MATH 332 - Multivariable Calculus Question
Bank
Question 1
Problem: Let F= (y2, x) be a vector field and Cbe the curve given by the
parabola y=x2from (0,0) to (1,1). Evaluate the line integral RCF·dr.
Step-by-Step Solution:
Step 1: Parametrize the curve C.
Since Cis the curve on the parabola y=x2from (0,0) to (1,1), we can
parametrize the curve by setting x=t,y=t2where tvaries from 0 to 1. Thus,
r(t) = (t, t2) with tin [0,1].
Step 2: Find the differential dr.
The derivative of r(t) is r′(t) = (1,2t). Thus, dr= (1,2t)dt.
Step 3: Calculate F(r(t)).
Substituting the parametric equations into F, we get:
F(t, t2)=(t4, t)
Step 4: Compute F(r(t)) ·dr(t).
The dot product F(r(t)) ·dr(t) = (t4, t)·(1,2t) = t4·1 + t·2t=t4+ 2t2.
Step 5: Evaluate the integral.
Calculate R1
0(t4+ 2t2)dt.
1. Integrate t4from 0 to 1:
Zt4dt =t5
5
1
0
=1
5−0 = 1
5
2. Integrate 2t2from 0 to 1:
Z2t2dt = 2 t3
3
1
0
= 2 1
3−0=2
3
Combine these results:
Z1
0
(t4+ 2t2)dt =1
5+2
3=3
15 +10
15 =13
15
1
Answer: The value of the line integral RCF·dris 13
15 .Question 1: Eval-
uating a Line Integral over a Vector Field
Problem: Let F = (y2, x)be a vector field and Cbe the curve given
by the parabola y=x2from (0,0) to (1,1). Evaluate the line integral
RCF·dr.
Step-by-Step Solution:
Step 1: Parametrize the curve C.
Since Cis the curve on the parabola y=x2from (0,0) to (1,1), we
can parametrize the curve by setting x=t,y=t2where tvaries from
0 to 1. Thus, r(t) = (t, t2)with tin [0,1].
Step 2: Find the differential dr.
The derivative of r(t)is r′(t) = (1,2t). Thus, dr= (1,2t)dt.
Step 3: Calculate F(r(t)).
Substituting the parametric equations into F, we get:
F(t, t2)=(t4, t)
Step 4: Compute F(r(t)) ·dr(t).
The dot product F(r(t)) ·dr(t) = (t4, t)·(1,2t) = t4·1 + t·2t=t4+ 2t2.
Step 5: Evaluate the integral.
Calculate R1
0(t4+ 2t2)dt.
1. Integrate t4from 0 to 1:
Zt4dt =t5
5
1
0
=1
5−0 = 1
5
2. Integrate 2t2from 0 to 1:
Z2t2dt = 2 t3
3
1
0
= 2 1
3−0=2
3
Combine these results:
Z1
0
(t4+ 2t2)dt =1
5+2
3=3
15 +10
15 =13
15
Answer: The value of the line integral RCF·dr is 13
15 .
Question 2
Evaluate the double integral RRR(x2y−3xy2)dA over the rectangle
Rwhere R= [0,2] ×[0,1].
Step-by-Step Solution:
Step 1: Identify the limits of integration.
Given that R= [0,2]×[0,1], we can set up the integral with xvarying
from 0to 2and yvarying from 0to 1.
2
Step 2: Set up the double integral.
The integral can be set up as:
ZZR
(x2y−3xy2)dA =Z1
0Z2
0
(x2y−3xy2)dx dy.
Step 3: Integrating with respect to xfirst.
Start by integrating the inner integral with respect to x:
Z2
0
(x2y−3xy2)dx.
Expanding the terms inside the integral:
=Z2
0
x2y dx −Z2
0
3xy2dx =yZ2
0
x2dx −3y2Z2
0
x dx.
Calculate R2
0x2dx using the power rule ( xn+1
n+1 ):
=x3
32
0
=23
3=8
3.
And calculate R2
0x dx:
=x2
22
0
=22
2= 2.
Return to the equation:
=y8
3−3y2(2) = 8y
3−6y2.
Step 4: Integrating the result with respect to y.
The outer integral is now:
Z1
08y
3−6y2dy =8
3Z1
0
y dy −6Z1
0
y2dy.
