MATH 332 - ADVANCED CALCULUS
- Stokes’ Theorem
Question Bank - Set 2
Liberty University
Question 1
Question
Let F(x, y, z) = (x, y2, xy)be a vector field, and let Sbe the surface in R3
defined by z=x2+y2for 0≤z≤1. Calculate the surface integral RRSF·dS
using Stokes’ Theorem.
Solution
Step 1: Determine the boundary curve of surface Sin the (x, y)-plane. The
boundary curve can be found by setting z= 1 in the surface equation z=x2+y2:
1 = x2+y2
This represents a circle in the (x, y)-plane with radius 1 centered at the origin.
Step 2: Calculate the outward unit normal vector to surface S. The unit
normal vector is given by:
n=∇z
|∇z|=D∂z
∂x ,∂z
∂y ,−1E
r∂z
∂x 2+∂z
∂y 2+ 1
where z=x2+y2on surface S. Calculating the partial derivatives:
∂z
∂x = 2xand ∂z
∂y = 2y
Therefore, the unit normal vector is:
n=*2x
p4x2+ 4y2+ 1,2y
p4x2+ 4y2+ 1,−1
p4x2+ 4y2+ 1+
Step 3: Calculate the curl of vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x y2xy
=⟨0,0, y −x⟩
Step 4: Apply Stokes’ Theorem, which states that for a surface Swith
boundary curve ∂S and outward unit normal vector n, the surface integral of
the curl of Fover Sis equal to the line integral of Faround the boundary curve
∂S:ZZS∇ × F·dS=I∂S
F·dr
Substitute in the given values and simplify the line integral to find the final
answer.
Question 2
Question
Let Sbe the surface of the cone given by the equation z=px2+y2, bounded
by the plane z= 3 and oriented upwards. Let Fbe the vector field defined by
F(x, y, z) = (z2, x, y). Calculate the flux of Facross the surface Susing Stokes’
Theorem.
Solution
Let’s denote the boundary curve of the surface Sas C, defined by the intersection
of the cone and the plane z= 3. Then, using Stokes’ Theorem, we have:
Flux(F, S) = ZZ
S
curl(F)·ndS =I
C
F·dr
where curl(F)is the curl of the vector field F,nis the unit normal vector
to the surface S,dS is the area element on S, and dris the tangent vector to
the curve C.
Step 1: Find the curl of F The curl of Fis given by:
curl(F) = ∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
z2x y
=∂y
∂x −∂x
∂y i−∂z2
∂x −∂z2
∂y j+∂x
∂x −∂y
∂y k
= (−1−0)i−(0 −0)j+ (1 −1)k=−i
Step 2: Parameterize the boundary curve CThe intersection of the
cone and the plane z= 3 gives us the boundary curve Cas a circle of radius 3
centered at the origin. We can parameterize this circle as:
r(t) = (3 cos t, 3 sin t, 3),0≤t≤2π
2
Step 3: Calculate the line integral Substitute the parameterization r(t)
into Fto get F(r(t)):
F(r(t)) = (32,3 cos t, 3 sin t) = (9,3 cos t, 3 sin t)
Now we can calculate the line integral along C:
I
C
F·dr=Z2π
0
F(r(t)) ·r′(t)dt =Z2π
0
(9,3 cos t, 3 sin t)·(−3 sin t, 3 cos t, 0) dt
=Z2π
0−27 sin tcos t+ 9 cos tsin t dt = 0
Therefore, the flux of Facross the surface Sis 0.
Question 3
Question
Let
F(x, y, z) = (y2, z2, x2)be a vector field in R3. Let Sbe the surface defined
by z=x2+y2for 1≤z≤4. Use Stokes’ Theorem to evaluate the circulation
of
Faround the boundary of Sin the counterclockwise direction.
Solution
Step 1: First, we need to find the curl of
F. The curl of a vector field
F(x, y, z) =
(M, N, P )is given by:
∇ ×
F=
i j k
∂
∂x
∂
∂y
∂
∂z
M N P
In this case,
F(x, y, z)=(y2, z2, x2), so M=y2,N=z2, and P=x2. Calcu-
lating the curl gives:
∇ ×
F= (0,−2z, 2y)
Step 2: Next, we need to find the normal vector to the surface S. The
normal vector ˆnis the gradient of the scalar function z=x2+y2:
∇z= (2x, 2y, −1)
Step 3: Now, we can apply Stokes’ Theorem. Stokes’ Theorem states that
the circulation of a vector field
Faround a closed curve Cis equal to the surface
integral of the curl of
Fover the surface Sthat is bounded by C:
IC
F·dr =ZZS
(∇ ×
F)·ˆn dS
3
Step 4: Substituting our values, the circulation of
Faround the boundary
of Sis given by:
IC
F·dr =ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS
Step 5: Simplifying the dot product gives:
ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS =ZZS
(−4zy + 4yx)dS
Step 6: The surface Sis the region in the xy-plane bounded by 1≤z≤4.
Thus, we can convert the surface integral to a double integral over the region in
the xy-plane:
ZZS
(−4zy + 4yx)dS =ZZR
(−4y(x2+y2)+4x(1)) dA
Step 7: Evaluating the double integral over the region Rgives the circulation
of
Faround the boundary of S.
Question 4
Question
Let Sbe the surface of the portion of the plane 2x+ 3y+z= 6 that lies in the
first octant, oriented upward, and let F= (x2, y2, z2). Use Stokes’ Theorem to
evaluate the surface integral RRSF·dS.
Solution
Step 1: Find the normal vector nto the surface S.
The equation of the given plane can be written as z= 6 −2x−3y. We can
express the normal vector as n=∂z
∂x ,∂z
∂y ,−1= (−2,−3,−1).
Step 2: Calculate the curl of F.
The curl of a vector field F= (P, Q, R)is given by
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F= (x2, y2, z2), we have P=x2,Q=y2, and R=z2. Calculating the
curl, we get
∇ × F= (0,0,0)
Step 3: Apply Stokes’ Theorem.
Stokes’ Theorem states that for a vector field Fand a surface Swith bound-
ary ∂S oriented by the outward unit normal, the surface integral of ∇×Fover
Sis equal to the line integral of Faround ∂S. Since the curl of Fis zero, the
surface integral over Sis zero.
Therefore, RRSF·dS= 0.
4
Question 5
Question
Let Sbe the surface of the region bounded by the paraboloid z=x2+y2and
the plane z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,
where F(x, y, z) = (y2, x, z)is a vector field.
Solution
Step 1: Determine the boundary curve Cof the surface Sin the xy-plane.
Since Sis bounded by the paraboloid z=x2+y2and the plane z= 4, the
boundary curve Ccan be parametrized by r(t) = (t, 2−t2), where −1≤t≤1.
Step 2: Calculate the curl of F:∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2x z
= (1,−1,0).
Step 3: Determine the unit normal vector nto the surface S. Since S
is given by z=g(x, y) = x2+y2and is below the plane z= 4, we have
n=∇g
|∇g|=⟨−2x,−2y,1⟩
√1+4(x2+y2).
Step 4: Calculate the line integral of Falong the boundary curve Cusing
Stokes’ Theorem:
ZC
F·dr=ZZS∇ × F·dS=ZZS
(1,−1,0) ·ndA
Step 5: Express the surface integral in terms of xand y:
ZZS
(1,−1,0) ·ndA =ZZD⟨−2x, −2y, 1⟩·⟨1,1,0⟩dA =ZZD
(−2x−2y)dA
Step 6: Integrate over the region Din the xy-plane:
ZZD
(−2x−2y)dA =Z1
−1Z1
−1
(−2x−2y)dydx = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 6
Question
Let
F(x, y, z) = x2
i+y2
j+z2
kbe a vector field, and let Sbe the part of the
paraboloid z= 4 −x2−y2that lies inside the cylinder x2+y2= 1. Use Stokes’
Theorem to evaluate the surface integral RRS(∇ ×
F)·d
S.
5
Solution
Step 1: Let’s first find the curl of
F.
∇ ×
F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y)
i−(2z−2x)
j+ (2y−2x)
k
Step 2: Now, find the unit normal vector to the surface S. Since Sis the
surface of the paraboloid, we have r(x, y) = x
i+y
j+(4−x2−y2)
k. The normal
vector is given by
N=∇r
||∇r||.
∇r =−2x
i−2y
j+
k
||∇r|| =p(−2x)2+ (−2y)2+ 1 = p4x2+ 4y2+ 1
So, the unit normal vector is
N=−2x
i−2y
j+
k
√4x2+4y2+1 .
Step 3: Substitute the curl and normal vector into the surface integral.
ZZS
(∇ ×
F)·d
S=ZZS
((2z−2y)
i−(2z−2x)
j+ (2y−2x)
k)·
NdS
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS
Step 4: Now evaluate the surface integral using Stokes’ Theorem. The
boundary curve of Sis the unit circle in the xy-plane. Calculate the line integral
along this curve and apply Stokes’ Theorem to relate the two. Apply
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS =IC
F·dr
Question 7
Question
Let Sbe the part of the surface x2+y2+z2= 9 that lies above the plane z= 0
and oriented upwards. Calculate the flux of the vector field F= (x2+y2)i+
zj+ykacross the surface S.
Solution
Step 1: Find the outward unit normal nto S. Since Sis the part of the surface
x2+y2+z2= 9 that lies above the plane z= 0, and oriented upwards, we
can write the equation of Sas z=p9−x2−y2. Then, the gradient of zis
∇z= (−x
√9−x2−y2,−y
√9−x2−y2,1). Hence, the unit normal vector to Sis given
by
n=∇z
∥∇z∥=1
q1 + x2+y2
9−x2−y2
(−x
p9−x2−y2,−y
p9−x2−y2,1)
6
Step 2: Calculate the curl of F. The curl of Fis given by
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2z y
= (∂y
∂y −∂z
∂z )i−(∂(x2+y2)
∂x −∂y
∂z )j+( ∂
∂x (x2+y2)−∂z
∂y )k
= 0i−0j+ 2yk= 2yk
Step 3: Apply Stokes’ Theorem. The flux of Facross Sis given by
ZZ
S
∇ × F·dS=ZZ
S
(2yk)·ndS
=ZZ
D
(2y)·1
q1 + x2+y2
9−x2−y2
dA
where Dis the projection of Sonto the xy-plane.
Step 4: Evaluate the flux integral. We can switch to polar coordinates by
letting x=rcos(θ)and y=rsin(θ). Thus, dA =rdrdθ, and the integral
becomes 2π
Z
0
3
Z
0
2rsin(θ)·1
q1 + r2
9
drdθ
This integral can be evaluated to find the flux of Facross the surface S.
Question 8
Question
Let Sbe the part of the paraboloid z=x2+y2that lies above the plane z= 1
oriented upwards. Use Stokes’ Theorem to evaluate the circulation of the vector
field F= (x2, y2, z2)around the curve Cwhere Cis the curve of intersection of
Sand the plane y=x.
Solution
Step 1: First, we need to parameterize the curve C:
Notice that y=xand z=x2+y2= 2x2. Therefore, a parameterization
of the curve Ccan be given by r(t) = (t, t, 2t2)where tvaries over the interval
[0,1].
Step 2: Next, we need to compute the curl of the vector field F:
The curl of Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y, 2x−2z, 2y−2x)
7
Step 3: Now, we need to compute the surface normal to Sat each point
(x, y, z)lying on the paraboloid z=x2+y2oriented upwards. The surface
normal is given by (−∂z
∂x ,−∂z
∂y ,1).
Therefore, the surface normal is (−2x, −2y, 1).
Step 4: Now, we can compute the circulation of Faround the curve Cusing
Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·ndS
where nis the unit normal to the surface Sand dS is the surface element of
S.
Step 5: Substituting all the values we found into the equation for Stokes’
Theorem, we get:
IC
F·dr=ZZS
(2z−2y, 2x−2z, 2y−2x)·(−2x, −2y, 1) dS
Step 6: We can now simplify this expression and evaluate the double integral
to find the circulation of Faround the curve C.
Question 9
Question
Let Sbe the part of the plane z= 4 −x−ythat lies above the triangle in the
xy-plane with vertices at (0,0),(1,0), and (0,1). Calculate the surface integral
RRS∇ × F·dS, where F(x, y, z) = ⟨x2, y2, z2⟩.
Solution
Step 1: Find the outward unit normal to S.
The equation of the plane z= 4 −x−ycan be rewritten as x+y+z= 4, which
means the normal vector to the plane is n=⟨1,1,1⟩. To get the unit normal
vector, divide by the magnitude: nunit =
|n| = ⟨1,1,1⟩√12+12+12=1
√3⟨1,1,1⟩=⟨1
√3,1
√3,1
√3⟩.
Step 2: Calculate ∇ × F.
The curl of Fis ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
=⟨0,0,2y−2x⟩.
Step 3: Calculate the dot product ∇ × F·nunit.
∇ × F·nunit =⟨0,0,2y−2x⟩·⟨ 1
√3,1
√3,1
√3⟩=1
√3(2y−2x).
Step 4: Integrate over the surface S.
The area of the triangle in the xy-plane is 1
2, so we have:
RRS∇ × F·dS=RRD∇ × F·(nunit dS) = RRD∇ × F·nunit dA
=RRD
1
√3(2y−2x)dA
8
=1
√3R1
0R1−x
0(2y−2x)dy dx
=1
√3R1
0y2−2xy1−x
0dx
=1
√3R1
0(1 −x)2−2x(1 −x)dx
=1
√3R1
0(1 −2x+x2−2x+ 2x2)dx
=1
√3R1
0(1 −4x+ 3x2)dx
=1
√3x−2x2+x31
0
=1
√3(1 −2 + 1)
=1
√3
Question 10
Question
Let F(x, y, z) = (x2+y, y2+z, z2+x)be a vector field and let Sbe the surface
defined by z=px2+y2in the region 0≤z≤1. Use Stokes’ Theorem to
evaluate the line integral HCF·dr, where Cis the boundary curve of Soriented
counterclockwise when viewed from above.
Solution
Step 1: Calculate the curl of F. The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y y2+z z2+x
.
=∂(z2+x)
∂y −∂(y2+z)
∂z i−∂(z2+x)
∂x −∂(x2+y)
∂z j+∂(y2+z)
∂x −∂(x2+y)
∂y k.
= (0 −1)i−(2x−1)j+ (2z−1)k.
=−i−(2x−1)j+ (2z−1)k.
Step 2: Determine the unit normal vector to the surface S. The unit normal
vector to the surface Sis given by n=1
√1+( ∂z
∂x )2+( ∂z
∂y )2(−∂z
∂x ,−∂z
∂y ,1). Since
z=px2+y2, we have ∂z
∂x =x
√x2+y2and ∂z
∂y =y
√x2+y2. So, the unit normal
vector is n=1
√2(−x
√x2+y2,−y
√x2+y2,1).
