MATH 332 - ADVANCED CALCULUS Stokes' Theorem Question Bank Set 4

MATH 332 - ADVANCED CALCULUS

- Stokes’ Theorem

Question Bank - Set 4

Liberty University

Question 1

Question

Let Sbe the part of the plane z= 4 −x−2ythat lies above the triangle with

vertices (0, 0, 0), (2, 0, 0), and (0, 1, 0) oriented counterclockwise when viewed

from above. Let F= (x2+ 2y, y2, z2)be a vector field. Calculate the surface

integral RRS∇ × F·dSusing Stokes’ Theorem.

Solution

Step 1: Find the normal vector to the surface S.

The normal vector to the surface Sis given by n=∇h(x, y, z), where h(x, y, z) =

4−x−2yis the height function for the plane z= 4 −x−2y. Thus, we have

∇h(x, y, z) = (−∂h

∂x,−∂h

∂y ,1) = (−1,−2,1).

Step 2: Calculate ∇ × F.

We have ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+ 2y y2z2



=∂(z2)

∂y −∂(y2)

∂z i−∂(z2)

∂x −∂(x2+2y)

∂z j+∂(y2)

∂x −∂(x2+2y)

∂y k

= (0 −0)i−(0 −0)j+ (0 −1)k=−k.

Step 3: Calculate the boundary of the surface S.

The boundary of the surface Sis the triangle in the xy-plane with vertices (0,

0), (2, 0), and (0, 1). Let Cbe the curve representing the boundary where C1

is the line segment from (0, 0) to (2, 0), C2is the line segment from (2, 0) to

(0, 1), and C3is the line segment from (0, 1) to (0, 0).

Step 4: Calculate the circulation of ∇ × Faround the boundary of S.

The circulation of ∇ × Faround the boundary Cis given by

IC∇ × F·dr=ZZS∇ × F·dS

=IC1∇ × F·dr+IC2∇ × F·dr+IC3∇ × F·dr

=Z2

F(x, 0) ·T1dx +Z1

F(2, y)·T2dy +Z0

F(x, 1) ·T3dx

=Z2

(x2,0,(4

Question 2

Question

Let Sbe the part of the plane z=x+ythat lies above the square 0≤x≤

1,0≤y≤1. Use Stokes’ Theorem to evaluate the surface integral

ZZS

(∇ × F)·dS

where F(x, y, z) = (z2, y2, x2).

Solution

Step 1: To apply Stokes’ Theorem, we first need to compute the curl of F.

∇×F=



i j k

∂

∂x

∂

∂y

∂

∂z

z2y2x2



=∂

∂y (x2)−∂

∂z (y2),∂

∂z (z2)−∂

∂x(x2),∂

∂x(y2)−∂

∂y (z2)

= (−2y, 2z, 2x)

Step 2: The outward unit normal vector to the surface Scan be calculated

as n= ( ∂z

∂x ,∂z

∂y ,−1) = (1,1,−1).

Step 3: The surface integral can be rewritten as

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·ndA

where Dis the projection of Sonto the xy-plane, that is, the square 0≤x≤

1,0≤y≤1.

Step 4: Substituting the values of ∇×Fand n, the surface integral becomes

ZZD

((−2y, 2z, 2x)) ·(1,1,−1) dA =ZZD

(−2y+ 2z−2x)dA

Step 5: Now, we evaluate the double integral over the square D:

ZZD

(−2y+ 2z−2x)dA =Z1

0Z1

(−2y+ 2(x+y)−2x)dy dx

=Z1

0−2Z1

y dy + 2 Z1

(x+y)dy −2Z1

x dydx

Step 6: Solving the inner integrals and simplifying gives:

=Z1

(−1 + 2(x+1

2)−2x)dx =Z1

(1 −2x)dx =x−x21

0= 1 −1 = 0

Therefore, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 3

Question

Let Cbe the curve defined by x=t2,y=t3,z=t4for 0≤t≤1. Let Sbe the

surface bounded by Cand the plane z= 1. Use Stokes’ Theorem to evaluate

HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).

Solution

1. First, parameterize the curve C:

The parameterization of Cis given by r(t) = (t2, t3, t4)for 0≤t≤1.

2. Calculate the derivative of r(t):

r′(t) = (2t, 3t2,4t3).

3. Compute the curl of F:

The curl of Fis given by ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



Simplifying, we get ∇ × F= (−1,−1,−1).

4. Calculate the line integral using Stokes’ Theorem:

By Stokes’ Theorem, HCF·dr=RRS(∇ × F)·dS, where Sis the surface

bounded by Cand z= 1.

Since ∇ × F= (−1,−1,−1), the line integral simplifies to HCF·dr=

RRS(−1,−1,−1) ·ndS, where nis the normal vector to the surface S.

5. Determine the normal vector to the surface S:

The normal vector to the surface Sis n=±(−∂z

∂x ,−∂z

∂y ,1) = ±(0,0,1).

As the normal vector should point outward from the bounded region, we

choose n= (0,0,1).

6. Compute the surface integral:

RRS(−1,−1,−1) ·(0,0,1) dS =−RRSdS =−RRR||ru×rv||dA, where R

is the parameterization region of the surface S.

As r(u, v) = (u, v, 1) and ∂r

∂u ×∂r

∂v = (−1,0,0), the surface integral simpli-

fies to −RRR1dA.

It follows that HCF·dr=−A, where Ais the area of the region R.

Question 4

Question

Let Sbe the part of the paraboloid z=x2+y2that lies below the plane

z= 1. Use Stokes’ Theorem to evaluate the circulation of the vector field

F=⟨yez, zex, xey⟩around the boundary curve of S, oriented counterclockwise

when viewed from above.

Solution

Step 1: First, let’s find the curl of F. The curl of a vector field F=⟨P, Q, R⟩

is defined as ∇ × F=∂R

∂y −∂Q

∂z i+∂P

∂z −∂R

∂x j+∂Q

∂x −∂P

∂y k. So, for our

vector field F=⟨yez, zex, xey⟩, we have:

∇×F=∂(xey)

∂y −∂(zex)

∂z i+∂(yez)

∂z −∂(xey)

∂x j+∂(zex)

∂x −∂(yez)

∂y k

= (xey−zex)i+ (yez−xey)j+ (zex−yez)k

Step 2: Next, we will find the outward unit normal vector to the surface S.

Since Sis the part of the paraboloid z=x2+y2below the plane z= 1, the unit

normal vector on Sis given by N=∇(z−x2−y2)

||∇(z−x2−y2)|| =−2xi

−2yj+kp1+4x2+ 4y2.

Step 3: Now, we can apply Stokes’ Theorem, which states that HCF·dr=

RRS(∇ × F)·NdS. Let’s first calculate RRS(∇ × F)·NdS:

ZZS

(∇×F)·NdS =ZZS

((xey−zex)i+(yez−xey)j+(zex−yez)k)·−2xi

−2yj+kp1+4x2+ 4y2dS

Step 4: To evaluate the surface integral, we need to find a parameterization

of S. Let’s use the parameterization r(u, v) = ⟨u, v, u2+v2⟩for 0≤u≤1and

−√1−u2≤v≤√1−u2. The normal vector to the surface can be calculated

as the cross product of ruand rv.

Step 5: After parameterizing S, we can compute the surface integral over S

using the parameterization. Finally, we substitute the results back into Stokes’

Theorem to find the circulation of Faround the boundary curve of S.

Question 5

Question

Let Sbe the surface given by z=x2+y2for 0≤z≤1oriented with outward

normal. Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS,

where F(x, y, z) = (x2+y2,2xy, z).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y22xy z



= (0,0,2y−2x)

Step 2: Determine the unit normal vector nto the surface S. Since Sis

given by z=x2+y2, the normal vector can be written as:

n=∇z

|∇z|=⟨−2x, −2y, 1⟩

p1+4x2+ 4y2

Step 3: Calculate the outward unit normal vector dS. The outward unit

normal vector dSis given by dS=ndS =−2x

√1+4x2+4y2,−2y

√1+4x2+4y2,1

√1+4x2+4y2dA

Step 4: Setup the surface integral using Stokes’ Theorem. Stokes’ Theorem

states that the surface integral of ∇ × Fover Sis equal to the line integral of

Falong the boundary of S, which in this case is the circle x2+y2= 1 in the

xy-plane.

Therefore, the surface integral RRS(∇ × F)·dScan be rewritten as the line

integral HCF·draround the boundary curve Cof S.

Step 5: Parameterize the curve C. Parameterize the circle Cin the xy-plane

with x= cos t,y= sin t,0≤t≤2π.

Step 6: Evaluate the line integral. Substitute the parameterization of the

curve into Fand the differential dr, then calculate F·dr.

Step 7: Compute the final result. Calculate the line integral over Cto find

the value of the surface integral using Stokes’ Theorem.

Question 6

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to calculate the flux of the curl of the vector

field F(x, y, z) = (xz, yz, x2)across the surface S, where the direction of positive

orientation is the outward normal.

Solution

1. First, we calculate the curl of the vector field F(x, y, z):

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

xz yz x2



=∂(x2)

∂y −∂(yz)

∂z i−∂(xz)

∂x −∂(x2)

∂z j+∂(yz)

∂x −∂(xz)

∂y k

= (0 −y)i−(z−0)j+ (z−z)k=−yi−zj

2. Then, we calculate the outward unit normal vector to the surface S, which

is given by

n=∇g

|∇g|=∇z

|∇z|=i+j+k

√12+ 12+ 12=1

√3(i+j+k)

3. The surface integral of the curl of Fover Scan be calculated using Stokes’

Theorem: ZZS

(∇ × F)·dS=IC

F·dr

=Z3

1Z2π

(−y, −z)·(1

√3cos(ϕ),1

√3sin(ϕ),1

√3)dϕdz

4. Performing the dot product and integrating, we get:

1Z2π

0−y

√3cos(ϕ)−z

√3sin(ϕ) + 1

√3dzdϕ

=Z3

1Z2π

0−z

√3sin(ϕ) + 1

√3dzdϕ

5. Further evaluating the integral, we obtain:

=Z3

1−2πz

√3dz =−2πz2

√33

=−2π32−1

√3=−2π8

√3

Therefore, the flux of the curl of the vector field F(x, y, z)across the surface

Sis −8π

√3.

Question 8

Question

Let Sbe the portion of the plane z=x+ythat lies inside the cylinder x2+y2=

1oriented with outward-pointing normal. Use Stokes’ Theorem to evaluate

RRSF·dS, where F=yzi+xzj+xyk.

Solution

Step 1: First, find the curl of F: Let F=Pi+Qj+Rk. Then, the curl of Fis

given by

curl(F) = ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



=∂R

∂y −∂Q

∂z i−∂R

∂x −∂P

∂z j+∂Q

∂x −∂P

∂y k

For F=yzi+xzj+xyk, we have P=yz,Q=xz,R=xy. Thus, the curl of

Fis

curl(F) = (z−z)i−(y−y)j+ (x−x)k=0

Step 2: By Stokes’ Theorem, we have

ZZS

F·dS=ZZ∂S

F·dr

Since the curl of Fis 0, the line integral vanishes. Therefore, RRSF·dS= 0.

Question 9

Question

Let Sbe the part of the paraboloid z= 4−x2−y2that lies above the plane z= 1.

Orient Swith outward normal. Use Stokes’ Theorem to evaluate HCF·drwhere

F(x, y, z) = xi+yj+zkand Cis the boundary of Soriented counterclockwise

when viewed from above.

Solution

Step 1: Find the curl of F. Step 2: Compute the surface integral of the curl

of Fover the region Sto apply Stokes’ Theorem. Step 3: Parameterize the

boundary curve Cand compute the line integral of Fover C.

Let’s start with these steps.

Question 10

Question

Let F=x2i+y2j+z2kbe a vector field, and let Sbe the surface enclosed by

the cylinder x2+y2= 1 and the planes z= 0 and z= 2. Use Stokes’ Theorem

to evaluate the line integral HCF·draround the boundary curve Cof S.

Solution

Step 1: First, we need to parameterize the boundary curve C, which consists of

the circle x2+y2= 1 on the plane z= 0 and the same circle on the plane z= 2.

We will parameterize the circle on the plane z= 0 as r1(t) = cos(t)i+ sin(t)j

for 0≤t≤2π. The circle on the plane z= 2 can be parameterized similarly as

r2(t) = cos(t)i+ sin(t)j+ 2k.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2y2z2



= (2z−2y)i−(2z−2x)j+ 0k= 2(z−y)i−2(z−x)j

Step 3: We can now apply Stokes’ Theorem to relate the line integral over

Cto the surface integral over S. Using Stokes’ Theorem, we have:

F·dr=ZZS

(∇ × F)·ndS

where nis the outward unit normal to S. Since the surface Sis bounded by

C, the outward unit normal npoints outward from the cylinder.

Step 4: We will calculate the surface integral over S:

ZZS

(∇ × F)·ndS =ZZS

(2(z−y)i−2(z−x)j)·kdS

Since the normal vector nis in the kdirection, the dot product simplifies

to −2(z−x). To find the limits of integration, we note that the projection of

Sonto the xy-plane is the circle x2+y2= 1. Thus, the limits for xand yare

x2+y2≤1, and for z, the limits are 0≤z≤2.