Calculate R1
0y dy and R1
0y2dy respectively:
=8
3y2
21
0
−6y3
31
0
=8
3·1
2−6·1
3.
Simplifying:
=4
3−2 = −2
3.
Step 5: Conclude with the final result.
3
Thus, the value of the double integral RRR(x2y−3xy2)dA over the
region Ris −2
3.
The steps illustrate the method of iterating through integrals in
multiple variables, a fundamental concept in multivariable calculus.
Question 2: Evaluating a Double Integral Over a Rectangular Region
Evaluate the double integral RRR(x2y−3xy2)dA over the rectangle
Rwhere R= [0,2] ×[0,1].
Step-by-Step Solution:
Step 1: Identify the limits of integration.
Given that R= [0,2]×[0,1], we can set up the integral with xvarying
from 0to 2and yvarying from 0to 1.
Step 2: Set up the double integral.
The integral can be set up as:
ZZR
(x2y−3xy2)dA =Z1
0Z2
0
(x2y−3xy2)dx dy.
Step 3: Integrating with respect to xfirst.
Start by integrating the inner integral with respect to x:
Z2
0
(x2y−3xy2)dx.
Expanding the terms inside the integral:
=Z2
0
x2y dx −Z2
0
3xy2dx =yZ2
0
x2dx −3y2Z2
0
x dx.
Calculate R2
0x2dx using the power rule ( xn+1
n+1 ):
=x3
32
0
=23
3=8
3.
And calculate R2
0x dx:
=x2
22
0
=22
2= 2.
Return to the equation:
=y8
3−3y2(2) = 8y
3−6y2.
Step 4: Integrating the result with respect to y.
The outer integral is now:
Z1
08y
3−6y2dy =8
3Z1
0
y dy −6Z1
0
y2dy.
4
Calculate R1
0y dy and R1
0y2dy respectively:
=8
3y2
21
0
−6y3
31
0
=8
3·1
2−6·1
3.
Simplifying:
=4
3−2 = −2
3.
Step 5: Conclude with the final result.
Thus, the value of the double integral RRR(x2y−3xy2)dA over the
region Ris −2
3.
The steps illustrate the method of iterating through integrals in
multiple variables, a fundamental concept in multivariable calculus.
Question 3
Problem Statement: Consider the function f(x, y) = x2y+exy +
sin(x+y). Find ∂f
∂x and ∂f
∂y .
Step-by-Step Solution for ∂f
∂x
Step 1: Apply the partial derivative with respect to x. To find ∂f
∂x ,
we differentiate f(x, y)with respect to x, treating yas a constant.
Step 2: Differentiate each term separately. - The first term is x2y.
Differentiating this with respect to xgives:
∂
∂x(x2y)=2xy (using the power rule and treating yas a constant)
- The second term is exy. Differentiating this with respect to x
using the chain rule:
∂
∂x(exy ) = exy ·∂
∂x(xy) = exy ·y
- The third term is sin(x+y). Differentiate using the chain rule:
∂
∂x(sin(x+y)) = cos(x+y)·∂
∂x(x+y) = cos(x+y)·1 = cos(x+y)
Step 3: Sum up the derivatives. Combining all these, we get:
∂f
∂x = 2xy +yexy + cos(x+y)
Step-by-Step Solution for ∂f
∂y
Step 1: Apply the partial derivative with respect to y. To find ∂f
∂y ,
we differentiate f(x, y)with respect to y, treating xas a constant.
5
Step 2: Differentiate each term separately. - The first term is x2y.
Differentiating this with respect to ygives:
∂
∂y (x2y) = x2(using the constant multiple rule)
- The second term is exy. Differentiating this with respect to y
using the chain rule:
∂
∂y (exy) = exy ·∂
∂y (xy) = exy ·x
- The third term is sin(x+y). Differentiate using the chain rule:
∂
∂y (sin(x+y)) = cos(x+y)·∂
∂y (x+y) = cos(x+y)·1 = cos(x+y)
Step 3: Sum up the derivatives. Combining these derivatives, we
get: ∂f
∂y =x2+xexy + cos(x+y)
Answers: - ∂f
∂x = 2xy +yexy + cos(x+y)-∂f
∂y =x2+xexy + cos(x+y)
Question 3: Multivariable Calculus - Finding Partial Derivatives
Problem Statement: Consider the function f(x, y) = x2y+exy +
sin(x+y). Find ∂f
∂x and ∂f
∂y .
