MATH 332 - ADVANCED CALCULUS
- Double integrals over rectangular and
general regions
Question Bank - Set 8
Liberty University
Question 1
Question
Evaluate the double integral RRRexy dA over the region Rbounded by the curves
y=x2,y= 4,x= 0, and x= 2.
Solution
To evaluate the double integral over the region R, we first need to determine
the bounds of integration.
• Consider the limits of integration for x: We see that xranges from 0 to 2.
• For each value of xin this range, yranges from x2to 4.
Step 1: Setup the double integral
ZZR
exy dA =Z2
0Z4
x2
exy dy dx
Step 2: Evaluate the inner integral with respect to y
Z4
x2
exy dy =exy
x4
x2
=e4x
x−ex3
x
Step 3: Evaluate the outer integral with respect to x
Z2
0 e4x
x−ex3
x!dx
Step 4: Integrate the first term
Z2
0
e4x
xdx =1
4e4x2
0
=e8
4−1
4
Step 5: Integrate the second term
Z2
0
ex3
xdx =1
3ex32
0
=e8
3−1
3
Step 6: Combine the results
e8
4−1
4−e8
3−1
3=e8
12 −1
6
Therefore, the value of the double integral RRRexy dA is e8
12 −1
6.
Question 2
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Sketch the region R
The region Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the
points of intersection, we set the curves equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the x-values where y=x2and y= 2xintersect. Thus,
the region Ris bounded by x= 1,y=x2, and y= 2x, as shown below:
xy2;2y= 2x
x= 1
Step 2: Set up the double integral
The given double integral can be set up as:
ZZR
ey2dA =Zx=2
x=1 Zy=2x
y=x2
ey2dy dx
2
Step 3: Evaluate the inner integral
Integrating with respect to yfirst, we get:
Zey2dy =Zey2dy
=1
2√πerf(y)
Therefore, the inner integral becomes:
Zy=2x
y=x2
ey2dy =1
2√πerf(4x)−erf(x2)
Step 4: Evaluate the outer integral
Now we integrate the above expression with respect to x:
Zx=2
x=1
1
2√πerf(4x)−erf(x2)dx
This integral may not have a straightforward closed form solution and might
require numerical methods for evaluation.
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x,x= 0, and x= 2.
Solution
Step 1: First, sketch the region Rto understand its shape and boundaries.
Step 2: To evaluate the given double integral, we will convert it into an
iterated integral and choose the order of integration. Since the region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 2, it is better to
integrate with respect to yfirst and then with respect to x. Thus, the iterated
integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to yfirst:
=Z2
0x2y+y3
32x
x2
dx
Step 4: Simplify and evaluate the integral:
=Z2
02x3−x4+8x
3−8x3
3dx
3
=Z2
0−x4−5x3
3+8x
3dx
=−x5
5−5x4
12 +4x2
32
0
=−32
5−80
12 +16
3
=−32
5−160
12 +64
12
=−32
5−96
12
=−32
5−8
=−32
5−40
5
=−72
5
Therefore, the value of the given double integral over the region Ris −72
5.
Question 4
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Draw a sketch of the region Rto visualize the bounds of integration.
The region Ris bound by the curves y=x2,y= 1,x= 0, and x= 1. It is a
triangle with vertices at (0,0),(1,0), and (1,1).
Step 2: Identify the bounds of integration for xand y. The region Rcan be
expressed as 0≤x≤1,0≤y≤x2.
Step 3: Set up the double integral with the given function and region. We
have RRRey2dA =R1
0Rx2
0ey2dy dx.
Step 4: Integrate with respect to yfirst. Rx2
0ey2dy =ey2
x2
0=ex4−e0=
ex4−1.
Step 5: Now, integrate with respect to x.R1
0(ex4−1) dx =1
4ex4−x
1
0=
1
4e−1.
4
Step 6: Therefore, the value of the double integral RRRey2dA over the region
Ris 1
4e−1.
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the given double integral over the region R, we will first determine
the bounds for xand yin terms of the region.
We observe that the region Ris enclosed between the curves y=x2and
y= 2xin the first quadrant. Therefore, we can express the bounds for xas
0≤x≤2and the bounds for yas x2≤y≤2x.
Now, we can rewrite the given double integral as:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydydx
Step 1: Evaluate the inner integral with respect to y:
Z2x
x2
ex+ydy =ex+y2x
x2
=e3x−ex2
Step 2: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z2
0
(e3x−ex2)dx ="e3x
3−ex2
2#2
0
=e6
3−e4
2−1
2
Therefore, the value of the double integral over the region Ris e6
3−e4
2−1
2.
Question 6
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
We begin by finding the points of intersection of the curves y=x2and y= 2x.
Step 1: Find the points of intersection. Setting the two equations equal to
each other gives: x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration. Since y= 2xis the larger
curve, the limits of integration on ywill be from x2to 2x, and the limits on x
will be from 0 to 2.
Step 3: Evaluate the double integral. The given double integral is:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx.
Step 4: Integrate with respect to y. Integrating with respect to yfirst, we
get:
Z2
0x2y+y3
32x
x2
dx
=Z2
0x2(2x) + (2x)3
3−(x2·x2+(x2)3
3)dx
=Z2
02x3+8x3
3−(x4+x6
3)dx
=Z2
0x3+8x3
3−x4−x6
3dx.
Step 5: Integrate with respect to x. Integrating the above expression with
respect to xgives:
x4
4+ 2x4−x5
5−x7
21 2
0
=24
4+ 2 ·24−25
5−27
21−04
4+ 2 ·04−05
5−07
21
=4 + 32 −32
5−128
21 −0
=3967
105 .
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 3967
105 .
Question 7
Question
Evaluate the double integral RRDex2+y2dA, where Dis the region bounded by
the curves y=x2and y= 2x.
6
Solution
To evaluate the double integral over the region D, we need to first determine
the limits of integration by finding the points of intersection between the two
curves y=x2and y= 2x.
Step 1: Determine the points of intersection: Setting x2= 2xgives
x2−2x= 0 ⇒x(x−2) = 0 which implies x= 0 and x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the integral: Since the region Dis bounded by y=x2
and y= 2x, the limits of integration for yare x2and 2x, and for xare 0and 2.
Thus, the integral becomes
Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral: Integrating with respect to y, we
have
Zex2+y2dy =Zex2+y2
y=2x
y=x2
=ex2+(2x)2
−ex2+x4
Step 4: Evaluate the double integral: Now, we evaluate the double
integral by integrating the result from Step 3 with respect to xover the limits
0to 2:
Z2
0
(ex2+(2x)2
−ex2+x4)dx =Z2
0
(e5x2
−ex4)dx
We can now solve this integral to obtain the final result.
Question 8
Question
Evaluate the double integral RRD(x2−y2)dA where Dis the region in the first
quadrant bounded by the curves y=x2and x=y2.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Find the limits of integration for y:The curves y=x2
and x=y2intersect at the points (0,0) and (1,1). Therefore, the limits of
integration for yare 0≤y≤x2.
Step 2: Find the limits of integration for x:To find the limits of
integration for x, we need to determine the range of xvalues within the region
D. This can be done by setting x=y2and y=x2equal to each other and
solving for xto find the intersection points.
7
Setting x=y2and y=x2equal to each other gives y2=xand x=y2,
which implies y4=y. So, y= 0 or y= 1.
Thus, the limits of integration for xare 0≤x≤y2where 0≤y≤1.
Step 3: Evaluate the double integral:
ZZD
(x2−y2)dA
=Z1
0Zy2
0
(x2−y2)dx dy
=Z1
0x3
3−xy2y2
0
dy
=Z1
0(y2)3
3−y4dy
=Z1
0y6
3−y4dy
=Z1
01
3−y2dy
=y
3−y3
31
0
=1
3−1
3−0+0
= 0.
Therefore, RRD(x2−y2)dA = 0.
Question 9
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and y= 1.
Solution
To solve this problem, we need to find the limits of integration by determining
the intersection points of the given curves.
Step 1: Find the intersection points.
Setting x2= 2x, we find the intersection point to be x= 0 and x= 2.
Therefore, the limits of integration for xare from 0 to 2.
Step 2: Set up the integral with respect to y.
The curves y=x2and y= 1 bound the region Rwith xvarying from 0 to
1. Therefore, the limits of integration for yare from x2to 1.
Step 3: Evaluate the double integral.
8
ZZR
(x+y)dA
=Z2
0Z1
x2
(x+y)dydx
=Z2
0xy +y2
21
x2
dx
=Z2
0x+1
2−x3−x4
2dx
=x2
2+x−x4
4−x5
10 2
0
=1
2(2)2+ 2 −1
4(2)4−1
10(2)5
= 2 + 2 −4−32
10
= 4 −16
5
=4
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 4
5.
Question 10
Question
Evaluate the double integral RRRex2+y2dx dy, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of in-
tegration by finding the points of intersection of the two curves y=x2and
y= 2x.
