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MATH 117 - ELEMENTS OF MATHEMATICS Gradient, divergence, and curl Question Bank Set 10

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MATH 117 - ELEMENTS OF
MATHEMATICS - Gradient,
divergence, and curl
Question Bank - Set 10
Liberty University
Question 1
Question
Let F(x, y, z) = x2yi+zj+ezkbe a vector field. Determine 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 scalar function
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 2: In this case, F(x, y, z) = x2yi+zj+ezk. So, P(x, y, z) = x2y,
Q(x, y, z) = z, and R(x, y, z) = ez.
Step 3: Now, let’s find the partial derivatives of P,Q, and R:
∂P
∂x = 2xy, ∂Q
∂y = 0,∂R
∂z =ez.
Step 4: Finally, we can calculate the divergence of Fusing the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy +0+ez= 2xy +ez.
Question 2
Question
Let F(x, y, z)=(x2y, y2z, z2x) be a vector field in 3-dimensional space. Com-
pute the divergence of F.
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
In this case, F(x, y, z) = (x2y, y2z, z2x), so P=x2y,Q=y2z, and R=z2x.
Step 1: Compute the partial derivative of Pwith respect to x:
∂P
∂x =∂
∂x (x2y) = 2xy
Step 2: Compute the partial derivative of Qwith respect to y:
∂Q
∂y =∂
∂y (y2z) = 2yz
Step 3: Compute the partial derivative of Rwith respect to z:
∂R
∂z =∂
∂z (z2x)=2zx
Step 4: Now, we can calculate the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy + 2yz + 2zx = 2xy + 2yz + 2zx
Therefore, the divergence of the vector field F(x, y, z)=(x2y, y2z, z2x) is
2xy + 2yz + 2zx.
Question 3
Question
Let F(x, y, z) = (2xy +z2)i+ (x2+z2)j+ (y2+z2)kbe a vector field in R3.
Compute the gradient, divergence, and curl of F.
Solution
Given F(x, y, z) = (2xy +z2)i+(x2+z2)j+(y2+z2)k, we can find the gradient,
divergence, and curl of F.
Step 1: Gradient of F The gradient of a scalar field f(x, y, z) = P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
∇f=∂f
∂x i+∂f
∂y j+∂f
∂z k
2
Applying this to F, we have:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·2xy +z2, x2+z2, y2+z2
= 2x+ 0 + 0 = 2xi
Step 2: Divergence of F The divergence of a vector field F=P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
div F=∂P
∂x +∂Q
∂y +∂R
∂z
For F, we get:
div F=∂(2xy +z2)
∂x +∂(x2+z2)
∂y +∂(y2+z2)
∂z
= 2y+ 0 + 2y= 4y
So, the divergence of Fis 4y.
Step 3: Curl of F The curl of a vector field F=P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by:
curl F=∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
For F, we get:
curl F= (2y−0)i+ (0 −2x)j+ (0 −2y)k
= 2yi−2xj−2yk
Question 4
Question
Let F(x, y, z) = x2yzi+xyz2j+x2y2kbe a vector field. Compute the divergence
of Fat the point (2,−1,3).
Solution
To compute the divergence of the vector field F(x, y, z) = x2yzi+xyz2j+x2y2k
at the point (2,−1,3), we will use the formula for divergence:
div(F) = ∂F1
∂x +∂F2
∂y +∂F3
∂z
3
where F(x, y, z) = F1(x, y, z)i+F2(x, y, z)j+F3(x, y, z)k.
Step 1: Calculate the partial derivatives of each component of F:
∂F1
∂x = 2xyz, ∂F2
∂y =xz2,∂F3
∂z = 2x2y
Step 2: Substitute (x, y, z) = (2,−1,3) into each partial derivative to find
the values at the point:
∂F1
∂x (2,−1,3) = 2(−1)(3) = −6
∂F2
∂y (2,−1,3) = (2)(32) = 18
∂F3
∂z (2,−1,3) = 2(2)2(−1) = −8
Step 3: Add the values of the partial derivatives calculated in Step 2 to
find the divergence of Fat the point (2,−1,3):
div(F)(2,−1,3) = −6 + 18 −8=4
Therefore, the divergence of the vector field Fat the point (2,−1,3) is 4.
Question 5
Question
Let F(x, y, z) = (yz2, xz2, xy2) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
F1F2F3
where F= (F1, F2, F3).
Step 1: Find the partial derivatives of F. We have:
∂F1
∂x =z2,∂F2
∂y =xz2,∂F3
∂z = 2xy
Step 2: Evaluate the curl of Fat the point (1,2,3). Substitute the point
(1,2,3) into the partial derivatives:
∂F1
∂x = 32= 9,∂F2
∂y = 1 ×32= 9,∂F3
∂z = 2 ×1×2=4
4
Step 3: Calculate the curl of F. Using the formula for the curl, we get:
∇ × F=
i j k
9 0 0
0 9 4
= (0 −0)i−(0 −0)j+ (9 −0)k= 9k
Therefore, the curl of Fat the point (1,2,3) is 9k.
Question 6
Question
Let F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x)be a vector field in R3. Find
the divergence of F.
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
functions of x,y, and z, respectively, we use the formula
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Given F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x), we identify P=x2y+
sin(z), Q=xz +y2, and R=ey+ cos(x).
Step 1: Find ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 2xy,
∂Q
∂y = 2y,
∂R
∂z = 0.
Step 2: Calculate the divergence.
) = ∂P ∂x+∂Q
∂y +∂R
∂z =2xy+2y+0=2y(x+1).
Therefore, the divergence of the vector field Fis 2y(x+ 1).
Question 7
Question
Consider the vector field F(x, y, z) = (x2+y2+z2)i+xyzj+ (xy −z2)k. Find
the curl of F.
5
Solution
To find 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,
where P,Q, and Rare the component functions, we use the formula for the
curl:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Compute the partial derivatives of P,Q, and R:
∂P
∂y = 2y, ∂Q
∂z =x, ∂R
∂x = 0,∂P
∂z = 0,
∂Q
∂x =y, ∂R
∂y =x, ∂P
∂y = 2x, ∂Q
∂z =y.
Step 2: Plug these derivatives into the formula for the curl:
∇ × F= (x−y)i−xj+ (y−2x)k
Therefore, the curl of the vector field F(x, y, z) is (x−y)i−xj+ (y−2x)k.
Question 8
Question
Let F(x, y, z) = ⟨x2y, xyz, z2⟩. Compute the curl of F.
Solution
To find the curl of a vector field F=⟨P, Q, R⟩, where P,Q, and Rare functions
of x,y, and z, we use the formula:
curl F=∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k.
In this case, F(x, y, z) = ⟨x2y, xyz, z2⟩, so we have P(x, y, z) = x2y,Q(x, y, z) =
xyz, and R(x, y, z) = z2.
Step 1: Compute the partial derivatives:
∂P
∂y =x2,∂Q
∂z =xy, ∂R
∂x = 0,
∂R
∂y = 0,∂Q
∂x =yz, ∂P
∂z = 0.
Step 2: Plug the derivatives into the formula for the curl:
curl F= (0 −xy)i−(0 −0) j+ (yz −0) k.
Step 3: Simplify the expression:
curl F=−xyi+yzk.
Therefore, the curl of Fis curl F=−xyi+yzk.
6
Question 9
Question
Let F(x, y, z) = x2+y, y2+z, z2+xbe a vector field. Calculate the diver-
gence of Fat the point (1,2,3).
Solution
To find the divergence of Fat the point (1,2,3), we need to calculate the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) with Fand evaluate it at the
point (1,2,3). The divergence of vector field Fis given by:
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
Step 1: Calculate the partial derivatives of F.
∂
∂x (x2+y)=2x
∂
∂y (y2+z)=2y
∂
∂z (z2+x)=1
Step 2: Find the divergence of F.
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
= 2x+ 2y+ 1
Step 3: Evaluate the divergence at the point (1,2,3).
∇ · F(1,2,3) = 2(1) + 2(2) + 1
= 2 + 4 + 1
= 7
Therefore, the divergence of Fat the point (1,2,3) is 7 .
Question 10
Question
Let F(x, y, z) = yi+xj+zk. Calculate ∇ · ∇ × F.
7
Solution
Step 1: Let’s first find the curl of F.
Step 1: ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y x z
=∂z
∂y −∂y
∂z i−∂z
∂x −∂x
∂z j+∂y
∂x −∂x
∂y k=k−k+i=i
Step 2: Now, let’s find the divergence of i.
Step 2: ∇ · ∇ × F=∇ · i=∂
∂x (i)=0
Step 3: Therefore, ∇ · ∇ × F= 0 .