Step-by-Step Solution for ∂f
∂x
Step 1: Apply the partial derivative with respect to x. To find ∂f
∂x ,
we differentiate f(x, y)with respect to x, treating yas a constant.
Step 2: Differentiate each term separately. - The first term is x2y.
Differentiating this with respect to xgives:
∂
∂x(x2y)=2xy (using the power rule and treating yas a constant)
- The second term is exy. Differentiating this with respect to x
using the chain rule:
∂
∂x(exy ) = exy ·∂
∂x(xy) = exy ·y
- The third term is sin(x+y). Differentiate using the chain rule:
∂
∂x(sin(x+y)) = cos(x+y)·∂
∂x(x+y) = cos(x+y)·1 = cos(x+y)
Step 3: Sum up the derivatives. Combining all these, we get:
∂f
∂x = 2xy +yexy + cos(x+y)
6
Step-by-Step Solution for ∂f
∂y
Step 1: Apply the partial derivative with respect to y. To find ∂f
∂y ,
we differentiate f(x, y)with respect to y, treating xas a constant.
Step 2: Differentiate each term separately. - The first term is x2y.
Differentiating this with respect to ygives:
∂
∂y (x2y) = x2(using the constant multiple rule)
- The second term is exy. Differentiating this with respect to y
using the chain rule:
∂
∂y (exy) = exy ·∂
∂y (xy) = exy ·x
- The third term is sin(x+y). Differentiate using the chain rule:
∂
∂y (sin(x+y)) = cos(x+y)·∂
∂y (x+y) = cos(x+y)·1 = cos(x+y)
Step 3: Sum up the derivatives. Combining these derivatives, we
get: ∂f
∂y =x2+xexy + cos(x+y)
Answers: - ∂f
∂x = 2xy +yexy + cos(x+y)-∂f
∂y =x2+xexy + cos(x+y)
Question 4
Problem Statement: Evaluate the double integral
ZZD
(x2+y2)dx dy
where Dis the disk centered at the origin with radius 3.
Step-by-Step Solution:
Step 1: Convert the Integral to Polar Coordinates Since Dis a
disc centered at the origin with radius 3, it is convenient to use polar
coordinates. In polar coordinates, x=rcos θand y=rsin θ. The
area element dx dy is converted to polar coordinates as r dr dθ. The
expression x2+y2converts to r2.
The double integral becomes:
ZZD
(x2+y2)dx dy =Z2π
0Z3
0
r2·r dr dθ
Step 2: Evaluate the inner integral (over r)
Z3
0
r3dr
7
To integrate r3, use the power rule for integration:
Zr3dr =r4
4
3
0
=34
4−04
4=81
4
Step 3: Evaluate the outer integral (over θ) Now, substitute the
result of the inner integral back into the integral over θ:
Z2π
0
81
4dθ
This integral simplifies to multiplying 81
4by the length of the interval
over which θis integrated:
81
4×2π=81
2π
Final Answer: The value of the double integral RRD(x2+y2)dx dy
over the disk with radius 3 is 81
2π. Question 4: Evaluating a Double
Integral in Polar Coordinates
Problem Statement: Evaluate the double integral
ZZD
(x2+y2)dx dy
where Dis the disk centered at the origin with radius 3.
Step-by-Step Solution:
Step 1: Convert the Integral to Polar Coordinates Since Dis a
disc centered at the origin with radius 3, it is convenient to use polar
coordinates. In polar coordinates, x=rcos θand y=rsin θ. The
area element dx dy is converted to polar coordinates as r dr dθ. The
expression x2+y2converts to r2.
The double integral becomes:
ZZD
(x2+y2)dx dy =Z2π
0Z3
0
r2·r dr dθ
Step 2: Evaluate the inner integral (over r)
Z3
0
r3dr
To integrate r3, use the power rule for integration:
Zr3dr =r4
4
3
0
=34
4−04
4=81
4
Step 3: Evaluate the outer integral (over θ) Now, substitute the
result of the inner integral back into the integral over θ:
Z2π
0
81
4dθ
8
This integral simplifies to multiplying 81
4by the length of the interval
over which θis integrated:
81
4×2π=81
2π
Final Answer: The value of the double integral RRD(x2+y2)dx dy
over the disk with radius 3 is 81
2π.