Step 3: Calculate the surface integral of the curl of Fover the surface S.
The surface integral is given by
ZZS
(∇×F)·ndS =ZZS
(−1,−(2x−1),2z−1)·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
9
Since z=px2+y2, the surface integral simplifies to
ZZS
(−1,−(2x−1),2px2+y2−1) ·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
The rest of the solution would involve further simplification and integration.
Question 11
Question
Let Sbe the part of the plane z= 3xinside the cylinder x2+y2= 4 oriented
upward. Use Stokes’ Theorem to evaluate HCF·dr, where F(x, y, z) = xyzi+
(x+y+z)j+xykand Cis the boundary of S, where Cis oriented counter-
clockwise when viewed from above.
Solution
Step 1: First, let’s find the curl of F:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
xyz x +y+z xy
= (y−z)i−(z)j+ (1 + x)k
Step 2: Next, we calculate the surface area of S. Since Sis inside the
cylinder x2+y2= 4 and the plane z= 3x, we can use polar coordinates for
the parametric representation of S:r(r, θ) = rcos(θ)i+rsin(θ)j+ 3rcos(θ)k
for (r, θ)∈[0,2] ×[0,2π].
The magnitude of the cross product |rr×rθ|gives us the surface area element,
which is |rr×rθ|=|rr×rθ|=p1 + (3 cos(θ))2·r=p1 + 9 cos2(θ)·r.
Step 3: Now we can set up the double integral to calculate the flux of the
curl of Fover the surface Susing Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·NdS
where Nis the outward unit normal to S. In this case, N=1
√1+9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k).
Step 4: Thus, we have:
IC
F·dr=ZZS
curl(F)·NdS
=Z2π
0Z2
0
((y−z)i−(z)j+ (1 + x)k)·1
p1 + 9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k)p1 + 9 cos2(θ)drdθ
=Z2π
0Z2
0
(6 sin2(θ)−rsin(θ) + 3) drdθ
Now we can proceed with solving this double integral to find the flux.
10
Question 12
Question
Let Sbe the surface of the part of the paraboloid z= 4 −x2−y2that lies above
the xy-plane and oriented upwards. Let F= (x2, y2, z2)be a vector field in R3.
Calculate the surface integral RRSF·dSusing Stokes’ Theorem.
Solution
Step 1: Determine the boundary of the surface Son the xy-plane. The part
of the paraboloid z= 4 −x2−y2that lies above the xy-plane is the unit circle
centered at the origin. Thus, the boundary curve Cof Son the xy-plane is the
unit circle x2+y2= 1.
Step 2: Parameterize the boundary curve C. Let R(u) = (cos u, sin u, 0) for
0≤u≤2πbe a parameterization of the unit circle x2+y2= 1.
Step 3: Calculate the curl of the vector field F= (x2, y2, z2). The curl
∇ × Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0,0,−2z)
Step 4: Applying Stokes’ Theorem, the surface integral RRSF·dSis equal to
the line integral RCF·dr, where dr=⟨dx, dy, dz⟩. Using the parameterization
R(u)for C:
ZC
F·dr=Z2π
0
(cos2u, sin2u, 0)T·(−sin u, cos u, 0) du
Step 5: Evaluating the dot product for each component:
ZC
F·dr=Z2π
0
(−cos2usin u+ sin2ucos u)du
ZC
F·dr=Z2π
0
1
2sin(2u)du = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 13
Question
Let Sbe the portion of the plane z= 4 −x−yabove the region in the xy-
plane bounded by the curves y=x2and y= 2x, oriented with outward normal.
Compute the flux of the curl of the vector field F(x, y, z) = (−y, z, x)across the
surface S.
11
Solution
Step 1: Find the curl of the vector field F. The curl of a vector field F=
(P, Q, R)is given by the determinant:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
In this case, F= (−y, z, x), so P=−y,Q=z, and R=x. Calculating the
curl:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−y z x
= (∂x
∂y −∂z
∂z ,∂z
∂x −∂x
∂z ,∂y
∂z −∂y
∂x )
∇ × F= (0 −1,1−0,−1−(−1)) = (−1,1,0)
Step 2: Parameterize the surface S. Let x=t,y=t2, and z= 4 −t−t2.
Then the parametric equation of the surface Sis given by:
r(t) = (t, t2,4−t−t2)
Step 3: Calculate the normal vector to the surface. The normal vector to
the surface Sis given by the cross product of the partial derivatives of r(t)with
respect to t.
r′(t) = (dx
dt ,dy
dt ,dz
dt ) = (1,2t, −1)
The normal vector is then:
N=r′(t) = (1,2t, −1)
Step 4: Calculate the flux of the curl of Facross the surface S. The flux is
given by:
ZZS
(∇ × F)·NdS =ZZR
(∇ × F)·ˆ
NdA
where ˆ
Nis the unit normal vector to the surface. Since Nis already a unit
normal vector, ˆ
N=N= (1,2t, −1). The region Rin the xy-plane is the region
bounded by y=x2and y= 2x. Therefore, the limits of integration for xare
0≤x≤2, and for yare x2≤y≤2x.
Calculating the dot product and integrating over the region R:
ZZR
(∇ × F)·ˆ
NdA =ZZR
(−1,1,0) ·(1,2t, −1) dA
=ZZR−1+2t dA
Step
12
Question 14
Question
Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1
and z= 2. Use Stokes’ Theorem to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = (y2, z2, x2).
Solution
To use Stokes’ Theorem, we first need to find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂x2
∂y −∂z2
∂z i−∂x2
∂x −∂y2
∂z j+∂y2
∂x −∂y2
∂y k
= 0i−0j+ 2yk
= 2yk.
Now, we need to find the unit normal vector nto the surface S. Since the
cone z=px2+y2points outward, the normal vector npoints away from the
z-axis, so it is n=1
√2(x, y, −px2+y2).
The surface integral then becomes
ZZS
(∇ × F)·dS=ZZS
2yk·1
√2(x, y, −px2+y2)dS =ZZD
2y(−√2)dA,
where Drepresents the projection of Sonto the xy-plane.
Since the projection of Sonto the xy-plane is a circle of radius 1 centered
at the origin, we can use polar coordinates:
ZZD
2y(−√2)dA =Z2π
0Z1
0
2rsin(θ)(−√2)rdrdθ =−2√2Z2π
0Z1
0
r2sin(θ)drdθ.
Integrating with respect to rfirst, we get
Z1
0
r2sin(θ)dr =r3
31
0
sin(θ) = sin(θ)
3,
so the surface integral becomes
−2√2Z2π
0
sin(θ)
3dθ =−2√2
3Z2π
0
sin(θ)dθ = 0.
Therefore, the surface integral RRS(∇ × F)·dSis 0.
13
Question 15
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤1.
Use Stokes’ Theorem to evaluate RRScurl F ·dS, where F(x, y, z) = (x2+
y2, ez, z).
Solution
Step 1: Compute the curl of F, which is given by
curl F =∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2ezz
.
This simplifies to
curl F = (0 −1,1−0,2x+ 0) = (−1,1,2x).
Step 2: Find the unit normal vector to the surface S. Let
g(x, y, z) = z−(x2+y2).
Then, ∇g= (−2x, −2y, 1) is parallel to the normal vector to S. Normalizing,
we find the unit normal vector to be
n=1
p1 + 4(x2+y2)(−2x, −2y, 1).
Step 3: Calculate the outward-pointing unit normal vector dSon the surface
S:
dS=ndS =1
p1 + 4(x2+y2)(−2x, −2y, 1) dxdy.
Step 4: Stoke’s Theorem states that
ZZS
curl F ·dS=ZZD
(curl F)·ndA,
where Dis the projection of Sonto the xy-plane.
Step 5: The projection of Sonto the xy-plane is the disc x2+y2≤1. To
evaluate the double integral, we need to find (curl F)·n.
(curl F)·n= (−1,1,2x)·1
p1 + 4(x2+y2)(−2x, −2y, 1).
This simplifies to
−2x+1
p1 + 4(x2+y2).
Step 6: Finally, we can evaluate the given integral using the double integral
over the disc x2+y2≤1:
ZZD
(−2x+1
p1 + 4(x2+y2))dA.
14
Question 16
Question
Let Sbe the portion of the sphere x2+y2+z2= 4 with z≥0and z≤
p4−x2−y2. Use Stokes’ Theorem to evaluate the surface integral of F=
(2y, x, −z)over S.
Solution
Step 1: Compute the curl of F.
Curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2y x −z
= (0,0,2)
Step 2: Compute the unit normal vector to S. Since Sis the portion of the
sphere x2+y2+z2= 4 in the upper half plane, the outward unit normal vector
is n=∇g
||∇g||, where g(x, y, z) = z−p4−x2−y2.
∇g= (−x
p4−x2−y2,−y
p4−x2−y2,1)
||∇g|| =sx2+y2
4−x2−y2+ 1
So, the unit normal vector is
n=(−x, −y, 4−x2−y2)
px2+y2+ 4 −x2−y2= (−x
p4−x2−y2,−y
p4−x2−y2,2
p4−x2−y2)
Step 3: Compute the surface integral of Fover Susing Stokes’ Theorem.
ZZS
F·dS=ZZS
F·ndS =ZZD
Curl(F)·ndA
Where Dis the projection of Sonto the xy-plane. Since Curl(F)·n= (0,0,2) ·
(−x
√4−x2−y2,−y
√4−x2−y2,2
√4−x2−y2) = 4
√4−x2−y2, we have
ZZS
F·dS=ZZD
4
p4−x2−y2dA
Step 4: Switch to polar coordinates. In polar coordinates, the domain Dis
0≤r≤2and 0≤θ≤2π. The Jacobian is r, so dA =rdrdθ.
Step 5: Evaluate the integral.
ZZS
F·dS=Z2π
0Z2
0
4
√4−r2·r drdθ
This can be computed by first integrating with respect to r, which gives
R2π
0[4√4−r2]2
0dθ.
Further computation yields the value of the surface integral.
15
Question 17
Question
Let Sbe the part of the cone z=px2+y2that lies below the plane z= 3,
with outward orientation. Evaluate the surface integral HSF·dS, where F=
(2x, y, z2), using Stokes’ Thejsonhirbn.
Solution
Step 1: First, let’s find the normal vector to the surface S. The unit normal
vector is given by
n=∇g
||∇g||
where g(x, y, z) = z−px2+y2is a scalar function defining the surface. Thus,
∇g=∂g
∂x ,∂g
∂y ,∂g
∂z = −x
px2+y2,−y
px2+y2,1!
and
||∇g|| =v
u
u
t −x
px2+y2!2
+ −y
px2+y2!2
+ 12=√2.
Therefore, the unit normal vector to the surface Sis
n= −x
p2(x2+y2),−y
p2(x2+y2),1
√2!.
Step 2: The surface integral HSF·dScan be evaluated using Stokes’ Theorem,
which states that IS
F·dS=ZZR
(∇ × F)·ndA
where Ris the projection of the surface Sonto the xy-plane.
Step 3: Compute the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2x y z2
= (0,−2z, −1)
Step 4: Since npoints upward for the cone and z= 3 on the cone, we have
n= (0,0,1).
Step 5: The projection of the surface Sonto the xy-plane is the disk x2+y2≤
9. Thus, the surface integral becomes
ZZx2+y2≤9
(0,−2z, −1) ·(0,0,1) dA =−ZZx2+y2≤9
dA =−π32=−9π.
Therefore, the value of the surface integral is −9π.
16
Question 18
Question
Let Cbe the curve in the plane given by y=x2+ 1, where xranges from 0
to 2. Let Sbe the surface of the part of the paraboloid z=x2+y2that lies
above C. Use Stokes’ Theorem to evaluate the line integral HCF·dr, where
F(x, y, z) = (x+y)i+xzj+yk.
Solution
Step 1: First, we parameterize the curve C. Since y=x2+ 1, the parameteri-
zation of Cis given by r(t) = (t, t2+ 1,0) where 0≤t≤2.
Step 2: Calculate the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x+y xz y
= (0 −z)i−(1)j+ (1 −x)k=−zi−j+ (1 −x)k
Step 3: Calculate the surface integral of ∇ × Fover the surface Susing
Stokes’ Theorem: ZZS
(∇ × F)·ndS =IC
F·dr
where nis the unit normal to the surface.
Step 4: Compute the unit normal vector to the surface S: Since the surface
is the part of the paraboloid z=x2+y2lying above C, the unit normal vector to
the surface pointing upwards is given by n=−∇h
|∇h|where h(x, y, z) = z−x2−y2.
∇h= (−2x, −2y, 1) ⇒ |∇h|=p4x2+ 4y2+ 1
So, n=2x
√4x2+4y2+1 ,2y
√4x2+4y2+1 ,−1
√4x2+4y2+1 .
Step 5: Calculate the line integral over the curve C:
IC
F·dr=ZZS
(∇ × F)·ndS
=Z2
0Zt2+1
t2
(−t) 2t
p4t2+ 4(t2+1)+1!+(1−t2) 2(t2+ 1)
p4t2+ 4(t2+1)+1!dt
After calculating this double integral, we can find the value of the line inte-
gral using Stokes’ Theorem.
17
Question 19
Question
Let Sbe the part of the plane z= 4 −x−2ythat lies inside the cylinder
x2+y2= 1 oriented upward. Use Stokes’ theorem to evaluate the integral
ZZS
(∇ × F)·ˆn dS,
where F(x, y, z) = (y2z, xz, xyz).
Solution
Step 1: Determine the boundary curve of S.Since Slies inside the
cylinder x2+y2= 1, the boundary curve of Sis the circle Cin the xy-plane
given by x2+y2= 1.
Step 2: Determine the unit normal vector ˆnto the surface S.Since
Sis oriented upward, the direction of the unit normal vector ˆnto the surface S
points in the positive zdirection. Therefore, ˆn=⟨0,0,1⟩.
Step 3: Compute the curl of F.The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z xz xyz
=−y−x−xz.
Step 4: Compute the surface integral using Stokes’ theorem. By
Stokes’ theorem, we have
ZZS
(∇ × F)·ˆn dS =IC
F·dr,
where Cis the boundary curve of S, which is the circle C:x2+y2= 1 oriented
counterclockwise.
Parametrize Cusing x= cos tand y= sin t, where 0≤t≤2π. Then,
dr=⟨−sin t, cos t, 0⟩dt.