Question 11
Question
Let F(x, y, z) = ⟨x3y2,−2xyz, z2⟩be a vector field in R3. Compute ∇ · F, where
∇is the gradient operator.
Solution
Step 1: Compute ∇ · Fusing the formula ∇ · F=∂
∂x (Fx) + ∂
∂y (Fy) + ∂
∂z (Fz)
where F(x, y, z) = ⟨Fx, Fy, Fz⟩.
∇ · F=∂
∂x (x3y2) + ∂
∂y (−2xyz) + ∂
∂z (z2)
Step 2: Compute each partial derivative.
∂
∂x (x3y2)=3x2y2
∂
∂y (−2xyz) = −2xz
∂
∂z (z2) = 2z
Step 3: Put it all together.
∇ · F= 3x2y2−2xz + 2z
Therefore, the divergence of Fis 3x2y2−2xz + 2z.
8
Question 12
Question
Let F(x, y, z) = zcos(xyz)i+xcos(xyz)j+ycos(xyz)k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
zcos(xyz)xcos(xyz)ycos(xyz)
=∂(ycos(xyz))
∂y −∂(xcos(xyz))
∂z i−∂(zcos(xyz))
∂x −∂(ycos(xyz))
∂z j+∂(xcos(xyz))
∂x −∂(zcos(xyz))
∂y k
= (cos(xyz)−yz sin(xyz)) i−(−cos(xyz) + xz sin(xyz)) j+(−cos(xyz) + xy sin(xyz)) k
Step 2: Compute ∇ · (∇ × F)
∇·(∇ × F) = ∂
∂x (cos(xyz)−yz sin(xyz))+ ∂
∂y (−cos(xyz)+xz sin(xyz))+ ∂
∂z (−cos(xyz)+xy sin(xyz))
=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz)
Therefore, ∇ · ∇ × F=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz).
Question 13
Question
Let F(x, y, z)=(x2, y2, z2) be a vector field in R3. Compute ∇ · F.
Solution
1. Find the components of ∇ · Fusing the divergence operator:
∇ · F=∂
∂x (x2) + ∂
∂y (y2) + ∂
∂z (z2)
2. Calculate the partial derivatives:
∂
∂x (x2)=2x
∂
∂y (y2)=2y
∂
∂z (z2)=2z
9
3. Substitute the partial derivatives back into the expression for ∇ · F:
∇ · F= 2x+ 2y+ 2z= 2(x+y+z)
4. Therefore, the divergence of the vector field F(x, y, z) = (x2, y2, z2) is
∇ · F= 2(x+y+z).
Question 14
Question
Let F(x, y, z)=(x2+y2,2y, z) be a vector field in R3. Compute the curl of F.
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 1: Calculate the partial derivatives of F:
∂F1
∂x = 2x, ∂F2
∂y = 2,∂F3
∂z = 1
Step 2: Use the formula for the curl of a vector field:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 3: Substitute the partial derivatives into the formula:
curl(F) = (0 −0,0−2x, 2−2) = (0,−2x, 0)
Therefore, the curl of Fis (0,−2x, 0) .
Question 15
Question
Let F(x, y, z)=(xz2, yz2, xy2). Compute ∇ · F.
10
Solution
To compute ∇·F, we need to find the divergence of the vector field F= (P, Q, R),
where P=xz2,Q=yz2, and R=xy2.
Step 1: Find ∇ · Fusing the formula for divergence:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: Calculate the partial derivatives:
∂P
∂x =z2
∂Q
∂y =z2
∂R
∂z = 2xy
Step 3: Add up the partial derivatives:
∇ · F=z2+z2+ 2xy = 2z2+ 2xy
Therefore, the divergence of the vector field F(x, y, z)=(xz2, yz2, xy2) is
2z2+ 2xy .
Question 16
Question
Let F(x, y, z)=(x2+y2+z2)i+ (xy +yz)j+ (xz2)k. Calculate ∇ · F.
Solution
Step 1: Let F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k. Then,
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: In this case, P(x, y, z) = x2+y2+z2,Q(x, y, z) = xy +yz, and
R(x, y, z) = xz2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x, ∂Q
∂y =x+z, ∂R
∂z = 2xz
Step 4: Finally, find the divergence:
∇ · F= 2x+ (x+z)+2xz = 3x+z+ 2xz
Therefore, ∇ · F= 3x+z+ 2xz .
11
Question 17
Question
Let F(x, y, z) = (2xy +z)i+ (x2+z2)j+ (3xz −y)kbe a vector field in R3.
Calculate ∇ · F, where ∇=∂
∂x ,∂
∂y ,∂
∂z is the del operator.
Solution
To calculate the divergence of F, we use the formula ∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z .
Step 1: Calculate ∂Fx
∂x ,∂Fy
∂y , and ∂Fz
∂z .
∂Fx
∂x =∂
∂x (2xy +z) = 2y
∂Fy
∂y =∂
∂y (x2+z2)=0
∂Fz
∂z =∂
∂z (3xz −y)=3x
Step 2: Calculate ∇ · Fusing the formula.
∇ · F= 2y+ 0 + 3x= 3x+ 2y
Therefore, the divergence of the vector field Fis 3x+ 2y.
Question 18
Question
Let F(x, y, z) = yzi+xzj−xykbe a vector field in R3. Compute the curl of F.
Solution
To compute the curl of a vector field F=Pi+Qj+Rk(P,Q, and Rare the
component functions of F), we use the formula:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
12
Here, P=yz,Q=xz, and R=−xy. Thus, we have:
curl F=∇ × F
=
i j k
∂
∂x
∂
∂y
∂
∂z
yz xz −xy
=∂(−xy)
∂y −∂(xz)
∂z i−∂(yz)
∂x −∂(−xy)
∂z j+∂(xz)
∂x −∂(yz)
∂y k
= (−x−x)i−(z+z)j+ (z−y)k
=−2xi−2zj+ (z−y)k
Question 19
Question
Let F(x, y, z) = x2yi+xyzj+yz2kbe a vector field. Find the divergence of F.
Solution
To find the divergence of F, we need to apply the following formula:
div F=∇ · F=∂
∂x Fx+∂
∂y Fy+∂
∂z Fz
Step 1: Calculate ∂
∂x Fx
∂
∂x (x2y)=2xy
Step 2: Calculate ∂
∂y Fy
∂
∂y (xyz) = xz
Step 3: Calculate ∂
∂z Fz
∂
∂z (yz2) = 2yz
Step 4: Sum up the partial derivatives and find the divergence
div F= 2xy +xz + 2yz
Question 20
Question
Let Fbe a vector field in R3defined as F(x, y, z) = (x2y, yz, z2). Determine
the divergence of F.
13
Solution
To find the divergence of F, we use the formula div(F) = ∇ · F, where ∇is the
del operator.
Step 1: Compute the del operator ∇:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Write the vector field Fas a sum of its component functions:
F(x, y, z)=(x2y, yz, z2)
Step 3: Compute the components of the divergence using the dot product:
div(F) = ∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, yz, z2)
=∂
∂x (x2y) + ∂
∂y (yz) + ∂
∂z (z2)
Step 4: Differentiate each component function with respect to its corre-
sponding variable:
∂
∂x (x2y)=2xy
∂
∂y (yz) = z
∂
∂z (z2) = 2z
Step 5: Substitute the derivatives back into the divergence formula:
div(F)=2xy +z+ 2z
Step 6: Simplify the expression:
div(F)=2xy + 3z
Question 21
Question
Let F(x, y, z) = x2i+y2j+z2kbe a vector field in R3. Determine the curl of
F.
14
Solution
To find the curl of F, we first write Fin component form:
F(x, y, z) = x2i+y2j+z2k= (x2, y2, z2).
Step 1: Compute the curl of Fusing the formula:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
.
Step 2: Expand the determinant:
curl(F)=(∂
∂y (z2)−∂
∂z (y2))i−(∂
∂x (z2)−∂
∂z (x2))j+ ( ∂
∂x (y2)−∂
∂y (x2))k.
Step 3: Calculate the partial derivatives:
curl(F) = (0 −0)i−(0 −0)j+ (0 −0)k.
Step 4: Simplify the expression to find the curl of F:
curl(F) = 0.
Question 22
Question
Let Fbe a vector field in R3given by F(x, y, z) = ⟨x2y, y2z, zx⟩. Calculate the
divergence of Fat the point (1,−1,2).
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩at a point (x0, y0, z0), we
use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z) = ⟨x2y, y2z, zx⟩, we have P=x2y,Q=y2z, and
R=zx. Therefore, we need to calculate the partial derivatives of P,Q, and R
with respect to x,y, and z.
