MATH 117 - ELEMENTS OF
MATHEMATICS - Gradient,
divergence, and curl
Question Bank - Set 5
Liberty University
Question 1
Question
Let F(x, y, z) = x
x2+y2+z2i+y
x2+y2+z2j+z
x2+y2+z2k. Calculate the divergence
of F.
Solution
Step 1: The divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by the following formula:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: Identify P,Q, and Rin F(x, y, z). Here, P(x, y, z) = x
x2+y2+z2,
Q(x, y, z) = y
x2+y2+z2, and R(x, y, z) = z
x2+y2+z2.
Step 3: Calculate the partial derivatives of P,Q, and R.
∂P
∂x =∂
∂x x
x2+y2+z2=(x2+y2+z2)−x(2x)
(x2+y2+z2)2
∂Q
∂y =∂
∂y y
x2+y2+z2=(x2+y2+z2)−y(2y)
(x2+y2+z2)2
∂R
∂z =∂
∂z z
x2+y2+z2=(x2+y2+z2)−z(2z)
(x2+y2+z2)2
Step 4: Calculate the divergence of F.
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
∇·F=(x2+y2+z2)−x(2x)
(x2+y2+z2)2+(x2+y2+z2)−y(2y)
(x2+y2+z2)2+(x2+y2+z2)−z(2z)
(x2+y2+z2)2
∇ · F=3(x2+y2+z2)−2x2−2y2−2z2
(x2+y2+z2)2
∇ · F=x2+y2+z2
(x2+y2+z2)2
∇ · F=1
x2+y2+z2
Question 2
Question
Let F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzkbe a vector field. Calculate the
divergence of Fat the point (1,−1,2).
Solution
Step 1: The divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: For the given vector field F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzk, we
have:
P(x, y, z) = x2y, Q(x, y, z) = xz2−y, R(x, y, z)=3xyz
Step 3: Now, calculate the partial derivatives of P,Q, and Rwith respect to
x,y, and zrespectively. Then, substitute the point (1,−1,2) into these partial
derivatives. ∂P
∂x = 2xy, ∂Q
∂y =−1,∂R
∂z = 3xy
Step 4: Substitute the point (1,−1,2) into the partial derivatives:
∂P
∂x (1,−1,2) = 2(−1)(1) = −2
∂Q
∂y (1,−1,2) = −1
∂R
∂z (1,−1,2) = 3(−1)(2) = −6
Step 5: Finally, calculate the divergence of Fat the point (1,−1,2):
div(F)(1,−1,2) = ∂P
∂x (1,−1,2)+ ∂Q
∂y (1,−1,2)+ ∂R
∂z (1,−1,2) = −2−1−6 = −9
2
Question 3
Question
Let F(x, y, z) = ⟨2xy, x2+y2,3z⟩be a vector field in R3. Determine the diver-
gence of F.
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩in R3, where P,Q, and
Rare the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Given F(x, y, z) = ⟨2xy, x2+y2,3z⟩, we have P(x, y, z)=2xy,
Q(x, y, z) = x2+y2, and R(x, y, z)=3z.
Step 2: Calculate the partial derivatives:
∂P
∂x = 2y, ∂Q
∂y = 2y, ∂R
∂z = 3
Step 3: Now, find the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2y+ 2y+ 3 = 4y+ 3
Therefore, the divergence of the vector field Fis div(F)=4y+ 3.
Question 4
Question
Let F(x, y, z)=(x2y, xy2, xyz). Find the divergence of F.
Solution
To find the divergence of F, denoted as ∇ · F, we need to compute the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) and F.
Step 1: Compute the del operator:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Compute the components of F:
F(x, y, z)=(x2y, xy2, xyz)
3
Step 3: Find the divergence of F:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, xy2, xyz)
Step 4: Compute the dot product:
∇ · F=∂
∂x (x2y) + ∂
∂y (xy2) + ∂
∂z (xyz)
Step 5: Compute the partial derivatives:
∂
∂x (x2y)=2xy, ∂
∂y (xy2) = x2,∂
∂z (xyz) = xy
Step 6: Sum up the components:
∇ · F= 2xy +x2+xy
Step 7: Simplify the expression:
∇ · F=x2+ 3xy
Therefore, the divergence of Fis x2+ 3xy.
Question 5
Question
Let F(x, y, z) = yi+xj−zkbe a vector field in R3. Calculate the gradient,
divergence, and curl of F.
Solution
Step 1: Calculate the gradient of F. The gradient of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x i+∂N
∂y j+∂P
∂z k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂
∂x (y)i+∂
∂y (x)j+∂
∂z (−z)k.
Calculating the partial derivatives, we get:
∇ · F=j−k
Step 2: Calculate the divergence of F. The divergence of a vector field
F(x, y, z) = M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z
4
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂y
∂x +∂x
∂y +∂(−z)
∂z .
Calculating the partial derivatives, we get:
∇ · F= 0
Step 3: Calculate the curl of F. The curl of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ × F=∂P
∂y −∂N
∂z i−∂P
∂x −∂M
∂z j+∂N
∂x −∂M
∂y k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ × F=∂(−z)
∂y −∂x
∂z i−∂(−z)
∂x −∂y
∂z j+∂x
∂x −∂y
∂y k.
Calculating the partial derivatives, we get:
∇ × F=j+i
Question 6
Question
Let F(x, y, z) = x2yi+ 2xyzj+xy2k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2xyz xy2
∇ × F= (0 −0) i−(0 −2z)j+ (2y−2y)k
∇ × F=−2zj
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · (−2zj) = ∂
∂x (0) + ∂
∂y (−2z) + ∂
∂z (0)
∇ · ∇ × F=−2∂z
∂y =−2
Question 7
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field. Compute the curl of F.
5
Solution
To compute the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and Rfrom F(x, y, z). Here, P(x, y, z) = x2y,
Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Compute the partial derivatives necessary for the curl formula.
∂P
∂y =x2,∂Q
∂z =xy,
∂R
∂x = 2xz, ∂P
∂z = 0,
∂Q
∂x =yz, ∂P
∂y = 2xy.
Step 3: Substitute the partial derivatives into the formula for the curl.
∇ × F= (0 −xy)i−(2xz −0) j+ (yz −2xy)k
Step 4: Simplify the components of the curl.
∇ × F=−xyi−2xzj+ (yz −2xy)k
Therefore, the curl of the vector field F(x, y, z) is ∇ × F=−xyi−2xzj+
(yz −2xy)k.
Question 8
Question
Let
F=⟨x2, yz, z3⟩. Compute ∇ · ∇ ·
F.
Solution
Step 1: Compute ∇ ·
F. The divergence of a vector field
F=⟨P, Q, R⟩is given
by ∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z . In this case,
F=⟨x2, yz, z3⟩, so:
∇ ·
F=∂
∂x (x2) + ∂
∂y (yz) + ∂
∂z (z3)
∇ ·
F= 2x+z
6
Step 2: Compute ∇·∇·
F. The diverence of a vector field
G=∇·
Fis again
given by ∇ ·
G=∂
∂x (∇ ·
F) + ∂
∂y (∇ ·
F) + ∂
∂z (∇ ·
F). Substitute ∇ ·
F= 2x+z
into the formula:
∇ · ∇ ·
F=∂
∂x (2x+z) + ∂
∂y (2x+z) + ∂
∂z (2x+z)
∇ · ∇ ·
F= 2 + 0 + 1
∇ · ∇ ·
F= 3
Therefore, the value of ∇ · ∇ ·
Fis 3 .
Question 9
Question
Let F(x, y, z) = x
x2+y2i+y
x2+y2j+zk. Evaluate ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x
x2+y2
y
x2+y2z
=∂
∂y (z)−∂
∂z y
x2+y2i−∂
∂x (z)−∂
∂z x
x2+y2j+∂
∂x y
x2+y2−∂
∂y x
x2+y2k
=−2yx
(x2+y2)2i+2xy
(x2+y2)2j+ 0k
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∂
∂x −2yx
(x2+y2)2+∂
∂y 2xy
(x2+y2)2
=2y(x2−y2)
(x2+y2)3+2x(y2−x2)
(x2+y2)3
=2xy(x2−y2)−2yx(y2−x2)
(x2+y2)3
= 0
Therefore, ∇ · ∇ × F= 0 .
7
Question 10
Question
Let F(x, y, z) = xzi+yzj+xykbe a vector field in R3. Compute ∇ · (∇ × F),
where ∇represents the gradient operator and ×represents the cross product.
Solution
Step 1: Compute ∇ × F.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
xz yz xy
=∂(xy)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(xy)
∂z j+∂(yz)
∂x −∂(xz)
∂y k
= (x−y)i−(z−x)j+ (y−z)k
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (x−y) + ∂
∂y (−z+x) + ∂
∂z (y−z)
= 1 −1 + 1 = 1
Therefore, ∇ · (∇ × F) = 1 .
Question 11
Question
Let F(x, y, z)=(x2yz, xz2, y2z) be a vector field. Find the divergence of Fat
the point (1,2,3).
Solution
To find the divergence of a vector field F(x, y, z)=(P(x, y, z), Q(x, y, z), R(x, y, z)),
we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z)=(x2yz, xz2, y2z), we have P(x, y, z) = x2yz,Q(x, y, z) =
xz2, and R(x, y, z) = y2z.
Step 1: Compute the partial derivatives:
∂P
∂x = 2xyz, ∂Q
∂y = 0,∂R
∂z =y2
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂P
∂x
(1,2,3)
= 2(1)(2)(3) = 12,
8
∂Q
∂y
(1,2,3)
= 0,
∂R
∂z
(1,2,3)
= (2)2= 4.
Step 3: Find the divergence of Fat the point (1,2,3):
div(F)
(1,2,3)
= 12 + 0 + 4 = 16.
Therefore, the divergence of Fat the point (1,2,3) is 16.
Question 12
Question
Let F(x, y, z)=(ey+z2)i+ (xez−2y)j+ (2z+x2)kbe a vector field. Compute
∇ · ∇ × F.