Substitute the parametrization and Finto the line integral formula:
IC
F·dr
=Z2π
0⟨sin2tcos t, cos t, sin tcos t⟩ · ⟨−sin t, cos t, 0⟩dt
=Z2π
0
(−sin3tcos t+ cos t+ 0)dt
=Z2π
0
(−sin3tcos t+ cos t)dt
=1
4(sin4t) + sin t2π
0
=π
2.
18
Therefore, the value of the given integral is π
2.
Question 20
Question
Let Sbe the surface of the ellipsoid given by x2
a2+y2
b2+z2
c2= 1, oriented outward.
Use Stokes’ Theorem to evaluate
ZZS
F·dS
where F(x, y, z) = ⟨−z, x, y⟩.
Solution
Step 1: Compute the curl of F:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−z x y
=⟨1,1,1⟩
Step 2: Determine the unit normal vector to S: The unit normal vector to
the ellipsoid Sis given by
n=±∇g
|∇g|=±⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 3: Calculate the outward unit normal vector n: We choose the positive
sign for the outward orientation, so
n=⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 4: Calculate the surface area element dS: The surface area element dS
is given by
dS =|n|dA =p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
Step 5: Compute the normal component of F: Since n=⟨2x/a2,2y/b2,2z/c2⟩/p(2x/a2)2+ (2y/b2)2+ (2z/c2)2,
the normal component of Fis
F·n=⟨−z, x, y⟩ · ⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2=2x2
a2−z2+2y2
b2
19
Step 6: Express the given double integral in terms of dx dy: Since RRSF·dS=
RRSF·ndS, we have
ZZS
F·dS=ZZD
(F·n)p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
where Dis the projection of Sonto the xy-plane.
Step 7: Evaluate the double integral over D: Now we can evaluate the double
integral over Dusing the parameterization x=asin ϕcos θ,y=bsin ϕsin θ
where 0≤ϕ≤π,0≤θ≤2π.
Therefore, the final result is
Question 21
Question
Let Sbe the surface of the hemisphere x2+y2+z2= 1,z≥0. Calculate the
flux of the vector field F(x, y, z) = (xy, z, x)across Susing Stokes’ Theorem.
Solution
Step 1: Determine the boundary curve Cof the surface S. The boundary curve
Cof the hemisphere Sis the intersection of the hemisphere with the xy-plane.
This curve is a unit circle centered at the origin, given by x2+y2= 1.
Step 2: Parameterize the boundary curve C. We can parameterize the
boundary curve Cusing polar coordinates as x= cos t,y= sin t, where 0≤t≤
2π.
Step 3: Calculate the circulation of Faround C. The circulation of Faround
Ccan be calculated as:
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
Step 4: Simplify and evaluate the circulation integral.
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
=Z2π
0−cos tsin2t−cos2tdt
=Z2π
0−1
2sin(2t)−1 + cos(2t)
2dt
=−1
4cos(2t)−t
2−1
2sin(2t)2π
0
=−1
2π
20
Step 5: Apply Stokes’ Theorem. By Stokes’ Theorem, the flux of Facross
Sis equal to the circulation of Faround C, which is −1
2π. Thus, the flux of F
across Sis −1
2π.
Question 22
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤4
in the xy-plane, with upward orientation. Use Stoke’s Theorem to evaluate
RRS(∇ × F)·dSwhere F=⟨y2,−xz, xy⟩.
(You may assume that the outward unit normal to Sis n=⟨−2x, −2y, 1⟩.)
Solution
Step 1: We begin by calculating ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2−xz xy
=⟨0−z, 0−0,∂(xy)
∂x −∂(−xz)
∂y ⟩=⟨−z, 0, y +x⟩=⟨−z, 0, x +y⟩
Step 2: Next, we need to calculate the outward unit normal to S, denoted
by n, which is ⟨−2x, −2y, 1⟩. Therefore, dS=ndS =⟨−2x, −2y, 1⟩ · dSk=
−2x dSk.
Step 3: Now, let’s parameterize the surface Susing cylindrical coordinates.
Let x=rcos θ,y=rsin θ, where 0≤r≤2and 0≤θ≤2π. The normal
vector in cylindrical coordinates is ⟨−2rcos θ, −2rsin θ, 1⟩. We also have dS =
|−∇z|drdθ =|2r|drdθ = 2r drdθ.
Step 4: Now we evaluate the surface integral using Stokes’ Theorem:
ZZS
(∇ × F)·dS=ZZD
(∇ × F)·ndA
=ZZD⟨−z, 0, x +y⟩ · ⟨−2rcos θ, −2rsin θ, 1⟩ · 2r drdθ
=ZZD
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
0Z2
0
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
02
3r3cos θ+r2sin θ
2
0
dθ
21
=Z2π
016
3cos θ+ 4 sin θdθ
=16
3sin θ−4 cos θ
2π
0
=64
3
Therefore, RRS(∇ × F)·dS=64
3.
Question 23
Question
Let Sbe the part of the paraboloid z=x2+y2that lies inside the cylinder
x2+y2= 1. Use Stokes’ Theorem to evaluate RRS(∇×F)·dS, where F(x, y, z) =
(y2, z2, x2).
Solution
Step 1: Compute the curl of F:
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂
∂y (x2)−∂
∂z (z2),∂
∂z (y2)−∂
∂x (x2),∂
∂x (z2)−∂
∂y (y2)
= (0 −2z, 2y−0,2x−0) = (−2z, 2y, 2x)
Step 2: Calculate the unit normal vector nto the surface S: Since S
lies on the paraboloid z=x2+y2, the unit normal vector is given by n=
∂z
∂x ,∂z
∂y ,−1= (2x, 2y, −1).
Step 3: Compute the dot product of ∇ × Fand n:
(∇ × F)·n= (−2z, 2y, 2x)·(2x, 2y, −1) = −4zx + 4zy −2x
Step 4: Find the magnitude of the projection of the curl onto the surface
S: Since Slies inside the cylinder x2+y2= 1,Scan be parameterized by
r(r, θ)=(rcos θ, r sin θ, r2), where 0≤r≤1and 0≤θ≤2π. Then, the
magnitude of the projection is given by:
ZZS
(∇ × F)·dS=ZZD−4r3cos θsin θ+ 4r3cos θsin θ−2r dA
=ZZD−2r dA
Step 5: Evaluate the double integral over the region D:
ZZD−2r dA =−2Z2π
0Z1
0
r dr dθ =−2Z2π
0r2
21
0
dθ
=−2Z2π
0
1
2dθ =−πZ2π
0
dθ =−π(2π−0) = −2π2
Therefore, RRS(∇ × F)·dS=−2π2.
22
Question 24
Question
Let Sbe the part of the plane z= 4 −2x−ythat lies above the square Rin the
xy-plane with vertices at (0,0),(1,0),(1,1), and (0,1). Use Stokes’ Theorem
to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = x2i+xyzj+z2k.
Solution
Step 1: Find the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2xyz z2
=∂(z2)
∂y −∂(xyz)
∂z i−∂(z2)
∂x −∂(x2)
∂z j+∂(x2)
∂y −∂(z2)
∂x k
= (0 −0)i−(0 −0)j+ (0 −0)k=0
Step 2: Determine the orientation of Sand R. Since Sis above Rin the
xy-plane and the square Rgoes from (0,0) to (1,1) counterclockwise, we use
the upward-pointing normal vector for S,N=⟨2,−1,−1⟩.
Step 3: Evaluate the surface integral using Stokes’ Theorem.
ZZS
(∇ × F)·dS=ZZR
F·dr
where r(x, y) = xi+yj+ (4 −2x−y)k.
Step 4: Compute F·dr.
F·dr= (x2i+xyzj+z2k)·(dxi+dyj+dzk)
=x2dx +xydy +z2dz =x2dx +xydy + (4 −2x−y)2dz
Step 5: Parametrize the square R. Let x=uand y=v, where u, v ∈[0,1].
Then dz =−2dx −dy.
Step 6: Evaluate the integral over R.
Z1
0Z1
0
(u2du +uvdv + (4 −2u−v)2(−2du −dv))
=Z1
0Z1
0
((u2+ 4 −2u−v)2−2uv)dvdu
23
=Z1
0Z1
0
(u4+ 16 + v2−4u3−4u2v−4v+ 16u−4uv −2uv)dvdu
=Z1
0u4+ 16 + v+v2−4u3−16u2−4udvdu
=
Question 25
Question
Let Sbe the portion of the plane z= 1−x−ythat lies above the region bounded
by the parabola y=x2and the line y= 0. Let F(x, y, z) = x2i+y2j+z2k.
Find the flux of Facross the surface Sin the positive z-direction using Stokes’
Theorem.
Solution
Step 1: First, we need to find the normal vector to the surface S. Since Slies on
the plane z= 1−x−y, the normal vector to the surface Sis n=⟨∂z
∂x ,∂z
∂y ,−1⟩=
⟨−1,−1,−1⟩.
Step 2: Next, we need to find the curl of the vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0 −0) i−(0 −0) j+ (2y−2x)k= (2y−2x)k
Step 3: Now, we can find the flux of Facross the surface Susing Stokes’
Theorem:
Flux =ZZ
S
(∇ × F)·dS=ZZ
D
(∇ × F)·ndA
where Dis the projection of Sonto the xy-plane.
Step 4: The projection of the surface Sonto the xy-plane is the region
bounded by the parabola y=x2and the line y= 0, which can be described as
D={(x, y)|0≤x≤1,0≤y≤x2}.
Step 5: Evaluating the flux integral, we have:
Flux =ZZ
D
(2y−2x)· ⟨−1,−1,−1⟩ · dA =ZZ
D
(−2y+ 2x)dA
Step 6: Changing to polar coordinates for D, we have:
Flux =Z1
0Zx2
0
(−2rsin(θ)+2rcos(θ))rdrdθ
24
Step 7: Integrating with respect to rand θ, we get:
Flux =Z1
0−2r3sin(θ)
3+2r3cos(θ)
3x2
0
dθ
Step 8: Simplifying further:
Flux =Zπ
2
0−2x6sin(θ)
3+2x6cos(θ)
3dθ =2
3x6 Zπ
2
0
cos(θ)dθ −Zπ
2
0
sin(θ)dθ!
Step 9: Finally, integrating cos(θ)and sin(θ)over [0,π
2]gives:
Flux =
25
Step 3: Calculate the curl of vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x y2xy
=⟨0,0, y −x⟩
Step 4: Apply Stokes’ Theorem, which states that for a surface Swith
boundary curve ∂S and outward unit normal vector n, the surface integral of
the curl of Fover Sis equal to the line integral of Faround the boundary curve
∂S:ZZS∇ × F·dS=I∂S
F·dr
Substitute in the given values and simplify the line integral to find the final
answer.
Question 2
Question
Let Sbe the surface of the cone given by the equation z=px2+y2, bounded
by the plane z= 3 and oriented upwards. Let Fbe the vector field defined by
F(x, y, z) = (z2, x, y). Calculate the flux of Facross the surface Susing Stokes’
Theorem.
Solution
Let’s denote the boundary curve of the surface Sas C, defined by the intersection
of the cone and the plane z= 3. Then, using Stokes’ Theorem, we have:
Flux(F, S) = ZZ
S
curl(F)·ndS =I
C
F·dr
where curl(F)is the curl of the vector field F,nis the unit normal vector
to the surface S,dS is the area element on S, and dris the tangent vector to
the curve C.
Step 1: Find the curl of F The curl of Fis given by:
curl(F) = ∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
z2x y
=∂y
∂x −∂x
∂y i−∂z2
∂x −∂z2
∂y j+∂x
∂x −∂y
∂y k
= (−1−0)i−(0 −0)j+ (1 −1)k=−i
Step 2: Parameterize the boundary curve CThe intersection of the
cone and the plane z= 3 gives us the boundary curve Cas a circle of radius 3
centered at the origin. We can parameterize this circle as:
r(t) = (3 cos t, 3 sin t, 3),0≤t≤2π
2
Step 3: Calculate the line integral Substitute the parameterization r(t)
into Fto get F(r(t)):
F(r(t)) = (32,3 cos t, 3 sin t) = (9,3 cos t, 3 sin t)
Now we can calculate the line integral along C:
I
C
F·dr=Z2π
0
F(r(t)) ·r′(t)dt =Z2π
0
(9,3 cos t, 3 sin t)·(−3 sin t, 3 cos t, 0) dt
=Z2π
0−27 sin tcos t+ 9 cos tsin t dt = 0
Therefore, the flux of Facross the surface Sis 0.
Question 3
Question
Let
F(x, y, z) = (y2, z2, x2)be a vector field in R3. Let Sbe the surface defined
by z=x2+y2for 1≤z≤4. Use Stokes’ Theorem to evaluate the circulation
of
Faround the boundary of Sin the counterclockwise direction.
Solution
Step 1: First, we need to find the curl of
F. The curl of a vector field
F(x, y, z) =
(M, N, P )is given by:
∇ ×
F=
i j k
∂
∂x
∂
∂y
∂
∂z
M N P
In this case,
F(x, y, z)=(y2, z2, x2), so M=y2,N=z2, and P=x2. Calcu-
lating the curl gives:
∇ ×
F= (0,−2z, 2y)
Step 2: Next, we need to find the normal vector to the surface S. The
normal vector ˆnis the gradient of the scalar function z=x2+y2:
∇z= (2x, 2y, −1)
Step 3: Now, we can apply Stokes’ Theorem. Stokes’ Theorem states that
the circulation of a vector field
Faround a closed curve Cis equal to the surface
integral of the curl of
Fover the surface Sthat is bounded by C:
IC
F·dr =ZZS
(∇ ×
F)·ˆn dS
3
Step 4: Substituting our values, the circulation of
Faround the boundary
of Sis given by:
IC
F·dr =ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS
Step 5: Simplifying the dot product gives:
ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS =ZZS
(−4zy + 4yx)dS
Step 6: The surface Sis the region in the xy-plane bounded by 1≤z≤4.
Thus, we can convert the surface integral to a double integral over the region in
the xy-plane:
ZZS
(−4zy + 4yx)dS =ZZR
(−4y(x2+y2)+4x(1)) dA
Step 7: Evaluating the double integral over the region Rgives the circulation
of
Faround the boundary of S.
Question 4
Question
Let Sbe the surface of the portion of the plane 2x+ 3y+z= 6 that lies in the
first octant, oriented upward, and let F= (x2, y2, z2). Use Stokes’ Theorem to
evaluate the surface integral RRSF·dS.
Solution
Step 1: Find the normal vector nto the surface S.
The equation of the given plane can be written as z= 6 −2x−3y. We can
express the normal vector as n=∂z
∂x ,∂z
∂y ,−1= (−2,−3,−1).
Step 2: Calculate the curl of F.