Step 1: Calculate ∂P ∂x
∂P
∂x = 2xy
15
Step 2: Calculate ∂Q ∂y
∂Q
∂y = 2yz
Step 3: Calculate ∂R ∂z
∂R
∂z =x
Step 4: Find the divergence of F at the point (1,−1,2) Now, we can
find the divergence of Fat (1,−1,2) by evaluating the sum:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= 2(1)(−1) + 2(−1)(2) + 1
=−2−4+1
=−5
Therefore, the divergence of Fat the point (1,−1,2) is −5 .
Question 23
Question
Let F=x2yi+y2zj+z2xkbe a vector field in R3. Compute the curl of F.
Solution
Step 1: The curl of a vector field F=Pi+Qj+Rkin R3is given by the
determinant
curl F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
where P, Q, R are the components of F.
Step 2: For F=x2yi+y2zj+z2xk, we have P=x2y,Q=y2z, and
R=z2x.
Step 3: Compute the partial derivatives of P, Q, R with respect to x,y, and
z.
∂P
∂x = 2xy, ∂Q
∂y = 2yz, ∂R
∂z = 2zx
Step 4: Substitute the computed partial derivatives into the determinant
formula for the curl.
curl F=
i j k
2xy 2yz 2zx
x2y y2z z2x
16
Step 5: Expand the determinant to find the curl of F.
curl F= (2zx −2yz)i−(2zx −2xy)j+ (2xy −2yz)k
Step 6: Simplify the expression to get the final answer.
curl F= (2zx −2yz)i+ (2xy −2zx)j+ (2xy −2yz)k
Question 24
Question
Let F(x, y, z) = x2yi+z3yj+xy2zkbe a vector field. Compute ∇ · F.
Solution
To compute the divergence of F, we use the formula ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F=Pi+Qj+Rk.
Step 1: Identify P,Q, and R. Here, P=x2y,Q=z3y, and R=xy2z.
Step 2: Compute the partial derivatives.
∂P
∂x = 2xy, ∂Q
∂y =z3,∂R
∂z =xy2
Step 3: Find the divergence.
∇ · F= 2xy +z3+xy2
Therefore, the divergence of Fis 2xy +z3+xy2.
Question 25
Question
Let F=ex+yi+ey+zj+ez+xkbe a vector field in R3. Compute ∇ · (∇ × F).
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
ex+yey+zez+x
=∂ez+x
∂y −∂ey+z
∂z i−∂ez+x
∂x −∂ex+y
∂z j+∂ex+y
∂x −∂ey+z
∂y k
= (−ey+z)i−(−ex+y)j+ex+yk=−ey+zi+ex+yj+ex+yk
17
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (−ey+z) + ∂
∂y (ex+y) + ∂
∂z (ex+y)
= 0 + ex+y+ex+y= 2ex+y
Therefore, ∇ · (∇ × F) = 2ex+y.
Question 26
Question
Let F(x, y, z) = 2yz + sin(x), xz +ey, y2+ cos(z)be a vector field in R3.
Compute the divergence of F.
Solution
To find the divergence of F, we will use the formula for the divergence of a
vector field F= (P, Q, R):
div(F) = ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Compute the partial derivatives of P,Q, and R.
∂P
∂x =∂
∂x (2yz + sin(x)) = 0 + cos(x),
∂Q
∂y =∂
∂y (xz +ey) = 0 + ey,
∂R
∂z =∂
∂z y2+ cos(z)= 0 −sin(z).
Step 2: Add up the partial derivatives to find the divergence of F.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= cos(x) + ey−sin(z).
Therefore, the divergence of the vector field Fis cos(x) + ey−sin(z).
Question 27
Question
Let F(x, y, z) = yezi+zexj+xeykbe a vector field. Compute ∇ · (∇ × F).
18
Solution
Step 1: First, we find ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
yezzexxey
Step 2: Calculate the determinant:
∇ × F=∂(xey)
∂y −∂(zex)
∂z i−∂(yez)
∂x −∂(xey)
∂z j+∂(zex)
∂x −∂(yez)
∂y k
= (xey)i−(−zex)j+ (yez)k
=xeyi+zexj+yezk
Step 3: Next, we find ∇ · (∇ × F):
∇ · (∇ × F) = ∂
∂x (xey) + ∂
∂y (zex) + ∂
∂z (yez)
=ey+zex+yez
=ey+zex+xey
Therefore, ∇ · (∇ × F) = ey+zex+xey.
Question 28
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field in R3. Calculate 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, where P,Q, and Rare the vector field’s component functions, we
use the formula div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Identify the component functions of F:F(x, y, z) = x2yi+xyzj+
xz2k, where P(x, y, z) = x2y,Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Find the partial derivatives:
∂P
∂x = 2xy, ∂Q
∂y =xz, and ∂R
∂z = 2xz.
Step 3: Compute the divergence:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy +xz + 2xz = 2xy + 3xz.
Therefore, the divergence of the vector field Fis 2xy + 3xz.
19
Question 29
Question
Let F(x, y, z) = x2i+y3j+z4k. Compute ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y3z4
∇ × F= (∂z4
∂y −0)i−(∂z4
∂x −0)j+ ( ∂x2
∂y −∂y3
∂x )k
∇ × F= 0i−0j+ 0k=0
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · 0= 0
Therefore, ∇ · ∇ × F=0.
Question 30
Question
Let
F(x, y, z)=(x2+y2)ˆ
i+ (y2+z2)ˆ
j+ (z2+x2)ˆ
k. Compute ∇ ·
F.
Solution
Step 1: The divergence 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
kis given by the formula
∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z .
Step 2: In this case, P(x, y, z) = x2+y2,Q(x, y, z) = y2+z2, and R(x, y, z) =
z2+x2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x,
∂Q
∂y = 2y,
∂R
∂z = 2z.
20
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
In this case, F(x, y, z) = (x2y, y2z, z2x), so P=x2y,Q=y2z, and R=z2x.
Step 1: Compute the partial derivative of Pwith respect to x:
∂P
∂x =∂
∂x (x2y) = 2xy
Step 2: Compute the partial derivative of Qwith respect to y:
∂Q
∂y =∂
∂y (y2z) = 2yz
Step 3: Compute the partial derivative of Rwith respect to z:
∂R
∂z =∂
∂z (z2x)=2zx
Step 4: Now, we can calculate the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy + 2yz + 2zx = 2xy + 2yz + 2zx
Therefore, the divergence of the vector field F(x, y, z)=(x2y, y2z, z2x) is
2xy + 2yz + 2zx.
Question 3
Question
Let F(x, y, z) = (2xy +z2)i+ (x2+z2)j+ (y2+z2)kbe a vector field in R3.
Compute the gradient, divergence, and curl of F.
Solution
Given F(x, y, z) = (2xy +z2)i+(x2+z2)j+(y2+z2)k, we can find the gradient,
divergence, and curl of F.
Step 1: Gradient of F The gradient of a scalar field f(x, y, z) = P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
∇f=∂f
∂x i+∂f
∂y j+∂f
∂z k
2
Applying this to F, we have:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·2xy +z2, x2+z2, y2+z2
= 2x+ 0 + 0 = 2xi
Step 2: Divergence of F The divergence of a vector field F=P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
div F=∂P
∂x +∂Q
∂y +∂R
∂z
For F, we get:
div F=∂(2xy +z2)
∂x +∂(x2+z2)
∂y +∂(y2+z2)
∂z
= 2y+ 0 + 2y= 4y
So, the divergence of Fis 4y.
Step 3: Curl of F The curl of a vector field F=P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by:
curl F=∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
For F, we get:
curl F= (2y−0)i+ (0 −2x)j+ (0 −2y)k
= 2yi−2xj−2yk
Question 4
Question
Let F(x, y, z) = x2yzi+xyz2j+x2y2kbe a vector field. Compute the divergence
of Fat the point (2,−1,3).
Solution
To compute the divergence of the vector field F(x, y, z) = x2yzi+xyz2j+x2y2k
at the point (2,−1,3), we will use the formula for divergence:
div(F) = ∂F1
∂x +∂F2
∂y +∂F3
∂z
3
where F(x, y, z) = F1(x, y, z)i+F2(x, y, z)j+F3(x, y, z)k.
Step 1: Calculate the partial derivatives of each component of F:
∂F1
∂x = 2xyz, ∂F2
∂y =xz2,∂F3
∂z = 2x2y
Step 2: Substitute (x, y, z) = (2,−1,3) into each partial derivative to find
the values at the point:
∂F1
∂x (2,−1,3) = 2(−1)(3) = −6
∂F2
∂y (2,−1,3) = (2)(32) = 18
∂F3
∂z (2,−1,3) = 2(2)2(−1) = −8
Step 3: Add the values of the partial derivatives calculated in Step 2 to
find the divergence of Fat the point (2,−1,3):
div(F)(2,−1,3) = −6 + 18 −8=4
Therefore, the divergence of the vector field Fat the point (2,−1,3) is 4.