Step 1: Find the points of intersection Set x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Step 2: Determine the limits of integration Since y=x2is below
y= 2xin the region R, the limits of integration will be:
0≤x≤2
9
x2≤y≤2x
Step 3: Evaluate the double integral
ZZR
ex2+y2dx dy =Z2
0Z2x
x2
ex2+y2dy dx
Step 4: Integrate with respect to yfirst For the inner integral:
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2=ex2+(2x)2
−ex2+x2=e5x2
−e2x2
Step 5: Integrate with respect to xNow integrate the above expression
from 0 to 2:
Z2
0
(e5x2
−e2x2)dx ="e5x2
5−e2x2
2#2
0
=e20
5−e4
2−1
5−1
2
Therefore, the value of the double integral RRRex2+y2dx dy over the region
Ris e20
5−e4
2−3
10 .
Question 11
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region in the first
quadrant bounded by x= 0,y= 0,x= 3, and x+y= 4.
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by x= 0,y= 0,x= 3, and x+y= 4. To find the limits of
integration for x, we need to consider the intersection points of the lines: x= 0,
x= 3, and x+y= 4. Solving x= 0 and x+y= 4, we get y= 4. Solving x= 3
and x+y= 4, we get y= 1. So, the limits of integration for xare from 0 to
3. Similarly, to find the limits of integration for y, we consider the intersection
points of the lines: y= 0,x+y= 4, and x= 3. Solving y= 0 and x+y= 4,
we get x= 4. Solving x= 3 and x+y= 4, we get y= 1. So, the limits of
integration for yare from 0 to 4−x.
Step 2: Rewrite the given integral with the limits of integration. The double
integral becomes:
Z3
0Z4−x
0
(x2+y2)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Z4−x
0
(x2+y2)dy =x2y+y3
3
4−x
0
=x2(4 −x) + (4 −x)3
3
Step 4: Evaluate the outer integral with respect to x.
Z3
0x2(4 −x) + (4 −x)3
3dx =Z3
0
(4x2−x3+64
3−24x
3+12x2
3−x3
3)dx
=Z3
0
(3x2−2x3+64
3−8x+ 4x2−x3
3)dx
=Z3
0
(−2x3+ 7x2−8x+64
3)dx
Integrating term by term, we get:
−1
2x4+7
3x3−4x2+64
3x
3
0
=−81
2+ 21 −36 + 64 = 7
2
Thus, the value of the given double integral is 7
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 2.
Solution
To evaluate the double integral RRRxy dA, we need to set up the limits of inte-
gration for xand yby determining the region Rgeometrically.
Step 1: Determine the limits of integration for xand y
The region Ris bounded by the curves y=x2,y=x,x= 0, and x= 2.
Visually, Rlooks like a region between the curve y=x2and the line y=x,
where xranges from 0 to 2.
To find the limits of integration for x, we set y=x2and y=xequal to each
other:
x2=x=⇒x2−x= 0 =⇒x(x−1) = 0.
This gives us x= 0 and x= 1 as the values where the two curves intersect.
Therefore, the limits of integration for xare 0≤x≤1.
Next, in terms of y,xranges from x2to x, so the limits of integration for y
are x2≤y≤x.
Step 2: Evaluate the double integral
11
Now, we can rewrite the double integral over the region Ras:
ZZR
xy dA =Z1
0Zx
x2
xy dy dx.
Let’s now evaluate the double integral by first integrating with respect to y
and then with respect to x.
Z1
0Zx
x2
xy dy dx =Z1
01
2xy2x
x2
dx =Z1
01
2x3−1
2x5dx.
Integrating with respect to x, we get:
1
8x4−1
12x61
0
=1
8−1
12 =1
24.
Therefore, the value of the double integral RRRxy dA over the region Ris
1
24 .
Question 13
Question
Let Rbe the region bounded by the curves y= 2x2and y= 3x. Calculate the
double integral RRR4x+ 3y dA over the region R.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits
of integration for x, we first need to find the intersection points of the curves
y= 2x2and y= 3x. Setting them equal to each other:
2x2= 3x.
Solving for xgives us x= 0 and x= 3/2.
Therefore, the limits of integration for xare 0≤x≤3/2.
To find the limits of integration for y, we substitute the equations y= 2x2
and y= 3xinto the integral. This gives us 2x2≤y≤3x.
Step 2: Set up the double integral. The double integral to evaluate is:
ZZR
(4x+ 3y)dA =Z3/2
0Z3x
2x2
(4x+ 3y)dy dx.
Step 3: Evaluate the inner integral with respect to y.
Z3x
2x2
(4x+ 3y)dy =4xy +3
2y23x
2x2
.
12
Substitute the limits of integration:
= 4x(3x) + 3
2(3x)2−4x(2x2)−3
2(2x2)2.
Step 4: Simplify the expression.
= 12x2+27
2x2−8x3−6x2= 6x2−8x3+27
2x2.
Step 5: Evaluate the outer integral with respect to x.
=Z3/2
0
(6x2−8x3+27
2x2)dx =2x3−2x4+27
2x33/2
0
.
Substitute the limits of integration:
= 2 3
23
−23
24
+27
23
23
.
Step 6: Calculate the final answer.
= 9 −27
4+81
2=55
4.
Thus, the value of the double integral RRR(4x+ 3y)dA over the region Ris
55
4.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y= 2x.
Solution
Step 1: To identify the limits of integration, we need to find the points of
intersection of the curves y=x2and y= 2x. Setting the two equations equal
to each other gives us x2= 2x.
Step 2: Rearranging the equation, we get x2−2x= 0. Factoring out an x,
we have x(x−2) = 0. So, the solutions are x= 0 and x= 2.
Step 3: Next, we need to determine which curve is above the other in the
region R. Evaluating y=x2at x= 1 and y= 2xat x= 1, we find that y=x2
is above y= 2xin the region R.
Step 4: Therefore, the limits of integration for xwill be from 0to 1for the
inner integral, and the limits for ywill be from x2to 2xfor the outer integral.
13
Step 5: The double integral becomes:
Z1
0Z2x
x2
(x2+y2)dy dx
Step 6: Integrating with respect to yfirst, we get:
Z1
0x2y+y3
32x
x2
dx
Step 7: Substitute the limits of integration and simplify:
Z1
02x3+(2x)3
3−x4−x6
3dx
Step 8: Continuing to simplify gives:
2Z1
04x3
3−x4
3dx
Step 9: Integrating gives:
2x4
3−x5
15 1
0
Step 10: Evaluate the integral at the limits of integration:
21
3−1
15
Step 11: Finally, calculate the value to get the result:
24
15=8
15
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 8
15 .
Question 15
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2,y= 4,x= 0, and x= 2.
14
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded above by the line y= 4 and below by the curve y=x2.
The left boundary is x= 0 and the right boundary is x= 2. Thus, the limits
of integration are:
0≤x≤2
x2≤y≤4
Step 2: Now, we can set up and evaluate the double integral:
ZZR
(3x2+ 2y)dA =Z2
0Z4
x2
(3x2+ 2y)dydx
=Z2
03x2y+y24
x2dx
=Z2
03x2(4) + 42−3x2(x2)−x4dx
=Z2
0
(12x2+ 16 −3x4−x4)dx
=Z2
0
(−4x4+ 12x2+ 16) dx
=−4
5x5+ 4x3+ 16x2
0
=−4
5(2)5+ 4(2)3+ 16(2)−−4
5(0)5+ 4(0)3+ 16(0)
=−128
5+ 32 + 32−0
=−128
5+ 64
=192
5
Therefore, the value of the double integral is 192
5.
Question 16
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 4 and x= 0.
15
Solution
To evaluate the double integral RRRx2y dA, we first need to find the limits of
integration for xand y.
Step 1: Determine the limits of integration for y. The region Ris bounded
by y=x2,y= 4, and x= 0. Setting y=x2and y= 4 equal to each other gives
x2= 4, so x=±2. Therefore, the limits of integration for yare from x2to 4.
Step 2: Determine the limits of integration for x. Since Ris bounded by
x= 0 and x=±2, the limits of integration for xare from 0to 2.
Step 3: Setup the double integral. The given double integral can be written
as:
ZZR
x2y dA =Z2
0Z4
x2
x2y dy dx
Step 4: Evaluate the double integral.
Z2
0Z4
x2
x2y dy dx =Z2
0x2y2
24
x2
dx
=Z2
08x2−x6
2dx =Z2
0
8x2−x6
2dx
=8x3
3−x7
14 2
0
=64
3−128
14 =256
21
Hence, the value of the double integral RRRx2y dA over the region Ris 256
21 .
Question 17
Question
Evaluate the double integral ZZ
R
ex2+y2dA over the region Rbounded by the
curves y=x,y= 4x, and x= 1.
Solution
To evaluate the given double integral, we need to find the limits of integration
for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by y=xand y= 4x. Their intersection gives x=y= 0 and x=y= 1. So,
the limits of integration for xare from 0 to 1.