The curl of a vector field F= (P, Q, R)is given by
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F= (x2, y2, z2), we have P=x2,Q=y2, and R=z2. Calculating the
curl, we get
∇ × F= (0,0,0)
Step 3: Apply Stokes’ Theorem.
Stokes’ Theorem states that for a vector field Fand a surface Swith bound-
ary ∂S oriented by the outward unit normal, the surface integral of ∇×Fover
Sis equal to the line integral of Faround ∂S. Since the curl of Fis zero, the
surface integral over Sis zero.
Therefore, RRSF·dS= 0.
4
Question 5
Question
Let Sbe the surface of the region bounded by the paraboloid z=x2+y2and
the plane z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,
where F(x, y, z) = (y2, x, z)is a vector field.
Solution
Step 1: Determine the boundary curve Cof the surface Sin the xy-plane.
Since Sis bounded by the paraboloid z=x2+y2and the plane z= 4, the
boundary curve Ccan be parametrized by r(t) = (t, 2−t2), where −1≤t≤1.
Step 2: Calculate the curl of F:∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2x z
= (1,−1,0).
Step 3: Determine the unit normal vector nto the surface S. Since S
is given by z=g(x, y) = x2+y2and is below the plane z= 4, we have
n=∇g
|∇g|=⟨−2x,−2y,1⟩
√1+4(x2+y2).
Step 4: Calculate the line integral of Falong the boundary curve Cusing
Stokes’ Theorem:
ZC
F·dr=ZZS∇ × F·dS=ZZS
(1,−1,0) ·ndA
Step 5: Express the surface integral in terms of xand y:
ZZS
(1,−1,0) ·ndA =ZZD⟨−2x, −2y, 1⟩·⟨1,1,0⟩dA =ZZD
(−2x−2y)dA
Step 6: Integrate over the region Din the xy-plane:
ZZD
(−2x−2y)dA =Z1
−1Z1
−1
(−2x−2y)dydx = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 6
Question
Let
F(x, y, z) = x2
i+y2
j+z2
kbe a vector field, and let Sbe the part of the
paraboloid z= 4 −x2−y2that lies inside the cylinder x2+y2= 1. Use Stokes’
Theorem to evaluate the surface integral RRS(∇ ×
F)·d
S.
5
Solution
Step 1: Let’s first find the curl of
F.
∇ ×
F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y)
i−(2z−2x)
j+ (2y−2x)
k
Step 2: Now, find the unit normal vector to the surface S. Since Sis the
surface of the paraboloid, we have r(x, y) = x
i+y
j+(4−x2−y2)
k. The normal
vector is given by
N=∇r
||∇r||.
∇r =−2x
i−2y
j+
k
||∇r|| =p(−2x)2+ (−2y)2+ 1 = p4x2+ 4y2+ 1
So, the unit normal vector is
N=−2x
i−2y
j+
k
√4x2+4y2+1 .
Step 3: Substitute the curl and normal vector into the surface integral.
ZZS
(∇ ×
F)·d
S=ZZS
((2z−2y)
i−(2z−2x)
j+ (2y−2x)
k)·
NdS
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS
Step 4: Now evaluate the surface integral using Stokes’ Theorem. The
boundary curve of Sis the unit circle in the xy-plane. Calculate the line integral
along this curve and apply Stokes’ Theorem to relate the two. Apply
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS =IC
F·dr
Question 7
Question
Let Sbe the part of the surface x2+y2+z2= 9 that lies above the plane z= 0
and oriented upwards. Calculate the flux of the vector field F= (x2+y2)i+
zj+ykacross the surface S.
Solution
Step 1: Find the outward unit normal nto S. Since Sis the part of the surface
x2+y2+z2= 9 that lies above the plane z= 0, and oriented upwards, we
can write the equation of Sas z=p9−x2−y2. Then, the gradient of zis
∇z= (−x
√9−x2−y2,−y
√9−x2−y2,1). Hence, the unit normal vector to Sis given
by
n=∇z
∥∇z∥=1
q1 + x2+y2
9−x2−y2
(−x
p9−x2−y2,−y
p9−x2−y2,1)
6
Step 2: Calculate the curl of F. The curl of Fis given by
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2z y
= (∂y
∂y −∂z
∂z )i−(∂(x2+y2)
∂x −∂y
∂z )j+( ∂
∂x (x2+y2)−∂z
∂y )k
= 0i−0j+ 2yk= 2yk
Step 3: Apply Stokes’ Theorem. The flux of Facross Sis given by
ZZ
S
∇ × F·dS=ZZ
S
(2yk)·ndS
=ZZ
D
(2y)·1
q1 + x2+y2
9−x2−y2
dA
where Dis the projection of Sonto the xy-plane.
Step 4: Evaluate the flux integral. We can switch to polar coordinates by
letting x=rcos(θ)and y=rsin(θ). Thus, dA =rdrdθ, and the integral
becomes 2π
Z
0
3
Z
0
2rsin(θ)·1
q1 + r2
9
drdθ
This integral can be evaluated to find the flux of Facross the surface S.
Question 8
Question
Let Sbe the part of the paraboloid z=x2+y2that lies above the plane z= 1
oriented upwards. Use Stokes’ Theorem to evaluate the circulation of the vector
field F= (x2, y2, z2)around the curve Cwhere Cis the curve of intersection of
Sand the plane y=x.
Solution
Step 1: First, we need to parameterize the curve C:
Notice that y=xand z=x2+y2= 2x2. Therefore, a parameterization
of the curve Ccan be given by r(t) = (t, t, 2t2)where tvaries over the interval
[0,1].
Step 2: Next, we need to compute the curl of the vector field F:
The curl of Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y, 2x−2z, 2y−2x)
7
Step 3: Now, we need to compute the surface normal to Sat each point
(x, y, z)lying on the paraboloid z=x2+y2oriented upwards. The surface
normal is given by (−∂z
∂x ,−∂z
∂y ,1).
Therefore, the surface normal is (−2x, −2y, 1).
Step 4: Now, we can compute the circulation of Faround the curve Cusing
Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·ndS
where nis the unit normal to the surface Sand dS is the surface element of
S.
Step 5: Substituting all the values we found into the equation for Stokes’
Theorem, we get:
IC
F·dr=ZZS
(2z−2y, 2x−2z, 2y−2x)·(−2x, −2y, 1) dS
Step 6: We can now simplify this expression and evaluate the double integral
to find the circulation of Faround the curve C.
Question 9
Question
Let Sbe the part of the plane z= 4 −x−ythat lies above the triangle in the
xy-plane with vertices at (0,0),(1,0), and (0,1). Calculate the surface integral
RRS∇ × F·dS, where F(x, y, z) = ⟨x2, y2, z2⟩.
Solution
Step 1: Find the outward unit normal to S.
The equation of the plane z= 4 −x−ycan be rewritten as x+y+z= 4, which
means the normal vector to the plane is n=⟨1,1,1⟩. To get the unit normal
vector, divide by the magnitude: nunit =
|n| = ⟨1,1,1⟩√12+12+12=1
√3⟨1,1,1⟩=⟨1
√3,1
√3,1
√3⟩.
Step 2: Calculate ∇ × F.
The curl of Fis ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
=⟨0,0,2y−2x⟩.
Step 3: Calculate the dot product ∇ × F·nunit.
∇ × F·nunit =⟨0,0,2y−2x⟩·⟨ 1
√3,1
√3,1
√3⟩=1
√3(2y−2x).
Step 4: Integrate over the surface S.
The area of the triangle in the xy-plane is 1
2, so we have:
RRS∇ × F·dS=RRD∇ × F·(nunit dS) = RRD∇ × F·nunit dA
=RRD
1
√3(2y−2x)dA
8
=1
√3R1
0R1−x
0(2y−2x)dy dx
=1
√3R1
0y2−2xy1−x
0dx
=1
√3R1
0(1 −x)2−2x(1 −x)dx
=1
√3R1
0(1 −2x+x2−2x+ 2x2)dx
=1
√3R1
0(1 −4x+ 3x2)dx
=1
√3x−2x2+x31
0
=1
√3(1 −2 + 1)
=1
√3
Question 10
Question
Let F(x, y, z) = (x2+y, y2+z, z2+x)be a vector field and let Sbe the surface
defined by z=px2+y2in the region 0≤z≤1. Use Stokes’ Theorem to
evaluate the line integral HCF·dr, where Cis the boundary curve of Soriented
counterclockwise when viewed from above.
Solution
Step 1: Calculate the curl of F. The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y y2+z z2+x
.
=∂(z2+x)
∂y −∂(y2+z)
∂z i−∂(z2+x)
∂x −∂(x2+y)
∂z j+∂(y2+z)
∂x −∂(x2+y)
∂y k.
= (0 −1)i−(2x−1)j+ (2z−1)k.
=−i−(2x−1)j+ (2z−1)k.
Step 2: Determine the unit normal vector to the surface S. The unit normal
vector to the surface Sis given by n=1
√1+( ∂z
∂x )2+( ∂z
∂y )2(−∂z
∂x ,−∂z
∂y ,1). Since
z=px2+y2, we have ∂z
∂x =x
√x2+y2and ∂z
∂y =y
√x2+y2. So, the unit normal
vector is n=1
√2(−x
√x2+y2,−y
√x2+y2,1).
Step 3: Calculate the surface integral of the curl of Fover the surface S.
The surface integral is given by
ZZS
(∇×F)·ndS =ZZS
(−1,−(2x−1),2z−1)·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
9
Since z=px2+y2, the surface integral simplifies to
ZZS
(−1,−(2x−1),2px2+y2−1) ·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
The rest of the solution would involve further simplification and integration.
Question 11
Question
Let Sbe the part of the plane z= 3xinside the cylinder x2+y2= 4 oriented
upward. Use Stokes’ Theorem to evaluate HCF·dr, where F(x, y, z) = xyzi+
(x+y+z)j+xykand Cis the boundary of S, where Cis oriented counter-
clockwise when viewed from above.
Solution
Step 1: First, let’s find the curl of F:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
xyz x +y+z xy
= (y−z)i−(z)j+ (1 + x)k
Step 2: Next, we calculate the surface area of S. Since Sis inside the
cylinder x2+y2= 4 and the plane z= 3x, we can use polar coordinates for
the parametric representation of S:r(r, θ) = rcos(θ)i+rsin(θ)j+ 3rcos(θ)k
for (r, θ)∈[0,2] ×[0,2π].
The magnitude of the cross product |rr×rθ|gives us the surface area element,
which is |rr×rθ|=|rr×rθ|=p1 + (3 cos(θ))2·r=p1 + 9 cos2(θ)·r.
Step 3: Now we can set up the double integral to calculate the flux of the
curl of Fover the surface Susing Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·NdS
where Nis the outward unit normal to S. In this case, N=1
√1+9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k).
Step 4: Thus, we have:
IC
F·dr=ZZS
curl(F)·NdS
=Z2π
0Z2
0
((y−z)i−(z)j+ (1 + x)k)·1
p1 + 9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k)p1 + 9 cos2(θ)drdθ
=Z2π
0Z2
0
(6 sin2(θ)−rsin(θ) + 3) drdθ
Now we can proceed with solving this double integral to find the flux.
10
Question 12
Question
Let Sbe the surface of the part of the paraboloid z= 4 −x2−y2that lies above
the xy-plane and oriented upwards. Let F= (x2, y2, z2)be a vector field in R3.
Calculate the surface integral RRSF·dSusing Stokes’ Theorem.
Solution
Step 1: Determine the boundary of the surface Son the xy-plane. The part
of the paraboloid z= 4 −x2−y2that lies above the xy-plane is the unit circle
centered at the origin. Thus, the boundary curve Cof Son the xy-plane is the
unit circle x2+y2= 1.
Step 2: Parameterize the boundary curve C. Let R(u) = (cos u, sin u, 0) for
0≤u≤2πbe a parameterization of the unit circle x2+y2= 1.
Step 3: Calculate the curl of the vector field F= (x2, y2, z2). The curl
∇ × Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0,0,−2z)
Step 4: Applying Stokes’ Theorem, the surface integral RRSF·dSis equal to
the line integral RCF·dr, where dr=⟨dx, dy, dz⟩. Using the parameterization
R(u)for C:
ZC
F·dr=Z2π
0
(cos2u, sin2u, 0)T·(−sin u, cos u, 0) du
Step 5: Evaluating the dot product for each component:
ZC
F·dr=Z2π
0
(−cos2usin u+ sin2ucos u)du
ZC
F·dr=Z2π
0
1
2sin(2u)du = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 13
Question
Let Sbe the portion of the plane z= 4 −x−yabove the region in the xy-
plane bounded by the curves y=x2and y= 2x, oriented with outward normal.
Compute the flux of the curl of the vector field F(x, y, z) = (−y, z, x)across the
surface S.
11
Solution
Step 1: Find the curl of the vector field F. The curl of a vector field F=
(P, Q, R)is given by the determinant:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
In this case, F= (−y, z, x), so P=−y,Q=z, and R=x. Calculating the
curl:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−y z x
= (∂x
∂y −∂z
∂z ,∂z
∂x −∂x
∂z ,∂y
∂z −∂y
∂x )
∇ × F= (0 −1,1−0,−1−(−1)) = (−1,1,0)
Step 2: Parameterize the surface S. Let x=t,y=t2, and z= 4 −t−t2.
Then the parametric equation of the surface Sis given by:
r(t) = (t, t2,4−t−t2)
Step 3: Calculate the normal vector to the surface. The normal vector to
the surface Sis given by the cross product of the partial derivatives of r(t)with
respect to t.
r′(t) = (dx
dt ,dy
dt ,dz
dt ) = (1,2t, −1)
The normal vector is then:
N=r′(t) = (1,2t, −1)
Step 4: Calculate the flux of the curl of Facross the surface S. The flux is
given by:
ZZS
(∇ × F)·NdS =ZZR
(∇ × F)·ˆ
NdA
where ˆ
Nis the unit normal vector to the surface. Since Nis already a unit
normal vector, ˆ
N=N= (1,2t, −1). The region Rin the xy-plane is the region
bounded by y=x2and y= 2x. Therefore, the limits of integration for xare
0≤x≤2, and for yare x2≤y≤2x.
Calculating the dot product and integrating over the region R:
ZZR
(∇ × F)·ˆ
NdA =ZZR
(−1,1,0) ·(1,2t, −1) dA
=ZZR−1+2t dA
Step
12
Question 14
Question
Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1
and z= 2. Use Stokes’ Theorem to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = (y2, z2, x2).
Solution
To use Stokes’ Theorem, we first need to find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂x2
∂y −∂z2
∂z i−∂x2
∂x −∂y2
∂z j+∂y2
∂x −∂y2
∂y k
= 0i−0j+ 2yk
= 2yk.