Solution
Recall the identities ∇ · (∇ × F) = 0 for any vector field F.
We will first calculate ∇ × Fusing the formula
∇ × F=∂Fz
∂y −∂Fy
∂z i+∂Fx
∂z −∂Fz
∂x j+∂Fy
∂x −∂Fx
∂y k.
Step 1: Calculate ∇ × F
We have:
∇ × F=∂(2z+x2)
∂y −∂(xez−2y)
∂z i
+∂(ey+z2)
∂z −∂(2z+x2)
∂x j
+∂(xez−2y)
∂x −∂(ey+z2)
∂y k
= (0 −x)i+ (2z−0)j+ (ez−0)k
=−xi+ 2zj+ezk.
Step 2: Compute ∇ · (∇ × F)
Using the divergence of a vector field formula, we have:
∇ · (∇ × F) = ∂
∂x (−x) + ∂
∂y (2z) + ∂
∂z (ez).
∇ · (∇ × F) = −1 + 0 + ez=ez−1.
Therefore, ∇ · ∇ × F=ez−1 .
9
Question 13
Question
Let F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k. Compute the divergence
of F.
Solution
Step 1: The divergence of a vector field F=F1i+F2j+F3kis given by
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z .
Step 2: In this case, F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k.
Step 3: Let’s compute the partial derivatives of F1,F2, and F3:
∂F1
∂x =y2z2−2x2yz
(x2+y2+z2)2
∂F2
∂y =x2z2−2y2zx
(x2+y2+z2)2
∂F3
∂z =x2y2−2z2xy
(x2+y2+z2)2
Step 4: Now, we can compute the divergence of Fusing the formula:
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z
Step 5: Substitute the partial derivatives we calculated:
∇ · F=y2z2−2x2yz
(x2+y2+z2)2+x2z2−2y2zx
(x2+y2+z2)2+x2y2−2z2xy
(x2+y2+z2)2
Step 6: Simplifying further, we get:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Therefore, the divergence of Fis:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Question 14
Question
Let F(x, y, z)=(xez+ycos z, x sin z, xyz) be a vector field in R3. Determine
the curl of Fand evaluate it at the point (1, π, 2).
10
Solution
Step 1: Compute the curl of F.The curl of a vector field F= (P, Q, R) in R3
is given by:
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F(x, y, z)=(xez+ycos z, x sin z, xyz), the components are P=xez+ycos z,
Q=xsin z, and R=xyz.
Now, let’s compute the curl of F:
∇×F=∂(xyz)
∂y −∂(xsin z)
∂z ,∂(xez+ycos z)
∂z −∂(xyz)
∂x ,∂(xsin z)
∂x −∂(xez+ycos z)
∂y
Step 2: Compute the partial derivatives.
∂(xyz)
∂y =xz, ∂(xsin z)
∂z =xcos z
∂(xez+ycos z)
∂z =xez−ysin z, ∂(xyz)
∂x =yz
∂(xsin z)
∂x = sin z, ∂(xez+ycos z)
∂y = cos z
Step 3: Evaluate the curl at the point (1, π, 2).Substitute x= 1, y=π, and
z= 2 into the expressions we found:
∇ × F=2π−π, e2−πsin 2,sin 2 −e2= (π, e2−πsin 2,sin 2 −e2)
Therefore, the curl of Fat the point (1, π, 2) is (π, e2−πsin 2,sin 2 −e2).
Question 15
Question
Let F(x, y, z)=(x2y, y2z, z2x) be a vector field in R3. Calculate the divergence
of Fat the point (1,2,3).
Solution
To find the divergence of a vector field F= (M, N, P ) at a point (x0, y0, z0), we
use the formula
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z .
Step 1: Compute the partial derivatives of F: Let M=x2y,N=y2z, and
P=z2x.∂M
∂x = 2xy,
∂N
∂y = 2yz,
∂P
∂z = 2zx.
11
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂M
∂x
(1,2,3) = 2(1)(2) = 4,
∂N
∂y
(1,2,3) = 2(2)(3) = 12,
∂P
∂z
(1,2,3) = 2(3)(1) = 6.
Step 3: Calculate the divergence of Fat the point (1,2,3):
∇ · F
(1,2,3) =∂M
∂x
(1,2,3) +∂N
∂y
(1,2,3) +∂P
∂z
(1,2,3) = 4 + 12 + 6 = 22 .
Question 16
Question
Let F(x, y, z) = x2yi−z3j+y2kbe a vector field. Determine the divergence of
F.
Solution
To find the divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula for divergence: div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Find P,Q, and R. Here, P(x, y, z) = x2y,Q(x, y, z) = 0, and
R(x, y, z) = y2.
Step 2: Calculate the partial derivatives.
∂P
∂x = 2xy
∂Q
∂y = 0
∂R
∂z = 0
Step 3: Find the divergence.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy
Therefore, the divergence of Fis 2xy .
Question 17
Question
Let F(x, y, z)=(xz, x +y, yz) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
12
Solution
To find the curl of Fat the point (1,2,3), we first need to compute the Jacobian
matrix of Fand then evaluate it at the given point.
Step 1: Compute the Jacobian matrix of F The components of the
vector field F(x, y, z)=(xz, x +y, yz) are P(x, y, z) = xz,Q(x, y, z) = x+y,
and R(x, y, z) = yz. The Jacobian matrix of Fis given by:
JF=
∂R
∂y
∂P
∂z
∂Q
∂x
∂P
∂y
∂Q
∂z
∂R
∂x
∂Q
∂y
∂R
∂z
∂P
∂x
Calculating the partial derivatives:
∂P
∂y = 0,∂Q
∂z = 0,∂R
∂x =z
∂P
∂y =x, ∂Q
∂z = 0,∂R
∂x = 0
∂Q
∂y = 1,∂R
∂z =y, ∂P
∂x =z
Therefore, the Jacobian matrix is:
JF=
030
101
021
Step 2: Compute the curl of F at (1,2,3) The curl of Fis given by the
determinant of the Jacobian matrix:
∇ × F= det(JF)=0−3(1) = −3
Therefore, the curl of Fat the point (1,2,3) is −3 .
Question 18
Question
Let F(x, y, z) = ⟨x2y, xy2, xyz⟩be a vector field in R3. Compute the divergence
of Fat the point (1,−1,2).
Solution
To find the divergence of Fat the point (1,−1,2), we first need to compute the
divergence function div(F).
13
Step 1: Compute the divergence of F. The divergence of a vector field
F=⟨M, N, P ⟩is given by:
div(F) = ∂M
∂x +∂N
∂y +∂P
∂z
In this case, F(x, y, z) = ⟨x2y, xy2, xyz⟩. Therefore:
∂M
∂x = 2xy, ∂N
∂y = 2xy, ∂P
∂z =xy
Thus, the divergence of Fis:
div(F) = 2xy + 2xy +xy = 4xy +xy = 5xy
Step 2: Evaluate the divergence at (1,−1,2). Substitute the coordinates of
the point into the divergence function:
div(F)(1,−1,2) = 5(1)(−1) = −5
Therefore, the divergence of Fat the point (1,−1,2) is −5.
Question 19
Question
Let F(x, y, z) = (x2y, yz, xz) be a vector field in R3. Determine the divergence
and curl of F.
Solution
To find the divergence of F(x, y, z), we use the formula for divergence:
div(F) = ∇ · F=∂(x2y)
∂x +∂(yz)
∂y +∂(xz)
∂z
Step 1: Calculate the partial derivatives:
∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z, ∂(xz)
∂z =x
Step 2: Compute the divergence:
div(F)=2xy +z+x
Therefore, the divergence of Fis 2xy +z+x.
Next, we will find the curl of F(x, y, z) using the formula for curl:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y yz xz
14
Step 3: Calculate the determinant:
curl(F) = ∂(xz)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(x2y)
∂z j+∂(x2y)
∂x −∂(yz)
∂y k
Step 4: Calculate the partial derivatives:
∂(xz)
∂y = 0,∂(yz)
∂z =y, ∂(xz)
∂x =z,
∂(x2y)
∂z = 0,∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z
Step 5: Substitute the partial derivatives back into the curl expression:
curl(F) = (0 −y)i−(z−0)j+ (2xy −z)k
Therefore, the curl of Fis −yi−zj+ (2xy −z)k.
Question 20
Question
Let F(x, y, z)=(x2y+yz, xz2+y2, yz2+x2). Find the divergence of F.
Solution
To find the divergence of F, denoted by ∇ · F, we need to compute the dot
product of the gradient operator ∇with the vector field F.
∇ · F=∂
∂x (x2y+yz) + ∂
∂y (xz2+y2) + ∂
∂z (yz2+x2)
Step 1: Compute ∂
∂x (x2y+yz):
∂
∂x (x2y+yz)=2xy
Step 2: Compute ∂
∂y (xz2+y2):
∂
∂y (xz2+y2) = z2+ 2y
Step 3: Compute ∂
∂z (yz2+x2):
∂
∂z (yz2+x2) = 2yz
Therefore, the divergence of Fis:
∇ · F= 2xy +z2+ 2y+ 2yz
15
Question 21
Question
Let F(x, y, z) = x2yi+y2zj+z2xk. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y y2z z2x
= (∂z2x
∂y −∂y2z
∂z )i−(∂z2x
∂x −∂x2y
∂z )j+ ( ∂x2y
∂x −∂y2z
∂y )k
= (0 −2z)i−(z2−0)j+ (2xy −0)k=−2zi−z2j+ 2xyk
Step 2: Find ∇ · (−2zi−z2j+ 2xyk)
∇ · (−2zi−z2j+ 2xyk) = ∂
∂x (−2z) + ∂
∂y (−z2) + ∂
∂z (2xy)
= 0 + 0 + 2x= 2x
Therefore, ∇ · ∇ × F= 2x.
Question 22
Question
Let F(x, y, z) = x3yi+xz2j+xy2zk. Calculate ∇ · ∇ × F.