Step 5: We can now evaluate the surface integral:

ZZS

(∇ × F)·ndS

=Z2π

0Z1

0−2(2 −cos(t)) dr dθ

=−4Z2π

(2 −rcos(t))



0dθ

=−4Z2π

(2 −cos(t)−0) dθ

=−4·2π(2)

=−16π

Therefore, the line integral HCF·draround the boundary curve Cof Sis

−16π.

Question 11

Question

Let F(x, y, z) = (y2−z)i+ (z2−x)j+ (x2−y)kand let Sbe the portion of

the cone z=px2+y2that lies between the planes z= 1 and z= 3. Use

Stokes’ Theorem to evaluate HCF·dr, where Cis the counterclockwise-oriented

boundary curve of surface S.

Solution

Step 1: First, let’s parameterize the boundary curve Cof the surface S. The

boundary curve consists of two parts: the circle x2+y2= 1 at z= 1 and the

circle x2+y2= 9 at z= 3. We can parameterize these circles as follows: For

z= 1, we have the circle C1:x2+y2= 1. Parameterizing this circle gives us:

r1(t) = ⟨cos(t),sin(t),1⟩,where 0≤t≤2π.

For z= 3, we have the circle C2:x2+y2= 9. Parameterizing this circle gives

us:

r2(t) = ⟨3 cos(t),3 sin(t),3⟩,where 0≤t≤2π.

Step 2: Now, we need to compute the curl of F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y2−z z2−x x2−y



= (2y+ 1)i−(2z+ 1)j+ (2x+ 1)k.

Step 3: By Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS,

where nis the unit normal to S.

Step 4: To find the unit normal to S, we compute the gradient of z=

px2+y2as ∇z=⟨x

√x2+y2,y

√x2+y2,−1⟩.

Step 5: The unit normal to Sis then ˆ

n=∇z

||∇z|| =⟨x,y,−√x2+y2⟩

√x2+y2+1 .

Step 6: We can now evaluate the surface integral RRS(∇×F)·ndS using the

parameterizations of C1and C2from Step 1.

Question 12

Question

Let Sbe the surface of the portion of the plane z= 4 −x−2ythat lies above

the square 0≤x≤1,0≤y≤1, oriented upward. Calculate the flux of the

curl of the vector field F= 2xi+yj+zkacross S.

Solution

Step 1: Find the curl of the vector field F. The curl of a vector field F=

Pi+Qj+Rkis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



In this case, F= 2xi+yj+zk, so:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

2x y z



= (0 −0)i−(0 −1)j+ (1 −2)k=−j−k

Step 2: Use Stokes’ Theorem to find the flux of the curl of Facross S.

Stokes’ Theorem states:

ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary of the surface S.

Step 3: Find the boundary curve Cof the surface S. The boundary curve

Cof the surface Sis the square where 0≤x≤1,0≤y≤1. It consists of four

line segments: 1. y= 0,0≤x≤12. x= 1,0≤y≤13. y= 1,0≤x≤14.

x= 0,0≤y≤1

Step 4: Parameterize and calculate the line integral HCF·dr. We will

integrate over each of the four line segments of the boundary: 1. y= 0,0≤

x≤1:r(t) = ti,0≤t≤12. x= 1,0≤y≤1:r(t) = i+tj,0≤t≤13.

y= 1,0≤x≤1:r(t) = (1 −t)i+j,0≤t≤14. x= 0,0≤y≤1:r(t) = tj,

0≤t≤1

Step 5: Substitute the parameterization into Fand calculate the dot product

with dr. For each line segment, calculate: 1. HCF·dr=R1

0(2t, 0,4−t)·(1,0,0)dt

Question 13

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,

where F(x, y, z) = (z, x, y).

(Note: You may assume that a positively oriented unit normal vector to S

at each point is n=1

√2(−x, −y, √2).)

Solution

Step 1: Find the boundary curve Cof Swhere the orientation is consistent with

the orientation of Sgiven by the positively oriented unit normal vector n.

Since Slies between z= 1 and z= 4, the boundary curve Cconsists of the

intersection of the cone z=px2+y2and the planes z= 1 and z= 4. Solving

these equations, we get z= 1 and z= 4:1 = px2+y2and 4 = px2+y2.

Thus, the boundary curve Cis the circle x2+y2= 1.

Step 2: Parameterize the boundary curve C.

Let Cbe parameterized by r(t) = (cos t, sin t, 1) where 0≤t≤2π.

Step 3: Find the curl of F.

The curl of Fis given by ∇ × F=∂Fz

∂y −∂Fy

∂z ,∂Fx

∂z −∂Fz

∂x ,∂Fy

∂x −∂Fx

∂y .

Calculating the components of the curl, we have: ∇×F= (0 −1,1−0,1−0) =

(−1,1,1).

Step 4: Evaluate the line integral HCF·dr.

Using Stokes’ Theorem, we have: RRSF·dS=HCF·dr.

F·dr=Z2π

F(r(t)) ·r′(t)dt

Z2π

F(r(t)) ·r′(t)dt =Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt

Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt =Z2π

0−sin t+ cos2tdt

Z2π

0−sin t+ cos2tdt =−cos t+t

2+sin 2t

42π

= 0

Therefore, the surface integral RRSF·dSis 0.

Question 14

Question

Let Fbe a vector field defined by F(x, y, z) = (y2z+ cos x)i+ (xz2+ sin y)j+

(xy2+ez)k. Calculate the surface integral RRScurl(F)·dS, where Sis the

portion of the plane z= 4 −x−2ythat lies above the triangle in the xy-plane

with vertices at (0,0),(1,0), and (0,2).

Solution

Step 1: Calculate curl(F).

curl(F) = 



i j k

∂

∂x

∂

∂y

∂

∂z

y2z+ cos x xz2+ sin y xy2+ez



=∂(xy2+ez)

∂y −∂(xz2+ sin y)

∂z i−∂(y2z+ cos x)

∂x −∂(xy2+ez)

∂z j+∂(xz2+ sin y)

∂x −∂(y2z+ cos x)

∂y k

= (2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k

Step 2: Calculate the unit normal vector to the surface S. The surface S

is defined by z= 4 −x−2y. The normal vector to this surface is n=∇z=

−i−2j+k. Therefore, the unit normal vector is n=−1

√6i−2

√6j+1

√6k.

Step 3: Calculate the magnitude of dS. The magnitude of dSis given by

dS =|n|dxdy =q6

6dxdy =dxdy.

Step 4: Calculate the surface integral. The surface integral can be calculated

as:

ZZS

curl(F)·dS=ZZS(2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k·−1

√6i−2

√6j+1

√6kdxdy

Now, we substitute z= 4 −x−2yinto the expression above and integrate over

the region in the xy-plane:

ZZS

curl(F)·dS=ZZR(2xy −(4 −x−2y)2)−1

√6−(sin x−2xy)2

√6+ (2x(4 −x−2y)−y2)1

√6dxdy

where Ris the triangle in the xy-plane

Question 15

Question

Let Sbe the hemisphere x2+y2+z2= 1, z ≥0, and let F= (x2+y2, y2+

z2, z2+x2). Calculate the flux of Facross Susing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2+z2z2+x2



=∂(z2+x2)

∂y −∂(y2+z2)

∂z i−∂(z2+x2)

∂x −∂(x2+y2)

∂z j+∂(y2+z2)

∂x −∂(x2+y2)

∂y k

= (0 −2z)i−(2x−0)j+ (2y−0)k=−2zi−2xj+ 2yk

Step 2: Calculate the surface integral of Fover Susing Stokes’ Theorem.

(∇ × F)·dS=I

∂S

F·dr

Step 3: The boundary of Sis the circle x2+y2= 1, z = 0.

r(t) = cos(t)i+ sin(t)j,0≤t≤2π

Step 4: Parameterize the curve C.

r′(t) = −sin(t)i+ cos(t)j

r′(t)×k=−cos(t)k

Step 5: Calculate the line integral.

∂S

F·dr=Z2π

F(r(t)) ·r′(t)×kdt

=Z2π

(−sin(t))2+ cos(t)2+ (−sin(t))2dt =Z2π

2dt = 4π

Therefore, the flux of Facross Sis 4π.

Question 16

Question

Let Sbe the part of the plane x+ 2y+ 3z= 6 that lies inside the cylinder

x2+y2= 4. Use Stokes’ Theorem to evaluate

F·dr,

where F=⟨y, x, z⟩and Cis the boundary of Soriented counterclockwise as

viewed from above.

Solution

Step 1: Find the boundary curve Cof the surface S. This is the intersection of

the plane and the cylinder, so we have

x+ 2y+ 3z= 6

x2+y2= 4.

Solving these equations, we find that the boundary curve Cis the circle x2+y2=

4, which is the circle of radius 2 centered at the origin.

Step 2: Parameterize the boundary curve C. Let x= 2 cos t,y= 2 sin t,

where 0≤t≤2π. Then dr=⟨−2 sin t, 2 cos t, 0⟩dt.

Step 3: Compute F(r(t)).

F(r(t)) = ⟨2 sin t, 2 cos t, 2 sin t⟩.

Step 4: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y x z



=⟨1,1,−2⟩.

Step 5: Compute the line integral using Stokes’ Theorem.

F·dr=ZZS

(∇ × F)·ndS,

where nis the outward unit normal to S. Since the outward normal for the

plane x+ 2y+ 3z= 6 is n=⟨1,2,3⟩/√12+ 22+ 32=⟨1/√14,2/√14,3/√14⟩,

we have IC

F·dr=ZZS

(∇ × F)·ndS

=ZZS⟨1,1,−2⟩ · 1

√14⟨1,2,3⟩dS

√14 ZZS

(1 + 2 −6) dS

=−3

√14 ZZS

=−3

√14 ·Area(S).

Step 6: Find the area of the region S. This can be done by finding the

projection of Sonto the xy-plane and calculating the double integral over this

region.

Step 7: Calculate the final answer for the line integral.

Question 17

Question

Let Sbe the part of the plane z= 3 −x−ythat lies above the triangle with

vertices (0,0,0),(1,0,0), and (0,1,0), oriented counterclockwise when viewed

from above. Let Fbe the vector field F= (x2+y2, z, 2x−z). Calculate the

surface integral R RSF·dSusing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

curl(F) = ∂(2x−z)

∂y −∂(z)

∂z ,∂(x2+y2)

∂z −∂(2x−z)

∂x ,∂(z)

∂x −∂(x2+y2)

∂y 

= (−1−0,0−2,0−2) = (−1,−2,−2).

Step 2: Calculate the normal vector to the surface S. The normal vector to

the plane z= 3 −x−yis n= (−∂z

∂x ,−∂z

∂y ,1) = (−(−1),−(−1),1) = (1,1,1).

Step 3: Normalize the normal vector. The unit normal vector is n0=

|n| = (1, 1, 1)√12+12+12=(1

√3,1

√3).

Step 4: Calculate the surface integral using Stokes’ Theorem. Stokes’ Theo-

rem states that R RSF·dS=RRRVcurl(F)·ndV , where Vis the region enclosed

by S. Since Sis the part of the plane z= 3 −x−ythat lies above the given

triangle, the region Vis the triangle 0≤x≤1,0≤y≤1−x, 0≤z≤3−x−y.

The surface integral becomes:

Z ZS

F·dS=ZZZV

(−1,−2,−2) ·(1,1,1) dV

=ZZZV−5dV.

Step 5: Calculate the triple integral over region V. We have 0≤x≤1,0≤

y≤1−x, 0≤z≤3−x−y. Therefore, the triple integral becomes:

0Z1−x

0Z3−x−y

0−5dz dy dx

=Z1

0Z1−x

[−5z]3−x−y

0dy dx

=Z1

0Z1−x

0−5(3 −x−y)dy dx

=Z1

[−5(3y−xy −y2

2)]1−x

0dx

=Z1

0−53(1 −x)−x(1 −x)−(1 −x)2

2dx

=Z1

0−17

2x2+ 14x−15 dx

Question 18

Question

Let Sbe the part of the plane x+y+z= 1 that lies in the first octant.

Let F(x, y, z) = ⟨2xz, y2, x +z⟩. Use Stokes’ Theorem to evaluate the surface

integral

ZZS

curl F ·ndS

where nis the outward unit normal to S.

Solution

Step 1: First, we need to find the curl of F.

curl F=



i j k

∂

∂x

∂

∂y

∂

∂z

2xz y2x+z



=∂

∂y (x+z)−∂

∂z (y2),∂

∂z (2xz)−∂

∂x(x+z),∂

∂x(y2)−∂

∂y (2xz)

= (1 −0,2x−1,2z)

= (1,2x−1,2z)

Step 2: Parameterize the surface S. Since Sis the part of the plane x+y+z=

1in the first octant, we can parameterize Sas:

r(u, v) = ⟨u, v, 1−u−v⟩,0≤u≤1,0≤v≤1−u

Step 3: Calculate the normal vector n. The normal vector can be found by

taking the cross product of the partial derivatives of rwith respect to uand v.