Now, we need to find the unit normal vector nto the surface S. Since the
cone z=px2+y2points outward, the normal vector npoints away from the
z-axis, so it is n=1
√2(x, y, −px2+y2).
The surface integral then becomes
ZZS
(∇ × F)·dS=ZZS
2yk·1
√2(x, y, −px2+y2)dS =ZZD
2y(−√2)dA,
where Drepresents the projection of Sonto the xy-plane.
Since the projection of Sonto the xy-plane is a circle of radius 1 centered
at the origin, we can use polar coordinates:
ZZD
2y(−√2)dA =Z2π
0Z1
0
2rsin(θ)(−√2)rdrdθ =−2√2Z2π
0Z1
0
r2sin(θ)drdθ.
Integrating with respect to rfirst, we get
Z1
0
r2sin(θ)dr =r3
31
0
sin(θ) = sin(θ)
3,
so the surface integral becomes
−2√2Z2π
0
sin(θ)
3dθ =−2√2
3Z2π
0
sin(θ)dθ = 0.
Therefore, the surface integral RRS(∇ × F)·dSis 0.
13
Question 15
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤1.
Use Stokes’ Theorem to evaluate RRScurl F ·dS, where F(x, y, z) = (x2+
y2, ez, z).
Solution
Step 1: Compute the curl of F, which is given by
curl F =∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2ezz
.
This simplifies to
curl F = (0 −1,1−0,2x+ 0) = (−1,1,2x).
Step 2: Find the unit normal vector to the surface S. Let
g(x, y, z) = z−(x2+y2).
Then, ∇g= (−2x, −2y, 1) is parallel to the normal vector to S. Normalizing,
we find the unit normal vector to be
n=1
p1 + 4(x2+y2)(−2x, −2y, 1).
Step 3: Calculate the outward-pointing unit normal vector dSon the surface
S:
dS=ndS =1
p1 + 4(x2+y2)(−2x, −2y, 1) dxdy.
Step 4: Stoke’s Theorem states that
ZZS
curl F ·dS=ZZD
(curl F)·ndA,
where Dis the projection of Sonto the xy-plane.
Step 5: The projection of Sonto the xy-plane is the disc x2+y2≤1. To
evaluate the double integral, we need to find (curl F)·n.
(curl F)·n= (−1,1,2x)·1
p1 + 4(x2+y2)(−2x, −2y, 1).
This simplifies to
−2x+1
p1 + 4(x2+y2).
Step 6: Finally, we can evaluate the given integral using the double integral
over the disc x2+y2≤1:
ZZD
(−2x+1
p1 + 4(x2+y2))dA.
14
Question 16
Question
Let Sbe the portion of the sphere x2+y2+z2= 4 with z≥0and z≤
p4−x2−y2. Use Stokes’ Theorem to evaluate the surface integral of F=
(2y, x, −z)over S.
Solution
Step 1: Compute the curl of F.
Curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2y x −z
= (0,0,2)
Step 2: Compute the unit normal vector to S. Since Sis the portion of the
sphere x2+y2+z2= 4 in the upper half plane, the outward unit normal vector
is n=∇g
||∇g||, where g(x, y, z) = z−p4−x2−y2.
∇g= (−x
p4−x2−y2,−y
p4−x2−y2,1)
||∇g|| =sx2+y2
4−x2−y2+ 1
So, the unit normal vector is
n=(−x, −y, 4−x2−y2)
px2+y2+ 4 −x2−y2= (−x
p4−x2−y2,−y
p4−x2−y2,2
p4−x2−y2)
Step 3: Compute the surface integral of Fover Susing Stokes’ Theorem.
ZZS
F·dS=ZZS
F·ndS =ZZD
Curl(F)·ndA
Where Dis the projection of Sonto the xy-plane. Since Curl(F)·n= (0,0,2) ·
(−x
√4−x2−y2,−y
√4−x2−y2,2
√4−x2−y2) = 4
√4−x2−y2, we have
ZZS
F·dS=ZZD
4
p4−x2−y2dA
Step 4: Switch to polar coordinates. In polar coordinates, the domain Dis
0≤r≤2and 0≤θ≤2π. The Jacobian is r, so dA =rdrdθ.
Step 5: Evaluate the integral.
ZZS
F·dS=Z2π
0Z2
0
4
√4−r2·r drdθ
This can be computed by first integrating with respect to r, which gives
R2π
0[4√4−r2]2
0dθ.
Further computation yields the value of the surface integral.
15
Question 17
Question
Let Sbe the part of the cone z=px2+y2that lies below the plane z= 3,
with outward orientation. Evaluate the surface integral HSF·dS, where F=
(2x, y, z2), using Stokes’ Thejsonhirbn.
Solution
Step 1: First, let’s find the normal vector to the surface S. The unit normal
vector is given by
n=∇g
||∇g||
where g(x, y, z) = z−px2+y2is a scalar function defining the surface. Thus,
∇g=∂g
∂x ,∂g
∂y ,∂g
∂z = −x
px2+y2,−y
px2+y2,1!
and
||∇g|| =v
u
u
t −x
px2+y2!2
+ −y
px2+y2!2
+ 12=√2.
Therefore, the unit normal vector to the surface Sis
n= −x
p2(x2+y2),−y
p2(x2+y2),1
√2!.
Step 2: The surface integral HSF·dScan be evaluated using Stokes’ Theorem,
which states that IS
F·dS=ZZR
(∇ × F)·ndA
where Ris the projection of the surface Sonto the xy-plane.
Step 3: Compute the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2x y z2
= (0,−2z, −1)
Step 4: Since npoints upward for the cone and z= 3 on the cone, we have
n= (0,0,1).
Step 5: The projection of the surface Sonto the xy-plane is the disk x2+y2≤
9. Thus, the surface integral becomes
ZZx2+y2≤9
(0,−2z, −1) ·(0,0,1) dA =−ZZx2+y2≤9
dA =−π32=−9π.
Therefore, the value of the surface integral is −9π.
16
Question 18
Question
Let Cbe the curve in the plane given by y=x2+ 1, where xranges from 0
to 2. Let Sbe the surface of the part of the paraboloid z=x2+y2that lies
above C. Use Stokes’ Theorem to evaluate the line integral HCF·dr, where
F(x, y, z) = (x+y)i+xzj+yk.
Solution
Step 1: First, we parameterize the curve C. Since y=x2+ 1, the parameteri-
zation of Cis given by r(t) = (t, t2+ 1,0) where 0≤t≤2.
Step 2: Calculate the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x+y xz y
= (0 −z)i−(1)j+ (1 −x)k=−zi−j+ (1 −x)k
Step 3: Calculate the surface integral of ∇ × Fover the surface Susing
Stokes’ Theorem: ZZS
(∇ × F)·ndS =IC
F·dr
where nis the unit normal to the surface.
Step 4: Compute the unit normal vector to the surface S: Since the surface
is the part of the paraboloid z=x2+y2lying above C, the unit normal vector to
the surface pointing upwards is given by n=−∇h
|∇h|where h(x, y, z) = z−x2−y2.
∇h= (−2x, −2y, 1) ⇒ |∇h|=p4x2+ 4y2+ 1
So, n=2x
√4x2+4y2+1 ,2y
√4x2+4y2+1 ,−1
√4x2+4y2+1 .
Step 5: Calculate the line integral over the curve C:
IC
F·dr=ZZS
(∇ × F)·ndS
=Z2
0Zt2+1
t2
(−t) 2t
p4t2+ 4(t2+1)+1!+(1−t2) 2(t2+ 1)
p4t2+ 4(t2+1)+1!dt
After calculating this double integral, we can find the value of the line inte-
gral using Stokes’ Theorem.
17
Question 19
Question
Let Sbe the part of the plane z= 4 −x−2ythat lies inside the cylinder
x2+y2= 1 oriented upward. Use Stokes’ theorem to evaluate the integral
ZZS
(∇ × F)·ˆn dS,
where F(x, y, z) = (y2z, xz, xyz).
Solution
Step 1: Determine the boundary curve of S.Since Slies inside the
cylinder x2+y2= 1, the boundary curve of Sis the circle Cin the xy-plane
given by x2+y2= 1.
Step 2: Determine the unit normal vector ˆnto the surface S.Since
Sis oriented upward, the direction of the unit normal vector ˆnto the surface S
points in the positive zdirection. Therefore, ˆn=⟨0,0,1⟩.
Step 3: Compute the curl of F.The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z xz xyz
=−y−x−xz.
Step 4: Compute the surface integral using Stokes’ theorem. By
Stokes’ theorem, we have
ZZS
(∇ × F)·ˆn dS =IC
F·dr,
where Cis the boundary curve of S, which is the circle C:x2+y2= 1 oriented
counterclockwise.
Parametrize Cusing x= cos tand y= sin t, where 0≤t≤2π. Then,
dr=⟨−sin t, cos t, 0⟩dt.
Substitute the parametrization and Finto the line integral formula:
IC
F·dr
=Z2π
0⟨sin2tcos t, cos t, sin tcos t⟩ · ⟨−sin t, cos t, 0⟩dt
=Z2π
0
(−sin3tcos t+ cos t+ 0)dt
=Z2π
0
(−sin3tcos t+ cos t)dt
=1
4(sin4t) + sin t2π
0
=π
2.
18
Therefore, the value of the given integral is π
2.
Question 20
Question
Let Sbe the surface of the ellipsoid given by x2
a2+y2
b2+z2
c2= 1, oriented outward.
Use Stokes’ Theorem to evaluate
ZZS
F·dS
where F(x, y, z) = ⟨−z, x, y⟩.
Solution
Step 1: Compute the curl of F:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−z x y
=⟨1,1,1⟩
Step 2: Determine the unit normal vector to S: The unit normal vector to
the ellipsoid Sis given by
n=±∇g
|∇g|=±⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 3: Calculate the outward unit normal vector n: We choose the positive
sign for the outward orientation, so
n=⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 4: Calculate the surface area element dS: The surface area element dS
is given by
dS =|n|dA =p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
Step 5: Compute the normal component of F: Since n=⟨2x/a2,2y/b2,2z/c2⟩/p(2x/a2)2+ (2y/b2)2+ (2z/c2)2,
the normal component of Fis
F·n=⟨−z, x, y⟩ · ⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2=2x2
a2−z2+2y2
b2
19
Step 6: Express the given double integral in terms of dx dy: Since RRSF·dS=
RRSF·ndS, we have
ZZS
F·dS=ZZD
(F·n)p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
where Dis the projection of Sonto the xy-plane.
Step 7: Evaluate the double integral over D: Now we can evaluate the double
integral over Dusing the parameterization x=asin ϕcos θ,y=bsin ϕsin θ
where 0≤ϕ≤π,0≤θ≤2π.
Therefore, the final result is
Question 21
Question
Let Sbe the surface of the hemisphere x2+y2+z2= 1,z≥0. Calculate the
flux of the vector field F(x, y, z) = (xy, z, x)across Susing Stokes’ Theorem.
Solution
Step 1: Determine the boundary curve Cof the surface S. The boundary curve
Cof the hemisphere Sis the intersection of the hemisphere with the xy-plane.
This curve is a unit circle centered at the origin, given by x2+y2= 1.
Step 2: Parameterize the boundary curve C. We can parameterize the
boundary curve Cusing polar coordinates as x= cos t,y= sin t, where 0≤t≤
2π.
Step 3: Calculate the circulation of Faround C. The circulation of Faround
Ccan be calculated as:
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
Step 4: Simplify and evaluate the circulation integral.
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
=Z2π
0−cos tsin2t−cos2tdt
=Z2π
0−1
2sin(2t)−1 + cos(2t)
2dt
=−1
4cos(2t)−t
2−1
2sin(2t)2π
0
=−1
2π
20
Step 5: Apply Stokes’ Theorem. By Stokes’ Theorem, the flux of Facross
Sis equal to the circulation of Faround C, which is −1
2π. Thus, the flux of F
across Sis −1
2π.
Question 22
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤4
in the xy-plane, with upward orientation. Use Stoke’s Theorem to evaluate
RRS(∇ × F)·dSwhere F=⟨y2,−xz, xy⟩.
(You may assume that the outward unit normal to Sis n=⟨−2x, −2y, 1⟩.)
Solution
Step 1: We begin by calculating ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2−xz xy
=⟨0−z, 0−0,∂(xy)
∂x −∂(−xz)
∂y ⟩=⟨−z, 0, y +x⟩=⟨−z, 0, x +y⟩
Step 2: Next, we need to calculate the outward unit normal to S, denoted
by n, which is ⟨−2x, −2y, 1⟩. Therefore, dS=ndS =⟨−2x, −2y, 1⟩ · dSk=
−2x dSk.
Step 3: Now, let’s parameterize the surface Susing cylindrical coordinates.
Let x=rcos θ,y=rsin θ, where 0≤r≤2and 0≤θ≤2π. The normal
vector in cylindrical coordinates is ⟨−2rcos θ, −2rsin θ, 1⟩. We also have dS =
|−∇z|drdθ =|2r|drdθ = 2r drdθ.
Step 4: Now we evaluate the surface integral using Stokes’ Theorem:
ZZS
(∇ × F)·dS=ZZD
(∇ × F)·ndA
=ZZD⟨−z, 0, x +y⟩ · ⟨−2rcos θ, −2rsin θ, 1⟩ · 2r drdθ
=ZZD
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
0Z2
0
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
02
3r3cos θ+r2sin θ
2
0
dθ
21
=Z2π
016
3cos θ+ 4 sin θdθ
=16
3sin θ−4 cos θ
2π
0
=64
3
Therefore, RRS(∇ × F)·dS=64
3.
Question 23
Question
Let Sbe the part of the paraboloid z=x2+y2that lies inside the cylinder
x2+y2= 1. Use Stokes’ Theorem to evaluate RRS(∇×F)·dS, where F(x, y, z) =
(y2, z2, x2).
Solution
Step 1: Compute the curl of F:
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂
∂y (x2)−∂
∂z (z2),∂
∂z (y2)−∂
∂x (x2),∂
∂x (z2)−∂
∂y (y2)
= (0 −2z, 2y−0,2x−0) = (−2z, 2y, 2x)
Step 2: Calculate the unit normal vector nto the surface S: Since S
lies on the paraboloid z=x2+y2, the unit normal vector is given by n=
∂z
∂x ,∂z
∂y ,−1= (2x, 2y, −1).