Step 2: Find the limits of integration for y:The region Ris also
bounded by x= 1. From the equation x= 1, we see that the region Rends at
x= 1. Combining this with y=xand y= 4x, we can express the limits of y
as x≤y≤4x.
16
Step 3: Set up and evaluate the double integral: The double integral
can be written as:
ZZ
R
ex2+y2dA =Z1
0Z4x
x
ex2+y2dy dx
Now, we calculate:
Z1
0Z4x
x
ex2+y2dy dx =Z1
0hex2+y2iy=4x
y=xdx
=Z1
0hex2+16x2
−ex2+x2idx
=Z1
0he17x2
−e2x2idx
=1
17e17x2
−1
2e2x21
0
=1
17e17 −1
2e2−1
17 −1
2
=1
17e17 −1
2e2+15
34
Therefore, the value of the double integral is 1
17 e17 −1
2e2+15
34 .
Question 18
Question
Evaluate the double integral RRR(xcos y)dA, where Ris the region bounded by
the curves y= 0,y= 2π,x= 0, and x=π.
Solution
Step 1: Rewrite the double integral in terms of xand y:
ZZR
(xcos y)dA =Z2π
0Zπ
0
(xcos y)dx dy
Step 2: Integrate with respect to x:
Zπ
0
(xcos y)dx =1
2x2cos y
π
0
=1
2(π2−0) cos y=π2
2cos y
Step 3: Substitute the result of the previous step back into the double inte-
gral:
Z2π
0
π2
2cos y dy
17
Step 4: Integrate with respect to y:
Z2π
0
π2
2cos y dy =π2
2Z2π
0
cos y dy
Step 5: Apply integration property of cosine function:
π2
2Z2π
0
cos y dy =π2
2sin y
2π
0
=π2
2(sin 2π−sin 0)
Step 6: Simplify the expression:
π2
2(0 −0) = 0
Therefore, the value of the double integral RRR(xcos y)dA over the region R
is 0.
Question 19
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 4x−x2.
Solution
To evaluate the double integral over the region D, we first need to determine
the limits of integration.
Step 1: Find the intersection points of the two curves. The intersec-
tion points occur when x2= 4x−x2. Simplifying, we get 2x2−4x= 0, which
gives 2x(x−2) = 0 ⇒x= 0 or x= 2.
Step 2: Determine the limits of integration for xand yin the
double integral. Since the region Dis bounded by y=x2and y= 4x−x2,
the limits of integration for xare from 0to 2and for yare from x2to 4x−x2.
Step 3: Set up and evaluate the double integral. We have:
ZZD
(x2+y2)dA =Z2
0Z4x−x2
x2
(x2+y2)dydx
Now, substitute y=x2and y= 4x−x2into the integrand:
Z2
0Z4x−x2
x2
(x2+y2)dydx =Z2
0x2y+y3
34x−x2
x2
dx
Z2
0x2(4x−x2) + (4x−x2)3
3−x2(x2) + (x2)3
3dx
Now, simplify and evaluate the integral over the specified limits (from 0to
2) to get the final answer.
18
Question 20
Question
Evaluate the double integral RRR(x2+y2)dA over the region Renclosed by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over the given region R, we first need to deter-
mine the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at x= 0 and x= 1. Therefore, the limits of integration
for xare from 0 to 1.
Step 2: Determine the limits of integration for y:For each xbetween
0 and 1, the corresponding yvalues are between x2and 2x. Thus, the limits of
integration for yare from x2to 2x.
Step 3: Setting up the double integral: The given double integral can
be expressed as:
ZZR
(x2+y2)dA =Z1
0Z2x
x2
(x2+y2)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
Step 5: Simplify the expression:
= 2x3+8x3
3−x4−x6
3
Step 6: Evaluate the outer integral with respect to x:Now, we
integrate the result from Step 5 with respect to x:
Z1
0
(2x3+8x3
3−x4−x6
3)dx
Step 7: Simplify and compute the final answer:
=2x4
4+8x4
12 −x5
5−x7
21
1
0
=1
2+2
3−1
5−1
21
19
=105
42 +28
42 −10
42 −2
42
=121
42
Thus, the value of the double integral over the region Ris 121
42 .
Question 21
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 1,x= 1, and x= 2.
Solution
To solve this problem, we first need to determine the limits of integration for x
and y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 1,x= 1, and x= 2. Therefore, the
limits of integration for xare 1≤x≤2, and the limits of integration for yare
x2≤y≤1.
Step 2: Set up the double integral The double integral of ex+yover
region Rcan be set up as follows:
ZZR
ex+ydA =Z2
1Z1
x2
ex+ydy dx
Step 3: Integrate with respect to yIntegrating ex+ywith respect to y:
Zex+ydy =ex·Zeydy =ex·ey=ex+y
Step 4: Evaluate the inner integral Substitute the limits of integration
for yinto the integral:
Z1
x2
ex+ydy =ex+1 −ex+x2
Step 5: Evaluate the outer integral Integrate the inner integral with
respect to x:
Z2
1
(ex+1 −ex+x2)dx
Step 6: Evaluate the integral Now, evaluate the integral:
"ex+1 −ex2+x
1+2#2
1
=e3−e2−e4
3+e5
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
e3−e2−e4
3+e5
3.
20
Question 22
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y= 2x.
Calculate the double integral RRR(2y−x)dA where dA =dxdy.
Solution
Step 1: Determine the intersection points of the curves y=x2and y= 2x.
Setting x2= 2x, we get x2−2x= 0, which simplifies to x(x−2) = 0. So, x= 0
or x= 2.
Therefore, the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration for xand y. The region Ris
bounded by the curves y=x2and y= 2x. The limits of integration for yare
x2and 2x, and for xare from 0 to 2.
Step 3: Set up the integral. The given double integral becomes:
ZZR
(2y−x)dA =Z2
0Z2x
x2
(2y−x)dydx
Step 4: Evaluate the inner integral with respect to y.
Z2x
x2
(2y−x)dy =y2−xy2x
x2= [(2x)2−2x·2x]−[x2−x·x2]
= (4x2−4x)−(x2−x3) = 3x2−3x
Step 5: Evaluate the outer integral with respect to x.
Z2
0
3x2−3x dx =x3−3
2x22
0
= [23−3
2·22]−[0 −0]
= (8 −6) −0 = 2
Therefore, the value of the double integral over the region Ris 2.
Question 23
Question
Evaluate the double integral RRRe−x2
−y2dx dy, where Ris the region bounded
by the curves y=x2and y= 4.
21
Solution
To evaluate the double integral over R, we first need to determine the limits of
integration.
Step 1: Find the limits of integration for x:The curves y=x2and
y= 4 intersect when x2= 4, so x=±2. Therefore, the limits of integration for
xare −2and 2.
Step 2: Find the limits of integration for y:The curve y=x2intersects
y= 4 at y= 4. Therefore, the limits of integration for yare x2and 4.
Thus, the double integral becomes:
ZZR
e−x2
−y2dx dy =Z2
−2Z4
x2
e−x2
−y2dy dx
Step 3: Evaluate the integral: Start by evaluating the inner integral
with respect to y:
Z4
x2
e−x2
−y2dy =h−e−x2
−y2i4
x2
=−e−x2
−42+e−x2
−x2
=−e−x2
−16 +e−2x2
Now, integrate this result with respect to x:
Z2
−2−e−x2
−16 +e−2x2dx
=16e−x2
−16 −1
2e−2x22
−2
= 16e−4−16 −1
2e−8−16e−4−16 −1
2e−8
= 2e−20 −e−8
Therefore, the value of the double integral is 2e−20 −e−8.
Question 24
Question
Evaluate the double integral RRD(2x+y)dA, where Dis the region bounded by
the curves y=x2,y= 4,x=−2, and x= 2.
22
Solution
Step 1: Draw the region Ddescribed in the question. It is the region bounded
by the curves y=x2,y= 4,x=−2, and x= 2. This region is a rectangle with
a parabolic cutout along one side.
Step 2: Set up the double integral in terms of xand y:
ZZD
(2x+y)dA =Z2
−2Z4
x2
(2x+y)dy dx
Step 3: Integrate with respect to yfirst:
Z2
−2Z4
x2
(2x+y)dy dx =Z2
−22xy +1
2y24
x2
dx
Step 4: Substitute the limits of integration and simplify the expression:
=Z2
−28x+ 8 −2x3−x4dx
Step 5: Integrate with respect to x:
=4x2+ 8x−1
2x4−1
5x52
−2
Step 6: Evaluate the integral at the limits of integration:
= (16 + 16 −8−32
5)−(16 −16+8+32
5)
Step 7: Simplify the expression to find the final answer:
=192
5
Therefore, the value of the double integral RRD(2x+y)dA over the region
Dis 192
5.
Question 25
Question
Evaluate the double integral RRDex2
−y2dx dy, where Dis the region bounded
by y=x2and y= 2x−x2.
23
Solution
To evaluate the given double integral, we need to find the bounds of integration
for xand yby determining the region D.