∂u =⟨1,0,−1⟩,

∂v =⟨0,1,−1⟩So, n=

∂u×

∂v =⟨1,1,1⟩

Step 4: Evaluate the surface integral. Using Stokes’ Theorem, we have:

ZZS

curl F·ndS =ZZD

(curl F)·(ru×rv)dA

where Dis the region in the uv-plane corresponding to the surface S.

=ZZD

(1,2u−1,2(1 −u−v)) ·(1,1,1) dA

=ZZD

(1 + 2u−1−2v)dA

=ZZD

(2u−2v)dA

=Z1

0Z1−u

(2u−2v)dv du

=Z1

02uv −v2v=1−u

v=0 du

=Z1

(2u−(1 −u)2)du

=Z1

(3u2−2u+ 1) du

=u3−u2+u1

= 1 −1+1

= 1

Therefore, the surface integral is equal to 1.

Question 20

Question

Let Sbe the part of the plane z= 4 −x−2ythat lies above the region in the

xy-plane bounded by the rectangle with vertices (0,0),(4,0),(4,3), and (0,3).

Use Stokes’ Theorem to evaluate the surface integral

ZZS

F·dS

where F= (y−z, x +z, x +y).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y−z x +z x +y



=∂

∂y (x+y)−∂

∂z (x+z)i−∂

∂x(y−z)−∂

∂z (y+z)j+∂

∂x(x+z)−∂

∂y (y+z)k

=i−j−k

Step 2: Determine the unit normal vector nto the surface S. Given that

Sis the plane z= 4 −x−2y, the unit normal vector to the surface Sis

n=∇z

|∇z|=D−∂z

∂x ,−∂z

∂y ,1E=⟨−(−1),−(−2),1⟩=⟨1,2,1⟩.

Step 3: Calculate the curl of F·n. The curl of F·nis ∇ × (F·n) =

∇ × (x−y−z) = ∇ × (−3) = 0.

Step 4: Apply Stokes’ Theorem. Stokes’ Theorem states that

ZZS

F·dS=ZC

F·dr

where Cis the boundary of the surface Soriented counterclockwise when viewed

from above.

The boundary of the region in the xy-plane is the rectangle with vertices

(0,0),(4,0),(4,3), and (0,3). Parameterize this rectangle using r(u, v) =

⟨u, v, 4−u−2v⟩where 0≤u≤4and 0≤v≤3.

The line integral RCF·drbecomes

(0 −2(4) + 4)dv +Z4

(3 + 4)du =−20 + 12 = −8

Therefore, the surface integral evaluates to −8.

Question 21

Question

Let Cbe the curve of intersection of the plane z= 0 and the cylinder x2+y2= 1

oriented counterclockwise when viewed from above. Let F= (y+z, x +z, x +y)

be a vector field. Calculate the flux of Facross the curve Cusing Stokes’

Theorem.

Solution

Step 1: Parameterize the curve C.

The curve Cis simply the circle x2+y2= 1 in the xy-plane. We can

parameterize this curve by r(t) = (cos t, sin t, 0) for 0≤t≤2π.

Step 2: Find the curl of F.

The curl of Fis given by ∇ × F=



iˆ

jˆ

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



= (1,1,1).

Step 3: Calculate the line integral of Fover C.

By Stokes’ Theorem, the flux of Facross Cis equal to the line integral of F

over the curve C, which is given by

F·dr=Z2π

F(r(t))·r′(t)dt =Z2π

(0,0,1)·(−sin t, cos t, 0)dt =Z2π

0dt = 0.

Therefore, the flux of Facross the curve Cis 0.

Question 22

Question

Let Sbe the surface given by z=x2−y2for 0≤x≤2and 0≤y≤2. Use

Stokes’ Theorem to evaluate the line integral

(x2+ 2y)dx + (z2−y)dy + (2x−2z)dz

where Cis the boundary curve of Swith counterclockwise orientation when

viewed from above.

Solution

Step 1: First, we need to parameterize the surface S.

The surface Sis given by z=x2−y2,0≤x≤2, and 0≤y≤2.

Let’s define the parameterization r(u, v) = ⟨u, v, u2−v2⟩, where (u, v)∈

D={(u, v)|0≤u≤2,0≤v≤2}.

Step 2: Next, we need to calculate the partial derivatives of rwith respect

to uand v.

∂r

∂u =⟨1,0,2u⟩and ∂r

∂v =⟨0,1,−2v⟩

Step 3: Now, we calculate the curl of the vector field F=⟨x2+ 2y, z2−

y, 2x−2z⟩.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+ 2y z2−y2x−2z



=⟨−1,0,2+2⟩=⟨−1,0,4⟩

Step 4: Using Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS

where nis the unit normal vector to the surface S.

Step 5: The surface integral RRS(∇ × F)·ndS simplifies to

ZZD⟨−1,0,4⟩ · ∂r

∂u ×∂r

∂v du dv

=ZZD



1 0 2u

0 1 −2v

−1 0 4



du dv

Step 6: Computing the determinant and performing the dot product gives

=ZZD

(4 + 4u)du dv

Step 7: Evaluating the double integral over the region D= [0,2]×[0,2] gives

the final answer.

Question 23

Question

Let Sbe the part of the cone z2=x2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to evaluate the surface integral RRS(∇×F)·dS,

where F(x, y, z) = (x2+y2)i+ (y2−z2)j+ (x2−y2)k.

Solution

To apply Stokes’ Theorem, we need to compute the curl of Fand the normal

vector nto the surface S.

Step 1: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2−z2x2−y2



=∂(x2−y2)

∂y −∂(y2−z2)

∂z i−∂(x2+y2)

∂x −∂(x2−y2)

∂z j+∂(x2+y2)

∂y −∂(y2−z2)

∂x k

=−2yi−2xj+ 2zk

Step 2: Find the unit normal vector n.The unit normal vector to the

surface Sis n=1

√2(−i−j+k)since the normal vector points outward.

Step 3: Evaluate the surface integral using Stokes’ Theorem. Stokes’

Theorem states that ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary curve of S.

The boundary curve Cof Sis a circle in the xy-plane with radius 3parametrized

by r(t) = 3 cos(t)i+ 3 sin(t)jfor 0≤t≤2π.

Therefore, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

((9 cos2(t) + 9 sin2(t))(−3 sin(t)) + (9 sin2(t)−9 cos2(t))(3 cos(t))) dt

=Z2π

(−27 sin(t) cos(t) + 27 sin(t) cos(t)) dt

=Z2π

0dt = 0

Hence, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 24

Question

Let Sbe the surface of the cone z=px2+y2bounded by the plane z= 3.

Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS, where

F(x, y, z) = xi+yj+zk.

Solution

Step 1: Calculate the curl of F: The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x y z



=∂z

∂y −∂y

∂z i−∂z

∂x −∂x

∂z j+∂y

∂x −∂x

∂y k

=i+j+k

Step 2: Find the unit normal vector to the surface S: The unit normal vector

to the surface Sis given by n=(−x,−y,√x2+y2)

√x2+y2+1 .

Step 3: Calculate the outward unit normal vector: Since the cone is upward-

facing, the outward unit normal vector is the negation of the unit normal vector

calculated in Step 2. Therefore, n=(x,y,−√x2+y2)

√x2+y2+1 .

Step 4: Calculate dS: The surface element dSis given by dS=ndS, where

dS =q1+(∂z

∂x )2+ ( ∂z

∂y )2dA =√2dA.

Step 5: Evaluate the surface integral:

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·n√2dA

=ZZD

(i+j+k)·(x, y, −px2+y2)

px2+y2+ 1

√2dA

=ZZD

(−z)√2dA

=−√2ZZD

zdA

The integral RRDzdA represents the flux of zthrough the region D, which

in this case is the projection of Sonto the xy-plane. However, the surface of

the cone Sdoes not intersect the xy-plane, making the flux integral through D

equal to zero. Hence, the surface integral RRS(∇ × F)·dSis also equal to 0.

Question 25

Question

Let Sbe the part of the plane z= 4 −x−ythat lies above the region in the

xy-plane bounded by the curve x2+y2= 1. Calculate the flux of the vector

field F(x, y, z) = ⟨yz, xz, xy⟩across the surface S.

Solution

Step 1: First, we find the unit normal vector to the surface S, which is given by

n=∂z

∂x ,∂z

∂y ,−1.

In this case, ∂z

∂x =−1,∂z

∂y =−1, so n=⟨−1,−1,−1⟩.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

yz xz xy



=⟨z−x, −z, y⟩.

Step 3: Now, we can apply Stokes’ theorem, which states:

ZZS

(∇ × F)·dS=IC

F·dr,

where Cis the boundary curve of the surface Sand dSis the outward-pointing

normal vector on S.

Step 4: Since Slies above the unit circle in the xy-plane, the boundary

curve Cis simply the unit circle x2+y2= 1. Therefore, we can parameterize

the curve Cas r(t) = ⟨cos t, sin t, 4−cos t−sin t⟩for 0≤t≤2π.

Step 5: Calculating the line integral HCF·dr, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

0⟨sin t(4 −cos t−sin t),cos t(4 −cos t−sin t),cos tsin t⟩ · ⟨−sin t, cos t, −sin t−cos t⟩dt

=Z2π

(−sin2t(4 −cos t−sin t)−cos2t(4 −cos t−sin t) + cos tsin t(sin t+ cos t)) dt

=Z2π

(−4 + 5 sin t+ 5 cos t)dt

= 0.

Step 6: Therefore, the flux of the vector field Facross the surface Sis zero.

Step 4: Calculate the circulation of ∇ × Faround the boundary of S.

The circulation of ∇ × Faround the boundary Cis given by

IC∇ × F·dr=ZZS∇ × F·dS

=IC1∇ × F·dr+IC2∇ × F·dr+IC3∇ × F·dr

=Z2

F(x, 0) ·T1dx +Z1

F(2, y)·T2dy +Z0

F(x, 1) ·T3dx

=Z2

(x2,0,(4

Question 2

Question

Let Sbe the part of the plane z=x+ythat lies above the square 0≤x≤

1,0≤y≤1. Use Stokes’ Theorem to evaluate the surface integral

ZZS

(∇ × F)·dS

where F(x, y, z) = (z2, y2, x2).

Solution

Step 1: To apply Stokes’ Theorem, we first need to compute the curl of F.

∇×F=



i j k

∂

∂x

∂

∂y

∂

∂z

z2y2x2



=∂

∂y (x2)−∂

∂z (y2),∂

∂z (z2)−∂

∂x(x2),∂

∂x(y2)−∂

∂y (z2)

= (−2y, 2z, 2x)

Step 2: The outward unit normal vector to the surface Scan be calculated

as n= ( ∂z

∂x ,∂z

∂y ,−1) = (1,1,−1).

Step 3: The surface integral can be rewritten as

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·ndA

where Dis the projection of Sonto the xy-plane, that is, the square 0≤x≤

1,0≤y≤1.

Step 4: Substituting the values of ∇×Fand n, the surface integral becomes

ZZD

((−2y, 2z, 2x)) ·(1,1,−1) dA =ZZD

(−2y+ 2z−2x)dA

Step 5: Now, we evaluate the double integral over the square D:

ZZD

(−2y+ 2z−2x)dA =Z1

0Z1

(−2y+ 2(x+y)−2x)dy dx

=Z1

0−2Z1

y dy + 2 Z1

(x+y)dy −2Z1

x dydx

Step 6: Solving the inner integrals and simplifying gives:

=Z1

(−1 + 2(x+1

2)−2x)dx =Z1

(1 −2x)dx =x−x21

0= 1 −1 = 0

Therefore, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 3

Question

Let Cbe the curve defined by x=t2,y=t3,z=t4for 0≤t≤1. Let Sbe the

surface bounded by Cand the plane z= 1. Use Stokes’ Theorem to evaluate

HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).

Solution

1. First, parameterize the curve C:

The parameterization of Cis given by r(t) = (t2, t3, t4)for 0≤t≤1.

2. Calculate the derivative of r(t):

r′(t) = (2t, 3t2,4t3).

3. Compute the curl of F:

The curl of Fis given by ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



Simplifying, we get ∇ × F= (−1,−1,−1).

4. Calculate the line integral using Stokes’ Theorem:

By Stokes’ Theorem, HCF·dr=RRS(∇ × F)·dS, where Sis the surface

bounded by Cand z= 1.

Since ∇ × F= (−1,−1,−1), the line integral simplifies to HCF·dr=

RRS(−1,−1,−1) ·ndS, where nis the normal vector to the surface S.

5. Determine the normal vector to the surface S:

The normal vector to the surface Sis n=±(−∂z

∂x ,−∂z

∂y ,1) = ±(0,0,1).

As the normal vector should point outward from the bounded region, we

choose n= (0,0,1).

6. Compute the surface integral:

RRS(−1,−1,−1) ·(0,0,1) dS =−RRSdS =−RRR||ru×rv||dA, where R

is the parameterization region of the surface S.

As r(u, v) = (u, v, 1) and ∂r

∂u ×∂r

∂v = (−1,0,0), the surface integral simpli-

fies to −RRR1dA.

It follows that HCF·dr=−A, where Ais the area of the region R.