Question 5
Question
Let F(x, y, z) = (yz2, xz2, xy2) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
F1F2F3
where F= (F1, F2, F3).
Step 1: Find the partial derivatives of F. We have:
∂F1
∂x =z2,∂F2
∂y =xz2,∂F3
∂z = 2xy
Step 2: Evaluate the curl of Fat the point (1,2,3). Substitute the point
(1,2,3) into the partial derivatives:
∂F1
∂x = 32= 9,∂F2
∂y = 1 ×32= 9,∂F3
∂z = 2 ×1×2=4
4
Step 3: Calculate the curl of F. Using the formula for the curl, we get:
∇ × F=
i j k
9 0 0
0 9 4
= (0 −0)i−(0 −0)j+ (9 −0)k= 9k
Therefore, the curl of Fat the point (1,2,3) is 9k.
Question 6
Question
Let F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x)be a vector field in R3. Find
the divergence of F.
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
functions of x,y, and z, respectively, we use the formula
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Given F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x), we identify P=x2y+
sin(z), Q=xz +y2, and R=ey+ cos(x).
Step 1: Find ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 2xy,
∂Q
∂y = 2y,
∂R
∂z = 0.
Step 2: Calculate the divergence.
) = ∂P ∂x+∂Q
∂y +∂R
∂z =2xy+2y+0=2y(x+1).
Therefore, the divergence of the vector field Fis 2y(x+ 1).
Question 7
Question
Consider the vector field F(x, y, z) = (x2+y2+z2)i+xyzj+ (xy −z2)k. Find
the curl of F.
5
Solution
To find 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,
where P,Q, and Rare the component functions, we use the formula for the
curl:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Compute the partial derivatives of P,Q, and R:
∂P
∂y = 2y, ∂Q
∂z =x, ∂R
∂x = 0,∂P
∂z = 0,
∂Q
∂x =y, ∂R
∂y =x, ∂P
∂y = 2x, ∂Q
∂z =y.
Step 2: Plug these derivatives into the formula for the curl:
∇ × F= (x−y)i−xj+ (y−2x)k
Therefore, the curl of the vector field F(x, y, z) is (x−y)i−xj+ (y−2x)k.
Question 8
Question
Let F(x, y, z) = ⟨x2y, xyz, z2⟩. Compute the curl of F.
Solution
To find the curl of a vector field F=⟨P, Q, R⟩, where P,Q, and Rare functions
of x,y, and z, we use the formula:
curl F=∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k.
In this case, F(x, y, z) = ⟨x2y, xyz, z2⟩, so we have P(x, y, z) = x2y,Q(x, y, z) =
xyz, and R(x, y, z) = z2.
Step 1: Compute the partial derivatives:
∂P
∂y =x2,∂Q
∂z =xy, ∂R
∂x = 0,
∂R
∂y = 0,∂Q
∂x =yz, ∂P
∂z = 0.
Step 2: Plug the derivatives into the formula for the curl:
curl F= (0 −xy)i−(0 −0) j+ (yz −0) k.
Step 3: Simplify the expression:
curl F=−xyi+yzk.
Therefore, the curl of Fis curl F=−xyi+yzk.
6
Question 9
Question
Let F(x, y, z) = x2+y, y2+z, z2+xbe a vector field. Calculate the diver-
gence of Fat the point (1,2,3).
Solution
To find the divergence of Fat the point (1,2,3), we need to calculate the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) with Fand evaluate it at the
point (1,2,3). The divergence of vector field Fis given by:
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
Step 1: Calculate the partial derivatives of F.
∂
∂x (x2+y)=2x
∂
∂y (y2+z)=2y
∂
∂z (z2+x)=1
Step 2: Find the divergence of F.
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
= 2x+ 2y+ 1
Step 3: Evaluate the divergence at the point (1,2,3).
∇ · F(1,2,3) = 2(1) + 2(2) + 1
= 2 + 4 + 1
= 7
Therefore, the divergence of Fat the point (1,2,3) is 7 .
Question 10
Question
Let F(x, y, z) = yi+xj+zk. Calculate ∇ · ∇ × F.
7
Solution
Step 1: Let’s first find the curl of F.
Step 1: ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y x z
=∂z
∂y −∂y
∂z i−∂z
∂x −∂x
∂z j+∂y
∂x −∂x
∂y k=k−k+i=i
Step 2: Now, let’s find the divergence of i.
Step 2: ∇ · ∇ × F=∇ · i=∂
∂x (i)=0
Step 3: Therefore, ∇ · ∇ × F= 0 .
Question 11
Question
Let F(x, y, z) = ⟨x3y2,−2xyz, z2⟩be a vector field in R3. Compute ∇ · F, where
∇is the gradient operator.
Solution
Step 1: Compute ∇ · Fusing the formula ∇ · F=∂
∂x (Fx) + ∂
∂y (Fy) + ∂
∂z (Fz)
where F(x, y, z) = ⟨Fx, Fy, Fz⟩.
∇ · F=∂
∂x (x3y2) + ∂
∂y (−2xyz) + ∂
∂z (z2)
Step 2: Compute each partial derivative.
∂
∂x (x3y2)=3x2y2
∂
∂y (−2xyz) = −2xz
∂
∂z (z2) = 2z
Step 3: Put it all together.
∇ · F= 3x2y2−2xz + 2z
Therefore, the divergence of Fis 3x2y2−2xz + 2z.
8
Question 12
Question
Let F(x, y, z) = zcos(xyz)i+xcos(xyz)j+ycos(xyz)k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
zcos(xyz)xcos(xyz)ycos(xyz)
=∂(ycos(xyz))
∂y −∂(xcos(xyz))
∂z i−∂(zcos(xyz))
∂x −∂(ycos(xyz))
∂z j+∂(xcos(xyz))
∂x −∂(zcos(xyz))
∂y k
= (cos(xyz)−yz sin(xyz)) i−(−cos(xyz) + xz sin(xyz)) j+(−cos(xyz) + xy sin(xyz)) k
Step 2: Compute ∇ · (∇ × F)
∇·(∇ × F) = ∂
∂x (cos(xyz)−yz sin(xyz))+ ∂
∂y (−cos(xyz)+xz sin(xyz))+ ∂
∂z (−cos(xyz)+xy sin(xyz))
=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz)
Therefore, ∇ · ∇ × F=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz).
Question 13
Question
Let F(x, y, z)=(x2, y2, z2) be a vector field in R3. Compute ∇ · F.
Solution
1. Find the components of ∇ · Fusing the divergence operator:
∇ · F=∂
∂x (x2) + ∂
∂y (y2) + ∂
∂z (z2)
2. Calculate the partial derivatives:
∂
∂x (x2)=2x
∂
∂y (y2)=2y
∂
∂z (z2)=2z
9
3. Substitute the partial derivatives back into the expression for ∇ · F:
∇ · F= 2x+ 2y+ 2z= 2(x+y+z)
4. Therefore, the divergence of the vector field F(x, y, z) = (x2, y2, z2) is
∇ · F= 2(x+y+z).
Question 14
Question
Let F(x, y, z)=(x2+y2,2y, z) be a vector field in R3. Compute the curl of F.
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 1: Calculate the partial derivatives of F:
∂F1
∂x = 2x, ∂F2
∂y = 2,∂F3
∂z = 1
Step 2: Use the formula for the curl of a vector field:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 3: Substitute the partial derivatives into the formula:
curl(F) = (0 −0,0−2x, 2−2) = (0,−2x, 0)
Therefore, the curl of Fis (0,−2x, 0) .
Question 15
Question
Let F(x, y, z)=(xz2, yz2, xy2). Compute ∇ · F.
10
Solution
To compute ∇·F, we need to find the divergence of the vector field F= (P, Q, R),
where P=xz2,Q=yz2, and R=xy2.
Step 1: Find ∇ · Fusing the formula for divergence:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: Calculate the partial derivatives:
∂P
∂x =z2
∂Q
∂y =z2
∂R
∂z = 2xy
Step 3: Add up the partial derivatives:
∇ · F=z2+z2+ 2xy = 2z2+ 2xy
Therefore, the divergence of the vector field F(x, y, z)=(xz2, yz2, xy2) is
2z2+ 2xy .
Question 16
Question
Let F(x, y, z)=(x2+y2+z2)i+ (xy +yz)j+ (xz2)k. Calculate ∇ · F.
Solution
Step 1: Let F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k. Then,
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: In this case, P(x, y, z) = x2+y2+z2,Q(x, y, z) = xy +yz, and
R(x, y, z) = xz2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x, ∂Q
∂y =x+z, ∂R
∂z = 2xz
Step 4: Finally, find the divergence:
∇ · F= 2x+ (x+z)+2xz = 3x+z+ 2xz
Therefore, ∇ · F= 3x+z+ 2xz .