Step 1: Determine the bounds of integration for xand y
First, find the points of intersection of the curves y=x2and y= 2x−x2:
Setting x2= 2x−x2, we get 2x−2x2= 0 =⇒x(1 −2x) = 0. Thus, x= 0
or x=1
2.
Now, by observing the graph of the curves, we see that y= 2x−x2is above
y=x2in the interval 0≤x≤1
2and below y=x2in the interval 1
2≤x≤1.
So, the bounds of integration for xare 0≤x≤1
2.
For y, the bounds are given by the curves:
• Lower bound: y=x2
• Upper bound: y= 2x−x2
Thus, the bounds of integration for yare x2≤y≤2x−x2.
Step 2: Evaluate the integral
The double integral can now be expressed as:
ZZD
ex2
−y2dx dy =Z1
2
0Z2x−x2
x2
ex2
−y2dy dx
Let’s compute this integral step by step.
Step 3: Integrate with respect to y
Z2x−x2
x2
ex2
−y2dy =h−ex2
−y2i2x−x2
x2
=−ex2
−(2x−x2)2+ex2
−x2
=−ex2
−4x2+4x4+e0
=−e4x4
−3x2+ 1
Step 4: Integrate with respect to x
Z1
2
0−e4x4
−3x2+ 1 dx =−Z1
2
0
e4x4
−3x2dx +Z1
2
0
1dx
This integral may not have a closed-form solution and would typically be
calculated by numerical methods.
Therefore, the double integral RRDex2
−y2dx dy over the region Dbounded
by y=x2and y= 2x−x2needs to be evaluated numerically.
24
Step 4: Integrate the first term
Z2
0
e4x
xdx =1
4e4x2
0
=e8
4−1
4
Step 5: Integrate the second term
Z2
0
ex3
xdx =1
3ex32
0
=e8
3−1
3
Step 6: Combine the results
e8
4−1
4−e8
3−1
3=e8
12 −1
6
Therefore, the value of the double integral RRRexy dA is e8
12 −1
6.
Question 2
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Sketch the region R
The region Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the
points of intersection, we set the curves equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the x-values where y=x2and y= 2xintersect. Thus,
the region Ris bounded by x= 1,y=x2, and y= 2x, as shown below:
xy2;2y= 2x
x= 1
Step 2: Set up the double integral
The given double integral can be set up as:
ZZR
ey2dA =Zx=2
x=1 Zy=2x
y=x2
ey2dy dx
2
Step 3: Evaluate the inner integral
Integrating with respect to yfirst, we get:
Zey2dy =Zey2dy
=1
2√πerf(y)
Therefore, the inner integral becomes:
Zy=2x
y=x2
ey2dy =1
2√πerf(4x)−erf(x2)
Step 4: Evaluate the outer integral
Now we integrate the above expression with respect to x:
Zx=2
x=1
1
2√πerf(4x)−erf(x2)dx
This integral may not have a straightforward closed form solution and might
require numerical methods for evaluation.
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x,x= 0, and x= 2.
Solution
Step 1: First, sketch the region Rto understand its shape and boundaries.
Step 2: To evaluate the given double integral, we will convert it into an
iterated integral and choose the order of integration. Since the region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 2, it is better to
integrate with respect to yfirst and then with respect to x. Thus, the iterated
integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to yfirst:
=Z2
0x2y+y3
32x
x2
dx
Step 4: Simplify and evaluate the integral:
=Z2
02x3−x4+8x
3−8x3
3dx
3
=Z2
0−x4−5x3
3+8x
3dx
=−x5
5−5x4
12 +4x2
32
0
=−32
5−80
12 +16
3
=−32
5−160
12 +64
12
=−32
5−96
12
=−32
5−8
=−32
5−40
5
=−72
5
Therefore, the value of the given double integral over the region Ris −72
5.
Question 4
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Draw a sketch of the region Rto visualize the bounds of integration.
The region Ris bound by the curves y=x2,y= 1,x= 0, and x= 1. It is a
triangle with vertices at (0,0),(1,0), and (1,1).
Step 2: Identify the bounds of integration for xand y. The region Rcan be
expressed as 0≤x≤1,0≤y≤x2.
Step 3: Set up the double integral with the given function and region. We
have RRRey2dA =R1
0Rx2
0ey2dy dx.
Step 4: Integrate with respect to yfirst. Rx2
0ey2dy =ey2
x2
0=ex4−e0=
ex4−1.
Step 5: Now, integrate with respect to x.R1
0(ex4−1) dx =1
4ex4−x
1
0=
1
4e−1.
4
Step 6: Therefore, the value of the double integral RRRey2dA over the region
Ris 1
4e−1.
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the given double integral over the region R, we will first determine
the bounds for xand yin terms of the region.
We observe that the region Ris enclosed between the curves y=x2and
y= 2xin the first quadrant. Therefore, we can express the bounds for xas
0≤x≤2and the bounds for yas x2≤y≤2x.
Now, we can rewrite the given double integral as:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydydx
Step 1: Evaluate the inner integral with respect to y:
Z2x
x2
ex+ydy =ex+y2x
x2
=e3x−ex2
Step 2: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z2
0
(e3x−ex2)dx ="e3x
3−ex2
2#2
0
=e6
3−e4
2−1
2
Therefore, the value of the double integral over the region Ris e6
3−e4
2−1
2.
Question 6
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
We begin by finding the points of intersection of the curves y=x2and y= 2x.
Step 1: Find the points of intersection. Setting the two equations equal to
each other gives: x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration. Since y= 2xis the larger
curve, the limits of integration on ywill be from x2to 2x, and the limits on x
will be from 0 to 2.
Step 3: Evaluate the double integral. The given double integral is:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx.
Step 4: Integrate with respect to y. Integrating with respect to yfirst, we
get:
Z2
0x2y+y3
32x
x2
dx
=Z2
0x2(2x) + (2x)3
3−(x2·x2+(x2)3
3)dx
=Z2
02x3+8x3
3−(x4+x6
3)dx
=Z2
0x3+8x3
3−x4−x6
3dx.
Step 5: Integrate with respect to x. Integrating the above expression with
respect to xgives:
x4
4+ 2x4−x5
5−x7
21 2
0
=24
4+ 2 ·24−25
5−27
21−04
4+ 2 ·04−05
5−07
21
=4 + 32 −32
5−128
21 −0
=3967
105 .
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 3967
105 .
Question 7
Question
Evaluate the double integral RRDex2+y2dA, where Dis the region bounded by
the curves y=x2and y= 2x.
6
Solution
To evaluate the double integral over the region D, we need to first determine
the limits of integration by finding the points of intersection between the two
curves y=x2and y= 2x.
Step 1: Determine the points of intersection: Setting x2= 2xgives
x2−2x= 0 ⇒x(x−2) = 0 which implies x= 0 and x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the integral: Since the region Dis bounded by y=x2
and y= 2x, the limits of integration for yare x2and 2x, and for xare 0and 2.
Thus, the integral becomes
Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral: Integrating with respect to y, we
have
Zex2+y2dy =Zex2+y2
y=2x
y=x2
=ex2+(2x)2
−ex2+x4
Step 4: Evaluate the double integral: Now, we evaluate the double
integral by integrating the result from Step 3 with respect to xover the limits
0to 2:
Z2
0
(ex2+(2x)2
−ex2+x4)dx =Z2
0
(e5x2
−ex4)dx
We can now solve this integral to obtain the final result.
Question 8
Question
Evaluate the double integral RRD(x2−y2)dA where Dis the region in the first
quadrant bounded by the curves y=x2and x=y2.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Find the limits of integration for y:The curves y=x2
and x=y2intersect at the points (0,0) and (1,1). Therefore, the limits of
integration for yare 0≤y≤x2.
Step 2: Find the limits of integration for x:To find the limits of
integration for x, we need to determine the range of xvalues within the region
D. This can be done by setting x=y2and y=x2equal to each other and
solving for xto find the intersection points.
7
Setting x=y2and y=x2equal to each other gives y2=xand x=y2,
which implies y4=y. So, y= 0 or y= 1.
Thus, the limits of integration for xare 0≤x≤y2where 0≤y≤1.
Step 3: Evaluate the double integral:
ZZD
(x2−y2)dA
=Z1
0Zy2
0
(x2−y2)dx dy
=Z1
0x3
3−xy2y2
0
dy
=Z1
0(y2)3
3−y4dy
=Z1
0y6
3−y4dy
=Z1
01
3−y2dy
=y
3−y3
31
0
=1
3−1
3−0+0
= 0.
Therefore, RRD(x2−y2)dA = 0.
Question 9
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and y= 1.
Solution
To solve this problem, we need to find the limits of integration by determining
the intersection points of the given curves.
Step 1: Find the intersection points.
Setting x2= 2x, we find the intersection point to be x= 0 and x= 2.
Therefore, the limits of integration for xare from 0 to 2.
Step 2: Set up the integral with respect to y.
The curves y=x2and y= 1 bound the region Rwith xvarying from 0 to
1. Therefore, the limits of integration for yare from x2to 1.
Step 3: Evaluate the double integral.