Question 4

Question

Let Sbe the part of the paraboloid z=x2+y2that lies below the plane

z= 1. Use Stokes’ Theorem to evaluate the circulation of the vector field

F=⟨yez, zex, xey⟩around the boundary curve of S, oriented counterclockwise

when viewed from above.

Solution

Step 1: First, let’s find the curl of F. The curl of a vector field F=⟨P, Q, R⟩

is defined as ∇ × F=∂R

∂y −∂Q

∂z i+∂P

∂z −∂R

∂x j+∂Q

∂x −∂P

∂y k. So, for our

vector field F=⟨yez, zex, xey⟩, we have:

∇×F=∂(xey)

∂y −∂(zex)

∂z i+∂(yez)

∂z −∂(xey)

∂x j+∂(zex)

∂x −∂(yez)

∂y k

= (xey−zex)i+ (yez−xey)j+ (zex−yez)k

Step 2: Next, we will find the outward unit normal vector to the surface S.

Since Sis the part of the paraboloid z=x2+y2below the plane z= 1, the unit

normal vector on Sis given by N=∇(z−x2−y2)

||∇(z−x2−y2)|| =−2xi

−2yj+kp1+4x2+ 4y2.

Step 3: Now, we can apply Stokes’ Theorem, which states that HCF·dr=

RRS(∇ × F)·NdS. Let’s first calculate RRS(∇ × F)·NdS:

ZZS

(∇×F)·NdS =ZZS

((xey−zex)i+(yez−xey)j+(zex−yez)k)·−2xi

−2yj+kp1+4x2+ 4y2dS

Step 4: To evaluate the surface integral, we need to find a parameterization

of S. Let’s use the parameterization r(u, v) = ⟨u, v, u2+v2⟩for 0≤u≤1and

−√1−u2≤v≤√1−u2. The normal vector to the surface can be calculated

as the cross product of ruand rv.

Step 5: After parameterizing S, we can compute the surface integral over S

using the parameterization. Finally, we substitute the results back into Stokes’

Theorem to find the circulation of Faround the boundary curve of S.

Question 5

Question

Let Sbe the surface given by z=x2+y2for 0≤z≤1oriented with outward

normal. Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS,

where F(x, y, z) = (x2+y2,2xy, z).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y22xy z



= (0,0,2y−2x)

Step 2: Determine the unit normal vector nto the surface S. Since Sis

given by z=x2+y2, the normal vector can be written as:

n=∇z

|∇z|=⟨−2x, −2y, 1⟩

p1+4x2+ 4y2

Step 3: Calculate the outward unit normal vector dS. The outward unit

normal vector dSis given by dS=ndS =−2x

√1+4x2+4y2,−2y

√1+4x2+4y2,1

√1+4x2+4y2dA

Step 4: Setup the surface integral using Stokes’ Theorem. Stokes’ Theorem

states that the surface integral of ∇ × Fover Sis equal to the line integral of

Falong the boundary of S, which in this case is the circle x2+y2= 1 in the

xy-plane.

Therefore, the surface integral RRS(∇ × F)·dScan be rewritten as the line

integral HCF·draround the boundary curve Cof S.

Step 5: Parameterize the curve C. Parameterize the circle Cin the xy-plane

with x= cos t,y= sin t,0≤t≤2π.

Step 6: Evaluate the line integral. Substitute the parameterization of the

curve into Fand the differential dr, then calculate F·dr.

Step 7: Compute the final result. Calculate the line integral over Cto find

the value of the surface integral using Stokes’ Theorem.

Question 6

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to calculate the flux of the curl of the vector

field F(x, y, z) = (xz, yz, x2)across the surface S, where the direction of positive

orientation is the outward normal.

Solution

1. First, we calculate the curl of the vector field F(x, y, z):

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

xz yz x2



=∂(x2)

∂y −∂(yz)

∂z i−∂(xz)

∂x −∂(x2)

∂z j+∂(yz)

∂x −∂(xz)

∂y k

= (0 −y)i−(z−0)j+ (z−z)k=−yi−zj

2. Then, we calculate the outward unit normal vector to the surface S, which

is given by

n=∇g

|∇g|=∇z

|∇z|=i+j+k

√12+ 12+ 12=1

√3(i+j+k)

3. The surface integral of the curl of Fover Scan be calculated using Stokes’

Theorem: ZZS

(∇ × F)·dS=IC

F·dr

=Z3

1Z2π

(−y, −z)·(1

√3cos(ϕ),1

√3sin(ϕ),1

√3)dϕdz

4. Performing the dot product and integrating, we get:

1Z2π

0−y

√3cos(ϕ)−z

√3sin(ϕ) + 1

√3dzdϕ

=Z3

1Z2π

0−z

√3sin(ϕ) + 1

√3dzdϕ

5. Further evaluating the integral, we obtain:

=Z3

1−2πz

√3dz =−2πz2

√33

=−2π32−1

√3=−2π8

√3

Therefore, the flux of the curl of the vector field F(x, y, z)across the surface

Sis −8π

√3.

Question 8

Question

Let Sbe the portion of the plane z=x+ythat lies inside the cylinder x2+y2=

1oriented with outward-pointing normal. Use Stokes’ Theorem to evaluate

RRSF·dS, where F=yzi+xzj+xyk.

Solution

Step 1: First, find the curl of F: Let F=Pi+Qj+Rk. Then, the curl of Fis

given by

curl(F) = ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



=∂R

∂y −∂Q

∂z i−∂R

∂x −∂P

∂z j+∂Q

∂x −∂P

∂y k

For F=yzi+xzj+xyk, we have P=yz,Q=xz,R=xy. Thus, the curl of

Fis

curl(F) = (z−z)i−(y−y)j+ (x−x)k=0

Step 2: By Stokes’ Theorem, we have

ZZS

F·dS=ZZ∂S

F·dr

Since the curl of Fis 0, the line integral vanishes. Therefore, RRSF·dS= 0.

Question 9

Question

Let Sbe the part of the paraboloid z= 4−x2−y2that lies above the plane z= 1.

Orient Swith outward normal. Use Stokes’ Theorem to evaluate HCF·drwhere

F(x, y, z) = xi+yj+zkand Cis the boundary of Soriented counterclockwise

when viewed from above.

Solution

Step 1: Find the curl of F. Step 2: Compute the surface integral of the curl

of Fover the region Sto apply Stokes’ Theorem. Step 3: Parameterize the

boundary curve Cand compute the line integral of Fover C.

Let’s start with these steps.

Question 10

Question

Let F=x2i+y2j+z2kbe a vector field, and let Sbe the surface enclosed by

the cylinder x2+y2= 1 and the planes z= 0 and z= 2. Use Stokes’ Theorem

to evaluate the line integral HCF·draround the boundary curve Cof S.

Solution

Step 1: First, we need to parameterize the boundary curve C, which consists of

the circle x2+y2= 1 on the plane z= 0 and the same circle on the plane z= 2.

We will parameterize the circle on the plane z= 0 as r1(t) = cos(t)i+ sin(t)j

for 0≤t≤2π. The circle on the plane z= 2 can be parameterized similarly as

r2(t) = cos(t)i+ sin(t)j+ 2k.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2y2z2



= (2z−2y)i−(2z−2x)j+ 0k= 2(z−y)i−2(z−x)j

Step 3: We can now apply Stokes’ Theorem to relate the line integral over

Cto the surface integral over S. Using Stokes’ Theorem, we have:

F·dr=ZZS

(∇ × F)·ndS

where nis the outward unit normal to S. Since the surface Sis bounded by

C, the outward unit normal npoints outward from the cylinder.

Step 4: We will calculate the surface integral over S:

ZZS

(∇ × F)·ndS =ZZS

(2(z−y)i−2(z−x)j)·kdS

Since the normal vector nis in the kdirection, the dot product simplifies

to −2(z−x). To find the limits of integration, we note that the projection of

Sonto the xy-plane is the circle x2+y2= 1. Thus, the limits for xand yare

x2+y2≤1, and for z, the limits are 0≤z≤2.

Step 5: We can now evaluate the surface integral:

ZZS

(∇ × F)·ndS

=Z2π

0Z1

0−2(2 −cos(t)) dr dθ

=−4Z2π

(2 −rcos(t))



0dθ

=−4Z2π

(2 −cos(t)−0) dθ

=−4·2π(2)

=−16π

Therefore, the line integral HCF·draround the boundary curve Cof Sis

−16π.

Question 11

Question

Let F(x, y, z) = (y2−z)i+ (z2−x)j+ (x2−y)kand let Sbe the portion of

the cone z=px2+y2that lies between the planes z= 1 and z= 3. Use

Stokes’ Theorem to evaluate HCF·dr, where Cis the counterclockwise-oriented

boundary curve of surface S.

Solution

Step 1: First, let’s parameterize the boundary curve Cof the surface S. The

boundary curve consists of two parts: the circle x2+y2= 1 at z= 1 and the

circle x2+y2= 9 at z= 3. We can parameterize these circles as follows: For

z= 1, we have the circle C1:x2+y2= 1. Parameterizing this circle gives us:

r1(t) = ⟨cos(t),sin(t),1⟩,where 0≤t≤2π.

For z= 3, we have the circle C2:x2+y2= 9. Parameterizing this circle gives

us:

r2(t) = ⟨3 cos(t),3 sin(t),3⟩,where 0≤t≤2π.

Step 2: Now, we need to compute the curl of F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y2−z z2−x x2−y



= (2y+ 1)i−(2z+ 1)j+ (2x+ 1)k.

Step 3: By Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS,

where nis the unit normal to S.

Step 4: To find the unit normal to S, we compute the gradient of z=

px2+y2as ∇z=⟨x

√x2+y2,y

√x2+y2,−1⟩.

Step 5: The unit normal to Sis then ˆ

n=∇z

||∇z|| =⟨x,y,−√x2+y2⟩

√x2+y2+1 .

Step 6: We can now evaluate the surface integral RRS(∇×F)·ndS using the

parameterizations of C1and C2from Step 1.

Question 12

Question

Let Sbe the surface of the portion of the plane z= 4 −x−2ythat lies above

the square 0≤x≤1,0≤y≤1, oriented upward. Calculate the flux of the

curl of the vector field F= 2xi+yj+zkacross S.

Solution

Step 1: Find the curl of the vector field F. The curl of a vector field F=

Pi+Qj+Rkis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



In this case, F= 2xi+yj+zk, so:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

2x y z



= (0 −0)i−(0 −1)j+ (1 −2)k=−j−k

Step 2: Use Stokes’ Theorem to find the flux of the curl of Facross S.

Stokes’ Theorem states:

ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary of the surface S.

Step 3: Find the boundary curve Cof the surface S. The boundary curve

Cof the surface Sis the square where 0≤x≤1,0≤y≤1. It consists of four

line segments: 1. y= 0,0≤x≤12. x= 1,0≤y≤13. y= 1,0≤x≤14.

x= 0,0≤y≤1

Step 4: Parameterize and calculate the line integral HCF·dr. We will

integrate over each of the four line segments of the boundary: 1. y= 0,0≤

x≤1:r(t) = ti,0≤t≤12. x= 1,0≤y≤1:r(t) = i+tj,0≤t≤13.

y= 1,0≤x≤1:r(t) = (1 −t)i+j,0≤t≤14. x= 0,0≤y≤1:r(t) = tj,

0≤t≤1

Step 5: Substitute the parameterization into Fand calculate the dot product

with dr. For each line segment, calculate: 1. HCF·dr=R1

0(2t, 0,4−t)·(1,0,0)dt

Question 13

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,

where F(x, y, z) = (z, x, y).

(Note: You may assume that a positively oriented unit normal vector to S

at each point is n=1

√2(−x, −y, √2).)

Solution

Step 1: Find the boundary curve Cof Swhere the orientation is consistent with

the orientation of Sgiven by the positively oriented unit normal vector n.

Since Slies between z= 1 and z= 4, the boundary curve Cconsists of the

intersection of the cone z=px2+y2and the planes z= 1 and z= 4. Solving

these equations, we get z= 1 and z= 4:1 = px2+y2and 4 = px2+y2.

Thus, the boundary curve Cis the circle x2+y2= 1.

Step 2: Parameterize the boundary curve C.

Let Cbe parameterized by r(t) = (cos t, sin t, 1) where 0≤t≤2π.

Step 3: Find the curl of F.

The curl of Fis given by ∇ × F=∂Fz

∂y −∂Fy

∂z ,∂Fx

∂z −∂Fz

∂x ,∂Fy

∂x −∂Fx

∂y .

Calculating the components of the curl, we have: ∇×F= (0 −1,1−0,1−0) =

(−1,1,1).

Step 4: Evaluate the line integral HCF·dr.

Using Stokes’ Theorem, we have: RRSF·dS=HCF·dr.

F·dr=Z2π

F(r(t)) ·r′(t)dt

Z2π

F(r(t)) ·r′(t)dt =Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt

Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt =Z2π

0−sin t+ cos2tdt

Z2π

0−sin t+ cos2tdt =−cos t+t

2+sin 2t

42π

= 0

Therefore, the surface integral RRSF·dSis 0.