Solution
Step 1: First, calculate ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x3y xz2xy2z
Step 2: Expand the determinant and compute the cross product:
∇×F=∂(xy2z)
∂y −∂(xz2)
∂z i−∂(x3y)
∂x −∂(xy2z)
∂z j+∂(xz2)
∂x −∂(x3y)
∂y k
= (x2z−2xyz)i−(3x2y−xz)j+ (z2)k
16
Step 3: Next, calculate ∇ · ∇ × F.
∇ · ∇ × F=∂(x2z−2xyz)
∂x +∂(−3x2y+xz)
∂y +∂(z2)
∂z
= (2xz −2yz)+(−3x2+z)+0
= 2xz −2yz −3x2+z
So, ∇ · ∇ × F= 2xz −2yz −3x2+z.
Question 23
Question
Let F(x, y, z) = −yi+xj+2zk
x2+y2+z2. Calculate the curl of F.
Solution
To find the curl of F, we will use the formula for the curl of a vector field
F=Pi+Qj+Rk:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and R
Here, P=−y,Q=x, and R= 2z.
Step 2: Calculate the partial derivatives
We need to find the partial derivatives of P,Q, and R:
∂P
∂y =−1,∂Q
∂z = 0
∂R
∂x = 0,∂P
∂z = 0
∂Q
∂x = 1,∂P
∂y =−1
Step 3: Compute ∇ × F
Plugging these values into the formula for the curl, we get:
∇ × F= (0 −0)i−(0 −0)j+ (1 −(−1))k
∇ × F= 2k
Therefore, the curl of Fis 0i +0j + 2kor simply 2k.
17
Question 24
Question
Let F(x, y, z)=(x2+yz)i+ (y2+xz)j+ (z2+xy)k. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+yz y2+xz z2+xy
=∂(z2+xy)
∂y −∂(y2+xz)
∂z i−∂(z2+xy)
∂x −∂(x2+yz)
∂z j+∂(y2+xz)
∂x −∂(x2+yz)
∂y k
= (x−x)i−(z−z)j+ (y−y)k=0
Step 2: Find ∇ · ∇ × F.
∇ · 0= 0
Therefore, ∇ · ∇ × F= 0 .
Question 25
Question
Let F(x, y, z)=(xy2, xz2,−2yz) be a vector field. Compute the curl of F.
Solution
To compute the curl of F, we need to find the determinant of the following
matrix:
curl(F) =
i j k
∂
∂x
∂
∂y
∂
∂z
xy2xz2−2yz
Step 1: Compute the partial derivatives:
∂
∂x (xy2) = y2,∂
∂y (xz2) = 0 ,∂
∂z (−2yz) = −2y
So, rewrite the determinant with the computed derivatives:
curl(F) =
i j k
y20−2y
Step 2: Expand the determinant to find the curl:
curl(F) = ∂
∂y (−2y)−0i−∂
∂x (−2y)−y2j+y2−0k
18
Step 3: Simplify the curl to get the final answer:
curl(F)=2i+y2j+y2k
Therefore, the curl of Fis F= 2i+y2j+y2k.
Question 26
Question
Let
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k. Compute ∇ · ∇ ×
F.
Solution
Step 1: Compute ∇ ×
F.
The curl of a vector field
F(x, y, z) = P(x, y, z)ˆ
i+Q(x, y, z)ˆ
j+R(x, y, z)ˆ
k
is given by:
∇ ×
F=∂R
∂y −∂Q
∂z ˆ
i+∂P
∂z −∂R
∂x ˆ
j+∂Q
∂x −∂P
∂y ˆ
k
Given
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k, we have:
∇×
F=∂(xey+yz)
∂y −∂(zex+yz)
∂z ˆ
i+∂(yez+xz)
∂z −∂(xey+yz)
∂x ˆ
j+∂(zex+yz)
∂x −∂(yez+xz)
∂y ˆ
k
∇ ×
F= (xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k
Step 2: Compute ∇ · ∇ ×
F.
The divergence of a vector field
G(x, y, z) = M(x, y, z)ˆ
i+N(x, y, z)ˆ
j+
P(x, y, z)ˆ
kis given by:
∇ ·
G=∂M
∂x +∂N
∂y +∂P
∂z
Considering
G(x, y, z)=(xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k, we have:
∇ · ∇ ×
F=∂(xey−z)
∂x +∂(zex+x)
∂y +∂(yez−y)
∂z
∇ · ∇ ×
F= (ey+ 1) + zex−1 + yez
∇ · ∇ ×
F=ey+zex+yez
19
Question 27
Question
Let
F(x, y, z)=(exsin y, cos z, xyz). Calculate the curl of
F.
Solution
To calculate the curl of a vector field
F(x, y, z)=(P, Q, R), we use the formula:
curl
F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
Step 1: Identify P,Q, and Rin
F. Here P=exsin y,Q= cos z, and
R=xyz.
Step 2: Find ∂R
∂y ,∂Q
∂z ,∂P
∂z ,∂R
∂x ,∂Q
∂x , and ∂P
∂y .
∂R
∂y =x
∂Q
∂z = 0
∂P
∂z = 0
∂R
∂x =yz
∂Q
∂x = 0
∂P
∂y =excos y
Step 3: Calculate the curl of
F.
curl
F= (x−0,0−yz, 0−excos y)
curl
F= (x, −yz, −excos y)
Therefore, the curl of the vector field
F(x, y, z) = (exsin y, cos z, xyz) is
(x, −yz, −excos y).
Question 28
Question
Let F(x, y, z) = ⟨y2, xz, xyz⟩be a vector field. Calculate the divergence of F.
20
Solution
To find the divergence of F, denoted as ∇ · F, we will use the formula for the
divergence of a vector field in Cartesian coordinates:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F(x, y, z) = ⟨P, Q, R⟩. In this case, P=y2,Q=xz, and R=xyz.
Step 1: Calculate ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 0,
∂Q
∂y = 0,
∂R
∂z =xy.
Step 2: Sum the partial derivatives to find the divergence of F.
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
=0+0+xy
=xy .
Question 29
Question
Let F(x, y, z) = ⟨x2y, e4z, xy2⟩. Calculate the curl of Fat the point (1,0,2).
Solution
To calculate the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
curl(F) = ∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
Given F(x, y, z) = ⟨x2y, e4z, xy2⟩, we have P(x, y, z) = x2y,Q(x, y, z) = e4z,
and R(x, y, z) = xy2.
Let’s calculate the partial derivatives and evaluate at the point (1,0,2):
∂P
∂z = 0,∂Q
∂x = 0,∂R
∂y = 2x
∂Q
∂z = 4e4z,∂R
∂x =y2,∂P
∂y =x2
21
∇·F=(x2+y2+z2)−x(2x)
(x2+y2+z2)2+(x2+y2+z2)−y(2y)
(x2+y2+z2)2+(x2+y2+z2)−z(2z)
(x2+y2+z2)2
∇ · F=3(x2+y2+z2)−2x2−2y2−2z2
(x2+y2+z2)2
∇ · F=x2+y2+z2
(x2+y2+z2)2
∇ · F=1
x2+y2+z2
Question 2
Question
Let F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzkbe a vector field. Calculate the
divergence of Fat the point (1,−1,2).
Solution
Step 1: The divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: For the given vector field F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzk, we
have:
P(x, y, z) = x2y, Q(x, y, z) = xz2−y, R(x, y, z)=3xyz
Step 3: Now, calculate the partial derivatives of P,Q, and Rwith respect to
x,y, and zrespectively. Then, substitute the point (1,−1,2) into these partial
derivatives. ∂P
∂x = 2xy, ∂Q
∂y =−1,∂R
∂z = 3xy
Step 4: Substitute the point (1,−1,2) into the partial derivatives:
∂P
∂x (1,−1,2) = 2(−1)(1) = −2
∂Q
∂y (1,−1,2) = −1
∂R
∂z (1,−1,2) = 3(−1)(2) = −6
Step 5: Finally, calculate the divergence of Fat the point (1,−1,2):
div(F)(1,−1,2) = ∂P
∂x (1,−1,2)+ ∂Q
∂y (1,−1,2)+ ∂R
∂z (1,−1,2) = −2−1−6 = −9
2
Question 3
Question
Let F(x, y, z) = ⟨2xy, x2+y2,3z⟩be a vector field in R3. Determine the diver-
gence of F.
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩in R3, where P,Q, and
Rare the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Given F(x, y, z) = ⟨2xy, x2+y2,3z⟩, we have P(x, y, z)=2xy,
Q(x, y, z) = x2+y2, and R(x, y, z)=3z.
Step 2: Calculate the partial derivatives:
∂P
∂x = 2y, ∂Q
∂y = 2y, ∂R
∂z = 3
Step 3: Now, find the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2y+ 2y+ 3 = 4y+ 3
Therefore, the divergence of the vector field Fis div(F)=4y+ 3.
Question 4
Question
Let F(x, y, z)=(x2y, xy2, xyz). Find the divergence of F.
Solution
To find the divergence of F, denoted as ∇ · F, we need to compute the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) and F.
Step 1: Compute the del operator:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Compute the components of F:
F(x, y, z)=(x2y, xy2, xyz)
3
Step 3: Find the divergence of F:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, xy2, xyz)
Step 4: Compute the dot product:
∇ · F=∂
∂x (x2y) + ∂
∂y (xy2) + ∂
∂z (xyz)
Step 5: Compute the partial derivatives:
∂
∂x (x2y)=2xy, ∂
∂y (xy2) = x2,∂
∂z (xyz) = xy
Step 6: Sum up the components:
∇ · F= 2xy +x2+xy
Step 7: Simplify the expression:
∇ · F=x2+ 3xy
Therefore, the divergence of Fis x2+ 3xy.
Question 5
Question
Let F(x, y, z) = yi+xj−zkbe a vector field in R3. Calculate the gradient,
divergence, and curl of F.