8
ZZR
(x+y)dA
=Z2
0Z1
x2
(x+y)dydx
=Z2
0xy +y2
21
x2
dx
=Z2
0x+1
2−x3−x4
2dx
=x2
2+x−x4
4−x5
10 2
0
=1
2(2)2+ 2 −1
4(2)4−1
10(2)5
= 2 + 2 −4−32
10
= 4 −16
5
=4
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 4
5.
Question 10
Question
Evaluate the double integral RRRex2+y2dx dy, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of in-
tegration by finding the points of intersection of the two curves y=x2and
y= 2x.
Step 1: Find the points of intersection Set x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Step 2: Determine the limits of integration Since y=x2is below
y= 2xin the region R, the limits of integration will be:
0≤x≤2
9
x2≤y≤2x
Step 3: Evaluate the double integral
ZZR
ex2+y2dx dy =Z2
0Z2x
x2
ex2+y2dy dx
Step 4: Integrate with respect to yfirst For the inner integral:
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2=ex2+(2x)2
−ex2+x2=e5x2
−e2x2
Step 5: Integrate with respect to xNow integrate the above expression
from 0 to 2:
Z2
0
(e5x2
−e2x2)dx ="e5x2
5−e2x2
2#2
0
=e20
5−e4
2−1
5−1
2
Therefore, the value of the double integral RRRex2+y2dx dy over the region
Ris e20
5−e4
2−3
10 .
Question 11
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region in the first
quadrant bounded by x= 0,y= 0,x= 3, and x+y= 4.
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by x= 0,y= 0,x= 3, and x+y= 4. To find the limits of
integration for x, we need to consider the intersection points of the lines: x= 0,
x= 3, and x+y= 4. Solving x= 0 and x+y= 4, we get y= 4. Solving x= 3
and x+y= 4, we get y= 1. So, the limits of integration for xare from 0 to
3. Similarly, to find the limits of integration for y, we consider the intersection
points of the lines: y= 0,x+y= 4, and x= 3. Solving y= 0 and x+y= 4,
we get x= 4. Solving x= 3 and x+y= 4, we get y= 1. So, the limits of
integration for yare from 0 to 4−x.
Step 2: Rewrite the given integral with the limits of integration. The double
integral becomes:
Z3
0Z4−x
0
(x2+y2)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Z4−x
0
(x2+y2)dy =x2y+y3
3
4−x
0
=x2(4 −x) + (4 −x)3
3
Step 4: Evaluate the outer integral with respect to x.
Z3
0x2(4 −x) + (4 −x)3
3dx =Z3
0
(4x2−x3+64
3−24x
3+12x2
3−x3
3)dx
=Z3
0
(3x2−2x3+64
3−8x+ 4x2−x3
3)dx
=Z3
0
(−2x3+ 7x2−8x+64
3)dx
Integrating term by term, we get:
−1
2x4+7
3x3−4x2+64
3x
3
0
=−81
2+ 21 −36 + 64 = 7
2
Thus, the value of the given double integral is 7
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 2.
Solution
To evaluate the double integral RRRxy dA, we need to set up the limits of inte-
gration for xand yby determining the region Rgeometrically.
Step 1: Determine the limits of integration for xand y
The region Ris bounded by the curves y=x2,y=x,x= 0, and x= 2.
Visually, Rlooks like a region between the curve y=x2and the line y=x,
where xranges from 0 to 2.
To find the limits of integration for x, we set y=x2and y=xequal to each
other:
x2=x=⇒x2−x= 0 =⇒x(x−1) = 0.
This gives us x= 0 and x= 1 as the values where the two curves intersect.
Therefore, the limits of integration for xare 0≤x≤1.
Next, in terms of y,xranges from x2to x, so the limits of integration for y
are x2≤y≤x.
Step 2: Evaluate the double integral
11
Now, we can rewrite the double integral over the region Ras:
ZZR
xy dA =Z1
0Zx
x2
xy dy dx.
Let’s now evaluate the double integral by first integrating with respect to y
and then with respect to x.
Z1
0Zx
x2
xy dy dx =Z1
01
2xy2x
x2
dx =Z1
01
2x3−1
2x5dx.
Integrating with respect to x, we get:
1
8x4−1
12x61
0
=1
8−1
12 =1
24.
Therefore, the value of the double integral RRRxy dA over the region Ris
1
24 .
Question 13
Question
Let Rbe the region bounded by the curves y= 2x2and y= 3x. Calculate the
double integral RRR4x+ 3y dA over the region R.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits
of integration for x, we first need to find the intersection points of the curves
y= 2x2and y= 3x. Setting them equal to each other:
2x2= 3x.
Solving for xgives us x= 0 and x= 3/2.
Therefore, the limits of integration for xare 0≤x≤3/2.
To find the limits of integration for y, we substitute the equations y= 2x2
and y= 3xinto the integral. This gives us 2x2≤y≤3x.
Step 2: Set up the double integral. The double integral to evaluate is:
ZZR
(4x+ 3y)dA =Z3/2
0Z3x
2x2
(4x+ 3y)dy dx.
Step 3: Evaluate the inner integral with respect to y.
Z3x
2x2
(4x+ 3y)dy =4xy +3
2y23x
2x2
.
12
Substitute the limits of integration:
= 4x(3x) + 3
2(3x)2−4x(2x2)−3
2(2x2)2.
Step 4: Simplify the expression.
= 12x2+27
2x2−8x3−6x2= 6x2−8x3+27
2x2.
Step 5: Evaluate the outer integral with respect to x.
=Z3/2
0
(6x2−8x3+27
2x2)dx =2x3−2x4+27
2x33/2
0
.
Substitute the limits of integration:
= 2 3
23
−23
24
+27
23
23
.
Step 6: Calculate the final answer.
= 9 −27
4+81
2=55
4.
Thus, the value of the double integral RRR(4x+ 3y)dA over the region Ris
55
4.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y= 2x.
Solution
Step 1: To identify the limits of integration, we need to find the points of
intersection of the curves y=x2and y= 2x. Setting the two equations equal
to each other gives us x2= 2x.
Step 2: Rearranging the equation, we get x2−2x= 0. Factoring out an x,
we have x(x−2) = 0. So, the solutions are x= 0 and x= 2.
Step 3: Next, we need to determine which curve is above the other in the
region R. Evaluating y=x2at x= 1 and y= 2xat x= 1, we find that y=x2
is above y= 2xin the region R.
Step 4: Therefore, the limits of integration for xwill be from 0to 1for the
inner integral, and the limits for ywill be from x2to 2xfor the outer integral.
13
Step 5: The double integral becomes:
Z1
0Z2x
x2
(x2+y2)dy dx
Step 6: Integrating with respect to yfirst, we get:
Z1
0x2y+y3
32x
x2
dx
Step 7: Substitute the limits of integration and simplify:
Z1
02x3+(2x)3
3−x4−x6
3dx
Step 8: Continuing to simplify gives:
2Z1
04x3
3−x4
3dx
Step 9: Integrating gives:
2x4
3−x5
15 1
0
Step 10: Evaluate the integral at the limits of integration:
21
3−1
15
Step 11: Finally, calculate the value to get the result:
24
15=8
15
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 8
15 .
Question 15
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2,y= 4,x= 0, and x= 2.
14
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded above by the line y= 4 and below by the curve y=x2.
The left boundary is x= 0 and the right boundary is x= 2. Thus, the limits
of integration are:
0≤x≤2
x2≤y≤4
Step 2: Now, we can set up and evaluate the double integral:
ZZR
(3x2+ 2y)dA =Z2
0Z4
x2
(3x2+ 2y)dydx
=Z2
03x2y+y24
x2dx
=Z2
03x2(4) + 42−3x2(x2)−x4dx
=Z2
0
(12x2+ 16 −3x4−x4)dx
=Z2
0
(−4x4+ 12x2+ 16) dx
=−4
5x5+ 4x3+ 16x2
0
=−4
5(2)5+ 4(2)3+ 16(2)−−4
5(0)5+ 4(0)3+ 16(0)
=−128
5+ 32 + 32−0
=−128
5+ 64
=192
5
Therefore, the value of the double integral is 192
5.
Question 16
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 4 and x= 0.
15
Solution
To evaluate the double integral RRRx2y dA, we first need to find the limits of
integration for xand y.
Step 1: Determine the limits of integration for y. The region Ris bounded
by y=x2,y= 4, and x= 0. Setting y=x2and y= 4 equal to each other gives
x2= 4, so x=±2. Therefore, the limits of integration for yare from x2to 4.
Step 2: Determine the limits of integration for x. Since Ris bounded by
x= 0 and x=±2, the limits of integration for xare from 0to 2.
Step 3: Setup the double integral. The given double integral can be written
as:
ZZR
x2y dA =Z2
0Z4
x2
x2y dy dx
Step 4: Evaluate the double integral.