Question 14

Question

Let Fbe a vector field defined by F(x, y, z) = (y2z+ cos x)i+ (xz2+ sin y)j+

(xy2+ez)k. Calculate the surface integral RRScurl(F)·dS, where Sis the

portion of the plane z= 4 −x−2ythat lies above the triangle in the xy-plane

with vertices at (0,0),(1,0), and (0,2).

Solution

Step 1: Calculate curl(F).

curl(F) = 



i j k

∂

∂x

∂

∂y

∂

∂z

y2z+ cos x xz2+ sin y xy2+ez



=∂(xy2+ez)

∂y −∂(xz2+ sin y)

∂z i−∂(y2z+ cos x)

∂x −∂(xy2+ez)

∂z j+∂(xz2+ sin y)

∂x −∂(y2z+ cos x)

∂y k

= (2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k

Step 2: Calculate the unit normal vector to the surface S. The surface S

is defined by z= 4 −x−2y. The normal vector to this surface is n=∇z=

−i−2j+k. Therefore, the unit normal vector is n=−1

√6i−2

√6j+1

√6k.

Step 3: Calculate the magnitude of dS. The magnitude of dSis given by

dS =|n|dxdy =q6

6dxdy =dxdy.

Step 4: Calculate the surface integral. The surface integral can be calculated

as:

ZZS

curl(F)·dS=ZZS(2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k·−1

√6i−2

√6j+1

√6kdxdy

Now, we substitute z= 4 −x−2yinto the expression above and integrate over

the region in the xy-plane:

ZZS

curl(F)·dS=ZZR(2xy −(4 −x−2y)2)−1

√6−(sin x−2xy)2

√6+ (2x(4 −x−2y)−y2)1

√6dxdy

where Ris the triangle in the xy-plane

Question 15

Question

Let Sbe the hemisphere x2+y2+z2= 1, z ≥0, and let F= (x2+y2, y2+

z2, z2+x2). Calculate the flux of Facross Susing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2+z2z2+x2



=∂(z2+x2)

∂y −∂(y2+z2)

∂z i−∂(z2+x2)

∂x −∂(x2+y2)

∂z j+∂(y2+z2)

∂x −∂(x2+y2)

∂y k

= (0 −2z)i−(2x−0)j+ (2y−0)k=−2zi−2xj+ 2yk

Step 2: Calculate the surface integral of Fover Susing Stokes’ Theorem.

(∇ × F)·dS=I

∂S

F·dr

Step 3: The boundary of Sis the circle x2+y2= 1, z = 0.

r(t) = cos(t)i+ sin(t)j,0≤t≤2π

Step 4: Parameterize the curve C.

r′(t) = −sin(t)i+ cos(t)j

r′(t)×k=−cos(t)k

Step 5: Calculate the line integral.

∂S

F·dr=Z2π

F(r(t)) ·r′(t)×kdt

=Z2π

(−sin(t))2+ cos(t)2+ (−sin(t))2dt =Z2π

2dt = 4π

Therefore, the flux of Facross Sis 4π.

Question 16

Question

Let Sbe the part of the plane x+ 2y+ 3z= 6 that lies inside the cylinder

x2+y2= 4. Use Stokes’ Theorem to evaluate

F·dr,

where F=⟨y, x, z⟩and Cis the boundary of Soriented counterclockwise as

viewed from above.

Solution

Step 1: Find the boundary curve Cof the surface S. This is the intersection of

the plane and the cylinder, so we have

x+ 2y+ 3z= 6

x2+y2= 4.

Solving these equations, we find that the boundary curve Cis the circle x2+y2=

4, which is the circle of radius 2 centered at the origin.

Step 2: Parameterize the boundary curve C. Let x= 2 cos t,y= 2 sin t,

where 0≤t≤2π. Then dr=⟨−2 sin t, 2 cos t, 0⟩dt.

Step 3: Compute F(r(t)).

F(r(t)) = ⟨2 sin t, 2 cos t, 2 sin t⟩.

Step 4: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y x z



=⟨1,1,−2⟩.

Step 5: Compute the line integral using Stokes’ Theorem.

F·dr=ZZS

(∇ × F)·ndS,

where nis the outward unit normal to S. Since the outward normal for the

plane x+ 2y+ 3z= 6 is n=⟨1,2,3⟩/√12+ 22+ 32=⟨1/√14,2/√14,3/√14⟩,

we have IC

F·dr=ZZS

(∇ × F)·ndS

=ZZS⟨1,1,−2⟩ · 1

√14⟨1,2,3⟩dS

√14 ZZS

(1 + 2 −6) dS

=−3

√14 ZZS

=−3

√14 ·Area(S).

Step 6: Find the area of the region S. This can be done by finding the

projection of Sonto the xy-plane and calculating the double integral over this

region.

Step 7: Calculate the final answer for the line integral.

Question 17

Question

Let Sbe the part of the plane z= 3 −x−ythat lies above the triangle with

vertices (0,0,0),(1,0,0), and (0,1,0), oriented counterclockwise when viewed

from above. Let Fbe the vector field F= (x2+y2, z, 2x−z). Calculate the

surface integral R RSF·dSusing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

curl(F) = ∂(2x−z)

∂y −∂(z)

∂z ,∂(x2+y2)

∂z −∂(2x−z)

∂x ,∂(z)

∂x −∂(x2+y2)

∂y 

= (−1−0,0−2,0−2) = (−1,−2,−2).

Step 2: Calculate the normal vector to the surface S. The normal vector to

the plane z= 3 −x−yis n= (−∂z

∂x ,−∂z

∂y ,1) = (−(−1),−(−1),1) = (1,1,1).

Step 3: Normalize the normal vector. The unit normal vector is n0=

|n| = (1, 1, 1)√12+12+12=(1

√3,1

√3).

Step 4: Calculate the surface integral using Stokes’ Theorem. Stokes’ Theo-

rem states that R RSF·dS=RRRVcurl(F)·ndV , where Vis the region enclosed

by S. Since Sis the part of the plane z= 3 −x−ythat lies above the given

triangle, the region Vis the triangle 0≤x≤1,0≤y≤1−x, 0≤z≤3−x−y.

The surface integral becomes:

Z ZS

F·dS=ZZZV

(−1,−2,−2) ·(1,1,1) dV

=ZZZV−5dV.

Step 5: Calculate the triple integral over region V. We have 0≤x≤1,0≤

y≤1−x, 0≤z≤3−x−y. Therefore, the triple integral becomes:

0Z1−x

0Z3−x−y

0−5dz dy dx

=Z1

0Z1−x

[−5z]3−x−y

0dy dx

=Z1

0Z1−x

0−5(3 −x−y)dy dx

=Z1

[−5(3y−xy −y2

2)]1−x

0dx

=Z1

0−53(1 −x)−x(1 −x)−(1 −x)2

2dx

=Z1

0−17

2x2+ 14x−15 dx

Question 18

Question

Let Sbe the part of the plane x+y+z= 1 that lies in the first octant.

Let F(x, y, z) = ⟨2xz, y2, x +z⟩. Use Stokes’ Theorem to evaluate the surface

integral

ZZS

curl F ·ndS

where nis the outward unit normal to S.

Solution

Step 1: First, we need to find the curl of F.

curl F=



i j k

∂

∂x

∂

∂y

∂

∂z

2xz y2x+z



=∂

∂y (x+z)−∂

∂z (y2),∂

∂z (2xz)−∂

∂x(x+z),∂

∂x(y2)−∂

∂y (2xz)

= (1 −0,2x−1,2z)

= (1,2x−1,2z)

Step 2: Parameterize the surface S. Since Sis the part of the plane x+y+z=

1in the first octant, we can parameterize Sas:

r(u, v) = ⟨u, v, 1−u−v⟩,0≤u≤1,0≤v≤1−u

Step 3: Calculate the normal vector n. The normal vector can be found by

taking the cross product of the partial derivatives of rwith respect to uand v.

∂u =⟨1,0,−1⟩,

∂v =⟨0,1,−1⟩So, n=

∂u×

∂v =⟨1,1,1⟩

Step 4: Evaluate the surface integral. Using Stokes’ Theorem, we have:

ZZS

curl F·ndS =ZZD

(curl F)·(ru×rv)dA

where Dis the region in the uv-plane corresponding to the surface S.

=ZZD

(1,2u−1,2(1 −u−v)) ·(1,1,1) dA

=ZZD

(1 + 2u−1−2v)dA

=ZZD

(2u−2v)dA

=Z1

0Z1−u

(2u−2v)dv du

=Z1

02uv −v2v=1−u

v=0 du

=Z1

(2u−(1 −u)2)du

=Z1

(3u2−2u+ 1) du

=u3−u2+u1

= 1 −1+1

= 1

Therefore, the surface integral is equal to 1.

Question 20

Question

Let Sbe the part of the plane z= 4 −x−2ythat lies above the region in the

xy-plane bounded by the rectangle with vertices (0,0),(4,0),(4,3), and (0,3).

Use Stokes’ Theorem to evaluate the surface integral

ZZS

F·dS

where F= (y−z, x +z, x +y).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y−z x +z x +y



=∂

∂y (x+y)−∂

∂z (x+z)i−∂

∂x(y−z)−∂

∂z (y+z)j+∂

∂x(x+z)−∂

∂y (y+z)k

=i−j−k

Step 2: Determine the unit normal vector nto the surface S. Given that

Sis the plane z= 4 −x−2y, the unit normal vector to the surface Sis

n=∇z

|∇z|=D−∂z

∂x ,−∂z

∂y ,1E=⟨−(−1),−(−2),1⟩=⟨1,2,1⟩.

Step 3: Calculate the curl of F·n. The curl of F·nis ∇ × (F·n) =

∇ × (x−y−z) = ∇ × (−3) = 0.

Step 4: Apply Stokes’ Theorem. Stokes’ Theorem states that

ZZS

F·dS=ZC

F·dr

where Cis the boundary of the surface Soriented counterclockwise when viewed

from above.

The boundary of the region in the xy-plane is the rectangle with vertices

(0,0),(4,0),(4,3), and (0,3). Parameterize this rectangle using r(u, v) =

⟨u, v, 4−u−2v⟩where 0≤u≤4and 0≤v≤3.

The line integral RCF·drbecomes

(0 −2(4) + 4)dv +Z4

(3 + 4)du =−20 + 12 = −8

Therefore, the surface integral evaluates to −8.

Question 21

Question

Let Cbe the curve of intersection of the plane z= 0 and the cylinder x2+y2= 1

oriented counterclockwise when viewed from above. Let F= (y+z, x +z, x +y)

be a vector field. Calculate the flux of Facross the curve Cusing Stokes’

Theorem.

Solution

Step 1: Parameterize the curve C.

The curve Cis simply the circle x2+y2= 1 in the xy-plane. We can

parameterize this curve by r(t) = (cos t, sin t, 0) for 0≤t≤2π.

Step 2: Find the curl of F.

The curl of Fis given by ∇ × F=



iˆ

jˆ

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



= (1,1,1).

Step 3: Calculate the line integral of Fover C.

By Stokes’ Theorem, the flux of Facross Cis equal to the line integral of F

over the curve C, which is given by

F·dr=Z2π

F(r(t))·r′(t)dt =Z2π

(0,0,1)·(−sin t, cos t, 0)dt =Z2π

0dt = 0.

Therefore, the flux of Facross the curve Cis 0.

Question 22

Question

Let Sbe the surface given by z=x2−y2for 0≤x≤2and 0≤y≤2. Use

Stokes’ Theorem to evaluate the line integral

(x2+ 2y)dx + (z2−y)dy + (2x−2z)dz

where Cis the boundary curve of Swith counterclockwise orientation when

viewed from above.

Solution

Step 1: First, we need to parameterize the surface S.

The surface Sis given by z=x2−y2,0≤x≤2, and 0≤y≤2.

Let’s define the parameterization r(u, v) = ⟨u, v, u2−v2⟩, where (u, v)∈

D={(u, v)|0≤u≤2,0≤v≤2}.

Step 2: Next, we need to calculate the partial derivatives of rwith respect

to uand v.

∂r

∂u =⟨1,0,2u⟩and ∂r

∂v =⟨0,1,−2v⟩

Step 3: Now, we calculate the curl of the vector field F=⟨x2+ 2y, z2−

y, 2x−2z⟩.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+ 2y z2−y2x−2z



=⟨−1,0,2+2⟩=⟨−1,0,4⟩

Step 4: Using Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS

where nis the unit normal vector to the surface S.

Step 5: The surface integral RRS(∇ × F)·ndS simplifies to

ZZD⟨−1,0,4⟩ · ∂r

∂u ×∂r

∂v du dv

=ZZD



1 0 2u

0 1 −2v

−1 0 4



du dv

Step 6: Computing the determinant and performing the dot product gives

=ZZD

(4 + 4u)du dv

Step 7: Evaluating the double integral over the region D= [0,2]×[0,2] gives

the final answer.

Question 23

Question

Let Sbe the part of the cone z2=x2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to evaluate the surface integral RRS(∇×F)·dS,

where F(x, y, z) = (x2+y2)i+ (y2−z2)j+ (x2−y2)k.

Solution

To apply Stokes’ Theorem, we need to compute the curl of Fand the normal

vector nto the surface S.