Question 5
Consider the function f(x, y, z) = x2y+ezsin x.
Task: Determine the gradient vector of the function at the point
(1,−1,0).
—
Solution:
Step 1: Understand the Gradient The gradient of a function f
with variables x, y, z is a vector consisting of the partial derivatives of
fwith respect to each variable. It is denoted as ∇fand calculated
as:
∇f=∂f
∂x ,∂f
∂y ,∂f
∂z
Step 2: Calculate the Partial Derivative with respect to xGiven
the function f(x, y, z) = x2y+ezsin x,
∂f
∂x =∂
∂x(x2y) + ∂
∂x(ezsin x)
= 2xy +ezcos x
Step 3: Calculate the Partial Derivative with respect to y
∂f
∂y =∂
∂y (x2y) + ∂
∂y (ezsin x)
=x2
Step 4: Calculate the Partial Derivative with respect to z
∂f
∂z =∂
∂z (x2y) + ∂
∂z (ezsin x)
=ezsin x
Step 5: Evaluate the Gradient at the Point (1,−1,0) Substitute
x= 1,y=−1, and z= 0 into the partial derivatives:
∂f
∂x = 2(1)(−1) + e0cos 1 = −2 + cos 1
9
∂f
∂y = (1)2= 1
∂f
∂z =e0sin 1 = sin 1
Step 6: Form the Gradient Vector The gradient vector ∇fat the
point (1,−1,0) is,
∇f(1,−1,0) = (−2 + cos 1,1,sin 1)
Result: The gradient vector of the function f(x, y, z)at the point
(1,−1,0) is:
∇f(1,−1,0) = (−2 + cos 1,1,sin 1)
This represents the direction in which the function fincreases
most rapidly at the point (1,−1,0). Question 5: Find the Gradient
Vector
Consider the function f(x, y, z) = x2y+ezsin x.
Task: Determine the gradient vector of the function at the point
(1,−1,0).
—
Solution:
Step 1: Understand the Gradient The gradient of a function f
with variables x, y, z is a vector consisting of the partial derivatives of
fwith respect to each variable. It is denoted as ∇fand calculated
as:
∇f=∂f
∂x ,∂f
∂y ,∂f
∂z
Step 2: Calculate the Partial Derivative with respect to xGiven
the function f(x, y, z) = x2y+ezsin x,
∂f
∂x =∂
∂x(x2y) + ∂
∂x(ezsin x)
= 2xy +ezcos x
Step 3: Calculate the Partial Derivative with respect to y
∂f
∂y =∂
∂y (x2y) + ∂
∂y (ezsin x)
=x2
Step 4: Calculate the Partial Derivative with respect to z
∂f
∂z =∂
∂z (x2y) + ∂
∂z (ezsin x)
=ezsin x
10
Step 5: Evaluate the Gradient at the Point (1,−1,0) Substitute
x= 1,y=−1, and z= 0 into the partial derivatives:
∂f
∂x = 2(1)(−1) + e0cos 1 = −2 + cos 1
∂f
∂y = (1)2= 1
∂f
∂z =e0sin 1 = sin 1
Step 6: Form the Gradient Vector The gradient vector ∇fat the
point (1,−1,0) is,
∇f(1,−1,0) = (−2 + cos 1,1,sin 1)
Result: The gradient vector of the function f(x, y, z)at the point
(1,−1,0) is:
∇f(1,−1,0) = (−2 + cos 1,1,sin 1)
This represents the direction in which the function fincreases
most rapidly at the point (1,−1,0).
Question 6
Problem: Evaluate the surface integral of the scalar field f(x, y, z) =
x2+y2over the surface Sdefined by the cylinder x2+y2= 9 from z= 0
to z= 5.
Solution:
Step 1: Set up the Surface Integral The surface integral over a
scalar field is given by:
Z ZS
f(x, y, z)dS
Step 2: Parameterize the Surface Since the surface is a vertical
cylinder, a common parametrization in cylindrical coordinates is use-
ful: - Parameterizing the lateral surface of the cylinder, we can use
x= 3 cos θ,y= 3 sin θ, and z=z. - 0≤θ≤2π(complete circle around
the cylinder). - 0≤z≤5(height of the cylinder).