Solution
Step 1: Calculate the gradient of F. The gradient of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x i+∂N
∂y j+∂P
∂z k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂
∂x (y)i+∂
∂y (x)j+∂
∂z (−z)k.
Calculating the partial derivatives, we get:
∇ · F=j−k
Step 2: Calculate the divergence of F. The divergence of a vector field
F(x, y, z) = M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z
4
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂y
∂x +∂x
∂y +∂(−z)
∂z .
Calculating the partial derivatives, we get:
∇ · F= 0
Step 3: Calculate the curl of F. The curl of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ × F=∂P
∂y −∂N
∂z i−∂P
∂x −∂M
∂z j+∂N
∂x −∂M
∂y k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ × F=∂(−z)
∂y −∂x
∂z i−∂(−z)
∂x −∂y
∂z j+∂x
∂x −∂y
∂y k.
Calculating the partial derivatives, we get:
∇ × F=j+i
Question 6
Question
Let F(x, y, z) = x2yi+ 2xyzj+xy2k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2xyz xy2
∇ × F= (0 −0) i−(0 −2z)j+ (2y−2y)k
∇ × F=−2zj
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · (−2zj) = ∂
∂x (0) + ∂
∂y (−2z) + ∂
∂z (0)
∇ · ∇ × F=−2∂z
∂y =−2
Question 7
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field. Compute the curl of F.
5
Solution
To compute the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and Rfrom F(x, y, z). Here, P(x, y, z) = x2y,
Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Compute the partial derivatives necessary for the curl formula.
∂P
∂y =x2,∂Q
∂z =xy,
∂R
∂x = 2xz, ∂P
∂z = 0,
∂Q
∂x =yz, ∂P
∂y = 2xy.
Step 3: Substitute the partial derivatives into the formula for the curl.
∇ × F= (0 −xy)i−(2xz −0) j+ (yz −2xy)k
Step 4: Simplify the components of the curl.
∇ × F=−xyi−2xzj+ (yz −2xy)k
Therefore, the curl of the vector field F(x, y, z) is ∇ × F=−xyi−2xzj+
(yz −2xy)k.
Question 8
Question
Let
F=⟨x2, yz, z3⟩. Compute ∇ · ∇ ·
F.
Solution
Step 1: Compute ∇ ·
F. The divergence of a vector field
F=⟨P, Q, R⟩is given
by ∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z . In this case,
F=⟨x2, yz, z3⟩, so:
∇ ·
F=∂
∂x (x2) + ∂
∂y (yz) + ∂
∂z (z3)
∇ ·
F= 2x+z
6
Step 2: Compute ∇·∇·
F. The diverence of a vector field
G=∇·
Fis again
given by ∇ ·
G=∂
∂x (∇ ·
F) + ∂
∂y (∇ ·
F) + ∂
∂z (∇ ·
F). Substitute ∇ ·
F= 2x+z
into the formula:
∇ · ∇ ·
F=∂
∂x (2x+z) + ∂
∂y (2x+z) + ∂
∂z (2x+z)
∇ · ∇ ·
F= 2 + 0 + 1
∇ · ∇ ·
F= 3
Therefore, the value of ∇ · ∇ ·
Fis 3 .
Question 9
Question
Let F(x, y, z) = x
x2+y2i+y
x2+y2j+zk. Evaluate ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x
x2+y2
y
x2+y2z
=∂
∂y (z)−∂
∂z y
x2+y2i−∂
∂x (z)−∂
∂z x
x2+y2j+∂
∂x y
x2+y2−∂
∂y x
x2+y2k
=−2yx
(x2+y2)2i+2xy
(x2+y2)2j+ 0k
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∂
∂x −2yx
(x2+y2)2+∂
∂y 2xy
(x2+y2)2
=2y(x2−y2)
(x2+y2)3+2x(y2−x2)
(x2+y2)3
=2xy(x2−y2)−2yx(y2−x2)
(x2+y2)3
= 0
Therefore, ∇ · ∇ × F= 0 .
7
Question 10
Question
Let F(x, y, z) = xzi+yzj+xykbe a vector field in R3. Compute ∇ · (∇ × F),
where ∇represents the gradient operator and ×represents the cross product.
Solution
Step 1: Compute ∇ × F.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
xz yz xy
=∂(xy)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(xy)
∂z j+∂(yz)
∂x −∂(xz)
∂y k
= (x−y)i−(z−x)j+ (y−z)k
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (x−y) + ∂
∂y (−z+x) + ∂
∂z (y−z)
= 1 −1 + 1 = 1
Therefore, ∇ · (∇ × F) = 1 .
Question 11
Question
Let F(x, y, z)=(x2yz, xz2, y2z) be a vector field. Find the divergence of Fat
the point (1,2,3).
Solution
To find the divergence of a vector field F(x, y, z)=(P(x, y, z), Q(x, y, z), R(x, y, z)),
we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z)=(x2yz, xz2, y2z), we have P(x, y, z) = x2yz,Q(x, y, z) =
xz2, and R(x, y, z) = y2z.
Step 1: Compute the partial derivatives:
∂P
∂x = 2xyz, ∂Q
∂y = 0,∂R
∂z =y2
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂P
∂x
(1,2,3)
= 2(1)(2)(3) = 12,
8
∂Q
∂y
(1,2,3)
= 0,
∂R
∂z
(1,2,3)
= (2)2= 4.
Step 3: Find the divergence of Fat the point (1,2,3):
div(F)
(1,2,3)
= 12 + 0 + 4 = 16.
Therefore, the divergence of Fat the point (1,2,3) is 16.
Question 12
Question
Let F(x, y, z)=(ey+z2)i+ (xez−2y)j+ (2z+x2)kbe a vector field. Compute
∇ · ∇ × F.
Solution
Recall the identities ∇ · (∇ × F) = 0 for any vector field F.
We will first calculate ∇ × Fusing the formula
∇ × F=∂Fz
∂y −∂Fy
∂z i+∂Fx
∂z −∂Fz
∂x j+∂Fy
∂x −∂Fx
∂y k.
Step 1: Calculate ∇ × F
We have:
∇ × F=∂(2z+x2)
∂y −∂(xez−2y)
∂z i
+∂(ey+z2)
∂z −∂(2z+x2)
∂x j
+∂(xez−2y)
∂x −∂(ey+z2)
∂y k
= (0 −x)i+ (2z−0)j+ (ez−0)k
=−xi+ 2zj+ezk.
Step 2: Compute ∇ · (∇ × F)
Using the divergence of a vector field formula, we have:
∇ · (∇ × F) = ∂
∂x (−x) + ∂
∂y (2z) + ∂
∂z (ez).
∇ · (∇ × F) = −1 + 0 + ez=ez−1.
Therefore, ∇ · ∇ × F=ez−1 .
9
Question 13
Question
Let F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k. Compute the divergence
of F.
Solution
Step 1: The divergence of a vector field F=F1i+F2j+F3kis given by
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z .
Step 2: In this case, F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k.
Step 3: Let’s compute the partial derivatives of F1,F2, and F3:
∂F1
∂x =y2z2−2x2yz
(x2+y2+z2)2
∂F2
∂y =x2z2−2y2zx
(x2+y2+z2)2
∂F3
∂z =x2y2−2z2xy
(x2+y2+z2)2
Step 4: Now, we can compute the divergence of Fusing the formula:
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z
Step 5: Substitute the partial derivatives we calculated:
∇ · F=y2z2−2x2yz
(x2+y2+z2)2+x2z2−2y2zx
(x2+y2+z2)2+x2y2−2z2xy
(x2+y2+z2)2
Step 6: Simplifying further, we get:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Therefore, the divergence of Fis:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Question 14
Question
Let F(x, y, z)=(xez+ycos z, x sin z, xyz) be a vector field in R3. Determine
the curl of Fand evaluate it at the point (1, π, 2).
10
Solution
Step 1: Compute the curl of F.The curl of a vector field F= (P, Q, R) in R3
is given by:
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F(x, y, z)=(xez+ycos z, x sin z, xyz), the components are P=xez+ycos z,
Q=xsin z, and R=xyz.
Now, let’s compute the curl of F:
∇×F=∂(xyz)
∂y −∂(xsin z)
∂z ,∂(xez+ycos z)
∂z −∂(xyz)
∂x ,∂(xsin z)
∂x −∂(xez+ycos z)
∂y
Step 2: Compute the partial derivatives.
∂(xyz)
∂y =xz, ∂(xsin z)
∂z =xcos z
∂(xez+ycos z)
∂z =xez−ysin z, ∂(xyz)
∂x =yz
∂(xsin z)
∂x = sin z, ∂(xez+ycos z)
∂y = cos z
Step 3: Evaluate the curl at the point (1, π, 2).Substitute x= 1, y=π, and
z= 2 into the expressions we found:
∇ × F=2π−π, e2−πsin 2,sin 2 −e2= (π, e2−πsin 2,sin 2 −e2)
Therefore, the curl of Fat the point (1, π, 2) is (π, e2−πsin 2,sin 2 −e2).
Question 15
Question
Let F(x, y, z)=(x2y, y2z, z2x) be a vector field in R3. Calculate the divergence
of Fat the point (1,2,3).
Solution
To find the divergence of a vector field F= (M, N, P ) at a point (x0, y0, z0), we
use the formula
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z .
Step 1: Compute the partial derivatives of F: Let M=x2y,N=y2z, and
P=z2x.∂M
∂x = 2xy,
∂N
∂y = 2yz,
∂P
∂z = 2zx.
11
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂M
∂x
(1,2,3) = 2(1)(2) = 4,
∂N
∂y
(1,2,3) = 2(2)(3) = 12,
∂P
∂z
(1,2,3) = 2(3)(1) = 6.
Step 3: Calculate the divergence of Fat the point (1,2,3):
∇ · F
(1,2,3) =∂M
∂x
(1,2,3) +∂N
∂y
(1,2,3) +∂P
∂z
(1,2,3) = 4 + 12 + 6 = 22 .