Step 1: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2−z2x2−y2



=∂(x2−y2)

∂y −∂(y2−z2)

∂z i−∂(x2+y2)

∂x −∂(x2−y2)

∂z j+∂(x2+y2)

∂y −∂(y2−z2)

∂x k

=−2yi−2xj+ 2zk

Step 2: Find the unit normal vector n.The unit normal vector to the

surface Sis n=1

√2(−i−j+k)since the normal vector points outward.

Step 3: Evaluate the surface integral using Stokes’ Theorem. Stokes’

Theorem states that ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary curve of S.

The boundary curve Cof Sis a circle in the xy-plane with radius 3parametrized

by r(t) = 3 cos(t)i+ 3 sin(t)jfor 0≤t≤2π.

Therefore, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

((9 cos2(t) + 9 sin2(t))(−3 sin(t)) + (9 sin2(t)−9 cos2(t))(3 cos(t))) dt

=Z2π

(−27 sin(t) cos(t) + 27 sin(t) cos(t)) dt

=Z2π

0dt = 0

Hence, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 24

Question

Let Sbe the surface of the cone z=px2+y2bounded by the plane z= 3.

Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS, where

F(x, y, z) = xi+yj+zk.

Solution

Step 1: Calculate the curl of F: The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x y z



=∂z

∂y −∂y

∂z i−∂z

∂x −∂x

∂z j+∂y

∂x −∂x

∂y k

=i+j+k

Step 2: Find the unit normal vector to the surface S: The unit normal vector

to the surface Sis given by n=(−x,−y,√x2+y2)

√x2+y2+1 .

Step 3: Calculate the outward unit normal vector: Since the cone is upward-

facing, the outward unit normal vector is the negation of the unit normal vector

calculated in Step 2. Therefore, n=(x,y,−√x2+y2)

√x2+y2+1 .

Step 4: Calculate dS: The surface element dSis given by dS=ndS, where

dS =q1+(∂z

∂x )2+ ( ∂z

∂y )2dA =√2dA.

Step 5: Evaluate the surface integral:

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·n√2dA

=ZZD

(i+j+k)·(x, y, −px2+y2)

px2+y2+ 1

√2dA

=ZZD

(−z)√2dA

=−√2ZZD

zdA

The integral RRDzdA represents the flux of zthrough the region D, which

in this case is the projection of Sonto the xy-plane. However, the surface of

the cone Sdoes not intersect the xy-plane, making the flux integral through D

equal to zero. Hence, the surface integral RRS(∇ × F)·dSis also equal to 0.

Question 25

Question

Let Sbe the part of the plane z= 4 −x−ythat lies above the region in the

xy-plane bounded by the curve x2+y2= 1. Calculate the flux of the vector

field F(x, y, z) = ⟨yz, xz, xy⟩across the surface S.

Solution

Step 1: First, we find the unit normal vector to the surface S, which is given by

n=∂z

∂x ,∂z

∂y ,−1.

In this case, ∂z

∂x =−1,∂z

∂y =−1, so n=⟨−1,−1,−1⟩.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

yz xz xy



=⟨z−x, −z, y⟩.

Step 3: Now, we can apply Stokes’ theorem, which states:

ZZS

(∇ × F)·dS=IC

F·dr,

where Cis the boundary curve of the surface Sand dSis the outward-pointing

normal vector on S.

Step 4: Since Slies above the unit circle in the xy-plane, the boundary

curve Cis simply the unit circle x2+y2= 1. Therefore, we can parameterize

the curve Cas r(t) = ⟨cos t, sin t, 4−cos t−sin t⟩for 0≤t≤2π.

Step 5: Calculating the line integral HCF·dr, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

0⟨sin t(4 −cos t−sin t),cos t(4 −cos t−sin t),cos tsin t⟩ · ⟨−sin t, cos t, −sin t−cos t⟩dt

=Z2π

(−sin2t(4 −cos t−sin t)−cos2t(4 −cos t−sin t) + cos tsin t(sin t+ cos t)) dt

=Z2π

(−4 + 5 sin t+ 5 cos t)dt

= 0.

Step 6: Therefore, the flux of the vector field Facross the surface Sis zero.

Step 4: Calculate the circulation of ∇ × Faround the boundary of S.

The circulation of ∇ × Faround the boundary Cis given by

IC∇ × F·dr=ZZS∇ × F·dS

=IC1∇ × F·dr+IC2∇ × F·dr+IC3∇ × F·dr

=Z2

F(x, 0) ·T1dx +Z1

F(2, y)·T2dy +Z0

F(x, 1) ·T3dx

=Z2

(x2,0,(4

Question 2

Question

Let Sbe the part of the plane z=x+ythat lies above the square 0≤x≤

1,0≤y≤1. Use Stokes’ Theorem to evaluate the surface integral

ZZS

(∇ × F)·dS

where F(x, y, z) = (z2, y2, x2).

Solution

Step 1: To apply Stokes’ Theorem, we first need to compute the curl of F.

∇×F=



i j k

∂

∂x

∂

∂y

∂

∂z

z2y2x2



=∂

∂y (x2)−∂

∂z (y2),∂

∂z (z2)−∂

∂x(x2),∂

∂x(y2)−∂

∂y (z2)

= (−2y, 2z, 2x)

Step 2: The outward unit normal vector to the surface Scan be calculated

as n= ( ∂z

∂x ,∂z

∂y ,−1) = (1,1,−1).

Step 3: The surface integral can be rewritten as

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·ndA

where Dis the projection of Sonto the xy-plane, that is, the square 0≤x≤

1,0≤y≤1.

Step 4: Substituting the values of ∇×Fand n, the surface integral becomes

ZZD

((−2y, 2z, 2x)) ·(1,1,−1) dA =ZZD

(−2y+ 2z−2x)dA

Step 5: Now, we evaluate the double integral over the square D:

ZZD

(−2y+ 2z−2x)dA =Z1

0Z1

(−2y+ 2(x+y)−2x)dy dx

=Z1

0−2Z1

y dy + 2 Z1

(x+y)dy −2Z1

x dydx

Step 6: Solving the inner integrals and simplifying gives:

=Z1

(−1 + 2(x+1

2)−2x)dx =Z1

(1 −2x)dx =x−x21

0= 1 −1 = 0

Therefore, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 3

Question

Let Cbe the curve defined by x=t2,y=t3,z=t4for 0≤t≤1. Let Sbe the

surface bounded by Cand the plane z= 1. Use Stokes’ Theorem to evaluate

HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).

Solution

1. First, parameterize the curve C:

The parameterization of Cis given by r(t) = (t2, t3, t4)for 0≤t≤1.

2. Calculate the derivative of r(t):

r′(t) = (2t, 3t2,4t3).

3. Compute the curl of F:

The curl of Fis given by ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



Simplifying, we get ∇ × F= (−1,−1,−1).

4. Calculate the line integral using Stokes’ Theorem:

By Stokes’ Theorem, HCF·dr=RRS(∇ × F)·dS, where Sis the surface

bounded by Cand z= 1.

Since ∇ × F= (−1,−1,−1), the line integral simplifies to HCF·dr=

RRS(−1,−1,−1) ·ndS, where nis the normal vector to the surface S.

5. Determine the normal vector to the surface S:

The normal vector to the surface Sis n=±(−∂z

∂x ,−∂z

∂y ,1) = ±(0,0,1).

As the normal vector should point outward from the bounded region, we

choose n= (0,0,1).

6. Compute the surface integral:

RRS(−1,−1,−1) ·(0,0,1) dS =−RRSdS =−RRR||ru×rv||dA, where R

is the parameterization region of the surface S.

As r(u, v) = (u, v, 1) and ∂r

∂u ×∂r

∂v = (−1,0,0), the surface integral simpli-

fies to −RRR1dA.

It follows that HCF·dr=−A, where Ais the area of the region R.

Question 4

Question

Let Sbe the part of the paraboloid z=x2+y2that lies below the plane

z= 1. Use Stokes’ Theorem to evaluate the circulation of the vector field

F=⟨yez, zex, xey⟩around the boundary curve of S, oriented counterclockwise

when viewed from above.

Solution

Step 1: First, let’s find the curl of F. The curl of a vector field F=⟨P, Q, R⟩

is defined as ∇ × F=∂R

∂y −∂Q

∂z i+∂P

∂z −∂R

∂x j+∂Q

∂x −∂P

∂y k. So, for our

vector field F=⟨yez, zex, xey⟩, we have:

∇×F=∂(xey)

∂y −∂(zex)

∂z i+∂(yez)

∂z −∂(xey)

∂x j+∂(zex)

∂x −∂(yez)

∂y k

= (xey−zex)i+ (yez−xey)j+ (zex−yez)k

Step 2: Next, we will find the outward unit normal vector to the surface S.

Since Sis the part of the paraboloid z=x2+y2below the plane z= 1, the unit

normal vector on Sis given by N=∇(z−x2−y2)

||∇(z−x2−y2)|| =−2xi

−2yj+kp1+4x2+ 4y2.

Step 3: Now, we can apply Stokes’ Theorem, which states that HCF·dr=

RRS(∇ × F)·NdS. Let’s first calculate RRS(∇ × F)·NdS:

ZZS

(∇×F)·NdS =ZZS

((xey−zex)i+(yez−xey)j+(zex−yez)k)·−2xi

−2yj+kp1+4x2+ 4y2dS

Step 4: To evaluate the surface integral, we need to find a parameterization

of S. Let’s use the parameterization r(u, v) = ⟨u, v, u2+v2⟩for 0≤u≤1and

−√1−u2≤v≤√1−u2. The normal vector to the surface can be calculated

as the cross product of ruand rv.

Step 5: After parameterizing S, we can compute the surface integral over S

using the parameterization. Finally, we substitute the results back into Stokes’

Theorem to find the circulation of Faround the boundary curve of S.

Question 5

Question

Let Sbe the surface given by z=x2+y2for 0≤z≤1oriented with outward

normal. Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS,

where F(x, y, z) = (x2+y2,2xy, z).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y22xy z



= (0,0,2y−2x)

Step 2: Determine the unit normal vector nto the surface S. Since Sis

given by z=x2+y2, the normal vector can be written as:

n=∇z

|∇z|=⟨−2x, −2y, 1⟩

p1+4x2+ 4y2

Step 3: Calculate the outward unit normal vector dS. The outward unit

normal vector dSis given by dS=ndS =−2x

√1+4x2+4y2,−2y

√1+4x2+4y2,1

√1+4x2+4y2dA

Step 4: Setup the surface integral using Stokes’ Theorem. Stokes’ Theorem

states that the surface integral of ∇ × Fover Sis equal to the line integral of

Falong the boundary of S, which in this case is the circle x2+y2= 1 in the

xy-plane.

Therefore, the surface integral RRS(∇ × F)·dScan be rewritten as the line

integral HCF·draround the boundary curve Cof S.

Step 5: Parameterize the curve C. Parameterize the circle Cin the xy-plane

with x= cos t,y= sin t,0≤t≤2π.

Step 6: Evaluate the line integral. Substitute the parameterization of the

curve into Fand the differential dr, then calculate F·dr.

Step 7: Compute the final result. Calculate the line integral over Cto find

the value of the surface integral using Stokes’ Theorem.

Question 6

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to calculate the flux of the curl of the vector

field F(x, y, z) = (xz, yz, x2)across the surface S, where the direction of positive

orientation is the outward normal.

Solution

1. First, we calculate the curl of the vector field F(x, y, z):

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

xz yz x2



=∂(x2)

∂y −∂(yz)

∂z i−∂(xz)

∂x −∂(x2)

∂z j+∂(yz)

∂x −∂(xz)

∂y k

= (0 −y)i−(z−0)j+ (z−z)k=−yi−zj

2. Then, we calculate the outward unit normal vector to the surface S, which

is given by

n=∇g

|∇g|=∇z

|∇z|=i+j+k

√12+ 12+ 12=1

√3(i+j+k)

3. The surface integral of the curl of Fover Scan be calculated using Stokes’

Theorem: ZZS

(∇ × F)·dS=IC

F·dr

=Z3

1Z2π

(−y, −z)·(1

√3cos(ϕ),1

√3sin(ϕ),1

√3)dϕdz

4. Performing the dot product and integrating, we get:

1Z2π

0−y

√3cos(ϕ)−z

√3sin(ϕ) + 1

√3dzdϕ

=Z3

1Z2π

0−z

√3sin(ϕ) + 1

√3dzdϕ

5. Further evaluating the integral, we obtain:

=Z3

1−2πz

√3dz =−2πz2

√33

=−2π32−1

√3=−2π8

√3

Therefore, the flux of the curl of the vector field F(x, y, z)across the surface

Sis −8π

√3.

Question 8

Question

Let Sbe the portion of the plane z=x+ythat lies inside the cylinder x2+y2=

1oriented with outward-pointing normal. Use Stokes’ Theorem to evaluate

RRSF·dS, where F=yzi+xzj+xyk.