Step 3: Compute dS Using the parametrization from Step 2, the el-
ement of surface area dS when a surface is parameterized as (x(θ, z), y(θ, z), z)
is:
dS =|rθ×rz|dθ dz,
where: - rθ= (−3 sin θ, 3 cos θ, 0) - rz= (0,0,1) - rθ×rz= (3 cos θ, 3 sin θ, 0)
-|rθ×rz|= 3
Thus, dS = 3 dθ dz.
11
Step 4: Compute the Integral Plug everything into the integral:
Z5
z=0 Z2π
θ=0
(9 cos2θ+ 9 sin2θ) 3 dθ dz =Z5
z=0 Z2π
θ=0
27 dθ dz.
Step 5: Integrate with Respect to θand z
Z5
z=0
[27 ×(2π−0)] dz =Z5
z=0
54π dz = 54π×(5 −0) = 270π.
Final Answer: The surface integral of the scalar field f(x, y, z) =
x2+y2over the cylinder is 270π. Question 6: Surface Integral of a
Scalar Field
Problem: Evaluate the surface integral of the scalar field f(x, y, z) =
x2+y2over the surface Sdefined by the cylinder x2+y2= 9 from z= 0
to z= 5.
Solution:
Step 1: Set up the Surface Integral The surface integral over a
scalar field is given by:
Z ZS
f(x, y, z)dS
Step 2: Parameterize the Surface Since the surface is a vertical
cylinder, a common parametrization in cylindrical coordinates is use-
ful: - Parameterizing the lateral surface of the cylinder, we can use
x= 3 cos θ,y= 3 sin θ, and z=z. - 0≤θ≤2π(complete circle around
the cylinder). - 0≤z≤5(height of the cylinder).
Step 3: Compute dS Using the parametrization from Step 2, the el-
ement of surface area dS when a surface is parameterized as (x(θ, z), y(θ, z), z)
is:
dS =|rθ×rz|dθ dz,
where: - rθ= (−3 sin θ, 3 cos θ, 0) - rz= (0,0,1) - rθ×rz= (3 cos θ, 3 sin θ, 0)
-|rθ×rz|= 3
Thus, dS = 3 dθ dz.
Step 4: Compute the Integral Plug everything into the integral:
Z5
z=0 Z2π
θ=0
(9 cos2θ+ 9 sin2θ) 3 dθ dz =Z5
z=0 Z2π
θ=0
27 dθ dz.
Step 5: Integrate with Respect to θand z
Z5
z=0
[27 ×(2π−0)] dz =Z5
z=0
54π dz = 54π×(5 −0) = 270π.
Final Answer: The surface integral of the scalar field f(x, y, z) =
x2+y2over the cylinder is 270π.
12
Question 7
Problem: Evaluate the triple integral RRREz dx dy dz, where Eis
the cylindrical region defined by x2+y2≤9,0≤z≤5.
Step-by-Step Solution
Step 1: Understand the Region of Integration * The problem in-
volves a cylinder with a radius of 3 along the z-axis from z= 0 to
z= 5. * The circular cross-section in the x-y plane has x2+y2≤9.
Step 2: Convert to Cylindrical Coordinates * To simplify the in-
tegral, we convert the problem into cylindrical coordinates. Here,
x=rcos θ,y=rsin θ, and z=z. The differential element dx dy dz
becomes r dr dθ dz.
Step 3: Setup the Integral in Cylindrical Coordinates * The limits
for rare from 0 to 3 (the radius of the cylinder). * The limits for θ
are from 0 to 2π(since it is a full circle). * The limits for zare from
0 to 5 (the height of the cylinder).
* The integral in cylindrical coordinates hence is:
Z5
0Z2π
0Z3
0
z r dr dθ dz
Step 4: Evaluate the Integral * First, integrate with respect to r:
Z3
0
z r dr =z1
2r23
0
=z·9
2=9z
2
* Next, integrate with respect to θ:
Z2π
0
9z
2dθ =9z
2×2π= 9πz
* Finally, integrate with respect to z:
Z5
0
9πz dz = 9π1
2z25
0
= 9π·25
2=225π
2
Answer: The value of the triple integral RRREz dx dy dz over the
cylindrical region Eis 225π
2. Question 7: Evaluate the Triple Integral
Over a Cylinder
Problem: Evaluate the triple integral RRREz dx dy dz, where Eis
the cylindrical region defined by x2+y2≤9,0≤z≤5.