Question 16
Question
Let F(x, y, z) = x2yi−z3j+y2kbe a vector field. Determine the divergence of
F.
Solution
To find the divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula for divergence: div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Find P,Q, and R. Here, P(x, y, z) = x2y,Q(x, y, z) = 0, and
R(x, y, z) = y2.
Step 2: Calculate the partial derivatives.
∂P
∂x = 2xy
∂Q
∂y = 0
∂R
∂z = 0
Step 3: Find the divergence.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy
Therefore, the divergence of Fis 2xy .
Question 17
Question
Let F(x, y, z)=(xz, x +y, yz) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
12
Solution
To find the curl of Fat the point (1,2,3), we first need to compute the Jacobian
matrix of Fand then evaluate it at the given point.
Step 1: Compute the Jacobian matrix of F The components of the
vector field F(x, y, z)=(xz, x +y, yz) are P(x, y, z) = xz,Q(x, y, z) = x+y,
and R(x, y, z) = yz. The Jacobian matrix of Fis given by:
JF=
∂R
∂y
∂P
∂z
∂Q
∂x
∂P
∂y
∂Q
∂z
∂R
∂x
∂Q
∂y
∂R
∂z
∂P
∂x
Calculating the partial derivatives:
∂P
∂y = 0,∂Q
∂z = 0,∂R
∂x =z
∂P
∂y =x, ∂Q
∂z = 0,∂R
∂x = 0
∂Q
∂y = 1,∂R
∂z =y, ∂P
∂x =z
Therefore, the Jacobian matrix is:
JF=
030
101
021
Step 2: Compute the curl of F at (1,2,3) The curl of Fis given by the
determinant of the Jacobian matrix:
∇ × F= det(JF)=0−3(1) = −3
Therefore, the curl of Fat the point (1,2,3) is −3 .
Question 18
Question
Let F(x, y, z) = ⟨x2y, xy2, xyz⟩be a vector field in R3. Compute the divergence
of Fat the point (1,−1,2).
Solution
To find the divergence of Fat the point (1,−1,2), we first need to compute the
divergence function div(F).
13
Step 1: Compute the divergence of F. The divergence of a vector field
F=⟨M, N, P ⟩is given by:
div(F) = ∂M
∂x +∂N
∂y +∂P
∂z
In this case, F(x, y, z) = ⟨x2y, xy2, xyz⟩. Therefore:
∂M
∂x = 2xy, ∂N
∂y = 2xy, ∂P
∂z =xy
Thus, the divergence of Fis:
div(F) = 2xy + 2xy +xy = 4xy +xy = 5xy
Step 2: Evaluate the divergence at (1,−1,2). Substitute the coordinates of
the point into the divergence function:
div(F)(1,−1,2) = 5(1)(−1) = −5
Therefore, the divergence of Fat the point (1,−1,2) is −5.
Question 19
Question
Let F(x, y, z) = (x2y, yz, xz) be a vector field in R3. Determine the divergence
and curl of F.
Solution
To find the divergence of F(x, y, z), we use the formula for divergence:
div(F) = ∇ · F=∂(x2y)
∂x +∂(yz)
∂y +∂(xz)
∂z
Step 1: Calculate the partial derivatives:
∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z, ∂(xz)
∂z =x
Step 2: Compute the divergence:
div(F)=2xy +z+x
Therefore, the divergence of Fis 2xy +z+x.
Next, we will find the curl of F(x, y, z) using the formula for curl:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y yz xz
14
Step 3: Calculate the determinant:
curl(F) = ∂(xz)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(x2y)
∂z j+∂(x2y)
∂x −∂(yz)
∂y k
Step 4: Calculate the partial derivatives:
∂(xz)
∂y = 0,∂(yz)
∂z =y, ∂(xz)
∂x =z,
∂(x2y)
∂z = 0,∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z
Step 5: Substitute the partial derivatives back into the curl expression:
curl(F) = (0 −y)i−(z−0)j+ (2xy −z)k
Therefore, the curl of Fis −yi−zj+ (2xy −z)k.
Question 20
Question
Let F(x, y, z)=(x2y+yz, xz2+y2, yz2+x2). Find the divergence of F.
Solution
To find the divergence of F, denoted by ∇ · F, we need to compute the dot
product of the gradient operator ∇with the vector field F.
∇ · F=∂
∂x (x2y+yz) + ∂
∂y (xz2+y2) + ∂
∂z (yz2+x2)
Step 1: Compute ∂
∂x (x2y+yz):
∂
∂x (x2y+yz)=2xy
Step 2: Compute ∂
∂y (xz2+y2):
∂
∂y (xz2+y2) = z2+ 2y
Step 3: Compute ∂
∂z (yz2+x2):
∂
∂z (yz2+x2) = 2yz
Therefore, the divergence of Fis:
∇ · F= 2xy +z2+ 2y+ 2yz
15
Question 21
Question
Let F(x, y, z) = x2yi+y2zj+z2xk. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y y2z z2x
= (∂z2x
∂y −∂y2z
∂z )i−(∂z2x
∂x −∂x2y
∂z )j+ ( ∂x2y
∂x −∂y2z
∂y )k
= (0 −2z)i−(z2−0)j+ (2xy −0)k=−2zi−z2j+ 2xyk
Step 2: Find ∇ · (−2zi−z2j+ 2xyk)
∇ · (−2zi−z2j+ 2xyk) = ∂
∂x (−2z) + ∂
∂y (−z2) + ∂
∂z (2xy)
= 0 + 0 + 2x= 2x
Therefore, ∇ · ∇ × F= 2x.
Question 22
Question
Let F(x, y, z) = x3yi+xz2j+xy2zk. Calculate ∇ · ∇ × F.
Solution
Step 1: First, calculate ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x3y xz2xy2z
Step 2: Expand the determinant and compute the cross product:
∇×F=∂(xy2z)
∂y −∂(xz2)
∂z i−∂(x3y)
∂x −∂(xy2z)
∂z j+∂(xz2)
∂x −∂(x3y)
∂y k
= (x2z−2xyz)i−(3x2y−xz)j+ (z2)k
16
Step 3: Next, calculate ∇ · ∇ × F.
∇ · ∇ × F=∂(x2z−2xyz)
∂x +∂(−3x2y+xz)
∂y +∂(z2)
∂z
= (2xz −2yz)+(−3x2+z)+0
= 2xz −2yz −3x2+z
So, ∇ · ∇ × F= 2xz −2yz −3x2+z.
Question 23
Question
Let F(x, y, z) = −yi+xj+2zk
x2+y2+z2. Calculate the curl of F.
Solution
To find the curl of F, we will use the formula for the curl of a vector field
F=Pi+Qj+Rk:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and R
Here, P=−y,Q=x, and R= 2z.
Step 2: Calculate the partial derivatives
We need to find the partial derivatives of P,Q, and R:
∂P
∂y =−1,∂Q
∂z = 0
∂R
∂x = 0,∂P
∂z = 0
∂Q
∂x = 1,∂P
∂y =−1
Step 3: Compute ∇ × F
Plugging these values into the formula for the curl, we get:
∇ × F= (0 −0)i−(0 −0)j+ (1 −(−1))k
∇ × F= 2k
Therefore, the curl of Fis 0i +0j + 2kor simply 2k.
17
Question 24
Question
Let F(x, y, z)=(x2+yz)i+ (y2+xz)j+ (z2+xy)k. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+yz y2+xz z2+xy
=∂(z2+xy)
∂y −∂(y2+xz)
∂z i−∂(z2+xy)
∂x −∂(x2+yz)
∂z j+∂(y2+xz)
∂x −∂(x2+yz)
∂y k
= (x−x)i−(z−z)j+ (y−y)k=0
Step 2: Find ∇ · ∇ × F.
∇ · 0= 0
Therefore, ∇ · ∇ × F= 0 .
Question 25
Question
Let F(x, y, z)=(xy2, xz2,−2yz) be a vector field. Compute the curl of F.
Solution
To compute the curl of F, we need to find the determinant of the following
matrix:
curl(F) =
i j k
∂
∂x
∂
∂y
∂
∂z
xy2xz2−2yz
Step 1: Compute the partial derivatives:
∂
∂x (xy2) = y2,∂
∂y (xz2) = 0 ,∂
∂z (−2yz) = −2y
So, rewrite the determinant with the computed derivatives:
curl(F) =
i j k
y20−2y
Step 2: Expand the determinant to find the curl:
curl(F) = ∂
∂y (−2y)−0i−∂
∂x (−2y)−y2j+y2−0k
18
Step 3: Simplify the curl to get the final answer:
curl(F)=2i+y2j+y2k
Therefore, the curl of Fis F= 2i+y2j+y2k.
Question 26
Question
Let
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k. Compute ∇ · ∇ ×
F.
Solution
Step 1: Compute ∇ ×
F.
The curl of a vector field
F(x, y, z) = P(x, y, z)ˆ
i+Q(x, y, z)ˆ
j+R(x, y, z)ˆ
k
is given by:
∇ ×
F=∂R
∂y −∂Q
∂z ˆ
i+∂P
∂z −∂R
∂x ˆ
j+∂Q
∂x −∂P
∂y ˆ
k
Given
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k, we have:
∇×
F=∂(xey+yz)
∂y −∂(zex+yz)
∂z ˆ
i+∂(yez+xz)
∂z −∂(xey+yz)
∂x ˆ
j+∂(zex+yz)
∂x −∂(yez+xz)
∂y ˆ
k
∇ ×
F= (xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k
Step 2: Compute ∇ · ∇ ×
F.
The divergence of a vector field
G(x, y, z) = M(x, y, z)ˆ
i+N(x, y, z)ˆ
j+
P(x, y, z)ˆ
kis given by:
∇ ·
G=∂M
∂x +∂N
∂y +∂P
∂z
Considering
G(x, y, z)=(xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k, we have:
∇ · ∇ ×
F=∂(xey−z)
∂x +∂(zex+x)
∂y +∂(yez−y)
∂z
∇ · ∇ ×
F= (ey+ 1) + zex−1 + yez
∇ · ∇ ×
F=ey+zex+yez
19
Question 27
Question
Let
F(x, y, z)=(exsin y, cos z, xyz). Calculate the curl of
F.