11
Question 17
Question
Let F(x, y, z) = (2xy +z)i+ (x2+z2)j+ (3xz −y)kbe a vector field in R3.
Calculate ∇ · F, where ∇=∂
∂x ,∂
∂y ,∂
∂z is the del operator.
Solution
To calculate the divergence of F, we use the formula ∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z .
Step 1: Calculate ∂Fx
∂x ,∂Fy
∂y , and ∂Fz
∂z .
∂Fx
∂x =∂
∂x (2xy +z) = 2y
∂Fy
∂y =∂
∂y (x2+z2)=0
∂Fz
∂z =∂
∂z (3xz −y)=3x
Step 2: Calculate ∇ · Fusing the formula.
∇ · F= 2y+ 0 + 3x= 3x+ 2y
Therefore, the divergence of the vector field Fis 3x+ 2y.
Question 18
Question
Let F(x, y, z) = yzi+xzj−xykbe a vector field in R3. Compute the curl of F.
Solution
To compute the curl of a vector field F=Pi+Qj+Rk(P,Q, and Rare the
component functions of F), we use the formula:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
12
Here, P=yz,Q=xz, and R=−xy. Thus, we have:
curl F=∇ × F
=
i j k
∂
∂x
∂
∂y
∂
∂z
yz xz −xy
=∂(−xy)
∂y −∂(xz)
∂z i−∂(yz)
∂x −∂(−xy)
∂z j+∂(xz)
∂x −∂(yz)
∂y k
= (−x−x)i−(z+z)j+ (z−y)k
=−2xi−2zj+ (z−y)k
Question 19
Question
Let F(x, y, z) = x2yi+xyzj+yz2kbe a vector field. Find the divergence of F.
Solution
To find the divergence of F, we need to apply the following formula:
div F=∇ · F=∂
∂x Fx+∂
∂y Fy+∂
∂z Fz
Step 1: Calculate ∂
∂x Fx
∂
∂x (x2y)=2xy
Step 2: Calculate ∂
∂y Fy
∂
∂y (xyz) = xz
Step 3: Calculate ∂
∂z Fz
∂
∂z (yz2) = 2yz
Step 4: Sum up the partial derivatives and find the divergence
div F= 2xy +xz + 2yz
Question 20
Question
Let Fbe a vector field in R3defined as F(x, y, z) = (x2y, yz, z2). Determine
the divergence of F.
13
Solution
To find the divergence of F, we use the formula div(F) = ∇ · F, where ∇is the
del operator.
Step 1: Compute the del operator ∇:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Write the vector field Fas a sum of its component functions:
F(x, y, z)=(x2y, yz, z2)
Step 3: Compute the components of the divergence using the dot product:
div(F) = ∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, yz, z2)
=∂
∂x (x2y) + ∂
∂y (yz) + ∂
∂z (z2)
Step 4: Differentiate each component function with respect to its corre-
sponding variable:
∂
∂x (x2y)=2xy
∂
∂y (yz) = z
∂
∂z (z2) = 2z
Step 5: Substitute the derivatives back into the divergence formula:
div(F)=2xy +z+ 2z
Step 6: Simplify the expression:
div(F)=2xy + 3z
Question 21
Question
Let F(x, y, z) = x2i+y2j+z2kbe a vector field in R3. Determine the curl of
F.
14
Solution
To find the curl of F, we first write Fin component form:
F(x, y, z) = x2i+y2j+z2k= (x2, y2, z2).
Step 1: Compute the curl of Fusing the formula:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
.
Step 2: Expand the determinant:
curl(F)=(∂
∂y (z2)−∂
∂z (y2))i−(∂
∂x (z2)−∂
∂z (x2))j+ ( ∂
∂x (y2)−∂
∂y (x2))k.
Step 3: Calculate the partial derivatives:
curl(F) = (0 −0)i−(0 −0)j+ (0 −0)k.
Step 4: Simplify the expression to find the curl of F:
curl(F) = 0.
Question 22
Question
Let Fbe a vector field in R3given by F(x, y, z) = ⟨x2y, y2z, zx⟩. Calculate the
divergence of Fat the point (1,−1,2).
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩at a point (x0, y0, z0), we
use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z) = ⟨x2y, y2z, zx⟩, we have P=x2y,Q=y2z, and
R=zx. Therefore, we need to calculate the partial derivatives of P,Q, and R
with respect to x,y, and z.
Step 1: Calculate ∂P ∂x
∂P
∂x = 2xy
15
Step 2: Calculate ∂Q ∂y
∂Q
∂y = 2yz
Step 3: Calculate ∂R ∂z
∂R
∂z =x
Step 4: Find the divergence of F at the point (1,−1,2) Now, we can
find the divergence of Fat (1,−1,2) by evaluating the sum:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= 2(1)(−1) + 2(−1)(2) + 1
=−2−4+1
=−5
Therefore, the divergence of Fat the point (1,−1,2) is −5 .
Question 23
Question
Let F=x2yi+y2zj+z2xkbe a vector field in R3. Compute the curl of F.
Solution
Step 1: The curl of a vector field F=Pi+Qj+Rkin R3is given by the
determinant
curl F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
where P, Q, R are the components of F.
Step 2: For F=x2yi+y2zj+z2xk, we have P=x2y,Q=y2z, and
R=z2x.
Step 3: Compute the partial derivatives of P, Q, R with respect to x,y, and
z.
∂P
∂x = 2xy, ∂Q
∂y = 2yz, ∂R
∂z = 2zx
Step 4: Substitute the computed partial derivatives into the determinant
formula for the curl.
curl F=
i j k
2xy 2yz 2zx
x2y y2z z2x
16
Step 5: Expand the determinant to find the curl of F.
curl F= (2zx −2yz)i−(2zx −2xy)j+ (2xy −2yz)k
Step 6: Simplify the expression to get the final answer.
curl F= (2zx −2yz)i+ (2xy −2zx)j+ (2xy −2yz)k
Question 24
Question
Let F(x, y, z) = x2yi+z3yj+xy2zkbe a vector field. Compute ∇ · F.
Solution
To compute the divergence of F, we use the formula ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F=Pi+Qj+Rk.
Step 1: Identify P,Q, and R. Here, P=x2y,Q=z3y, and R=xy2z.
Step 2: Compute the partial derivatives.
∂P
∂x = 2xy, ∂Q
∂y =z3,∂R
∂z =xy2
Step 3: Find the divergence.
∇ · F= 2xy +z3+xy2
Therefore, the divergence of Fis 2xy +z3+xy2.
Question 25
Question
Let F=ex+yi+ey+zj+ez+xkbe a vector field in R3. Compute ∇ · (∇ × F).
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
ex+yey+zez+x
=∂ez+x
∂y −∂ey+z
∂z i−∂ez+x
∂x −∂ex+y
∂z j+∂ex+y
∂x −∂ey+z
∂y k
= (−ey+z)i−(−ex+y)j+ex+yk=−ey+zi+ex+yj+ex+yk
17
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (−ey+z) + ∂
∂y (ex+y) + ∂
∂z (ex+y)
= 0 + ex+y+ex+y= 2ex+y
Therefore, ∇ · (∇ × F) = 2ex+y.
Question 26
Question
Let F(x, y, z) = 2yz + sin(x), xz +ey, y2+ cos(z)be a vector field in R3.
Compute the divergence of F.
Solution
To find the divergence of F, we will use the formula for the divergence of a
vector field F= (P, Q, R):
div(F) = ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Compute the partial derivatives of P,Q, and R.
∂P
∂x =∂
∂x (2yz + sin(x)) = 0 + cos(x),
∂Q
∂y =∂
∂y (xz +ey) = 0 + ey,
∂R
∂z =∂
∂z y2+ cos(z)= 0 −sin(z).
Step 2: Add up the partial derivatives to find the divergence of F.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= cos(x) + ey−sin(z).
Therefore, the divergence of the vector field Fis cos(x) + ey−sin(z).
Question 27
Question
Let F(x, y, z) = yezi+zexj+xeykbe a vector field. Compute ∇ · (∇ × F).
18
Solution
Step 1: First, we find ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
yezzexxey
Step 2: Calculate the determinant:
∇ × F=∂(xey)
∂y −∂(zex)
∂z i−∂(yez)
∂x −∂(xey)
∂z j+∂(zex)
∂x −∂(yez)
∂y k
= (xey)i−(−zex)j+ (yez)k
=xeyi+zexj+yezk
Step 3: Next, we find ∇ · (∇ × F):
∇ · (∇ × F) = ∂
∂x (xey) + ∂
∂y (zex) + ∂
∂z (yez)
=ey+zex+yez
=ey+zex+xey
Therefore, ∇ · (∇ × F) = ey+zex+xey.