Z2
0Z4
x2
x2y dy dx =Z2
0x2y2
24
x2
dx
=Z2
08x2−x6
2dx =Z2
0
8x2−x6
2dx
=8x3
3−x7
14 2
0
=64
3−128
14 =256
21
Hence, the value of the double integral RRRx2y dA over the region Ris 256
21 .
Question 17
Question
Evaluate the double integral ZZ
R
ex2+y2dA over the region Rbounded by the
curves y=x,y= 4x, and x= 1.
Solution
To evaluate the given double integral, we need to find the limits of integration
for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by y=xand y= 4x. Their intersection gives x=y= 0 and x=y= 1. So,
the limits of integration for xare from 0 to 1.
Step 2: Find the limits of integration for y:The region Ris also
bounded by x= 1. From the equation x= 1, we see that the region Rends at
x= 1. Combining this with y=xand y= 4x, we can express the limits of y
as x≤y≤4x.
16
Step 3: Set up and evaluate the double integral: The double integral
can be written as:
ZZ
R
ex2+y2dA =Z1
0Z4x
x
ex2+y2dy dx
Now, we calculate:
Z1
0Z4x
x
ex2+y2dy dx =Z1
0hex2+y2iy=4x
y=xdx
=Z1
0hex2+16x2
−ex2+x2idx
=Z1
0he17x2
−e2x2idx
=1
17e17x2
−1
2e2x21
0
=1
17e17 −1
2e2−1
17 −1
2
=1
17e17 −1
2e2+15
34
Therefore, the value of the double integral is 1
17 e17 −1
2e2+15
34 .
Question 18
Question
Evaluate the double integral RRR(xcos y)dA, where Ris the region bounded by
the curves y= 0,y= 2π,x= 0, and x=π.
Solution
Step 1: Rewrite the double integral in terms of xand y:
ZZR
(xcos y)dA =Z2π
0Zπ
0
(xcos y)dx dy
Step 2: Integrate with respect to x:
Zπ
0
(xcos y)dx =1
2x2cos y
π
0
=1
2(π2−0) cos y=π2
2cos y
Step 3: Substitute the result of the previous step back into the double inte-
gral:
Z2π
0
π2
2cos y dy
17
Step 4: Integrate with respect to y:
Z2π
0
π2
2cos y dy =π2
2Z2π
0
cos y dy
Step 5: Apply integration property of cosine function:
π2
2Z2π
0
cos y dy =π2
2sin y
2π
0
=π2
2(sin 2π−sin 0)
Step 6: Simplify the expression:
π2
2(0 −0) = 0
Therefore, the value of the double integral RRR(xcos y)dA over the region R
is 0.
Question 19
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 4x−x2.
Solution
To evaluate the double integral over the region D, we first need to determine
the limits of integration.
Step 1: Find the intersection points of the two curves. The intersec-
tion points occur when x2= 4x−x2. Simplifying, we get 2x2−4x= 0, which
gives 2x(x−2) = 0 ⇒x= 0 or x= 2.
Step 2: Determine the limits of integration for xand yin the
double integral. Since the region Dis bounded by y=x2and y= 4x−x2,
the limits of integration for xare from 0to 2and for yare from x2to 4x−x2.
Step 3: Set up and evaluate the double integral. We have:
ZZD
(x2+y2)dA =Z2
0Z4x−x2
x2
(x2+y2)dydx
Now, substitute y=x2and y= 4x−x2into the integrand:
Z2
0Z4x−x2
x2
(x2+y2)dydx =Z2
0x2y+y3
34x−x2
x2
dx
Z2
0x2(4x−x2) + (4x−x2)3
3−x2(x2) + (x2)3
3dx
Now, simplify and evaluate the integral over the specified limits (from 0to
2) to get the final answer.
18
Question 20
Question
Evaluate the double integral RRR(x2+y2)dA over the region Renclosed by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over the given region R, we first need to deter-
mine the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at x= 0 and x= 1. Therefore, the limits of integration
for xare from 0 to 1.
Step 2: Determine the limits of integration for y:For each xbetween
0 and 1, the corresponding yvalues are between x2and 2x. Thus, the limits of
integration for yare from x2to 2x.
Step 3: Setting up the double integral: The given double integral can
be expressed as:
ZZR
(x2+y2)dA =Z1
0Z2x
x2
(x2+y2)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
Step 5: Simplify the expression:
= 2x3+8x3
3−x4−x6
3
Step 6: Evaluate the outer integral with respect to x:Now, we
integrate the result from Step 5 with respect to x:
Z1
0
(2x3+8x3
3−x4−x6
3)dx
Step 7: Simplify and compute the final answer:
=2x4
4+8x4
12 −x5
5−x7
21
1
0
=1
2+2
3−1
5−1
21
19
=105
42 +28
42 −10
42 −2
42
=121
42
Thus, the value of the double integral over the region Ris 121
42 .
Question 21
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 1,x= 1, and x= 2.
Solution
To solve this problem, we first need to determine the limits of integration for x
and y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 1,x= 1, and x= 2. Therefore, the
limits of integration for xare 1≤x≤2, and the limits of integration for yare
x2≤y≤1.
Step 2: Set up the double integral The double integral of ex+yover
region Rcan be set up as follows:
ZZR
ex+ydA =Z2
1Z1
x2
ex+ydy dx
Step 3: Integrate with respect to yIntegrating ex+ywith respect to y:
Zex+ydy =ex·Zeydy =ex·ey=ex+y
Step 4: Evaluate the inner integral Substitute the limits of integration
for yinto the integral:
Z1
x2
ex+ydy =ex+1 −ex+x2
Step 5: Evaluate the outer integral Integrate the inner integral with
respect to x:
Z2
1
(ex+1 −ex+x2)dx
Step 6: Evaluate the integral Now, evaluate the integral:
"ex+1 −ex2+x
1+2#2
1
=e3−e2−e4
3+e5
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
e3−e2−e4
3+e5
3.
20
Question 22
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y= 2x.
Calculate the double integral RRR(2y−x)dA where dA =dxdy.
Solution
Step 1: Determine the intersection points of the curves y=x2and y= 2x.
Setting x2= 2x, we get x2−2x= 0, which simplifies to x(x−2) = 0. So, x= 0
or x= 2.
Therefore, the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration for xand y. The region Ris
bounded by the curves y=x2and y= 2x. The limits of integration for yare
x2and 2x, and for xare from 0 to 2.
Step 3: Set up the integral. The given double integral becomes:
ZZR
(2y−x)dA =Z2
0Z2x
x2
(2y−x)dydx
Step 4: Evaluate the inner integral with respect to y.
Z2x
x2
(2y−x)dy =y2−xy2x
x2= [(2x)2−2x·2x]−[x2−x·x2]
= (4x2−4x)−(x2−x3) = 3x2−3x
Step 5: Evaluate the outer integral with respect to x.
Z2
0
3x2−3x dx =x3−3
2x22
0
= [23−3
2·22]−[0 −0]
= (8 −6) −0 = 2
Therefore, the value of the double integral over the region Ris 2.
Question 23
Question
Evaluate the double integral RRRe−x2
−y2dx dy, where Ris the region bounded
by the curves y=x2and y= 4.
21
Solution
To evaluate the double integral over R, we first need to determine the limits of
integration.
Step 1: Find the limits of integration for x:The curves y=x2and
y= 4 intersect when x2= 4, so x=±2. Therefore, the limits of integration for
xare −2and 2.
Step 2: Find the limits of integration for y:The curve y=x2intersects
y= 4 at y= 4. Therefore, the limits of integration for yare x2and 4.
Thus, the double integral becomes:
ZZR
e−x2
−y2dx dy =Z2
−2Z4
x2
e−x2
−y2dy dx
Step 3: Evaluate the integral: Start by evaluating the inner integral
with respect to y:
Z4
x2
e−x2
−y2dy =h−e−x2
−y2i4
x2
=−e−x2
−42+e−x2
−x2
=−e−x2
−16 +e−2x2
Now, integrate this result with respect to x:
Z2
−2−e−x2
−16 +e−2x2dx
=16e−x2
−16 −1
2e−2x22
−2
= 16e−4−16 −1
2e−8−16e−4−16 −1
2e−8
= 2e−20 −e−8
Therefore, the value of the double integral is 2e−20 −e−8.
Question 24
Question
Evaluate the double integral RRD(2x+y)dA, where Dis the region bounded by
the curves y=x2,y= 4,x=−2, and x= 2.
22
Solution
Step 1: Draw the region Ddescribed in the question. It is the region bounded
by the curves y=x2,y= 4,x=−2, and x= 2. This region is a rectangle with
a parabolic cutout along one side.
Step 2: Set up the double integral in terms of xand y:
ZZD
(2x+y)dA =Z2
−2Z4
x2
(2x+y)dy dx
Step 3: Integrate with respect to yfirst:
Z2
−2Z4
x2
(2x+y)dy dx =Z2
−22xy +1
2y24
x2
dx
Step 4: Substitute the limits of integration and simplify the expression:
=Z2
−28x+ 8 −2x3−x4dx
Step 5: Integrate with respect to x:
=4x2+ 8x−1
2x4−1
5x52
−2
Step 6: Evaluate the integral at the limits of integration:
= (16 + 16 −8−32
5)−(16 −16+8+32
5)
Step 7: Simplify the expression to find the final answer:
=192
5
Therefore, the value of the double integral RRD(2x+y)dA over the region
Dis 192
5.