Step-by-Step Solution
Step 1: Understand the Region of Integration * The problem in-
volves a cylinder with a radius of 3 along the z-axis from z= 0 to
z= 5. * The circular cross-section in the x-y plane has x2+y2≤9.
Step 2: Convert to Cylindrical Coordinates * To simplify the in-
tegral, we convert the problem into cylindrical coordinates. Here,
13
x=rcos θ,y=rsin θ, and z=z. The differential element dx dy dz
becomes r dr dθ dz.
Step 3: Setup the Integral in Cylindrical Coordinates * The limits
for rare from 0 to 3 (the radius of the cylinder). * The limits for θ
are from 0 to 2π(since it is a full circle). * The limits for zare from
0 to 5 (the height of the cylinder).
* The integral in cylindrical coordinates hence is:
Z5
0Z2π
0Z3
0
z r dr dθ dz
Step 4: Evaluate the Integral * First, integrate with respect to r:
Z3
0
z r dr =z1
2r23
0
=z·9
2=9z
2
* Next, integrate with respect to θ:
Z2π
0
9z
2dθ =9z
2×2π= 9πz
* Finally, integrate with respect to z:
Z5
0
9πz dz = 9π1
2z25
0
= 9π·25
2=225π
2
Answer: The value of the triple integral RRREz dx dy dz over the
cylindrical region Eis 225π
2.
Question 8
Problem:
Consider the function f(x, y) = x3+ 3xy2−15x. You are at the point
P(2,−1).
a) Find the gradient of fat P.
b) Determine the directional derivative of fat Pin the direction
of the vector v =⟨3,4⟩.
Solution:
Part (a): Finding the Gradient at Point P(2,−1)
Step 1: Calculate the partial derivatives of f. - The partial deriva-
tive of fwith respect to x, denoted as fx, is:
fx=∂
∂x(x3+ 3xy2−15x)=3x2+ 3y2−15.
- The partial derivative of fwith respect to y, denoted as fy, is:
fy=∂
∂y (x3+ 3xy2−15x)=6xy.
14
Step 2: Evaluate these derivatives at point P(2,−1): - fx(2,−1) =
3(2)2+ 3(−1)2−15 = 12 + 3 −15 = 0. - fy(2,−1) = 6(2)(−1) = −12.
Step 3: Compile the gradient vector: - ∇f(2,−1) = ⟨0,−12⟩.
Answer for part (a): f(2,−1) = ⟨0,−12⟩.
Part (b): Directional Derivative in the Direction of Vector v
Step 1: Normalize the vector v. - The magnitude |v|is p(3)2+ (4)2=
5. - The unit vector in the direction of v is u =3
5,4
5.
Step 2: Use the formula for the directional derivative: - The di-
rectional derivative of fat Pin the direction of u is given by ∇f·u:
Duf=⟨0,−12⟩ · 3
5,4
5= 0 ·3
5+ (−12) ·4
5=−48
5.
Answer for part (b): The directional derivative of fat P(2,−1)
in the direction of v is −48
5. Question 8: Multivariable Calculus -
Gradient and Directional Derivative
Problem:
Consider the function f(x, y) = x3+ 3xy2−15x. You are at the point
P(2,−1).
a) Find the gradient of fat P.
b) Determine the directional derivative of fat Pin the direction
of the vector v =⟨3,4⟩.
Solution:
Part (a): Finding the Gradient at Point P(2,−1)
Step 1: Calculate the partial derivatives of f. - The partial deriva-
tive of fwith respect to x, denoted as fx, is:
fx=∂
∂x(x3+ 3xy2−15x)=3x2+ 3y2−15.
- The partial derivative of fwith respect to y, denoted as fy, is:
fy=∂
∂y (x3+ 3xy2−15x)=6xy.
Step 2: Evaluate these derivatives at point P(2,−1): - fx(2,−1) =
3(2)2+ 3(−1)2−15 = 12 + 3 −15 = 0. - fy(2,−1) = 6(2)(−1) = −12.
Step 3: Compile the gradient vector: - ∇f(2,−1) = ⟨0,−12⟩.
Answer for part (a): f(2,−1) = ⟨0,−12⟩.
Part (b): Directional Derivative in the Direction of Vector v
Step 1: Normalize the vector v. - The magnitude |v|is p(3)2+ (4)2=
5. - The unit vector in the direction of v is u =3
5,4
5.