Question 28
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field in R3. Calculate 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, where P,Q, and Rare the vector field’s component functions, we
use the formula div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Identify the component functions of F:F(x, y, z) = x2yi+xyzj+
xz2k, where P(x, y, z) = x2y,Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Find the partial derivatives:
∂P
∂x = 2xy, ∂Q
∂y =xz, and ∂R
∂z = 2xz.
Step 3: Compute the divergence:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy +xz + 2xz = 2xy + 3xz.
Therefore, the divergence of the vector field Fis 2xy + 3xz.
19
Question 29
Question
Let F(x, y, z) = x2i+y3j+z4k. Compute ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y3z4
∇ × F= (∂z4
∂y −0)i−(∂z4
∂x −0)j+ ( ∂x2
∂y −∂y3
∂x )k
∇ × F= 0i−0j+ 0k=0
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · 0= 0
Therefore, ∇ · ∇ × F=0.
Question 30
Question
Let
F(x, y, z)=(x2+y2)ˆ
i+ (y2+z2)ˆ
j+ (z2+x2)ˆ
k. Compute ∇ ·
F.
Solution
Step 1: The divergence 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
kis given by the formula
∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z .
Step 2: In this case, P(x, y, z) = x2+y2,Q(x, y, z) = y2+z2, and R(x, y, z) =
z2+x2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x,
∂Q
∂y = 2y,
∂R
∂z = 2z.
20
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
the component functions, we use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
In this case, F(x, y, z) = (x2y, y2z, z2x), so P=x2y,Q=y2z, and R=z2x.
Step 1: Compute the partial derivative of Pwith respect to x:
∂P
∂x =∂
∂x (x2y) = 2xy
Step 2: Compute the partial derivative of Qwith respect to y:
∂Q
∂y =∂
∂y (y2z) = 2yz
Step 3: Compute the partial derivative of Rwith respect to z:
∂R
∂z =∂
∂z (z2x)=2zx
Step 4: Now, we can calculate the divergence of F:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy + 2yz + 2zx = 2xy + 2yz + 2zx
Therefore, the divergence of the vector field F(x, y, z)=(x2y, y2z, z2x) is
2xy + 2yz + 2zx.
Question 3
Question
Let F(x, y, z) = (2xy +z2)i+ (x2+z2)j+ (y2+z2)kbe a vector field in R3.
Compute the gradient, divergence, and curl of F.
Solution
Given F(x, y, z) = (2xy +z2)i+(x2+z2)j+(y2+z2)k, we can find the gradient,
divergence, and curl of F.
Step 1: Gradient of F The gradient of a scalar field f(x, y, z) = P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
∇f=∂f
∂x i+∂f
∂y j+∂f
∂z k
2
Applying this to F, we have:
∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·2xy +z2, x2+z2, y2+z2
= 2x+ 0 + 0 = 2xi
Step 2: Divergence of F The divergence of a vector field F=P(x, y, z)i+
Q(x, y, z)j+R(x, y, z)kis given by:
div F=∂P
∂x +∂Q
∂y +∂R
∂z
For F, we get:
div F=∂(2xy +z2)
∂x +∂(x2+z2)
∂y +∂(y2+z2)
∂z
= 2y+ 0 + 2y= 4y
So, the divergence of Fis 4y.
Step 3: Curl of F The curl of a vector field F=P(x, y, z)i+Q(x, y, z)j+
R(x, y, z)kis given by:
curl F=∂R
∂y −∂Q
∂z i+∂P
∂z −∂R
∂x j+∂Q
∂x −∂P
∂y k
For F, we get:
curl F= (2y−0)i+ (0 −2x)j+ (0 −2y)k
= 2yi−2xj−2yk
Question 4
Question
Let F(x, y, z) = x2yzi+xyz2j+x2y2kbe a vector field. Compute the divergence
of Fat the point (2,−1,3).
Solution
To compute the divergence of the vector field F(x, y, z) = x2yzi+xyz2j+x2y2k
at the point (2,−1,3), we will use the formula for divergence:
div(F) = ∂F1
∂x +∂F2
∂y +∂F3
∂z
3
where F(x, y, z) = F1(x, y, z)i+F2(x, y, z)j+F3(x, y, z)k.
Step 1: Calculate the partial derivatives of each component of F:
∂F1
∂x = 2xyz, ∂F2
∂y =xz2,∂F3
∂z = 2x2y
Step 2: Substitute (x, y, z) = (2,−1,3) into each partial derivative to find
the values at the point:
∂F1
∂x (2,−1,3) = 2(−1)(3) = −6
∂F2
∂y (2,−1,3) = (2)(32) = 18
∂F3
∂z (2,−1,3) = 2(2)2(−1) = −8
Step 3: Add the values of the partial derivatives calculated in Step 2 to
find the divergence of Fat the point (2,−1,3):
div(F)(2,−1,3) = −6 + 18 −8=4
Therefore, the divergence of the vector field Fat the point (2,−1,3) is 4.
Question 5
Question
Let F(x, y, z) = (yz2, xz2, xy2) be a vector field in R3. Calculate the curl of F
at the point (1,2,3).
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
F1F2F3
where F= (F1, F2, F3).
Step 1: Find the partial derivatives of F. We have:
∂F1
∂x =z2,∂F2
∂y =xz2,∂F3
∂z = 2xy
Step 2: Evaluate the curl of Fat the point (1,2,3). Substitute the point
(1,2,3) into the partial derivatives:
∂F1
∂x = 32= 9,∂F2
∂y = 1 ×32= 9,∂F3
∂z = 2 ×1×2=4
4
Step 3: Calculate the curl of F. Using the formula for the curl, we get:
∇ × F=
i j k
9 0 0
0 9 4
= (0 −0)i−(0 −0)j+ (9 −0)k= 9k
Therefore, the curl of Fat the point (1,2,3) is 9k.
Question 6
Question
Let F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x)be a vector field in R3. Find
the divergence of F.
Solution
To find the divergence of a vector field F= (P, Q, R), where P,Q, and Rare
functions of x,y, and z, respectively, we use the formula
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Given F(x, y, z) = x2y+ sin(z), xz +y2, ey+ cos(x), we identify P=x2y+
sin(z), Q=xz +y2, and R=ey+ cos(x).
Step 1: Find ∂P
∂x ,∂Q
∂y , and ∂R
∂z .
∂P
∂x = 2xy,
∂Q
∂y = 2y,
∂R
∂z = 0.
Step 2: Calculate the divergence.
) = ∂P ∂x+∂Q
∂y +∂R
∂z =2xy+2y+0=2y(x+1).
Therefore, the divergence of the vector field Fis 2y(x+ 1).
Question 7
Question
Consider the vector field F(x, y, z) = (x2+y2+z2)i+xyzj+ (xy −z2)k. Find
the curl of F.
5
Solution
To find 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,
where P,Q, and Rare the component functions, we use the formula for the
curl:
∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k
Step 1: Compute the partial derivatives of P,Q, and R:
∂P
∂y = 2y, ∂Q
∂z =x, ∂R
∂x = 0,∂P
∂z = 0,
∂Q
∂x =y, ∂R
∂y =x, ∂P
∂y = 2x, ∂Q
∂z =y.
Step 2: Plug these derivatives into the formula for the curl:
∇ × F= (x−y)i−xj+ (y−2x)k
Therefore, the curl of the vector field F(x, y, z) is (x−y)i−xj+ (y−2x)k.
Question 8
Question
Let F(x, y, z) = ⟨x2y, xyz, z2⟩. Compute the curl of F.
Solution
To find the curl of a vector field F=⟨P, Q, R⟩, where P,Q, and Rare functions
of x,y, and z, we use the formula:
curl F=∇ × F=∂R
∂y −∂Q
∂z i−∂R
∂x −∂P
∂z j+∂Q
∂x −∂P
∂y k.
In this case, F(x, y, z) = ⟨x2y, xyz, z2⟩, so we have P(x, y, z) = x2y,Q(x, y, z) =
xyz, and R(x, y, z) = z2.
Step 1: Compute the partial derivatives:
∂P
∂y =x2,∂Q
∂z =xy, ∂R
∂x = 0,
∂R
∂y = 0,∂Q
∂x =yz, ∂P
∂z = 0.
Step 2: Plug the derivatives into the formula for the curl:
curl F= (0 −xy)i−(0 −0) j+ (yz −0) k.
Step 3: Simplify the expression:
curl F=−xyi+yzk.
Therefore, the curl of Fis curl F=−xyi+yzk.
6
Question 9
Question
Let F(x, y, z) = x2+y, y2+z, z2+xbe a vector field. Calculate the diver-
gence of Fat the point (1,2,3).