Solution

Step 1: First, find the curl of F: Let F=Pi+Qj+Rk. Then, the curl of Fis

given by

curl(F) = ∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



=∂R

∂y −∂Q

∂z i−∂R

∂x −∂P

∂z j+∂Q

∂x −∂P

∂y k

For F=yzi+xzj+xyk, we have P=yz,Q=xz,R=xy. Thus, the curl of

Fis

curl(F) = (z−z)i−(y−y)j+ (x−x)k=0

Step 2: By Stokes’ Theorem, we have

ZZS

F·dS=ZZ∂S

F·dr

Since the curl of Fis 0, the line integral vanishes. Therefore, RRSF·dS= 0.

Question 9

Question

Let Sbe the part of the paraboloid z= 4−x2−y2that lies above the plane z= 1.

Orient Swith outward normal. Use Stokes’ Theorem to evaluate HCF·drwhere

F(x, y, z) = xi+yj+zkand Cis the boundary of Soriented counterclockwise

when viewed from above.

Solution

Step 1: Find the curl of F. Step 2: Compute the surface integral of the curl

of Fover the region Sto apply Stokes’ Theorem. Step 3: Parameterize the

boundary curve Cand compute the line integral of Fover C.

Let’s start with these steps.

Question 10

Question

Let F=x2i+y2j+z2kbe a vector field, and let Sbe the surface enclosed by

the cylinder x2+y2= 1 and the planes z= 0 and z= 2. Use Stokes’ Theorem

to evaluate the line integral HCF·draround the boundary curve Cof S.

Solution

Step 1: First, we need to parameterize the boundary curve C, which consists of

the circle x2+y2= 1 on the plane z= 0 and the same circle on the plane z= 2.

We will parameterize the circle on the plane z= 0 as r1(t) = cos(t)i+ sin(t)j

for 0≤t≤2π. The circle on the plane z= 2 can be parameterized similarly as

r2(t) = cos(t)i+ sin(t)j+ 2k.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2y2z2



= (2z−2y)i−(2z−2x)j+ 0k= 2(z−y)i−2(z−x)j

Step 3: We can now apply Stokes’ Theorem to relate the line integral over

Cto the surface integral over S. Using Stokes’ Theorem, we have:

F·dr=ZZS

(∇ × F)·ndS

where nis the outward unit normal to S. Since the surface Sis bounded by

C, the outward unit normal npoints outward from the cylinder.

Step 4: We will calculate the surface integral over S:

ZZS

(∇ × F)·ndS =ZZS

(2(z−y)i−2(z−x)j)·kdS

Since the normal vector nis in the kdirection, the dot product simplifies

to −2(z−x). To find the limits of integration, we note that the projection of

Sonto the xy-plane is the circle x2+y2= 1. Thus, the limits for xand yare

x2+y2≤1, and for z, the limits are 0≤z≤2.

Step 5: We can now evaluate the surface integral:

ZZS

(∇ × F)·ndS

=Z2π

0Z1

0−2(2 −cos(t)) dr dθ

=−4Z2π

(2 −rcos(t))



0dθ

=−4Z2π

(2 −cos(t)−0) dθ

=−4·2π(2)

=−16π

Therefore, the line integral HCF·draround the boundary curve Cof Sis

−16π.

Question 11

Question

Let F(x, y, z) = (y2−z)i+ (z2−x)j+ (x2−y)kand let Sbe the portion of

the cone z=px2+y2that lies between the planes z= 1 and z= 3. Use

Stokes’ Theorem to evaluate HCF·dr, where Cis the counterclockwise-oriented

boundary curve of surface S.

Solution

Step 1: First, let’s parameterize the boundary curve Cof the surface S. The

boundary curve consists of two parts: the circle x2+y2= 1 at z= 1 and the

circle x2+y2= 9 at z= 3. We can parameterize these circles as follows: For

z= 1, we have the circle C1:x2+y2= 1. Parameterizing this circle gives us:

r1(t) = ⟨cos(t),sin(t),1⟩,where 0≤t≤2π.

For z= 3, we have the circle C2:x2+y2= 9. Parameterizing this circle gives

us:

r2(t) = ⟨3 cos(t),3 sin(t),3⟩,where 0≤t≤2π.

Step 2: Now, we need to compute the curl of F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y2−z z2−x x2−y



= (2y+ 1)i−(2z+ 1)j+ (2x+ 1)k.

Step 3: By Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS,

where nis the unit normal to S.

Step 4: To find the unit normal to S, we compute the gradient of z=

px2+y2as ∇z=⟨x

√x2+y2,y

√x2+y2,−1⟩.

Step 5: The unit normal to Sis then ˆ

n=∇z

||∇z|| =⟨x,y,−√x2+y2⟩

√x2+y2+1 .

Step 6: We can now evaluate the surface integral RRS(∇×F)·ndS using the

parameterizations of C1and C2from Step 1.

Question 12

Question

Let Sbe the surface of the portion of the plane z= 4 −x−2ythat lies above

the square 0≤x≤1,0≤y≤1, oriented upward. Calculate the flux of the

curl of the vector field F= 2xi+yj+zkacross S.

Solution

Step 1: Find the curl of the vector field F. The curl of a vector field F=

Pi+Qj+Rkis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

P Q R



In this case, F= 2xi+yj+zk, so:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

2x y z



= (0 −0)i−(0 −1)j+ (1 −2)k=−j−k

Step 2: Use Stokes’ Theorem to find the flux of the curl of Facross S.

Stokes’ Theorem states:

ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary of the surface S.

Step 3: Find the boundary curve Cof the surface S. The boundary curve

Cof the surface Sis the square where 0≤x≤1,0≤y≤1. It consists of four

line segments: 1. y= 0,0≤x≤12. x= 1,0≤y≤13. y= 1,0≤x≤14.

x= 0,0≤y≤1

Step 4: Parameterize and calculate the line integral HCF·dr. We will

integrate over each of the four line segments of the boundary: 1. y= 0,0≤

x≤1:r(t) = ti,0≤t≤12. x= 1,0≤y≤1:r(t) = i+tj,0≤t≤13.

y= 1,0≤x≤1:r(t) = (1 −t)i+j,0≤t≤14. x= 0,0≤y≤1:r(t) = tj,

0≤t≤1

Step 5: Substitute the parameterization into Fand calculate the dot product

with dr. For each line segment, calculate: 1. HCF·dr=R1

0(2t, 0,4−t)·(1,0,0)dt

Question 13

Question

Let Sbe the part of the cone z=px2+y2that lies between the planes z= 1

and z= 4. Use Stokes’ Theorem to evaluate the surface integral RRSF·dS,

where F(x, y, z) = (z, x, y).

(Note: You may assume that a positively oriented unit normal vector to S

at each point is n=1

√2(−x, −y, √2).)

Solution

Step 1: Find the boundary curve Cof Swhere the orientation is consistent with

the orientation of Sgiven by the positively oriented unit normal vector n.

Since Slies between z= 1 and z= 4, the boundary curve Cconsists of the

intersection of the cone z=px2+y2and the planes z= 1 and z= 4. Solving

these equations, we get z= 1 and z= 4:1 = px2+y2and 4 = px2+y2.

Thus, the boundary curve Cis the circle x2+y2= 1.

Step 2: Parameterize the boundary curve C.

Let Cbe parameterized by r(t) = (cos t, sin t, 1) where 0≤t≤2π.

Step 3: Find the curl of F.

The curl of Fis given by ∇ × F=∂Fz

∂y −∂Fy

∂z ,∂Fx

∂z −∂Fz

∂x ,∂Fy

∂x −∂Fx

∂y .

Calculating the components of the curl, we have: ∇×F= (0 −1,1−0,1−0) =

(−1,1,1).

Step 4: Evaluate the line integral HCF·dr.

Using Stokes’ Theorem, we have: RRSF·dS=HCF·dr.

F·dr=Z2π

F(r(t)) ·r′(t)dt

Z2π

F(r(t)) ·r′(t)dt =Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt

Z2π

(1,cos t, sin t)·(−sin t, cos t, 0)dt =Z2π

0−sin t+ cos2tdt

Z2π

0−sin t+ cos2tdt =−cos t+t

2+sin 2t

42π

= 0

Therefore, the surface integral RRSF·dSis 0.

Question 14

Question

Let Fbe a vector field defined by F(x, y, z) = (y2z+ cos x)i+ (xz2+ sin y)j+

(xy2+ez)k. Calculate the surface integral RRScurl(F)·dS, where Sis the

portion of the plane z= 4 −x−2ythat lies above the triangle in the xy-plane

with vertices at (0,0),(1,0), and (0,2).

Solution

Step 1: Calculate curl(F).

curl(F) = 



i j k

∂

∂x

∂

∂y

∂

∂z

y2z+ cos x xz2+ sin y xy2+ez



=∂(xy2+ez)

∂y −∂(xz2+ sin y)

∂z i−∂(y2z+ cos x)

∂x −∂(xy2+ez)

∂z j+∂(xz2+ sin y)

∂x −∂(y2z+ cos x)

∂y k

= (2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k

Step 2: Calculate the unit normal vector to the surface S. The surface S

is defined by z= 4 −x−2y. The normal vector to this surface is n=∇z=

−i−2j+k. Therefore, the unit normal vector is n=−1

√6i−2

√6j+1

√6k.

Step 3: Calculate the magnitude of dS. The magnitude of dSis given by

dS =|n|dxdy =q6

6dxdy =dxdy.

Step 4: Calculate the surface integral. The surface integral can be calculated

as:

ZZS

curl(F)·dS=ZZS(2xy −z2)i−(sin x−2xy)j+ (2xz −y2)k·−1

√6i−2

√6j+1

√6kdxdy

Now, we substitute z= 4 −x−2yinto the expression above and integrate over

the region in the xy-plane:

ZZS

curl(F)·dS=ZZR(2xy −(4 −x−2y)2)−1

√6−(sin x−2xy)2

√6+ (2x(4 −x−2y)−y2)1

√6dxdy

where Ris the triangle in the xy-plane

Question 15

Question

Let Sbe the hemisphere x2+y2+z2= 1, z ≥0, and let F= (x2+y2, y2+

z2, z2+x2). Calculate the flux of Facross Susing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2+z2z2+x2



=∂(z2+x2)

∂y −∂(y2+z2)

∂z i−∂(z2+x2)

∂x −∂(x2+y2)

∂z j+∂(y2+z2)

∂x −∂(x2+y2)

∂y k

= (0 −2z)i−(2x−0)j+ (2y−0)k=−2zi−2xj+ 2yk

Step 2: Calculate the surface integral of Fover Susing Stokes’ Theorem.

(∇ × F)·dS=I

∂S

F·dr

Step 3: The boundary of Sis the circle x2+y2= 1, z = 0.

r(t) = cos(t)i+ sin(t)j,0≤t≤2π

Step 4: Parameterize the curve C.

r′(t) = −sin(t)i+ cos(t)j

r′(t)×k=−cos(t)k

Step 5: Calculate the line integral.

∂S

F·dr=Z2π

F(r(t)) ·r′(t)×kdt

=Z2π

(−sin(t))2+ cos(t)2+ (−sin(t))2dt =Z2π

2dt = 4π

Therefore, the flux of Facross Sis 4π.

Question 16

Question

Let Sbe the part of the plane x+ 2y+ 3z= 6 that lies inside the cylinder

x2+y2= 4. Use Stokes’ Theorem to evaluate

F·dr,

where F=⟨y, x, z⟩and Cis the boundary of Soriented counterclockwise as

viewed from above.

Solution

Step 1: Find the boundary curve Cof the surface S. This is the intersection of

the plane and the cylinder, so we have

x+ 2y+ 3z= 6

x2+y2= 4.

Solving these equations, we find that the boundary curve Cis the circle x2+y2=

4, which is the circle of radius 2 centered at the origin.

Step 2: Parameterize the boundary curve C. Let x= 2 cos t,y= 2 sin t,

where 0≤t≤2π. Then dr=⟨−2 sin t, 2 cos t, 0⟩dt.

Step 3: Compute F(r(t)).

F(r(t)) = ⟨2 sin t, 2 cos t, 2 sin t⟩.

Step 4: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y x z



=⟨1,1,−2⟩.

Step 5: Compute the line integral using Stokes’ Theorem.

F·dr=ZZS

(∇ × F)·ndS,

where nis the outward unit normal to S. Since the outward normal for the

plane x+ 2y+ 3z= 6 is n=⟨1,2,3⟩/√12+ 22+ 32=⟨1/√14,2/√14,3/√14⟩,

we have IC

F·dr=ZZS

(∇ × F)·ndS

=ZZS⟨1,1,−2⟩ · 1

√14⟨1,2,3⟩dS

√14 ZZS

(1 + 2 −6) dS

=−3

√14 ZZS

=−3

√14 ·Area(S).

Step 6: Find the area of the region S. This can be done by finding the

projection of Sonto the xy-plane and calculating the double integral over this

region.

Step 7: Calculate the final answer for the line integral.

Question 17

Question

Let Sbe the part of the plane z= 3 −x−ythat lies above the triangle with

vertices (0,0,0),(1,0,0), and (0,1,0), oriented counterclockwise when viewed

from above. Let Fbe the vector field F= (x2+y2, z, 2x−z). Calculate the

surface integral R RSF·dSusing Stokes’ Theorem.