Step 3: Compute the dot product of ∇ × Fand n:
(∇ × F)·n= (−2z, 2y, 2x)·(2x, 2y, −1) = −4zx + 4zy −2x
Step 4: Find the magnitude of the projection of the curl onto the surface
S: Since Slies inside the cylinder x2+y2= 1,Scan be parameterized by
r(r, θ)=(rcos θ, r sin θ, r2), where 0≤r≤1and 0≤θ≤2π. Then, the
magnitude of the projection is given by:
ZZS
(∇ × F)·dS=ZZD−4r3cos θsin θ+ 4r3cos θsin θ−2r dA
=ZZD−2r dA
Step 5: Evaluate the double integral over the region D:
ZZD−2r dA =−2Z2π
0Z1
0
r dr dθ =−2Z2π
0r2
21
0
dθ
=−2Z2π
0
1
2dθ =−πZ2π
0
dθ =−π(2π−0) = −2π2
Therefore, RRS(∇ × F)·dS=−2π2.
22
Question 24
Question
Let Sbe the part of the plane z= 4 −2x−ythat lies above the square Rin the
xy-plane with vertices at (0,0),(1,0),(1,1), and (0,1). Use Stokes’ Theorem
to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = x2i+xyzj+z2k.
Solution
Step 1: Find the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2xyz z2
=∂(z2)
∂y −∂(xyz)
∂z i−∂(z2)
∂x −∂(x2)
∂z j+∂(x2)
∂y −∂(z2)
∂x k
= (0 −0)i−(0 −0)j+ (0 −0)k=0
Step 2: Determine the orientation of Sand R. Since Sis above Rin the
xy-plane and the square Rgoes from (0,0) to (1,1) counterclockwise, we use
the upward-pointing normal vector for S,N=⟨2,−1,−1⟩.
Step 3: Evaluate the surface integral using Stokes’ Theorem.
ZZS
(∇ × F)·dS=ZZR
F·dr
where r(x, y) = xi+yj+ (4 −2x−y)k.
Step 4: Compute F·dr.
F·dr= (x2i+xyzj+z2k)·(dxi+dyj+dzk)
=x2dx +xydy +z2dz =x2dx +xydy + (4 −2x−y)2dz
Step 5: Parametrize the square R. Let x=uand y=v, where u, v ∈[0,1].
Then dz =−2dx −dy.
Step 6: Evaluate the integral over R.
Z1
0Z1
0
(u2du +uvdv + (4 −2u−v)2(−2du −dv))
=Z1
0Z1
0
((u2+ 4 −2u−v)2−2uv)dvdu
23
=Z1
0Z1
0
(u4+ 16 + v2−4u3−4u2v−4v+ 16u−4uv −2uv)dvdu
=Z1
0u4+ 16 + v+v2−4u3−16u2−4udvdu
=
Question 25
Question
Let Sbe the portion of the plane z= 1−x−ythat lies above the region bounded
by the parabola y=x2and the line y= 0. Let F(x, y, z) = x2i+y2j+z2k.
Find the flux of Facross the surface Sin the positive z-direction using Stokes’
Theorem.
Solution
Step 1: First, we need to find the normal vector to the surface S. Since Slies on
the plane z= 1−x−y, the normal vector to the surface Sis n=⟨∂z
∂x ,∂z
∂y ,−1⟩=
⟨−1,−1,−1⟩.
Step 2: Next, we need to find the curl of the vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0 −0) i−(0 −0) j+ (2y−2x)k= (2y−2x)k
Step 3: Now, we can find the flux of Facross the surface Susing Stokes’
Theorem:
Flux =ZZ
S
(∇ × F)·dS=ZZ
D
(∇ × F)·ndA
where Dis the projection of Sonto the xy-plane.
Step 4: The projection of the surface Sonto the xy-plane is the region
bounded by the parabola y=x2and the line y= 0, which can be described as
D={(x, y)|0≤x≤1,0≤y≤x2}.
Step 5: Evaluating the flux integral, we have:
Flux =ZZ
D
(2y−2x)· ⟨−1,−1,−1⟩ · dA =ZZ
D
(−2y+ 2x)dA
Step 6: Changing to polar coordinates for D, we have:
Flux =Z1
0Zx2
0
(−2rsin(θ)+2rcos(θ))rdrdθ
24
Step 7: Integrating with respect to rand θ, we get:
Flux =Z1
0−2r3sin(θ)
3+2r3cos(θ)
3x2
0
dθ
Step 8: Simplifying further:
Flux =Zπ
2
0−2x6sin(θ)
3+2x6cos(θ)
3dθ =2
3x6 Zπ
2
0
cos(θ)dθ −Zπ
2
0
sin(θ)dθ!
Step 9: Finally, integrating cos(θ)and sin(θ)over [0,π
2]gives:
Flux =
25
Step 3: Calculate the curl of vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x y2xy
=⟨0,0, y −x⟩
Step 4: Apply Stokes’ Theorem, which states that for a surface Swith
boundary curve ∂S and outward unit normal vector n, the surface integral of
the curl of Fover Sis equal to the line integral of Faround the boundary curve
∂S:ZZS∇ × F·dS=I∂S
F·dr
Substitute in the given values and simplify the line integral to find the final
answer.
Question 2
Question
Let Sbe the surface of the cone given by the equation z=px2+y2, bounded
by the plane z= 3 and oriented upwards. Let Fbe the vector field defined by
F(x, y, z) = (z2, x, y). Calculate the flux of Facross the surface Susing Stokes’
Theorem.
Solution
Let’s denote the boundary curve of the surface Sas C, defined by the intersection
of the cone and the plane z= 3. Then, using Stokes’ Theorem, we have:
Flux(F, S) = ZZ
S
curl(F)·ndS =I
C
F·dr
where curl(F)is the curl of the vector field F,nis the unit normal vector
to the surface S,dS is the area element on S, and dris the tangent vector to
the curve C.
Step 1: Find the curl of F The curl of Fis given by:
curl(F) = ∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
z2x y
=∂y
∂x −∂x
∂y i−∂z2
∂x −∂z2
∂y j+∂x
∂x −∂y
∂y k
= (−1−0)i−(0 −0)j+ (1 −1)k=−i
Step 2: Parameterize the boundary curve CThe intersection of the
cone and the plane z= 3 gives us the boundary curve Cas a circle of radius 3
centered at the origin. We can parameterize this circle as:
r(t) = (3 cos t, 3 sin t, 3),0≤t≤2π
2
Step 3: Calculate the line integral Substitute the parameterization r(t)
into Fto get F(r(t)):
F(r(t)) = (32,3 cos t, 3 sin t) = (9,3 cos t, 3 sin t)
Now we can calculate the line integral along C:
I
C
F·dr=Z2π
0
F(r(t)) ·r′(t)dt =Z2π
0
(9,3 cos t, 3 sin t)·(−3 sin t, 3 cos t, 0) dt
=Z2π
0−27 sin tcos t+ 9 cos tsin t dt = 0
Therefore, the flux of Facross the surface Sis 0.
Question 3
Question
Let
F(x, y, z) = (y2, z2, x2)be a vector field in R3. Let Sbe the surface defined
by z=x2+y2for 1≤z≤4. Use Stokes’ Theorem to evaluate the circulation
of
Faround the boundary of Sin the counterclockwise direction.
Solution
Step 1: First, we need to find the curl of
F. The curl of a vector field
F(x, y, z) =
(M, N, P )is given by:
∇ ×
F=
i j k
∂
∂x
∂
∂y
∂
∂z
M N P
In this case,
F(x, y, z)=(y2, z2, x2), so M=y2,N=z2, and P=x2. Calcu-
lating the curl gives:
∇ ×
F= (0,−2z, 2y)
Step 2: Next, we need to find the normal vector to the surface S. The
normal vector ˆnis the gradient of the scalar function z=x2+y2:
∇z= (2x, 2y, −1)
Step 3: Now, we can apply Stokes’ Theorem. Stokes’ Theorem states that
the circulation of a vector field
Faround a closed curve Cis equal to the surface
integral of the curl of
Fover the surface Sthat is bounded by C:
IC
F·dr =ZZS
(∇ ×
F)·ˆn dS
3
Step 4: Substituting our values, the circulation of
Faround the boundary
of Sis given by:
IC
F·dr =ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS
Step 5: Simplifying the dot product gives:
ZZS
(0,−2z, 2y)·(2x, 2y, −1) dS =ZZS
(−4zy + 4yx)dS
Step 6: The surface Sis the region in the xy-plane bounded by 1≤z≤4.
Thus, we can convert the surface integral to a double integral over the region in
the xy-plane:
ZZS
(−4zy + 4yx)dS =ZZR
(−4y(x2+y2)+4x(1)) dA
Step 7: Evaluating the double integral over the region Rgives the circulation
of
Faround the boundary of S.
Question 4
Question
Let Sbe the surface of the portion of the plane 2x+ 3y+z= 6 that lies in the
first octant, oriented upward, and let F= (x2, y2, z2). Use Stokes’ Theorem to
evaluate the surface integral RRSF·dS.
Solution
Step 1: Find the normal vector nto the surface S.
The equation of the given plane can be written as z= 6 −2x−3y. We can
express the normal vector as n=∂z
∂x ,∂z
∂y ,−1= (−2,−3,−1).
Step 2: Calculate the curl of F.
The curl of a vector field F= (P, Q, R)is given by
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F= (x2, y2, z2), we have P=x2,Q=y2, and R=z2. Calculating the
curl, we get
∇ × F= (0,0,0)
Step 3: Apply Stokes’ Theorem.
Stokes’ Theorem states that for a vector field Fand a surface Swith bound-
ary ∂S oriented by the outward unit normal, the surface integral of ∇×Fover
Sis equal to the line integral of Faround ∂S. Since the curl of Fis zero, the
surface integral over Sis zero.
Therefore, RRSF·dS= 0.
4
Question 5
Question
Let Sbe the surface of the region bounded by the paraboloid z=x2+y2and
the plane z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,
where F(x, y, z) = (y2, x, z)is a vector field.
Solution
Step 1: Determine the boundary curve Cof the surface Sin the xy-plane.
Since Sis bounded by the paraboloid z=x2+y2and the plane z= 4, the
boundary curve Ccan be parametrized by r(t) = (t, 2−t2), where −1≤t≤1.
Step 2: Calculate the curl of F:∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2x z
= (1,−1,0).
Step 3: Determine the unit normal vector nto the surface S. Since S
is given by z=g(x, y) = x2+y2and is below the plane z= 4, we have
n=∇g
|∇g|=⟨−2x,−2y,1⟩
√1+4(x2+y2).
Step 4: Calculate the line integral of Falong the boundary curve Cusing
Stokes’ Theorem:
ZC
F·dr=ZZS∇ × F·dS=ZZS
(1,−1,0) ·ndA
Step 5: Express the surface integral in terms of xand y:
ZZS
(1,−1,0) ·ndA =ZZD⟨−2x, −2y, 1⟩·⟨1,1,0⟩dA =ZZD
(−2x−2y)dA
Step 6: Integrate over the region Din the xy-plane:
ZZD
(−2x−2y)dA =Z1
−1Z1
−1
(−2x−2y)dydx = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 6
Question
Let
F(x, y, z) = x2
i+y2
j+z2
kbe a vector field, and let Sbe the part of the
paraboloid z= 4 −x2−y2that lies inside the cylinder x2+y2= 1. Use Stokes’
Theorem to evaluate the surface integral RRS(∇ ×
F)·d
S.
5
Solution
Step 1: Let’s first find the curl of
F.
∇ ×
F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y)
i−(2z−2x)
j+ (2y−2x)
k
Step 2: Now, find the unit normal vector to the surface S. Since Sis the
surface of the paraboloid, we have r(x, y) = x
i+y
j+(4−x2−y2)
k. The normal
vector is given by
N=∇r
||∇r||.
∇r =−2x
i−2y
j+
k
||∇r|| =p(−2x)2+ (−2y)2+ 1 = p4x2+ 4y2+ 1
So, the unit normal vector is
N=−2x
i−2y
j+
k
√4x2+4y2+1 .
Step 3: Substitute the curl and normal vector into the surface integral.
ZZS
(∇ ×
F)·d
S=ZZS
((2z−2y)
i−(2z−2x)
j+ (2y−2x)
k)·
NdS
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS
Step 4: Now evaluate the surface integral using Stokes’ Theorem. The
boundary curve of Sis the unit circle in the xy-plane. Calculate the line integral
along this curve and apply Stokes’ Theorem to relate the two. Apply
ZZS
(2z−2y)(−2x) + (2z−2x)(−2y) + (2y−2x)dS =IC
F·dr
Question 7
Question
Let Sbe the part of the surface x2+y2+z2= 9 that lies above the plane z= 0
and oriented upwards. Calculate the flux of the vector field F= (x2+y2)i+
zj+ykacross the surface S.
Solution
Step 1: Find the outward unit normal nto S. Since Sis the part of the surface
x2+y2+z2= 9 that lies above the plane z= 0, and oriented upwards, we
can write the equation of Sas z=p9−x2−y2. Then, the gradient of zis
∇z= (−x
√9−x2−y2,−y
√9−x2−y2,1). Hence, the unit normal vector to Sis given
by
n=∇z
∥∇z∥=1
q1 + x2+y2
9−x2−y2
(−x
p9−x2−y2,−y
p9−x2−y2,1)
6
Step 2: Calculate the curl of F. The curl of Fis given by
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2z y
= (∂y
∂y −∂z
∂z )i−(∂(x2+y2)
∂x −∂y
∂z )j+( ∂
∂x (x2+y2)−∂z
∂y )k
= 0i−0j+ 2yk= 2yk
Step 3: Apply Stokes’ Theorem. The flux of Facross Sis given by
ZZ
S
∇ × F·dS=ZZ
S
(2yk)·ndS
=ZZ
D
(2y)·1
q1 + x2+y2
9−x2−y2
dA
where Dis the projection of Sonto the xy-plane.
Step 4: Evaluate the flux integral. We can switch to polar coordinates by
letting x=rcos(θ)and y=rsin(θ). Thus, dA =rdrdθ, and the integral
becomes 2π
Z
0
3
Z
0
2rsin(θ)·1
q1 + r2
9
drdθ
This integral can be evaluated to find the flux of Facross the surface S.
Question 8
Question
Let Sbe the part of the paraboloid z=x2+y2that lies above the plane z= 1
oriented upwards. Use Stokes’ Theorem to evaluate the circulation of the vector
field F= (x2, y2, z2)around the curve Cwhere Cis the curve of intersection of
Sand the plane y=x.
Solution
Step 1: First, we need to parameterize the curve C:
Notice that y=xand z=x2+y2= 2x2. Therefore, a parameterization
of the curve Ccan be given by r(t) = (t, t, 2t2)where tvaries over the interval
[0,1].