Solution
To find the divergence of Fat the point (1,2,3), we need to calculate the dot
product of the del operator (∇=∂
∂x ,∂
∂y ,∂
∂z ) with Fand evaluate it at the
point (1,2,3). The divergence of vector field Fis given by:
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
Step 1: Calculate the partial derivatives of F.
∂
∂x (x2+y)=2x
∂
∂y (y2+z)=2y
∂
∂z (z2+x)=1
Step 2: Find the divergence of F.
∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z
= 2x+ 2y+ 1
Step 3: Evaluate the divergence at the point (1,2,3).
∇ · F(1,2,3) = 2(1) + 2(2) + 1
= 2 + 4 + 1
= 7
Therefore, the divergence of Fat the point (1,2,3) is 7 .
Question 10
Question
Let F(x, y, z) = yi+xj+zk. Calculate ∇ · ∇ × F.
7
Solution
Step 1: Let’s first find the curl of F.
Step 1: ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y x z
=∂z
∂y −∂y
∂z i−∂z
∂x −∂x
∂z j+∂y
∂x −∂x
∂y k=k−k+i=i
Step 2: Now, let’s find the divergence of i.
Step 2: ∇ · ∇ × F=∇ · i=∂
∂x (i)=0
Step 3: Therefore, ∇ · ∇ × F= 0 .
Question 11
Question
Let F(x, y, z) = ⟨x3y2,−2xyz, z2⟩be a vector field in R3. Compute ∇ · F, where
∇is the gradient operator.
Solution
Step 1: Compute ∇ · Fusing the formula ∇ · F=∂
∂x (Fx) + ∂
∂y (Fy) + ∂
∂z (Fz)
where F(x, y, z) = ⟨Fx, Fy, Fz⟩.
∇ · F=∂
∂x (x3y2) + ∂
∂y (−2xyz) + ∂
∂z (z2)
Step 2: Compute each partial derivative.
∂
∂x (x3y2)=3x2y2
∂
∂y (−2xyz) = −2xz
∂
∂z (z2) = 2z
Step 3: Put it all together.
∇ · F= 3x2y2−2xz + 2z
Therefore, the divergence of Fis 3x2y2−2xz + 2z.
8
Question 12
Question
Let F(x, y, z) = zcos(xyz)i+xcos(xyz)j+ycos(xyz)k. Compute ∇ · ∇ × F.
Solution
Step 1: Compute ∇ × F
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
zcos(xyz)xcos(xyz)ycos(xyz)
=∂(ycos(xyz))
∂y −∂(xcos(xyz))
∂z i−∂(zcos(xyz))
∂x −∂(ycos(xyz))
∂z j+∂(xcos(xyz))
∂x −∂(zcos(xyz))
∂y k
= (cos(xyz)−yz sin(xyz)) i−(−cos(xyz) + xz sin(xyz)) j+(−cos(xyz) + xy sin(xyz)) k
Step 2: Compute ∇ · (∇ × F)
∇·(∇ × F) = ∂
∂x (cos(xyz)−yz sin(xyz))+ ∂
∂y (−cos(xyz)+xz sin(xyz))+ ∂
∂z (−cos(xyz)+xy sin(xyz))
=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz)
Therefore, ∇ · ∇ × F=−y2zcos(xyz)−x2zcos(xyz)−x2ycos(xyz).
Question 13
Question
Let F(x, y, z)=(x2, y2, z2) be a vector field in R3. Compute ∇ · F.
Solution
1. Find the components of ∇ · Fusing the divergence operator:
∇ · F=∂
∂x (x2) + ∂
∂y (y2) + ∂
∂z (z2)
2. Calculate the partial derivatives:
∂
∂x (x2)=2x
∂
∂y (y2)=2y
∂
∂z (z2)=2z
9
3. Substitute the partial derivatives back into the expression for ∇ · F:
∇ · F= 2x+ 2y+ 2z= 2(x+y+z)
4. Therefore, the divergence of the vector field F(x, y, z) = (x2, y2, z2) is
∇ · F= 2(x+y+z).
Question 14
Question
Let F(x, y, z)=(x2+y2,2y, z) be a vector field in R3. Compute the curl of F.
Solution
To find the curl of a vector field F(x, y, z), we use the formula:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 1: Calculate the partial derivatives of F:
∂F1
∂x = 2x, ∂F2
∂y = 2,∂F3
∂z = 1
Step 2: Use the formula for the curl of a vector field:
curl(F) = ∂F3
∂y −∂F2
∂z ,∂F1
∂z −∂F3
∂x ,∂F2
∂x −∂F1
∂y
Step 3: Substitute the partial derivatives into the formula:
curl(F) = (0 −0,0−2x, 2−2) = (0,−2x, 0)
Therefore, the curl of Fis (0,−2x, 0) .
Question 15
Question
Let F(x, y, z)=(xz2, yz2, xy2). Compute ∇ · F.
10
Solution
To compute ∇·F, we need to find the divergence of the vector field F= (P, Q, R),
where P=xz2,Q=yz2, and R=xy2.
Step 1: Find ∇ · Fusing the formula for divergence:
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: Calculate the partial derivatives:
∂P
∂x =z2
∂Q
∂y =z2
∂R
∂z = 2xy
Step 3: Add up the partial derivatives:
∇ · F=z2+z2+ 2xy = 2z2+ 2xy
Therefore, the divergence of the vector field F(x, y, z)=(xz2, yz2, xy2) is
2z2+ 2xy .
Question 16
Question
Let F(x, y, z)=(x2+y2+z2)i+ (xy +yz)j+ (xz2)k. Calculate ∇ · F.
Solution
Step 1: Let F(x, y, z) = P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k. Then,
∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 2: In this case, P(x, y, z) = x2+y2+z2,Q(x, y, z) = xy +yz, and
R(x, y, z) = xz2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x, ∂Q
∂y =x+z, ∂R
∂z = 2xz
Step 4: Finally, find the divergence:
∇ · F= 2x+ (x+z)+2xz = 3x+z+ 2xz
Therefore, ∇ · F= 3x+z+ 2xz .
11
Question 17
Question
Let F(x, y, z) = (2xy +z)i+ (x2+z2)j+ (3xz −y)kbe a vector field in R3.
Calculate ∇ · F, where ∇=∂
∂x ,∂
∂y ,∂
∂z is the del operator.
Solution
To calculate the divergence of F, we use the formula ∇ · F=∂Fx
∂x +∂Fy
∂y +∂Fz
∂z .
Step 1: Calculate ∂Fx
∂x ,∂Fy
∂y , and ∂Fz
∂z .
∂Fx
∂x =∂
∂x (2xy +z) = 2y
∂Fy
∂y =∂
∂y (x2+z2)=0
∂Fz
∂z =∂
∂z (3xz −y)=3x
Step 2: Calculate ∇ · Fusing the formula.
∇ · F= 2y+ 0 + 3x= 3x+ 2y
Therefore, the divergence of the vector field Fis 3x+ 2y.
Question 18
Question
Let F(x, y, z) = yzi+xzj−xykbe a vector field in R3. Compute the curl of F.
Solution
To compute the curl of a vector field F=Pi+Qj+Rk(P,Q, and Rare the
component functions of F), we use the formula:
curl F=∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
12
Here, P=yz,Q=xz, and R=−xy. Thus, we have:
curl F=∇ × F
=
i j k
∂
∂x
∂
∂y
∂
∂z
yz xz −xy
=∂(−xy)
∂y −∂(xz)
∂z i−∂(yz)
∂x −∂(−xy)
∂z j+∂(xz)
∂x −∂(yz)
∂y k
= (−x−x)i−(z+z)j+ (z−y)k
=−2xi−2zj+ (z−y)k
Question 19
Question
Let F(x, y, z) = x2yi+xyzj+yz2kbe a vector field. Find the divergence of F.
Solution
To find the divergence of F, we need to apply the following formula:
div F=∇ · F=∂
∂x Fx+∂
∂y Fy+∂
∂z Fz
Step 1: Calculate ∂
∂x Fx
∂
∂x (x2y)=2xy
Step 2: Calculate ∂
∂y Fy
∂
∂y (xyz) = xz
Step 3: Calculate ∂
∂z Fz
∂
∂z (yz2) = 2yz
Step 4: Sum up the partial derivatives and find the divergence
div F= 2xy +xz + 2yz
Question 20
Question
Let Fbe a vector field in R3defined as F(x, y, z) = (x2y, yz, z2). Determine
the divergence of F.
13
Solution
To find the divergence of F, we use the formula div(F) = ∇ · F, where ∇is the
del operator.