Solution
To calculate the curl of a vector field
F(x, y, z)=(P, Q, R), we use the formula:
curl
F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
Step 1: Identify P,Q, and Rin
F. Here P=exsin y,Q= cos z, and
R=xyz.
Step 2: Find ∂R
∂y ,∂Q
∂z ,∂P
∂z ,∂R
∂x ,∂Q
∂x , and ∂P
∂y .
∂R
∂y =x
∂Q
∂z = 0
∂P
∂z = 0
∂R
∂x =yz
∂Q
∂x = 0
∂P
∂y =excos y
Step 3: Calculate the curl of
F.
curl
F= (x−0,0−yz, 0−excos y)
curl
F= (x, −yz, −excos y)
Therefore, the curl of the vector field
F(x, y, z) = (exsin y, cos z, xyz) is
(x, −yz, −excos y).
Question 28
Question
Let F(x, y, z) = ⟨y2, xz, xyz⟩be a vector field. Calculate the divergence of F.
20
Solution
To find the divergence of F, denoted as ∇ · F, we will use the formula for the
divergence of a vector field in Cartesian coordinates:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F(x, y, z) = ⟨P, Q, R⟩. In this case, P=y2,Q=xz, and R=xyz.
Step 1: Calculate ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 0,
∂Q
∂y = 0,
∂R
∂z =xy.
Step 2: Sum the partial derivatives to find the divergence of F.
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
=0+0+xy
=xy .
Question 29
Question
Let F(x, y, z) = ⟨x2y, e4z, xy2⟩. Calculate the curl of Fat the point (1,0,2).
Solution
To calculate the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
curl(F) = ∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
Given F(x, y, z) = ⟨x2y, e4z, xy2⟩, we have P(x, y, z) = x2y,Q(x, y, z) = e4z,
and R(x, y, z) = xy2.
Let’s calculate the partial derivatives and evaluate at the point (1,0,2):
∂P
∂z = 0,∂Q
∂x = 0,∂R
∂y = 2x
∂Q
∂z = 4e4z,∂R
∂x =y2,∂P
∂y =x2
21
∇·F=(x2+y2+z2)−x(2x)
(x2+y2+z2)2+(x2+y2+z2)−y(2y)
(x2+y2+z2)2+(x2+y2+z2)−z(2z)
(x2+y2+z2)2
∇ · F=3(x2+y2+z2)−2x2−2y2−2z2
(x2+y2+z2)2
∇ · F=x2+y2+z2
(x2+y2+z2)2
∇ · F=1
x2+y2+z2
Question 2
Question
Let F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzkbe a vector field. Calculate the
divergence of Fat the point (1,−1,2).
Solution
Step 1: The divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: For the given vector field F(x, y, z) = x2yi+ (xz2−y)j+ 3xyzk, we
have:
P(x, y, z) = x2y, Q(x, y, z) = xz2−y, R(x, y, z)=3xyz
Step 3: Now, calculate the partial derivatives of P,Q, and Rwith respect to
x,y, and zrespectively. Then, substitute the point (1,−1,2) into these partial
derivatives. ∂P
∂x = 2xy, ∂Q
∂y =−1,∂R
∂z = 3xy
Step 4: Substitute the point (1,−1,2) into the partial derivatives:
∂P
∂x (1,−1,2) = 2(−1)(1) = −2
∂Q
∂y (1,−1,2) = −1
∂R
∂z (1,−1,2) = 3(−1)(2) = −6
Step 5: Finally, calculate the divergence of Fat the point (1,−1,2):
div(F)(1,−1,2) = ∂P
∂x (1,−1,2)+ ∂Q
∂y (1,−1,2)+ ∂R
∂z (1,−1,2) = −2−1−6 = −9
2
Question 3
Question
Let F(x, y, z) = ⟨2xy, x2+y2,3z⟩be a vector field in R3. Determine the diver-
gence of F.
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩in R3, where P,Q, and
Rare the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Given F(x, y, z) = ⟨2xy, x2+y2,3z⟩, we have P(x, y, z)=2xy,
Q(x, y, z) = x2+y2, and R(x, y, z)=3z.
Step 2: Calculate the partial derivatives:
∂P
∂x = 2y, ∂Q
∂y = 2y, ∂R
∂z = 3
Step 3: Now, find the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2y+ 2y+ 3 = 4y+ 3
Therefore, the divergence of the vector field Fis div(F)=4y+ 3.
Question 4
Question
Let F(x, y, z)=(x2y, xy2, xyz). Find the divergence of F.
Solution
To find the divergence of F, denoted as ∇ · F, we need to compute the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) and F.
Step 1: Compute the del operator:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Compute the components of F:
F(x, y, z)=(x2y, xy2, xyz)
3
Step 3: Find the divergence of F:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, xy2, xyz)
Step 4: Compute the dot product:
∇ · F=∂
∂x (x2y) + ∂
∂y (xy2) + ∂
∂z (xyz)
Step 5: Compute the partial derivatives:
∂
∂x (x2y)=2xy, ∂
∂y (xy2) = x2,∂
∂z (xyz) = xy
Step 6: Sum up the components:
∇ · F= 2xy +x2+xy
Step 7: Simplify the expression:
∇ · F=x2+ 3xy
Therefore, the divergence of Fis x2+ 3xy.
Question 5
Question
Let F(x, y, z) = yi+xj−zkbe a vector field in R3. Calculate the gradient,
divergence, and curl of F.
Solution
Step 1: Calculate the gradient of F. The gradient of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x i+∂N
∂y j+∂P
∂z k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂
∂x (y)i+∂
∂y (x)j+∂
∂z (−z)k.
Calculating the partial derivatives, we get:
∇ · F=j−k
Step 2: Calculate the divergence of F. The divergence of a vector field
F(x, y, z) = M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z
4
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ · F=∂y
∂x +∂x
∂y +∂(−z)
∂z .
Calculating the partial derivatives, we get:
∇ · F= 0
Step 3: Calculate the curl of F. The curl of a vector field F(x, y, z) =
M(x, y, z)i+N(x, y, z)j+P(x, y, z)kis given by:
∇ × F=∂P
∂y −∂N
∂z i−∂P
∂x −∂M
∂z j+∂N
∂x −∂M
∂y k
In this case, F(x, y, z) = yi+xj−zk.
So, ∇ × F=∂(−z)
∂y −∂x
∂z i−∂(−z)
∂x −∂y
∂z j+∂x
∂x −∂y
∂y k.
Calculating the partial derivatives, we get:
∇ × F=j+i
Question 6
Question
Let F(x, y, z) = x2yi+ 2xyzj+xy2k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2xyz xy2
∇ × F= (0 −0) i−(0 −2z)j+ (2y−2y)k
∇ × F=−2zj
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · (−2zj) = ∂
∂x (0) + ∂
∂y (−2z) + ∂
∂z (0)
∇ · ∇ × F=−2∂z
∂y =−2
Question 7
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field. Compute the curl of F.
5
Solution
To compute the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and Rfrom F(x, y, z). Here, P(x, y, z) = x2y,
Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Compute the partial derivatives necessary for the curl formula.
∂P
∂y =x2,∂Q
∂z =xy,
∂R
∂x = 2xz, ∂P
∂z = 0,
∂Q
∂x =yz, ∂P
∂y = 2xy.
Step 3: Substitute the partial derivatives into the formula for the curl.
∇ × F= (0 −xy)i−(2xz −0) j+ (yz −2xy)k
Step 4: Simplify the components of the curl.
∇ × F=−xyi−2xzj+ (yz −2xy)k
Therefore, the curl of the vector field F(x, y, z) is ∇ × F=−xyi−2xzj+
(yz −2xy)k.
Question 8
Question
Let
F=⟨x2, yz, z3⟩. Compute ∇ · ∇ ·
F.
Solution
Step 1: Compute ∇ ·
F. The divergence of a vector field
F=⟨P, Q, R⟩is given
by ∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z . In this case,
F=⟨x2, yz, z3⟩, so:
∇ ·
F=∂
∂x (x2) + ∂
∂y (yz) + ∂
∂z (z3)
∇ ·
F= 2x+z
6
Step 2: Compute ∇·∇·
F. The diverence of a vector field
G=∇·
Fis again
given by ∇ ·
G=∂
∂x (∇ ·
F) + ∂
∂y (∇ ·
F) + ∂
∂z (∇ ·
F). Substitute ∇ ·
F= 2x+z
into the formula:
∇ · ∇ ·
F=∂
∂x (2x+z) + ∂
∂y (2x+z) + ∂
∂z (2x+z)
∇ · ∇ ·
F= 2 + 0 + 1
∇ · ∇ ·
F= 3
Therefore, the value of ∇ · ∇ ·
Fis 3 .
Question 9
Question
Let F(x, y, z) = x
x2+y2i+y
x2+y2j+zk. Evaluate ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x
x2+y2
y
x2+y2z
=∂
∂y (z)−∂
∂z y
x2+y2i−∂
∂x (z)−∂
∂z x
x2+y2j+∂
∂x y
x2+y2−∂
∂y x
x2+y2k
=−2yx
(x2+y2)2i+2xy
(x2+y2)2j+ 0k
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∂
∂x −2yx
(x2+y2)2+∂
∂y 2xy
(x2+y2)2
=2y(x2−y2)
(x2+y2)3+2x(y2−x2)
(x2+y2)3
=2xy(x2−y2)−2yx(y2−x2)
(x2+y2)3
= 0
Therefore, ∇ · ∇ × F= 0 .
7
Question 10
Question
Let F(x, y, z) = xzi+yzj+xykbe a vector field in R3. Compute ∇ · (∇ × F),
where ∇represents the gradient operator and ×represents the cross product.
Solution
Step 1: Compute ∇ × F.