Question 25
Question
Evaluate the double integral RRDex2
−y2dx dy, where Dis the region bounded
by y=x2and y= 2x−x2.
23
Solution
To evaluate the given double integral, we need to find the bounds of integration
for xand yby determining the region D.
Step 1: Determine the bounds of integration for xand y
First, find the points of intersection of the curves y=x2and y= 2x−x2:
Setting x2= 2x−x2, we get 2x−2x2= 0 =⇒x(1 −2x) = 0. Thus, x= 0
or x=1
2.
Now, by observing the graph of the curves, we see that y= 2x−x2is above
y=x2in the interval 0≤x≤1
2and below y=x2in the interval 1
2≤x≤1.
So, the bounds of integration for xare 0≤x≤1
2.
For y, the bounds are given by the curves:
• Lower bound: y=x2
• Upper bound: y= 2x−x2
Thus, the bounds of integration for yare x2≤y≤2x−x2.
Step 2: Evaluate the integral
The double integral can now be expressed as:
ZZD
ex2
−y2dx dy =Z1
2
0Z2x−x2
x2
ex2
−y2dy dx
Let’s compute this integral step by step.
Step 3: Integrate with respect to y
Z2x−x2
x2
ex2
−y2dy =h−ex2
−y2i2x−x2
x2
=−ex2
−(2x−x2)2+ex2
−x2
=−ex2
−4x2+4x4+e0
=−e4x4
−3x2+ 1
Step 4: Integrate with respect to x
Z1
2
0−e4x4
−3x2+ 1 dx =−Z1
2
0
e4x4
−3x2dx +Z1
2
0
1dx
This integral may not have a closed-form solution and would typically be
calculated by numerical methods.
Therefore, the double integral RRDex2
−y2dx dy over the region Dbounded
by y=x2and y= 2x−x2needs to be evaluated numerically.
24
Step 4: Integrate the first term
Z2
0
e4x
xdx =1
4e4x2
0
=e8
4−1
4
Step 5: Integrate the second term
Z2
0
ex3
xdx =1
3ex32
0
=e8
3−1
3
Step 6: Combine the results
e8
4−1
4−e8
3−1
3=e8
12 −1
6
Therefore, the value of the double integral RRRexy dA is e8
12 −1
6.
Question 2
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Sketch the region R
The region Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the
points of intersection, we set the curves equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the x-values where y=x2and y= 2xintersect. Thus,
the region Ris bounded by x= 1,y=x2, and y= 2x, as shown below:
xy2;2y= 2x
x= 1
Step 2: Set up the double integral
The given double integral can be set up as:
ZZR
ey2dA =Zx=2
x=1 Zy=2x
y=x2
ey2dy dx
2
Step 3: Evaluate the inner integral
Integrating with respect to yfirst, we get:
Zey2dy =Zey2dy
=1
2√πerf(y)
Therefore, the inner integral becomes:
Zy=2x
y=x2
ey2dy =1
2√πerf(4x)−erf(x2)
Step 4: Evaluate the outer integral
Now we integrate the above expression with respect to x:
Zx=2
x=1
1
2√πerf(4x)−erf(x2)dx
This integral may not have a straightforward closed form solution and might
require numerical methods for evaluation.
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x,x= 0, and x= 2.
Solution
Step 1: First, sketch the region Rto understand its shape and boundaries.
Step 2: To evaluate the given double integral, we will convert it into an
iterated integral and choose the order of integration. Since the region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 2, it is better to
integrate with respect to yfirst and then with respect to x. Thus, the iterated
integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to yfirst:
=Z2
0x2y+y3
32x
x2
dx
Step 4: Simplify and evaluate the integral:
=Z2
02x3−x4+8x
3−8x3
3dx
3
=Z2
0−x4−5x3
3+8x
3dx
=−x5
5−5x4
12 +4x2
32
0
=−32
5−80
12 +16
3
=−32
5−160
12 +64
12
=−32
5−96
12
=−32
5−8
=−32
5−40
5
=−72
5
Therefore, the value of the given double integral over the region Ris −72
5.
Question 4
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Draw a sketch of the region Rto visualize the bounds of integration.
The region Ris bound by the curves y=x2,y= 1,x= 0, and x= 1. It is a
triangle with vertices at (0,0),(1,0), and (1,1).
Step 2: Identify the bounds of integration for xand y. The region Rcan be
expressed as 0≤x≤1,0≤y≤x2.
Step 3: Set up the double integral with the given function and region. We
have RRRey2dA =R1
0Rx2
0ey2dy dx.
Step 4: Integrate with respect to yfirst. Rx2
0ey2dy =ey2
x2
0=ex4−e0=
ex4−1.
Step 5: Now, integrate with respect to x.R1
0(ex4−1) dx =1
4ex4−x
1
0=
1
4e−1.
4
Step 6: Therefore, the value of the double integral RRRey2dA over the region
Ris 1
4e−1.
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the given double integral over the region R, we will first determine
the bounds for xand yin terms of the region.
We observe that the region Ris enclosed between the curves y=x2and
y= 2xin the first quadrant. Therefore, we can express the bounds for xas
0≤x≤2and the bounds for yas x2≤y≤2x.
Now, we can rewrite the given double integral as:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydydx
Step 1: Evaluate the inner integral with respect to y:
Z2x
x2
ex+ydy =ex+y2x
x2
=e3x−ex2
Step 2: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z2
0
(e3x−ex2)dx ="e3x
3−ex2
2#2
0
=e6
3−e4
2−1
2
Therefore, the value of the double integral over the region Ris e6
3−e4
2−1
2.
Question 6
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
We begin by finding the points of intersection of the curves y=x2and y= 2x.
Step 1: Find the points of intersection. Setting the two equations equal to
each other gives: x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration. Since y= 2xis the larger
curve, the limits of integration on ywill be from x2to 2x, and the limits on x
will be from 0 to 2.
Step 3: Evaluate the double integral. The given double integral is:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx.
Step 4: Integrate with respect to y. Integrating with respect to yfirst, we
get:
Z2
0x2y+y3
32x
x2
dx
=Z2
0x2(2x) + (2x)3
3−(x2·x2+(x2)3
3)dx
=Z2
02x3+8x3
3−(x4+x6
3)dx
=Z2
0x3+8x3
3−x4−x6
3dx.
Step 5: Integrate with respect to x. Integrating the above expression with
respect to xgives:
x4
4+ 2x4−x5
5−x7
21 2
0
=24
4+ 2 ·24−25
5−27
21−04
4+ 2 ·04−05
5−07
21
=4 + 32 −32
5−128
21 −0
=3967
105 .
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 3967
105 .
Question 7
Question
Evaluate the double integral RRDex2+y2dA, where Dis the region bounded by
the curves y=x2and y= 2x.
6
Solution
To evaluate the double integral over the region D, we need to first determine
the limits of integration by finding the points of intersection between the two
curves y=x2and y= 2x.
Step 1: Determine the points of intersection: Setting x2= 2xgives
x2−2x= 0 ⇒x(x−2) = 0 which implies x= 0 and x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the integral: Since the region Dis bounded by y=x2
and y= 2x, the limits of integration for yare x2and 2x, and for xare 0and 2.
Thus, the integral becomes
Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral: Integrating with respect to y, we
have
Zex2+y2dy =Zex2+y2
y=2x
y=x2
=ex2+(2x)2
−ex2+x4
Step 4: Evaluate the double integral: Now, we evaluate the double
integral by integrating the result from Step 3 with respect to xover the limits
0to 2:
Z2
0
(ex2+(2x)2
−ex2+x4)dx =Z2
0
(e5x2
−ex4)dx
We can now solve this integral to obtain the final result.
Question 8
Question
Evaluate the double integral RRD(x2−y2)dA where Dis the region in the first
quadrant bounded by the curves y=x2and x=y2.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Find the limits of integration for y:The curves y=x2
and x=y2intersect at the points (0,0) and (1,1). Therefore, the limits of
integration for yare 0≤y≤x2.
Step 2: Find the limits of integration for x:To find the limits of
integration for x, we need to determine the range of xvalues within the region
D. This can be done by setting x=y2and y=x2equal to each other and
solving for xto find the intersection points.
7
Setting x=y2and y=x2equal to each other gives y2=xand x=y2,
which implies y4=y. So, y= 0 or y= 1.
Thus, the limits of integration for xare 0≤x≤y2where 0≤y≤1.
Step 3: Evaluate the double integral:
ZZD
(x2−y2)dA
=Z1
0Zy2
0
(x2−y2)dx dy
=Z1
0x3
3−xy2y2
0
dy
=Z1
0(y2)3
3−y4dy
=Z1
0y6
3−y4dy
=Z1
01
3−y2dy
=y
3−y3
31
0
=1
3−1
3−0+0
= 0.