Step 2: Use the formula for the directional derivative: - The di-
rectional derivative of fat Pin the direction of u is given by ∇f·u:
Duf=⟨0,−12⟩ · 3
5,4
5= 0 ·3
5+ (−12) ·4
5=−48
5.
Answer for part (b): The directional derivative of fat P(2,−1) in
the direction of v is −48
5.
15
Question 9
Evaluate the double integral of the function f(x, y) = x2yover the
region defined by 0≤y≤1and y≤x≤2.
Solution:
Step 1: Sketch the region of integration. - The region of integration
is determined by two conditions: 0≤y≤1and y≤x≤2. This is a
triangular region starting from the line y= 0 to y= 1 with xvalues
starting from the line x=yto x= 2.
Step 2: Set up the double integral. - Since the limits for yare
constants and the limits for xare functions of y, it’s easier to integrate
with respect to xfirst. The limits for xare from yto 2, and the limits
for yare from 0to 1. Therefore, the integral setup is:
Z1
0Z2
y
x2y dx dy
Step 3: Compute the inner integral with respect to x. - Integrate
x2ywith respect to x. Since ycan be treated as a constant while
integrating with respect to x, the integral becomes:
Z2
y
x2y dx =yZ2
y
x2dx =yx3
32
y
=y23
3−y3
3=y8
3−y3
3
=8y
3−y4
3
Step 4: Compute the outer integral with respect to y. - Integrate
the expression from the previous step with respect to y:
Z1
08y
3−y4
3dy =8
3Z1
0
y dy −1
3Z1
0
y4dy
Compute each integral:
Z1
0
y dy =y2
2
1
0
=1
2
Z1
0
y4dy =y5
5
1
0
=1
5
Therefore, the integral becomes:
8
3·1
2−1
3·1
5=4
3−1
15 =20
15 −1
15 =19
15
Step 5: Conclusion - The value of the integral is:
19
15
16
This result represents the volume under the surface described by
f(x, y) = x2yover the specified region in the xy-plane. Question 9:
Multivariable Calculus for Liberty University
Evaluate the double integral of the function f(x, y) = x2yover the
region defined by 0≤y≤1and y≤x≤2.
Solution:
Step 1: Sketch the region of integration. - The region of integration
is determined by two conditions: 0≤y≤1and y≤x≤2. This is a
triangular region starting from the line y= 0 to y= 1 with xvalues
starting from the line x=yto x= 2.
Step 2: Set up the double integral. - Since the limits for yare
constants and the limits for xare functions of y, it’s easier to integrate
with respect to xfirst. The limits for xare from yto 2, and the limits
for yare from 0to 1. Therefore, the integral setup is:
Z1
0Z2
y
x2y dx dy
Step 3: Compute the inner integral with respect to x. - Integrate
x2ywith respect to x. Since ycan be treated as a constant while
integrating with respect to x, the integral becomes:
Z2
y
x2y dx =yZ2
y
x2dx =yx3
32
y
=y23
3−y3
3=y8
3−y3
3
=8y
3−y4
3
Step 4: Compute the outer integral with respect to y. - Integrate
the expression from the previous step with respect to y:
Z1
08y
3−y4
3dy =8
3Z1
0
y dy −1
3Z1
0
y4dy
Compute each integral:
Z1
0
y dy =y2
2
1
0
=1
2
Z1
0
y4dy =y5
5
1
0
=1
5
Therefore, the integral becomes:
8
3·1
2−1
3·1
5=4
3−1
15 =20
15 −1
15 =19
15
Step 5: Conclusion - The value of the integral is:
19
15
This result represents the volume under the surface described by
f(x, y) = x2yover the specified region in the xy-plane.
17
Question 10
Problem Statement: Calculate the line integral of the vector field
F= (x2+y2,2xy)over the curve Cdefined by the quarter circle x2+y2=
4from (2,0) to (0,2), traversing counterclockwise.
Step-by-Step Solution:
Step 1: Parametrize the curve C.
Since Cis a quarter circle from (2,0) to (0,2), let’s parametrize it
using:
x= 2 cos(t), y = 2 sin(t)
where tranges from 0to π/2to cover the quarter circle in the coun-
tercycle direction.
Step 2: Compute the derivatives of the parametric equations.
Compute dx and dy:dx
dt =−2 sin(t),
dy
dt = 2 cos(t).