∇×F=
i j k
∂
∂x
∂
∂y
∂
∂z
xz yz xy
=∂(xy)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(xy)
∂z j+∂(yz)
∂x −∂(xz)
∂y k
= (x−y)i−(z−x)j+ (y−z)k
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (x−y) + ∂
∂y (−z+x) + ∂
∂z (y−z)
= 1 −1 + 1 = 1
Therefore, ∇ · (∇ × F) = 1 .
Question 11
Question
Let F(x, y, z)=(x2yz, xz2, y2z) be a vector field. Find the divergence of Fat
the point (1,2,3).
Solution
To find the divergence of a vector field F(x, y, z)=(P(x, y, z), Q(x, y, z), R(x, y, z)),
we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z)=(x2yz, xz2, y2z), we have P(x, y, z) = x2yz,Q(x, y, z) =
xz2, and R(x, y, z) = y2z.
Step 1: Compute the partial derivatives:
∂P
∂x = 2xyz, ∂Q
∂y = 0,∂R
∂z =y2
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂P
∂x
(1,2,3)
= 2(1)(2)(3) = 12,
8
∂Q
∂y
(1,2,3)
= 0,
∂R
∂z
(1,2,3)
= (2)2= 4.
Step 3: Find the divergence of Fat the point (1,2,3):
div(F)
(1,2,3)
= 12 + 0 + 4 = 16.
Therefore, the divergence of Fat the point (1,2,3) is 16.
Question 12
Question
Let F(x, y, z)=(ey+z2)i+ (xez−2y)j+ (2z+x2)kbe a vector field. Compute
∇ · ∇ × F.
Solution
Recall the identities ∇ · (∇ × F) = 0 for any vector field F.
We will first calculate ∇ × Fusing the formula
∇ × F=∂Fz
∂y −∂Fy
∂z i+∂Fx
∂z −∂Fz
∂x j+∂Fy
∂x −∂Fx
∂y k.
Step 1: Calculate ∇ × F
We have:
∇ × F=∂(2z+x2)
∂y −∂(xez−2y)
∂z i
+∂(ey+z2)
∂z −∂(2z+x2)
∂x j
+∂(xez−2y)
∂x −∂(ey+z2)
∂y k
= (0 −x)i+ (2z−0)j+ (ez−0)k
=−xi+ 2zj+ezk.
Step 2: Compute ∇ · (∇ × F)
Using the divergence of a vector field formula, we have:
∇ · (∇ × F) = ∂
∂x (−x) + ∂
∂y (2z) + ∂
∂z (ez).
∇ · (∇ × F) = −1 + 0 + ez=ez−1.
Therefore, ∇ · ∇ × F=ez−1 .
9
Question 13
Question
Let F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k. Compute the divergence
of F.
Solution
Step 1: The divergence of a vector field F=F1i+F2j+F3kis given by
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z .
Step 2: In this case, F(x, y, z) = xyz
x2+y2+z2i+yzx
x2+y2+z2j+zxy
x2+y2+z2k.
Step 3: Let’s compute the partial derivatives of F1,F2, and F3:
∂F1
∂x =y2z2−2x2yz
(x2+y2+z2)2
∂F2
∂y =x2z2−2y2zx
(x2+y2+z2)2
∂F3
∂z =x2y2−2z2xy
(x2+y2+z2)2
Step 4: Now, we can compute the divergence of Fusing the formula:
∇ · F=∂F1
∂x +∂F2
∂y +∂F3
∂z
Step 5: Substitute the partial derivatives we calculated:
∇ · F=y2z2−2x2yz
(x2+y2+z2)2+x2z2−2y2zx
(x2+y2+z2)2+x2y2−2z2xy
(x2+y2+z2)2
Step 6: Simplifying further, we get:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Therefore, the divergence of Fis:
∇ · F=3x2y2+ 3y2z2+ 3z2x2−6x2yz −6y2zx −6z2xy
(x2+y2+z2)2
Question 14
Question
Let F(x, y, z)=(xez+ycos z, x sin z, xyz) be a vector field in R3. Determine
the curl of Fand evaluate it at the point (1, π, 2).
10
Solution
Step 1: Compute the curl of F.The curl of a vector field F= (P, Q, R) in R3
is given by:
∇ × F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
For F(x, y, z)=(xez+ycos z, x sin z, xyz), the components are P=xez+ycos z,
Q=xsin z, and R=xyz.
Now, let’s compute the curl of F:
∇×F=∂(xyz)
∂y −∂(xsin z)
∂z ,∂(xez+ycos z)
∂z −∂(xyz)
∂x ,∂(xsin z)
∂x −∂(xez+ycos z)
∂y
Step 2: Compute the partial derivatives.
∂(xyz)
∂y =xz, ∂(xsin z)
∂z =xcos z
∂(xez+ycos z)
∂z =xez−ysin z, ∂(xyz)
∂x =yz
∂(xsin z)
∂x = sin z, ∂(xez+ycos z)
∂y = cos z
Step 3: Evaluate the curl at the point (1, π, 2).Substitute x= 1, y=π, and
z= 2 into the expressions we found:
∇ × F=2π−π, e2−πsin 2,sin 2 −e2= (π, e2−πsin 2,sin 2 −e2)
Therefore, the curl of Fat the point (1, π, 2) is (π, e2−πsin 2,sin 2 −e2).
Question 15
Question
Let F(x, y, z)=(x2y, y2z, z2x) be a vector field in R3. Calculate the divergence
of Fat the point (1,2,3).
Solution
To find the divergence of a vector field F= (M, N, P ) at a point (x0, y0, z0), we
use the formula
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z .
Step 1: Compute the partial derivatives of F: Let M=x2y,N=y2z, and
P=z2x.∂M
∂x = 2xy,
∂N
∂y = 2yz,
∂P
∂z = 2zx.
11
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂M
∂x
(1,2,3) = 2(1)(2) = 4,
∂N
∂y
(1,2,3) = 2(2)(3) = 12,
∂P
∂z
(1,2,3) = 2(3)(1) = 6.
Step 3: Calculate the divergence of Fat the point (1,2,3):
∇ · F
(1,2,3) =∂M
∂x
(1,2,3) +∂N
∂y
(1,2,3) +∂P
∂z
(1,2,3) = 4 + 12 + 6 = 22 .
Question 16
Question
Let F(x, y, z) = x2yi−z3j+y2kbe a vector field. Determine the divergence of
F.
Solution
To find the divergence of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula for divergence: div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Find P,Q, and R. Here, P(x, y, z) = x2y,Q(x, y, z) = 0, and
R(x, y, z) = y2.
Step 2: Calculate the partial derivatives.
∂P
∂x = 2xy
∂Q
∂y = 0
∂R
∂z = 0
Step 3: Find the divergence.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy
Therefore, the divergence of Fis 2xy .
Question 17
Question
Let F(x, y, z)=(xz, x +y, yz) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
12
Solution
To find the curl of Fat the point (1,2,3), we first need to compute the Jacobian
matrix of Fand then evaluate it at the given point.
Step 1: Compute the Jacobian matrix of F The components of the
vector field F(x, y, z)=(xz, x +y, yz) are P(x, y, z) = xz,Q(x, y, z) = x+y,
and R(x, y, z) = yz. The Jacobian matrix of Fis given by:
JF=
∂R
∂y
∂P
∂z
∂Q
∂x
∂P
∂y
∂Q
∂z
∂R
∂x
∂Q
∂y
∂R
∂z
∂P
∂x
Calculating the partial derivatives:
∂P
∂y = 0,∂Q
∂z = 0,∂R
∂x =z
∂P
∂y =x, ∂Q
∂z = 0,∂R
∂x = 0
∂Q
∂y = 1,∂R
∂z =y, ∂P
∂x =z
Therefore, the Jacobian matrix is:
JF=
030
101
021
Step 2: Compute the curl of F at (1,2,3) The curl of Fis given by the
determinant of the Jacobian matrix:
∇ × F= det(JF)=0−3(1) = −3
Therefore, the curl of Fat the point (1,2,3) is −3 .
Question 18
Question
Let F(x, y, z) = ⟨x2y, xy2, xyz⟩be a vector field in R3. Compute the divergence
of Fat the point (1,−1,2).
Solution
To find the divergence of Fat the point (1,−1,2), we first need to compute the
divergence function div(F).
13
Step 1: Compute the divergence of F. The divergence of a vector field
F=⟨M, N, P ⟩is given by:
div(F) = ∂M
∂x +∂N
∂y +∂P
∂z
In this case, F(x, y, z) = ⟨x2y, xy2, xyz⟩. Therefore:
∂M
∂x = 2xy, ∂N
∂y = 2xy, ∂P
∂z =xy
Thus, the divergence of Fis:
div(F) = 2xy + 2xy +xy = 4xy +xy = 5xy
Step 2: Evaluate the divergence at (1,−1,2). Substitute the coordinates of
the point into the divergence function:
div(F)(1,−1,2) = 5(1)(−1) = −5
Therefore, the divergence of Fat the point (1,−1,2) is −5.
Question 19
Question
Let F(x, y, z) = (x2y, yz, xz) be a vector field in R3. Determine the divergence
and curl of F.
Solution
To find the divergence of F(x, y, z), we use the formula for divergence:
div(F) = ∇ · F=∂(x2y)
∂x +∂(yz)
∂y +∂(xz)
∂z
Step 1: Calculate the partial derivatives:
∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z, ∂(xz)
∂z =x
Step 2: Compute the divergence:
div(F)=2xy +z+x
Therefore, the divergence of Fis 2xy +z+x.
Next, we will find the curl of F(x, y, z) using the formula for curl:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y yz xz
14
Step 3: Calculate the determinant:
curl(F) = ∂(xz)
∂y −∂(yz)
∂z i−∂(xz)
∂x −∂(x2y)
∂z j+∂(x2y)
∂x −∂(yz)
∂y k
Step 4: Calculate the partial derivatives:
∂(xz)
∂y = 0,∂(yz)
∂z =y, ∂(xz)
∂x =z,
∂(x2y)
∂z = 0,∂(x2y)
∂x = 2xy, ∂(yz)
∂y =z
Step 5: Substitute the partial derivatives back into the curl expression:
curl(F) = (0 −y)i−(z−0)j+ (2xy −z)k
Therefore, the curl of Fis −yi−zj+ (2xy −z)k.