Step 1: Compute the del operator ∇:
∇=∂
∂x ,∂
∂y ,∂
∂z
Step 2: Write the vector field Fas a sum of its component functions:
F(x, y, z)=(x2y, yz, z2)
Step 3: Compute the components of the divergence using the dot product:
div(F) = ∇ · F=∂
∂x ,∂
∂y ,∂
∂z ·(x2y, yz, z2)
=∂
∂x (x2y) + ∂
∂y (yz) + ∂
∂z (z2)
Step 4: Differentiate each component function with respect to its corre-
sponding variable:
∂
∂x (x2y)=2xy
∂
∂y (yz) = z
∂
∂z (z2) = 2z
Step 5: Substitute the derivatives back into the divergence formula:
div(F)=2xy +z+ 2z
Step 6: Simplify the expression:
div(F)=2xy + 3z
Question 21
Question
Let F(x, y, z) = x2i+y2j+z2kbe a vector field in R3. Determine the curl of
F.
14
Solution
To find the curl of F, we first write Fin component form:
F(x, y, z) = x2i+y2j+z2k= (x2, y2, z2).
Step 1: Compute the curl of Fusing the formula:
curl(F) = ∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y2z2
.
Step 2: Expand the determinant:
curl(F)=(∂
∂y (z2)−∂
∂z (y2))i−(∂
∂x (z2)−∂
∂z (x2))j+ ( ∂
∂x (y2)−∂
∂y (x2))k.
Step 3: Calculate the partial derivatives:
curl(F) = (0 −0)i−(0 −0)j+ (0 −0)k.
Step 4: Simplify the expression to find the curl of F:
curl(F) = 0.
Question 22
Question
Let Fbe a vector field in R3given by F(x, y, z) = ⟨x2y, y2z, zx⟩. Calculate the
divergence of Fat the point (1,−1,2).
Solution
To find the divergence of a vector field F=⟨P, Q, R⟩at a point (x0, y0, z0), we
use the formula:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
Given that F(x, y, z) = ⟨x2y, y2z, zx⟩, we have P=x2y,Q=y2z, and
R=zx. Therefore, we need to calculate the partial derivatives of P,Q, and R
with respect to x,y, and z.
Step 1: Calculate ∂P ∂x
∂P
∂x = 2xy
15
Step 2: Calculate ∂Q ∂y
∂Q
∂y = 2yz
Step 3: Calculate ∂R ∂z
∂R
∂z =x
Step 4: Find the divergence of F at the point (1,−1,2) Now, we can
find the divergence of Fat (1,−1,2) by evaluating the sum:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= 2(1)(−1) + 2(−1)(2) + 1
=−2−4+1
=−5
Therefore, the divergence of Fat the point (1,−1,2) is −5 .
Question 23
Question
Let F=x2yi+y2zj+z2xkbe a vector field in R3. Compute the curl of F.
Solution
Step 1: The curl of a vector field F=Pi+Qj+Rkin R3is given by the
determinant
curl F=
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
where P, Q, R are the components of F.
Step 2: For F=x2yi+y2zj+z2xk, we have P=x2y,Q=y2z, and
R=z2x.
Step 3: Compute the partial derivatives of P, Q, R with respect to x,y, and
z.
∂P
∂x = 2xy, ∂Q
∂y = 2yz, ∂R
∂z = 2zx
Step 4: Substitute the computed partial derivatives into the determinant
formula for the curl.
curl F=
i j k
2xy 2yz 2zx
x2y y2z z2x
16
Step 5: Expand the determinant to find the curl of F.
curl F= (2zx −2yz)i−(2zx −2xy)j+ (2xy −2yz)k
Step 6: Simplify the expression to get the final answer.
curl F= (2zx −2yz)i+ (2xy −2zx)j+ (2xy −2yz)k
Question 24
Question
Let F(x, y, z) = x2yi+z3yj+xy2zkbe a vector field. Compute ∇ · F.
Solution
To compute the divergence of F, we use the formula ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z ,
where F=Pi+Qj+Rk.
Step 1: Identify P,Q, and R. Here, P=x2y,Q=z3y, and R=xy2z.
Step 2: Compute the partial derivatives.
∂P
∂x = 2xy, ∂Q
∂y =z3,∂R
∂z =xy2
Step 3: Find the divergence.
∇ · F= 2xy +z3+xy2
Therefore, the divergence of Fis 2xy +z3+xy2.
Question 25
Question
Let F=ex+yi+ey+zj+ez+xkbe a vector field in R3. Compute ∇ · (∇ × F).
Solution
Step 1: Compute ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
ex+yey+zez+x
=∂ez+x
∂y −∂ey+z
∂z i−∂ez+x
∂x −∂ex+y
∂z j+∂ex+y
∂x −∂ey+z
∂y k
= (−ey+z)i−(−ex+y)j+ex+yk=−ey+zi+ex+yj+ex+yk
17
Step 2: Compute ∇ · (∇ × F).
∇ · (∇ × F) = ∂
∂x (−ey+z) + ∂
∂y (ex+y) + ∂
∂z (ex+y)
= 0 + ex+y+ex+y= 2ex+y
Therefore, ∇ · (∇ × F) = 2ex+y.
Question 26
Question
Let F(x, y, z) = 2yz + sin(x), xz +ey, y2+ cos(z)be a vector field in R3.
Compute the divergence of F.
Solution
To find the divergence of F, we will use the formula for the divergence of a
vector field F= (P, Q, R):
div(F) = ∇ · F=∂P
∂x +∂Q
∂y +∂R
∂z
Step 1: Compute the partial derivatives of P,Q, and R.
∂P
∂x =∂
∂x (2yz + sin(x)) = 0 + cos(x),
∂Q
∂y =∂
∂y (xz +ey) = 0 + ey,
∂R
∂z =∂
∂z y2+ cos(z)= 0 −sin(z).
Step 2: Add up the partial derivatives to find the divergence of F.
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z
= cos(x) + ey−sin(z).
Therefore, the divergence of the vector field Fis cos(x) + ey−sin(z).
Question 27
Question
Let F(x, y, z) = yezi+zexj+xeykbe a vector field. Compute ∇ · (∇ × F).
18
Solution
Step 1: First, we find ∇ × F:
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
yezzexxey
Step 2: Calculate the determinant:
∇ × F=∂(xey)
∂y −∂(zex)
∂z i−∂(yez)
∂x −∂(xey)
∂z j+∂(zex)
∂x −∂(yez)
∂y k
= (xey)i−(−zex)j+ (yez)k
=xeyi+zexj+yezk
Step 3: Next, we find ∇ · (∇ × F):
∇ · (∇ × F) = ∂
∂x (xey) + ∂
∂y (zex) + ∂
∂z (yez)
=ey+zex+yez
=ey+zex+xey
Therefore, ∇ · (∇ × F) = ey+zex+xey.
Question 28
Question
Let F(x, y, z) = x2yi+xyzj+xz2kbe a vector field in R3. Calculate 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, where P,Q, and Rare the vector field’s component functions, we
use the formula div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z .
Step 1: Identify the component functions of F:F(x, y, z) = x2yi+xyzj+
xz2k, where P(x, y, z) = x2y,Q(x, y, z) = xyz, and R(x, y, z) = xz2.
Step 2: Find the partial derivatives:
∂P
∂x = 2xy, ∂Q
∂y =xz, and ∂R
∂z = 2xz.
Step 3: Compute the divergence:
div(F) = ∂P
∂x +∂Q
∂y +∂R
∂z = 2xy +xz + 2xz = 2xy + 3xz.
Therefore, the divergence of the vector field Fis 2xy + 3xz.
19
Question 29
Question
Let F(x, y, z) = x2i+y3j+z4k. Compute ∇ · ∇ × F.
Solution
Step 1: Find ∇ × F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
x2y3z4
∇ × F= (∂z4
∂y −0)i−(∂z4
∂x −0)j+ ( ∂x2
∂y −∂y3
∂x )k
∇ × F= 0i−0j+ 0k=0
Step 2: Compute ∇ · ∇ × F.
∇ · ∇ × F=∇ · 0= 0
Therefore, ∇ · ∇ × F=0.
Question 30
Question
Let
F(x, y, z)=(x2+y2)ˆ
i+ (y2+z2)ˆ
j+ (z2+x2)ˆ
k. Compute ∇ ·
F.
Solution
Step 1: The divergence 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
kis given by the formula
∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z .
Step 2: In this case, P(x, y, z) = x2+y2,Q(x, y, z) = y2+z2, and R(x, y, z) =
z2+x2.
Step 3: Calculate the partial derivatives:
∂P
∂x = 2x,
∂Q
∂y = 2y,
∂R
∂z = 2z.
20
Step 4: Compute the divergence:
∇ ·
F=∂P
∂x +∂Q
∂y +∂R
∂z = 2x+ 2y+ 2z= 2(x+y+z).
Therefore, ∇ ·
F= 2(x+y+z) .
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