Step 2: Next, we need to compute the curl of the vector field F:
The curl of Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (2z−2y, 2x−2z, 2y−2x)
7
Step 3: Now, we need to compute the surface normal to Sat each point
(x, y, z)lying on the paraboloid z=x2+y2oriented upwards. The surface
normal is given by (−∂z
∂x ,−∂z
∂y ,1).
Therefore, the surface normal is (−2x, −2y, 1).
Step 4: Now, we can compute the circulation of Faround the curve Cusing
Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·ndS
where nis the unit normal to the surface Sand dS is the surface element of
S.
Step 5: Substituting all the values we found into the equation for Stokes’
Theorem, we get:
IC
F·dr=ZZS
(2z−2y, 2x−2z, 2y−2x)·(−2x, −2y, 1) dS
Step 6: We can now simplify this expression and evaluate the double integral
to find the circulation of Faround the curve C.
Question 9
Question
Let Sbe the part of the plane z= 4 −x−ythat lies above the triangle in the
xy-plane with vertices at (0,0),(1,0), and (0,1). Calculate the surface integral
RRS∇ × F·dS, where F(x, y, z) = ⟨x2, y2, z2⟩.
Solution
Step 1: Find the outward unit normal to S.
The equation of the plane z= 4 −x−ycan be rewritten as x+y+z= 4, which
means the normal vector to the plane is n=⟨1,1,1⟩. To get the unit normal
vector, divide by the magnitude: nunit =
|n| = ⟨1,1,1⟩√12+12+12=1
√3⟨1,1,1⟩=⟨1
√3,1
√3,1
√3⟩.
Step 2: Calculate ∇ × F.
The curl of Fis ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
=⟨0,0,2y−2x⟩.
Step 3: Calculate the dot product ∇ × F·nunit.
∇ × F·nunit =⟨0,0,2y−2x⟩·⟨ 1
√3,1
√3,1
√3⟩=1
√3(2y−2x).
Step 4: Integrate over the surface S.
The area of the triangle in the xy-plane is 1
2, so we have:
RRS∇ × F·dS=RRD∇ × F·(nunit dS) = RRD∇ × F·nunit dA
=RRD
1
√3(2y−2x)dA
8
=1
√3R1
0R1−x
0(2y−2x)dy dx
=1
√3R1
0y2−2xy1−x
0dx
=1
√3R1
0(1 −x)2−2x(1 −x)dx
=1
√3R1
0(1 −2x+x2−2x+ 2x2)dx
=1
√3R1
0(1 −4x+ 3x2)dx
=1
√3x−2x2+x31
0
=1
√3(1 −2 + 1)
=1
√3
Question 10
Question
Let F(x, y, z) = (x2+y, y2+z, z2+x)be a vector field and let Sbe the surface
defined by z=px2+y2in the region 0≤z≤1. Use Stokes’ Theorem to
evaluate the line integral HCF·dr, where Cis the boundary curve of Soriented
counterclockwise when viewed from above.
Solution
Step 1: Calculate the curl of F. The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y y2+z z2+x
.
=∂(z2+x)
∂y −∂(y2+z)
∂z i−∂(z2+x)
∂x −∂(x2+y)
∂z j+∂(y2+z)
∂x −∂(x2+y)
∂y k.
= (0 −1)i−(2x−1)j+ (2z−1)k.
=−i−(2x−1)j+ (2z−1)k.
Step 2: Determine the unit normal vector to the surface S. The unit normal
vector to the surface Sis given by n=1
√1+( ∂z
∂x )2+( ∂z
∂y )2(−∂z
∂x ,−∂z
∂y ,1). Since
z=px2+y2, we have ∂z
∂x =x
√x2+y2and ∂z
∂y =y
√x2+y2. So, the unit normal
vector is n=1
√2(−x
√x2+y2,−y
√x2+y2,1).
Step 3: Calculate the surface integral of the curl of Fover the surface S.
The surface integral is given by
ZZS
(∇×F)·ndS =ZZS
(−1,−(2x−1),2z−1)·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
9
Since z=px2+y2, the surface integral simplifies to
ZZS
(−1,−(2x−1),2px2+y2−1) ·1
√2(−x
px2+y2,−y
px2+y2,1)dS.
The rest of the solution would involve further simplification and integration.
Question 11
Question
Let Sbe the part of the plane z= 3xinside the cylinder x2+y2= 4 oriented
upward. Use Stokes’ Theorem to evaluate HCF·dr, where F(x, y, z) = xyzi+
(x+y+z)j+xykand Cis the boundary of S, where Cis oriented counter-
clockwise when viewed from above.
Solution
Step 1: First, let’s find the curl of F:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
xyz x +y+z xy
= (y−z)i−(z)j+ (1 + x)k
Step 2: Next, we calculate the surface area of S. Since Sis inside the
cylinder x2+y2= 4 and the plane z= 3x, we can use polar coordinates for
the parametric representation of S:r(r, θ) = rcos(θ)i+rsin(θ)j+ 3rcos(θ)k
for (r, θ)∈[0,2] ×[0,2π].
The magnitude of the cross product |rr×rθ|gives us the surface area element,
which is |rr×rθ|=|rr×rθ|=p1 + (3 cos(θ))2·r=p1 + 9 cos2(θ)·r.
Step 3: Now we can set up the double integral to calculate the flux of the
curl of Fover the surface Susing Stokes’ Theorem:
IC
F·dr=ZZS
(∇ × F)·NdS
where Nis the outward unit normal to S. In this case, N=1
√1+9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k).
Step 4: Thus, we have:
IC
F·dr=ZZS
curl(F)·NdS
=Z2π
0Z2
0
((y−z)i−(z)j+ (1 + x)k)·1
p1 + 9 cos2(θ)(−3 cos(θ)i,−3 sin(θ)j,1k)p1 + 9 cos2(θ)drdθ
=Z2π
0Z2
0
(6 sin2(θ)−rsin(θ) + 3) drdθ
Now we can proceed with solving this double integral to find the flux.
10
Question 12
Question
Let Sbe the surface of the part of the paraboloid z= 4 −x2−y2that lies above
the xy-plane and oriented upwards. Let F= (x2, y2, z2)be a vector field in R3.
Calculate the surface integral RRSF·dSusing Stokes’ Theorem.
Solution
Step 1: Determine the boundary of the surface Son the xy-plane. The part
of the paraboloid z= 4 −x2−y2that lies above the xy-plane is the unit circle
centered at the origin. Thus, the boundary curve Cof Son the xy-plane is the
unit circle x2+y2= 1.
Step 2: Parameterize the boundary curve C. Let R(u) = (cos u, sin u, 0) for
0≤u≤2πbe a parameterization of the unit circle x2+y2= 1.
Step 3: Calculate the curl of the vector field F= (x2, y2, z2). The curl
∇ × Fis given by:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0,0,−2z)
Step 4: Applying Stokes’ Theorem, the surface integral RRSF·dSis equal to
the line integral RCF·dr, where dr=⟨dx, dy, dz⟩. Using the parameterization
R(u)for C:
ZC
F·dr=Z2π
0
(cos2u, sin2u, 0)T·(−sin u, cos u, 0) du
Step 5: Evaluating the dot product for each component:
ZC
F·dr=Z2π
0
(−cos2usin u+ sin2ucos u)du
ZC
F·dr=Z2π
0
1
2sin(2u)du = 0
Therefore, the surface integral RRSF·dSis equal to 0.
Question 13
Question
Let Sbe the portion of the plane z= 4 −x−yabove the region in the xy-
plane bounded by the curves y=x2and y= 2x, oriented with outward normal.
Compute the flux of the curl of the vector field F(x, y, z) = (−y, z, x)across the
surface S.
11
Solution
Step 1: Find the curl of the vector field F. The curl of a vector field F=
(P, Q, R)is given by the determinant:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
In this case, F= (−y, z, x), so P=−y,Q=z, and R=x. Calculating the
curl:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−y z x
= (∂x
∂y −∂z
∂z ,∂z
∂x −∂x
∂z ,∂y
∂z −∂y
∂x )
∇ × F= (0 −1,1−0,−1−(−1)) = (−1,1,0)
Step 2: Parameterize the surface S. Let x=t,y=t2, and z= 4 −t−t2.
Then the parametric equation of the surface Sis given by:
r(t) = (t, t2,4−t−t2)
Step 3: Calculate the normal vector to the surface. The normal vector to
the surface Sis given by the cross product of the partial derivatives of r(t)with
respect to t.
r′(t) = (dx
dt ,dy
dt ,dz
dt ) = (1,2t, −1)
The normal vector is then:
N=r′(t) = (1,2t, −1)
Step 4: Calculate the flux of the curl of Facross the surface S. The flux is
given by:
ZZS
(∇ × F)·NdS =ZZR
(∇ × F)·ˆ
NdA
where ˆ
Nis the unit normal vector to the surface. Since Nis already a unit
normal vector, ˆ
N=N= (1,2t, −1). The region Rin the xy-plane is the region
bounded by y=x2and y= 2x. Therefore, the limits of integration for xare
0≤x≤2, and for yare x2≤y≤2x.
Calculating the dot product and integrating over the region R:
ZZR
(∇ × F)·ˆ
NdA =ZZR
(−1,1,0) ·(1,2t, −1) dA
=ZZR−1+2t dA
Step
12
Question 14
Question
Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1
and z= 2. Use Stokes’ Theorem to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = (y2, z2, x2).
Solution
To use Stokes’ Theorem, we first need to find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂x2
∂y −∂z2
∂z i−∂x2
∂x −∂y2
∂z j+∂y2
∂x −∂y2
∂y k
= 0i−0j+ 2yk
= 2yk.
Now, we need to find the unit normal vector nto the surface S. Since the
cone z=px2+y2points outward, the normal vector npoints away from the
z-axis, so it is n=1
√2(x, y, −px2+y2).
The surface integral then becomes
ZZS
(∇ × F)·dS=ZZS
2yk·1
√2(x, y, −px2+y2)dS =ZZD
2y(−√2)dA,
where Drepresents the projection of Sonto the xy-plane.
Since the projection of Sonto the xy-plane is a circle of radius 1 centered
at the origin, we can use polar coordinates:
ZZD
2y(−√2)dA =Z2π
0Z1
0
2rsin(θ)(−√2)rdrdθ =−2√2Z2π
0Z1
0
r2sin(θ)drdθ.
Integrating with respect to rfirst, we get
Z1
0
r2sin(θ)dr =r3
31
0
sin(θ) = sin(θ)
3,
so the surface integral becomes
−2√2Z2π
0
sin(θ)
3dθ =−2√2
3Z2π
0
sin(θ)dθ = 0.
Therefore, the surface integral RRS(∇ × F)·dSis 0.
13
Question 15
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤1.
Use Stokes’ Theorem to evaluate RRScurl F ·dS, where F(x, y, z) = (x2+
y2, ez, z).
Solution
Step 1: Compute the curl of F, which is given by
curl F =∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+y2ezz
.
This simplifies to
curl F = (0 −1,1−0,2x+ 0) = (−1,1,2x).
Step 2: Find the unit normal vector to the surface S. Let
g(x, y, z) = z−(x2+y2).
Then, ∇g= (−2x, −2y, 1) is parallel to the normal vector to S. Normalizing,
we find the unit normal vector to be
n=1
p1 + 4(x2+y2)(−2x, −2y, 1).
Step 3: Calculate the outward-pointing unit normal vector dSon the surface
S:
dS=ndS =1
p1 + 4(x2+y2)(−2x, −2y, 1) dxdy.
Step 4: Stoke’s Theorem states that
ZZS
curl F ·dS=ZZD
(curl F)·ndA,
where Dis the projection of Sonto the xy-plane.
Step 5: The projection of Sonto the xy-plane is the disc x2+y2≤1. To
evaluate the double integral, we need to find (curl F)·n.
(curl F)·n= (−1,1,2x)·1
p1 + 4(x2+y2)(−2x, −2y, 1).
This simplifies to
−2x+1
p1 + 4(x2+y2).
Step 6: Finally, we can evaluate the given integral using the double integral
over the disc x2+y2≤1:
ZZD
(−2x+1
p1 + 4(x2+y2))dA.
14
Question 16
Question
Let Sbe the portion of the sphere x2+y2+z2= 4 with z≥0and z≤
p4−x2−y2. Use Stokes’ Theorem to evaluate the surface integral of F=
(2y, x, −z)over S.
Solution
Step 1: Compute the curl of F.
Curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2y x −z
= (0,0,2)
Step 2: Compute the unit normal vector to S. Since Sis the portion of the
sphere x2+y2+z2= 4 in the upper half plane, the outward unit normal vector
is n=∇g
||∇g||, where g(x, y, z) = z−p4−x2−y2.
∇g= (−x
p4−x2−y2,−y
p4−x2−y2,1)
||∇g|| =sx2+y2
4−x2−y2+ 1
So, the unit normal vector is
n=(−x, −y, 4−x2−y2)
px2+y2+ 4 −x2−y2= (−x
p4−x2−y2,−y
p4−x2−y2,2
p4−x2−y2)
Step 3: Compute the surface integral of Fover Susing Stokes’ Theorem.
ZZS
F·dS=ZZS
F·ndS =ZZD
Curl(F)·ndA
Where Dis the projection of Sonto the xy-plane. Since Curl(F)·n= (0,0,2) ·
(−x
√4−x2−y2,−y
√4−x2−y2,2
√4−x2−y2) = 4
√4−x2−y2, we have
ZZS
F·dS=ZZD
4
p4−x2−y2dA
Step 4: Switch to polar coordinates. In polar coordinates, the domain Dis
0≤r≤2and 0≤θ≤2π. The Jacobian is r, so dA =rdrdθ.
Step 5: Evaluate the integral.
ZZS
F·dS=Z2π
0Z2
0
4
√4−r2·r drdθ
This can be computed by first integrating with respect to r, which gives
R2π
0[4√4−r2]2
0dθ.
Further computation yields the value of the surface integral.
15
Question 17
Question
Let Sbe the part of the cone z=px2+y2that lies below the plane z= 3,
with outward orientation. Evaluate the surface integral HSF·dS, where F=
(2x, y, z2), using Stokes’ Thejsonhirbn.
Solution
Step 1: First, let’s find the normal vector to the surface S. The unit normal
vector is given by
n=∇g
||∇g||
where g(x, y, z) = z−px2+y2is a scalar function defining the surface. Thus,
∇g=∂g
∂x ,∂g
∂y ,∂g
∂z = −x
px2+y2,−y
px2+y2,1!
and
||∇g|| =v
u
u
t −x
px2+y2!2
+ −y
px2+y2!2
+ 12=√2.