Question 20
Question
Let F(x, y, z)=(x2y+yz, xz2+y2, yz2+x2). Find the divergence of F.
Solution
To find the divergence of F, denoted by ∇ · F, we need to compute the dot
product of the gradient operator ∇with the vector field F.
∇ · F=∂
∂x (x2y+yz) + ∂
∂y (xz2+y2) + ∂
∂z (yz2+x2)
Step 1: Compute ∂
∂x (x2y+yz):
∂
∂x (x2y+yz)=2xy
Step 2: Compute ∂
∂y (xz2+y2):
∂
∂y (xz2+y2) = z2+ 2y
Step 3: Compute ∂
∂z (yz2+x2):
∂
∂z (yz2+x2) = 2yz
Therefore, the divergence of Fis:
∇ · F= 2xy +z2+ 2y+ 2yz
15
Question 21
Question
Let F(x, y, z) = x2yi+y2zj+z2xk. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y y2z z2x
= (∂z2x
∂y −∂y2z
∂z )i−(∂z2x
∂x −∂x2y
∂z )j+ ( ∂x2y
∂x −∂y2z
∂y )k
= (0 −2z)i−(z2−0)j+ (2xy −0)k=−2zi−z2j+ 2xyk
Step 2: Find ∇ · (−2zi−z2j+ 2xyk)
∇ · (−2zi−z2j+ 2xyk) = ∂
∂x (−2z) + ∂
∂y (−z2) + ∂
∂z (2xy)
= 0 + 0 + 2x= 2x
Therefore, ∇ · ∇ × F= 2x.
Question 22
Question
Let F(x, y, z) = x3yi+xz2j+xy2zk. Calculate ∇ · ∇ × F.
Solution
Step 1: First, calculate ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x3y xz2xy2z
Step 2: Expand the determinant and compute the cross product:
∇×F=∂(xy2z)
∂y −∂(xz2)
∂z i−∂(x3y)
∂x −∂(xy2z)
∂z j+∂(xz2)
∂x −∂(x3y)
∂y k
= (x2z−2xyz)i−(3x2y−xz)j+ (z2)k
16
Step 3: Next, calculate ∇ · ∇ × F.
∇ · ∇ × F=∂(x2z−2xyz)
∂x +∂(−3x2y+xz)
∂y +∂(z2)
∂z
= (2xz −2yz)+(−3x2+z)+0
= 2xz −2yz −3x2+z
So, ∇ · ∇ × F= 2xz −2yz −3x2+z.
Question 23
Question
Let F(x, y, z) = −yi+xj+2zk
x2+y2+z2. Calculate the curl of F.
Solution
To find the curl of F, we will use the formula for the curl of a vector field
F=Pi+Qj+Rk:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Identify P,Q, and R
Here, P=−y,Q=x, and R= 2z.
Step 2: Calculate the partial derivatives
We need to find the partial derivatives of P,Q, and R:
∂P
∂y =−1,∂Q
∂z = 0
∂R
∂x = 0,∂P
∂z = 0
∂Q
∂x = 1,∂P
∂y =−1
Step 3: Compute ∇ × F
Plugging these values into the formula for the curl, we get:
∇ × F= (0 −0)i−(0 −0)j+ (1 −(−1))k
∇ × F= 2k
Therefore, the curl of Fis 0i +0j + 2kor simply 2k.
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Question 24
Question
Let F(x, y, z)=(x2+yz)i+ (y2+xz)j+ (z2+xy)k. Calculate ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2+yz y2+xz z2+xy
=∂(z2+xy)
∂y −∂(y2+xz)
∂z i−∂(z2+xy)
∂x −∂(x2+yz)
∂z j+∂(y2+xz)
∂x −∂(x2+yz)
∂y k
= (x−x)i−(z−z)j+ (y−y)k=0
Step 2: Find ∇ · ∇ × F.
∇ · 0= 0
Therefore, ∇ · ∇ × F= 0 .
Question 25
Question
Let F(x, y, z)=(xy2, xz2,−2yz) be a vector field. Compute the curl of F.
Solution
To compute the curl of F, we need to find the determinant of the following
matrix:
curl(F) =
i j k
∂
∂x
∂
∂y
∂
∂z
xy2xz2−2yz
Step 1: Compute the partial derivatives:
∂
∂x (xy2) = y2,∂
∂y (xz2) = 0 ,∂
∂z (−2yz) = −2y
So, rewrite the determinant with the computed derivatives:
curl(F) =
i j k
y20−2y
Step 2: Expand the determinant to find the curl:
curl(F) = ∂
∂y (−2y)−0i−∂
∂x (−2y)−y2j+y2−0k
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Step 3: Simplify the curl to get the final answer:
curl(F)=2i+y2j+y2k
Therefore, the curl of Fis F= 2i+y2j+y2k.
Question 26
Question
Let
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k. Compute ∇ · ∇ ×
F.
Solution
Step 1: Compute ∇ ×
F.
The curl of a vector field
F(x, y, z) = P(x, y, z)ˆ
i+Q(x, y, z)ˆ
j+R(x, y, z)ˆ
k
is given by:
∇ ×
F=∂R
∂y −∂Q
∂z ˆ
i+∂P
∂z −∂R
∂x ˆ
j+∂Q
∂x −∂P
∂y ˆ
k
Given
F(x, y, z) = (yez+xz)ˆ
i+ (zex+yz)ˆ
j+ (xey+yz)ˆ
k, we have:
∇×
F=∂(xey+yz)
∂y −∂(zex+yz)
∂z ˆ
i+∂(yez+xz)
∂z −∂(xey+yz)
∂x ˆ
j+∂(zex+yz)
∂x −∂(yez+xz)
∂y ˆ
k
∇ ×
F= (xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k
Step 2: Compute ∇ · ∇ ×
F.
The divergence of a vector field
G(x, y, z) = M(x, y, z)ˆ
i+N(x, y, z)ˆ
j+
P(x, y, z)ˆ
kis given by:
∇ ·
G=∂M
∂x +∂N
∂y +∂P
∂z
Considering
G(x, y, z)=(xey−z)ˆ
i+ (zex+x)ˆ
j+ (yez−y)ˆ
k, we have:
∇ · ∇ ×
F=∂(xey−z)
∂x +∂(zex+x)
∂y +∂(yez−y)
∂z
∇ · ∇ ×
F= (ey+ 1) + zex−1 + yez
∇ · ∇ ×
F=ey+zex+yez
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Question 27
Question
Let
F(x, y, z)=(exsin y, cos z, xyz). Calculate the curl of
F.
Solution
To calculate the curl of a vector field
F(x, y, z)=(P, Q, R), we use the formula:
curl
F=∂R
∂y −∂Q
∂z ,∂P
∂z −∂R
∂x ,∂Q
∂x −∂P
∂y
Step 1: Identify P,Q, and Rin
F. Here P=exsin y,Q= cos z, and
R=xyz.
Step 2: Find ∂R
∂y ,∂Q
∂z ,∂P
∂z ,∂R
∂x ,∂Q
∂x , and ∂P
∂y .
∂R
∂y =x
∂Q
∂z = 0
∂P
∂z = 0
∂R
∂x =yz
∂Q
∂x = 0
∂P
∂y =excos y
Step 3: Calculate the curl of
F.
curl
F= (x−0,0−yz, 0−excos y)
curl
F= (x, −yz, −excos y)
Therefore, the curl of the vector field
F(x, y, z) = (exsin y, cos z, xyz) is
(x, −yz, −excos y).
Question 28
Question
Let F(x, y, z) = ⟨y2, xz, xyz⟩be a vector field. Calculate the divergence of F.
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Solution
To find the divergence of F, denoted as ∇ · F, we will use the formula for the
divergence of a vector field in Cartesian coordinates:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F(x, y, z) = ⟨P, Q, R⟩. In this case, P=y2,Q=xz, and R=xyz.
Step 1: Calculate ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 0,
∂Q
∂y = 0,
∂R
∂z =xy.
Step 2: Sum the partial derivatives to find the divergence of F.
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
=0+0+xy
=xy .
Question 29
Question
Let F(x, y, z) = ⟨x2y, e4z, xy2⟩. Calculate the curl of Fat the point (1,0,2).
Solution
To calculate the curl of a vector field F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)k, we use the formula:
curl(F) = ∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
Given F(x, y, z) = ⟨x2y, e4z, xy2⟩, we have P(x, y, z) = x2y,Q(x, y, z) = e4z,
and R(x, y, z) = xy2.
Let’s calculate the partial derivatives and evaluate at the point (1,0,2):
∂P
∂z = 0,∂Q
∂x = 0,∂R
∂y = 2x
∂Q
∂z = 4e4z,∂R
∂x =y2,∂P
∂y =x2
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Now, we evaluate the curl at (1,0,2):
curl(F) = (2x−0) i+0−y2j+0−x2k
curl(F)=2i+ 0j−k
Therefore, the curl of Fat the point (1,0,2) is ⟨2,0,−1⟩.
Question 30
Question
Let F(x, y, z)=(yexz +zsin y, xz cos y, y cos z). Compute ∇ · F.
Solution
To compute ∇ · F, where F(x, y, z)=(M(x, y, z), N(x, y, z), P (x, y, z)), we use
the formula:
∇ · F=∂M
∂x +∂N
∂y +∂P
∂z
Step 1: Compute ∂M
∂x ,∂N
∂y , and ∂P
∂z .
∂M
∂x =exz
∂N
∂y =−xz sin y
∂P
∂z =−ysin z
Step 2: Add the partial derivatives to find ∇ · F.
∇ · F=exz −xz sin y−ysin z
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