Step 3: Use these to write r′(t), the derivative of the position
vector:
r′(t)=(−2 sin(t),2 cos(t)).
Step 4: Plug the parametric equations into the vector field.
Replace xand yin the vector field:
F(x(t), y(t)) = ((2 cos(t))2+ (2 sin(t))2,2·2 cos(t)·2 sin(t))
F(x(t), y(t)) = (4cos2(t) + 4sin2(t),8 cos(t) sin(t)).
Use the identity cos2(t) + sin2(t)=1:
F(x(t), y(t)) = (4,8 cos(t) sin(t)).
Step 5: Compute the line integral.
Line integral is defined as:
ZC
F·dr=Zπ
2
0
F(x(t), y(t)) ·r′(t)dt
Compute the dot product F ·r′(t):
(4,8 cos(t) sin(t)) ·(−2 sin(t),2 cos(t)) = −8 sin(t) + 16 cos(t) sin(t) cos(t)
Utilize the identity cos(t) sin(t) = 1
2sin(2t):
F·r′(t) = −8 sin(t) + 8 sin(2t) cos(t)
18
Now, integrate from 0to π/2:
Zπ
2
0
(−8 sin(t) + 8 sin(2t) cos(t)) dt
Step 6: Evaluate the integral.
Separate and solve each integral:
Zπ
2
0
−8 sin(t)dt = [−8 cos(t)]
π
2
0=−8(cos(π/2) −cos(0)) = 8
Zπ
2
0
8 sin(2t) cos(t)dt = 4 Zπ
2
0
sin(2t)d(2t)=4×[−cos(2t)
2]
π
2
0= 0
Adding these up:
Zπ
2
0
F·r′(t)dt = 8 + 0 = 8
Answer: The line integral of the vector field F = (x2+y2,2xy)over
the quarter circle from (2,0) to (0,2) is 8. Question 10: Evaluating a
Line Integral
Problem Statement: Calculate the line integral of the vector field
F= (x2+y2,2xy)over the curve Cdefined by the quarter circle x2+y2=
4from (2,0) to (0,2), traversing counterclockwise.
Step-by-Step Solution:
Step 1: Parametrize the curve C.
Since Cis a quarter circle from (2,0) to (0,2), let’s parametrize it
using:
x= 2 cos(t), y = 2 sin(t)
where tranges from 0to π/2to cover the quarter circle in the coun-
tercycle direction.
Step 2: Compute the derivatives of the parametric equations.
Compute dx and dy:dx
dt =−2 sin(t),
dy
dt = 2 cos(t).
Step 3: Use these to write r′(t), the derivative of the position
vector:
r′(t)=(−2 sin(t),2 cos(t)).
Step 4: Plug the parametric equations into the vector field.
Replace xand yin the vector field:
F(x(t), y(t)) = ((2 cos(t))2+ (2 sin(t))2,2·2 cos(t)·2 sin(t))
F(x(t), y(t)) = (4cos2(t) + 4sin2(t),8 cos(t) sin(t)).
19
Use the identity cos2(t) + sin2(t)=1:
F(x(t), y(t)) = (4,8 cos(t) sin(t)).
Step 5: Compute the line integral.
Line integral is defined as:
ZC
F·dr=Zπ
2
0
F(x(t), y(t)) ·r′(t)dt
Compute the dot product F ·r′(t):
(4,8 cos(t) sin(t)) ·(−2 sin(t),2 cos(t)) = −8 sin(t) + 16 cos(t) sin(t) cos(t)
Utilize the identity cos(t) sin(t) = 1
2sin(2t):
F·r′(t) = −8 sin(t) + 8 sin(2t) cos(t)
Now, integrate from 0to π/2:
Zπ
2
0
(−8 sin(t) + 8 sin(2t) cos(t)) dt
Step 6: Evaluate the integral.
Separate and solve each integral:
Zπ
2
0
−8 sin(t)dt = [−8 cos(t)]
π
2
0=−8(cos(π/2) −cos(0)) = 8
Zπ
2
0
8 sin(2t) cos(t)dt = 4 Zπ
2
0
sin(2t)d(2t)=4×[−cos(2t)
2]
π
2
0= 0
Adding these up:
Zπ
2
0
F·r′(t)dt = 8 + 0 = 8
Answer: The line integral of the vector field F = (x2+y2,2xy)over
the quarter circle from (2,0) to (0,2) is 8.
20