Solution

Step 1: Calculate the curl of F. The curl of Fis given by:

curl(F) = ∂(2x−z)

∂y −∂(z)

∂z ,∂(x2+y2)

∂z −∂(2x−z)

∂x ,∂(z)

∂x −∂(x2+y2)

∂y 

= (−1−0,0−2,0−2) = (−1,−2,−2).

Step 2: Calculate the normal vector to the surface S. The normal vector to

the plane z= 3 −x−yis n= (−∂z

∂x ,−∂z

∂y ,1) = (−(−1),−(−1),1) = (1,1,1).

Step 3: Normalize the normal vector. The unit normal vector is n0=

|n| = (1, 1, 1)√12+12+12=(1

√3,1

√3).

Step 4: Calculate the surface integral using Stokes’ Theorem. Stokes’ Theo-

rem states that R RSF·dS=RRRVcurl(F)·ndV , where Vis the region enclosed

by S. Since Sis the part of the plane z= 3 −x−ythat lies above the given

triangle, the region Vis the triangle 0≤x≤1,0≤y≤1−x, 0≤z≤3−x−y.

The surface integral becomes:

Z ZS

F·dS=ZZZV

(−1,−2,−2) ·(1,1,1) dV

=ZZZV−5dV.

Step 5: Calculate the triple integral over region V. We have 0≤x≤1,0≤

y≤1−x, 0≤z≤3−x−y. Therefore, the triple integral becomes:

0Z1−x

0Z3−x−y

0−5dz dy dx

=Z1

0Z1−x

[−5z]3−x−y

0dy dx

=Z1

0Z1−x

0−5(3 −x−y)dy dx

=Z1

[−5(3y−xy −y2

2)]1−x

0dx

=Z1

0−53(1 −x)−x(1 −x)−(1 −x)2

2dx

=Z1

0−17

2x2+ 14x−15 dx

Question 18

Question

Let Sbe the part of the plane x+y+z= 1 that lies in the first octant.

Let F(x, y, z) = ⟨2xz, y2, x +z⟩. Use Stokes’ Theorem to evaluate the surface

integral

ZZS

curl F ·ndS

where nis the outward unit normal to S.

Solution

Step 1: First, we need to find the curl of F.

curl F=



i j k

∂

∂x

∂

∂y

∂

∂z

2xz y2x+z



=∂

∂y (x+z)−∂

∂z (y2),∂

∂z (2xz)−∂

∂x(x+z),∂

∂x(y2)−∂

∂y (2xz)

= (1 −0,2x−1,2z)

= (1,2x−1,2z)

Step 2: Parameterize the surface S. Since Sis the part of the plane x+y+z=

1in the first octant, we can parameterize Sas:

r(u, v) = ⟨u, v, 1−u−v⟩,0≤u≤1,0≤v≤1−u

Step 3: Calculate the normal vector n. The normal vector can be found by

taking the cross product of the partial derivatives of rwith respect to uand v.

∂u =⟨1,0,−1⟩,

∂v =⟨0,1,−1⟩So, n=

∂u×

∂v =⟨1,1,1⟩

Step 4: Evaluate the surface integral. Using Stokes’ Theorem, we have:

ZZS

curl F·ndS =ZZD

(curl F)·(ru×rv)dA

where Dis the region in the uv-plane corresponding to the surface S.

=ZZD

(1,2u−1,2(1 −u−v)) ·(1,1,1) dA

=ZZD

(1 + 2u−1−2v)dA

=ZZD

(2u−2v)dA

=Z1

0Z1−u

(2u−2v)dv du

=Z1

02uv −v2v=1−u

v=0 du

=Z1

(2u−(1 −u)2)du

=Z1

(3u2−2u+ 1) du

=u3−u2+u1

= 1 −1+1

= 1

Therefore, the surface integral is equal to 1.

Question 20

Question

Let Sbe the part of the plane z= 4 −x−2ythat lies above the region in the

xy-plane bounded by the rectangle with vertices (0,0),(4,0),(4,3), and (0,3).

Use Stokes’ Theorem to evaluate the surface integral

ZZS

F·dS

where F= (y−z, x +z, x +y).

Solution

Step 1: Calculate the curl of F. The curl of Fis given by

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

y−z x +z x +y



=∂

∂y (x+y)−∂

∂z (x+z)i−∂

∂x(y−z)−∂

∂z (y+z)j+∂

∂x(x+z)−∂

∂y (y+z)k

=i−j−k

Step 2: Determine the unit normal vector nto the surface S. Given that

Sis the plane z= 4 −x−2y, the unit normal vector to the surface Sis

n=∇z

|∇z|=D−∂z

∂x ,−∂z

∂y ,1E=⟨−(−1),−(−2),1⟩=⟨1,2,1⟩.

Step 3: Calculate the curl of F·n. The curl of F·nis ∇ × (F·n) =

∇ × (x−y−z) = ∇ × (−3) = 0.

Step 4: Apply Stokes’ Theorem. Stokes’ Theorem states that

ZZS

F·dS=ZC

F·dr

where Cis the boundary of the surface Soriented counterclockwise when viewed

from above.

The boundary of the region in the xy-plane is the rectangle with vertices

(0,0),(4,0),(4,3), and (0,3). Parameterize this rectangle using r(u, v) =

⟨u, v, 4−u−2v⟩where 0≤u≤4and 0≤v≤3.

The line integral RCF·drbecomes

(0 −2(4) + 4)dv +Z4

(3 + 4)du =−20 + 12 = −8

Therefore, the surface integral evaluates to −8.

Question 21

Question

Let Cbe the curve of intersection of the plane z= 0 and the cylinder x2+y2= 1

oriented counterclockwise when viewed from above. Let F= (y+z, x +z, x +y)

be a vector field. Calculate the flux of Facross the curve Cusing Stokes’

Theorem.

Solution

Step 1: Parameterize the curve C.

The curve Cis simply the circle x2+y2= 1 in the xy-plane. We can

parameterize this curve by r(t) = (cos t, sin t, 0) for 0≤t≤2π.

Step 2: Find the curl of F.

The curl of Fis given by ∇ × F=



iˆ

jˆ

∂

∂x

∂

∂y

∂

∂z

y+z x +z x +y



= (1,1,1).

Step 3: Calculate the line integral of Fover C.

By Stokes’ Theorem, the flux of Facross Cis equal to the line integral of F

over the curve C, which is given by

F·dr=Z2π

F(r(t))·r′(t)dt =Z2π

(0,0,1)·(−sin t, cos t, 0)dt =Z2π

0dt = 0.

Therefore, the flux of Facross the curve Cis 0.

Question 22

Question

Let Sbe the surface given by z=x2−y2for 0≤x≤2and 0≤y≤2. Use

Stokes’ Theorem to evaluate the line integral

(x2+ 2y)dx + (z2−y)dy + (2x−2z)dz

where Cis the boundary curve of Swith counterclockwise orientation when

viewed from above.

Solution

Step 1: First, we need to parameterize the surface S.

The surface Sis given by z=x2−y2,0≤x≤2, and 0≤y≤2.

Let’s define the parameterization r(u, v) = ⟨u, v, u2−v2⟩, where (u, v)∈

D={(u, v)|0≤u≤2,0≤v≤2}.

Step 2: Next, we need to calculate the partial derivatives of rwith respect

to uand v.

∂r

∂u =⟨1,0,2u⟩and ∂r

∂v =⟨0,1,−2v⟩

Step 3: Now, we calculate the curl of the vector field F=⟨x2+ 2y, z2−

y, 2x−2z⟩.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+ 2y z2−y2x−2z



=⟨−1,0,2+2⟩=⟨−1,0,4⟩

Step 4: Using Stokes’ Theorem, we have

F·dr=ZZS

(∇ × F)·ndS

where nis the unit normal vector to the surface S.

Step 5: The surface integral RRS(∇ × F)·ndS simplifies to

ZZD⟨−1,0,4⟩ · ∂r

∂u ×∂r

∂v du dv

=ZZD



1 0 2u

0 1 −2v

−1 0 4



du dv

Step 6: Computing the determinant and performing the dot product gives

=ZZD

(4 + 4u)du dv

Step 7: Evaluating the double integral over the region D= [0,2]×[0,2] gives

the final answer.

Question 23

Question

Let Sbe the part of the cone z2=x2+y2that lies between the planes z= 1

and z= 3. Use Stokes’ Theorem to evaluate the surface integral RRS(∇×F)·dS,

where F(x, y, z) = (x2+y2)i+ (y2−z2)j+ (x2−y2)k.

Solution

To apply Stokes’ Theorem, we need to compute the curl of Fand the normal

vector nto the surface S.

Step 1: Compute ∇ × F.

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x2+y2y2−z2x2−y2



=∂(x2−y2)

∂y −∂(y2−z2)

∂z i−∂(x2+y2)

∂x −∂(x2−y2)

∂z j+∂(x2+y2)

∂y −∂(y2−z2)

∂x k

=−2yi−2xj+ 2zk

Step 2: Find the unit normal vector n.The unit normal vector to the

surface Sis n=1

√2(−i−j+k)since the normal vector points outward.

Step 3: Evaluate the surface integral using Stokes’ Theorem. Stokes’

Theorem states that ZZS

(∇ × F)·dS=IC

F·dr

where Cis the boundary curve of S.

The boundary curve Cof Sis a circle in the xy-plane with radius 3parametrized

by r(t) = 3 cos(t)i+ 3 sin(t)jfor 0≤t≤2π.

Therefore, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

((9 cos2(t) + 9 sin2(t))(−3 sin(t)) + (9 sin2(t)−9 cos2(t))(3 cos(t))) dt

=Z2π

(−27 sin(t) cos(t) + 27 sin(t) cos(t)) dt

=Z2π

0dt = 0

Hence, the value of the surface integral RRS(∇ × F)·dSis 0.

Question 24

Question

Let Sbe the surface of the cone z=px2+y2bounded by the plane z= 3.

Use Stokes’ Theorem to evaluate the surface integral RRS(∇ × F)·dS, where

F(x, y, z) = xi+yj+zk.

Solution

Step 1: Calculate the curl of F: The curl of Fis given by:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

x y z



=∂z

∂y −∂y

∂z i−∂z

∂x −∂x

∂z j+∂y

∂x −∂x

∂y k

=i+j+k

Step 2: Find the unit normal vector to the surface S: The unit normal vector

to the surface Sis given by n=(−x,−y,√x2+y2)

√x2+y2+1 .

Step 3: Calculate the outward unit normal vector: Since the cone is upward-

facing, the outward unit normal vector is the negation of the unit normal vector

calculated in Step 2. Therefore, n=(x,y,−√x2+y2)

√x2+y2+1 .

Step 4: Calculate dS: The surface element dSis given by dS=ndS, where

dS =q1+(∂z

∂x )2+ ( ∂z

∂y )2dA =√2dA.

Step 5: Evaluate the surface integral:

ZZS

(∇ × F)·dS=ZZD

(∇ × F)·n√2dA

=ZZD

(i+j+k)·(x, y, −px2+y2)

px2+y2+ 1

√2dA

=ZZD

(−z)√2dA

=−√2ZZD

zdA

The integral RRDzdA represents the flux of zthrough the region D, which

in this case is the projection of Sonto the xy-plane. However, the surface of

the cone Sdoes not intersect the xy-plane, making the flux integral through D

equal to zero. Hence, the surface integral RRS(∇ × F)·dSis also equal to 0.

Question 25

Question

Let Sbe the part of the plane z= 4 −x−ythat lies above the region in the

xy-plane bounded by the curve x2+y2= 1. Calculate the flux of the vector

field F(x, y, z) = ⟨yz, xz, xy⟩across the surface S.

Solution

Step 1: First, we find the unit normal vector to the surface S, which is given by

n=∂z

∂x ,∂z

∂y ,−1.

In this case, ∂z

∂x =−1,∂z

∂y =−1, so n=⟨−1,−1,−1⟩.

Step 2: Next, we compute the curl of the vector field F:

∇ × F=



i j k

∂

∂x

∂

∂y

∂

∂z

yz xz xy



=⟨z−x, −z, y⟩.

Step 3: Now, we can apply Stokes’ theorem, which states:

ZZS

(∇ × F)·dS=IC

F·dr,

where Cis the boundary curve of the surface Sand dSis the outward-pointing

normal vector on S.

Step 4: Since Slies above the unit circle in the xy-plane, the boundary

curve Cis simply the unit circle x2+y2= 1. Therefore, we can parameterize

the curve Cas r(t) = ⟨cos t, sin t, 4−cos t−sin t⟩for 0≤t≤2π.

Step 5: Calculating the line integral HCF·dr, we have

F·dr=Z2π

F(r(t)) ·r′(t)dt

=Z2π

0⟨sin t(4 −cos t−sin t),cos t(4 −cos t−sin t),cos tsin t⟩ · ⟨−sin t, cos t, −sin t−cos t⟩dt

=Z2π

(−sin2t(4 −cos t−sin t)−cos2t(4 −cos t−sin t) + cos tsin t(sin t+ cos t)) dt

=Z2π

(−4 + 5 sin t+ 5 cos t)dt

= 0.

Step 6: Therefore, the flux of the vector field Facross the surface Sis zero.

MATH 332 - ADVANCED CALCULUS Stokes' Theorem Question Bank Set 4

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