Therefore, the unit normal vector to the surface Sis
n= −x
p2(x2+y2),−y
p2(x2+y2),1
√2!.
Step 2: The surface integral HSF·dScan be evaluated using Stokes’ Theorem,
which states that IS
F·dS=ZZR
(∇ × F)·ndA
where Ris the projection of the surface Sonto the xy-plane.
Step 3: Compute the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
2x y z2
= (0,−2z, −1)
Step 4: Since npoints upward for the cone and z= 3 on the cone, we have
n= (0,0,1).
Step 5: The projection of the surface Sonto the xy-plane is the disk x2+y2≤
9. Thus, the surface integral becomes
ZZx2+y2≤9
(0,−2z, −1) ·(0,0,1) dA =−ZZx2+y2≤9
dA =−π32=−9π.
Therefore, the value of the surface integral is −9π.
16
Question 18
Question
Let Cbe the curve in the plane given by y=x2+ 1, where xranges from 0
to 2. Let Sbe the surface of the part of the paraboloid z=x2+y2that lies
above C. Use Stokes’ Theorem to evaluate the line integral HCF·dr, where
F(x, y, z) = (x+y)i+xzj+yk.
Solution
Step 1: First, we parameterize the curve C. Since y=x2+ 1, the parameteri-
zation of Cis given by r(t) = (t, t2+ 1,0) where 0≤t≤2.
Step 2: Calculate the curl of F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x+y xz y
= (0 −z)i−(1)j+ (1 −x)k=−zi−j+ (1 −x)k
Step 3: Calculate the surface integral of ∇ × Fover the surface Susing
Stokes’ Theorem: ZZS
(∇ × F)·ndS =IC
F·dr
where nis the unit normal to the surface.
Step 4: Compute the unit normal vector to the surface S: Since the surface
is the part of the paraboloid z=x2+y2lying above C, the unit normal vector to
the surface pointing upwards is given by n=−∇h
|∇h|where h(x, y, z) = z−x2−y2.
∇h= (−2x, −2y, 1) ⇒ |∇h|=p4x2+ 4y2+ 1
So, n=2x
√4x2+4y2+1 ,2y
√4x2+4y2+1 ,−1
√4x2+4y2+1 .
Step 5: Calculate the line integral over the curve C:
IC
F·dr=ZZS
(∇ × F)·ndS
=Z2
0Zt2+1
t2
(−t) 2t
p4t2+ 4(t2+1)+1!+(1−t2) 2(t2+ 1)
p4t2+ 4(t2+1)+1!dt
After calculating this double integral, we can find the value of the line inte-
gral using Stokes’ Theorem.
17
Question 19
Question
Let Sbe the part of the plane z= 4 −x−2ythat lies inside the cylinder
x2+y2= 1 oriented upward. Use Stokes’ theorem to evaluate the integral
ZZS
(∇ × F)·ˆn dS,
where F(x, y, z) = (y2z, xz, xyz).
Solution
Step 1: Determine the boundary curve of S.Since Slies inside the
cylinder x2+y2= 1, the boundary curve of Sis the circle Cin the xy-plane
given by x2+y2= 1.
Step 2: Determine the unit normal vector ˆnto the surface S.Since
Sis oriented upward, the direction of the unit normal vector ˆnto the surface S
points in the positive zdirection. Therefore, ˆn=⟨0,0,1⟩.
Step 3: Compute the curl of F.The curl of Fis given by
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z xz xyz
=−y−x−xz.
Step 4: Compute the surface integral using Stokes’ theorem. By
Stokes’ theorem, we have
ZZS
(∇ × F)·ˆn dS =IC
F·dr,
where Cis the boundary curve of S, which is the circle C:x2+y2= 1 oriented
counterclockwise.
Parametrize Cusing x= cos tand y= sin t, where 0≤t≤2π. Then,
dr=⟨−sin t, cos t, 0⟩dt.
Substitute the parametrization and Finto the line integral formula:
IC
F·dr
=Z2π
0⟨sin2tcos t, cos t, sin tcos t⟩ · ⟨−sin t, cos t, 0⟩dt
=Z2π
0
(−sin3tcos t+ cos t+ 0)dt
=Z2π
0
(−sin3tcos t+ cos t)dt
=1
4(sin4t) + sin t2π
0
=π
2.
18
Therefore, the value of the given integral is π
2.
Question 20
Question
Let Sbe the surface of the ellipsoid given by x2
a2+y2
b2+z2
c2= 1, oriented outward.
Use Stokes’ Theorem to evaluate
ZZS
F·dS
where F(x, y, z) = ⟨−z, x, y⟩.
Solution
Step 1: Compute the curl of F:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
−z x y
=⟨1,1,1⟩
Step 2: Determine the unit normal vector to S: The unit normal vector to
the ellipsoid Sis given by
n=±∇g
|∇g|=±⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 3: Calculate the outward unit normal vector n: We choose the positive
sign for the outward orientation, so
n=⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
Step 4: Calculate the surface area element dS: The surface area element dS
is given by
dS =|n|dA =p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
Step 5: Compute the normal component of F: Since n=⟨2x/a2,2y/b2,2z/c2⟩/p(2x/a2)2+ (2y/b2)2+ (2z/c2)2,
the normal component of Fis
F·n=⟨−z, x, y⟩ · ⟨2x/a2,2y/b2,2z/c2⟩
p(2x/a2)2+ (2y/b2)2+ (2z/c2)2=2x2
a2−z2+2y2
b2
19
Step 6: Express the given double integral in terms of dx dy: Since RRSF·dS=
RRSF·ndS, we have
ZZS
F·dS=ZZD
(F·n)p(2x/a2)2+ (2y/b2)2+ (2z/c2)2
abc dx dy
where Dis the projection of Sonto the xy-plane.
Step 7: Evaluate the double integral over D: Now we can evaluate the double
integral over Dusing the parameterization x=asin ϕcos θ,y=bsin ϕsin θ
where 0≤ϕ≤π,0≤θ≤2π.
Therefore, the final result is
Question 21
Question
Let Sbe the surface of the hemisphere x2+y2+z2= 1,z≥0. Calculate the
flux of the vector field F(x, y, z) = (xy, z, x)across Susing Stokes’ Theorem.
Solution
Step 1: Determine the boundary curve Cof the surface S. The boundary curve
Cof the hemisphere Sis the intersection of the hemisphere with the xy-plane.
This curve is a unit circle centered at the origin, given by x2+y2= 1.
Step 2: Parameterize the boundary curve C. We can parameterize the
boundary curve Cusing polar coordinates as x= cos t,y= sin t, where 0≤t≤
2π.
Step 3: Calculate the circulation of Faround C. The circulation of Faround
Ccan be calculated as:
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
Step 4: Simplify and evaluate the circulation integral.
IC
F·dr=Z2π
0
(cos tsin t, 0,cos t)·(−sin t, cos t)dt
=Z2π
0−cos tsin2t−cos2tdt
=Z2π
0−1
2sin(2t)−1 + cos(2t)
2dt
=−1
4cos(2t)−t
2−1
2sin(2t)2π
0
=−1
2π
20
Step 5: Apply Stokes’ Theorem. By Stokes’ Theorem, the flux of Facross
Sis equal to the circulation of Faround C, which is −1
2π. Thus, the flux of F
across Sis −1
2π.
Question 22
Question
Let Sbe the part of the surface z=x2+y2that lies above the disk x2+y2≤4
in the xy-plane, with upward orientation. Use Stoke’s Theorem to evaluate
RRS(∇ × F)·dSwhere F=⟨y2,−xz, xy⟩.
(You may assume that the outward unit normal to Sis n=⟨−2x, −2y, 1⟩.)
Solution
Step 1: We begin by calculating ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2−xz xy
=⟨0−z, 0−0,∂(xy)
∂x −∂(−xz)
∂y ⟩=⟨−z, 0, y +x⟩=⟨−z, 0, x +y⟩
Step 2: Next, we need to calculate the outward unit normal to S, denoted
by n, which is ⟨−2x, −2y, 1⟩. Therefore, dS=ndS =⟨−2x, −2y, 1⟩ · dSk=
−2x dSk.
Step 3: Now, let’s parameterize the surface Susing cylindrical coordinates.
Let x=rcos θ,y=rsin θ, where 0≤r≤2and 0≤θ≤2π. The normal
vector in cylindrical coordinates is ⟨−2rcos θ, −2rsin θ, 1⟩. We also have dS =
|−∇z|drdθ =|2r|drdθ = 2r drdθ.
Step 4: Now we evaluate the surface integral using Stokes’ Theorem:
ZZS
(∇ × F)·dS=ZZD
(∇ × F)·ndA
=ZZD⟨−z, 0, x +y⟩ · ⟨−2rcos θ, −2rsin θ, 1⟩ · 2r drdθ
=ZZD
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
0Z2
0
(2r2cos θ+ 2rsin θ)drdθ
=Z2π
02
3r3cos θ+r2sin θ
2
0
dθ
21
=Z2π
016
3cos θ+ 4 sin θdθ
=16
3sin θ−4 cos θ
2π
0
=64
3
Therefore, RRS(∇ × F)·dS=64
3.
Question 23
Question
Let Sbe the part of the paraboloid z=x2+y2that lies inside the cylinder
x2+y2= 1. Use Stokes’ Theorem to evaluate RRS(∇×F)·dS, where F(x, y, z) =
(y2, z2, x2).
Solution
Step 1: Compute the curl of F:
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2z2x2
=∂
∂y (x2)−∂
∂z (z2),∂
∂z (y2)−∂
∂x (x2),∂
∂x (z2)−∂
∂y (y2)
= (0 −2z, 2y−0,2x−0) = (−2z, 2y, 2x)
Step 2: Calculate the unit normal vector nto the surface S: Since S
lies on the paraboloid z=x2+y2, the unit normal vector is given by n=
∂z
∂x ,∂z
∂y ,−1= (2x, 2y, −1).
Step 3: Compute the dot product of ∇ × Fand n:
(∇ × F)·n= (−2z, 2y, 2x)·(2x, 2y, −1) = −4zx + 4zy −2x
Step 4: Find the magnitude of the projection of the curl onto the surface
S: Since Slies inside the cylinder x2+y2= 1,Scan be parameterized by
r(r, θ)=(rcos θ, r sin θ, r2), where 0≤r≤1and 0≤θ≤2π. Then, the
magnitude of the projection is given by:
ZZS
(∇ × F)·dS=ZZD−4r3cos θsin θ+ 4r3cos θsin θ−2r dA
=ZZD−2r dA
Step 5: Evaluate the double integral over the region D:
ZZD−2r dA =−2Z2π
0Z1
0
r dr dθ =−2Z2π
0r2
21
0
dθ
=−2Z2π
0
1
2dθ =−πZ2π
0
dθ =−π(2π−0) = −2π2
Therefore, RRS(∇ × F)·dS=−2π2.
22
Question 24
Question
Let Sbe the part of the plane z= 4 −2x−ythat lies above the square Rin the
xy-plane with vertices at (0,0),(1,0),(1,1), and (0,1). Use Stokes’ Theorem
to evaluate the surface integral
ZZS
(∇ × F)·dS,
where F(x, y, z) = x2i+xyzj+z2k.
Solution
Step 1: Find the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2xyz z2
=∂(z2)
∂y −∂(xyz)
∂z i−∂(z2)
∂x −∂(x2)
∂z j+∂(x2)
∂y −∂(z2)
∂x k
= (0 −0)i−(0 −0)j+ (0 −0)k=0
Step 2: Determine the orientation of Sand R. Since Sis above Rin the
xy-plane and the square Rgoes from (0,0) to (1,1) counterclockwise, we use
the upward-pointing normal vector for S,N=⟨2,−1,−1⟩.
Step 3: Evaluate the surface integral using Stokes’ Theorem.
ZZS
(∇ × F)·dS=ZZR
F·dr
where r(x, y) = xi+yj+ (4 −2x−y)k.
Step 4: Compute F·dr.
F·dr= (x2i+xyzj+z2k)·(dxi+dyj+dzk)
=x2dx +xydy +z2dz =x2dx +xydy + (4 −2x−y)2dz
Step 5: Parametrize the square R. Let x=uand y=v, where u, v ∈[0,1].
Then dz =−2dx −dy.
Step 6: Evaluate the integral over R.
Z1
0Z1
0
(u2du +uvdv + (4 −2u−v)2(−2du −dv))
=Z1
0Z1
0
((u2+ 4 −2u−v)2−2uv)dvdu
23
=Z1
0Z1
0
(u4+ 16 + v2−4u3−4u2v−4v+ 16u−4uv −2uv)dvdu
=Z1
0u4+ 16 + v+v2−4u3−16u2−4udvdu
=
Question 25
Question
Let Sbe the portion of the plane z= 1−x−ythat lies above the region bounded
by the parabola y=x2and the line y= 0. Let F(x, y, z) = x2i+y2j+z2k.
Find the flux of Facross the surface Sin the positive z-direction using Stokes’
Theorem.
Solution
Step 1: First, we need to find the normal vector to the surface S. Since Slies on
the plane z= 1−x−y, the normal vector to the surface Sis n=⟨∂z
∂x ,∂z
∂y ,−1⟩=
⟨−1,−1,−1⟩.
Step 2: Next, we need to find the curl of the vector field F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
= (0 −0) i−(0 −0) j+ (2y−2x)k= (2y−2x)k
Step 3: Now, we can find the flux of Facross the surface Susing Stokes’
Theorem:
Flux =ZZ
S
(∇ × F)·dS=ZZ
D
(∇ × F)·ndA
where Dis the projection of Sonto the xy-plane.
Step 4: The projection of the surface Sonto the xy-plane is the region
bounded by the parabola y=x2and the line y= 0, which can be described as
D={(x, y)|0≤x≤1,0≤y≤x2}.
Step 5: Evaluating the flux integral, we have:
Flux =ZZ
D
(2y−2x)· ⟨−1,−1,−1⟩ · dA =ZZ
D
(−2y+ 2x)dA
Step 6: Changing to polar coordinates for D, we have:
Flux =Z1
0Zx2
0
(−2rsin(θ)+2rcos(θ))rdrdθ
24
Step 7: Integrating with respect to rand θ, we get:
Flux =Z1
0−2r3sin(θ)
3+2r3cos(θ)
3x2
0
dθ
Step 8: Simplifying further:
Flux =Zπ
2
0−2x6sin(θ)
3+2x6cos(θ)
3dθ =2
3x6 Zπ
2
0
cos(θ)dθ −Zπ
2
0
sin(θ)dθ!
Step 9: Finally, integrating cos(θ)and sin(θ)over [0,π
2]gives:
Flux =
25