Therefore, RRD(x2−y2)dA = 0.
Question 9
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and y= 1.
Solution
To solve this problem, we need to find the limits of integration by determining
the intersection points of the given curves.
Step 1: Find the intersection points.
Setting x2= 2x, we find the intersection point to be x= 0 and x= 2.
Therefore, the limits of integration for xare from 0 to 2.
Step 2: Set up the integral with respect to y.
The curves y=x2and y= 1 bound the region Rwith xvarying from 0 to
1. Therefore, the limits of integration for yare from x2to 1.
Step 3: Evaluate the double integral.
8
ZZR
(x+y)dA
=Z2
0Z1
x2
(x+y)dydx
=Z2
0xy +y2
21
x2
dx
=Z2
0x+1
2−x3−x4
2dx
=x2
2+x−x4
4−x5
10 2
0
=1
2(2)2+ 2 −1
4(2)4−1
10(2)5
= 2 + 2 −4−32
10
= 4 −16
5
=4
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 4
5.
Question 10
Question
Evaluate the double integral RRRex2+y2dx dy, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of in-
tegration by finding the points of intersection of the two curves y=x2and
y= 2x.
Step 1: Find the points of intersection Set x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Step 2: Determine the limits of integration Since y=x2is below
y= 2xin the region R, the limits of integration will be:
0≤x≤2
9
x2≤y≤2x
Step 3: Evaluate the double integral
ZZR
ex2+y2dx dy =Z2
0Z2x
x2
ex2+y2dy dx
Step 4: Integrate with respect to yfirst For the inner integral:
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2=ex2+(2x)2
−ex2+x2=e5x2
−e2x2
Step 5: Integrate with respect to xNow integrate the above expression
from 0 to 2:
Z2
0
(e5x2
−e2x2)dx ="e5x2
5−e2x2
2#2
0
=e20
5−e4
2−1
5−1
2
Therefore, the value of the double integral RRRex2+y2dx dy over the region
Ris e20
5−e4
2−3
10 .
Question 11
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region in the first
quadrant bounded by x= 0,y= 0,x= 3, and x+y= 4.
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by x= 0,y= 0,x= 3, and x+y= 4. To find the limits of
integration for x, we need to consider the intersection points of the lines: x= 0,
x= 3, and x+y= 4. Solving x= 0 and x+y= 4, we get y= 4. Solving x= 3
and x+y= 4, we get y= 1. So, the limits of integration for xare from 0 to
3. Similarly, to find the limits of integration for y, we consider the intersection
points of the lines: y= 0,x+y= 4, and x= 3. Solving y= 0 and x+y= 4,
we get x= 4. Solving x= 3 and x+y= 4, we get y= 1. So, the limits of
integration for yare from 0 to 4−x.
Step 2: Rewrite the given integral with the limits of integration. The double
integral becomes:
Z3
0Z4−x
0
(x2+y2)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Z4−x
0
(x2+y2)dy =x2y+y3
3
4−x
0
=x2(4 −x) + (4 −x)3
3
Step 4: Evaluate the outer integral with respect to x.
Z3
0x2(4 −x) + (4 −x)3
3dx =Z3
0
(4x2−x3+64
3−24x
3+12x2
3−x3
3)dx
=Z3
0
(3x2−2x3+64
3−8x+ 4x2−x3
3)dx
=Z3
0
(−2x3+ 7x2−8x+64
3)dx
Integrating term by term, we get:
−1
2x4+7
3x3−4x2+64
3x
3
0
=−81
2+ 21 −36 + 64 = 7
2
Thus, the value of the given double integral is 7
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 2.
Solution
To evaluate the double integral RRRxy dA, we need to set up the limits of inte-
gration for xand yby determining the region Rgeometrically.
Step 1: Determine the limits of integration for xand y
The region Ris bounded by the curves y=x2,y=x,x= 0, and x= 2.
Visually, Rlooks like a region between the curve y=x2and the line y=x,
where xranges from 0 to 2.
To find the limits of integration for x, we set y=x2and y=xequal to each
other:
x2=x=⇒x2−x= 0 =⇒x(x−1) = 0.
This gives us x= 0 and x= 1 as the values where the two curves intersect.
Therefore, the limits of integration for xare 0≤x≤1.
Next, in terms of y,xranges from x2to x, so the limits of integration for y
are x2≤y≤x.
Step 2: Evaluate the double integral
11
Now, we can rewrite the double integral over the region Ras:
ZZR
xy dA =Z1
0Zx
x2
xy dy dx.
Let’s now evaluate the double integral by first integrating with respect to y
and then with respect to x.
Z1
0Zx
x2
xy dy dx =Z1
01
2xy2x
x2
dx =Z1
01
2x3−1
2x5dx.
Integrating with respect to x, we get:
1
8x4−1
12x61
0
=1
8−1
12 =1
24.
Therefore, the value of the double integral RRRxy dA over the region Ris
1
24 .
Question 13
Question
Let Rbe the region bounded by the curves y= 2x2and y= 3x. Calculate the
double integral RRR4x+ 3y dA over the region R.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits
of integration for x, we first need to find the intersection points of the curves
y= 2x2and y= 3x. Setting them equal to each other:
2x2= 3x.
Solving for xgives us x= 0 and x= 3/2.
Therefore, the limits of integration for xare 0≤x≤3/2.
To find the limits of integration for y, we substitute the equations y= 2x2
and y= 3xinto the integral. This gives us 2x2≤y≤3x.
Step 2: Set up the double integral. The double integral to evaluate is:
ZZR
(4x+ 3y)dA =Z3/2
0Z3x
2x2
(4x+ 3y)dy dx.
Step 3: Evaluate the inner integral with respect to y.
Z3x
2x2
(4x+ 3y)dy =4xy +3
2y23x
2x2
.
12
Substitute the limits of integration:
= 4x(3x) + 3
2(3x)2−4x(2x2)−3
2(2x2)2.
Step 4: Simplify the expression.
= 12x2+27
2x2−8x3−6x2= 6x2−8x3+27
2x2.
Step 5: Evaluate the outer integral with respect to x.
=Z3/2
0
(6x2−8x3+27
2x2)dx =2x3−2x4+27
2x33/2
0
.
Substitute the limits of integration:
= 2 3
23
−23
24
+27
23
23
.
Step 6: Calculate the final answer.
= 9 −27
4+81
2=55
4.
Thus, the value of the double integral RRR(4x+ 3y)dA over the region Ris
55
4.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y= 2x.
Solution
Step 1: To identify the limits of integration, we need to find the points of
intersection of the curves y=x2and y= 2x. Setting the two equations equal
to each other gives us x2= 2x.
Step 2: Rearranging the equation, we get x2−2x= 0. Factoring out an x,
we have x(x−2) = 0. So, the solutions are x= 0 and x= 2.
Step 3: Next, we need to determine which curve is above the other in the
region R. Evaluating y=x2at x= 1 and y= 2xat x= 1, we find that y=x2
is above y= 2xin the region R.
Step 4: Therefore, the limits of integration for xwill be from 0to 1for the
inner integral, and the limits for ywill be from x2to 2xfor the outer integral.
13
Step 5: The double integral becomes:
Z1
0Z2x
x2
(x2+y2)dy dx
Step 6: Integrating with respect to yfirst, we get:
Z1
0x2y+y3
32x
x2
dx
Step 7: Substitute the limits of integration and simplify:
Z1
02x3+(2x)3
3−x4−x6
3dx
Step 8: Continuing to simplify gives:
2Z1
04x3
3−x4
3dx
Step 9: Integrating gives:
2x4
3−x5
15 1
0
Step 10: Evaluate the integral at the limits of integration:
21
3−1
15
Step 11: Finally, calculate the value to get the result:
24
15=8
15
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 8
15 .
Question 15
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2,y= 4,x= 0, and x= 2.
14
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded above by the line y= 4 and below by the curve y=x2.
The left boundary is x= 0 and the right boundary is x= 2. Thus, the limits
of integration are:
0≤x≤2
x2≤y≤4
Step 2: Now, we can set up and evaluate the double integral:
ZZR
(3x2+ 2y)dA =Z2
0Z4
x2
(3x2+ 2y)dydx
=Z2
03x2y+y24
x2dx
=Z2
03x2(4) + 42−3x2(x2)−x4dx
=Z2
0
(12x2+ 16 −3x4−x4)dx
=Z2
0
(−4x4+ 12x2+ 16) dx
=−4
5x5+ 4x3+ 16x2
0
=−4
5(2)5+ 4(2)3+ 16(2)−−4
5(0)5+ 4(0)3+ 16(0)
=−128
5+ 32 + 32−0
=−128
5+ 64
=192
5
Therefore, the value of the double integral is 192
5.
Question 16
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 4 and x= 0.
15
Solution
To evaluate the double integral RRRx2y dA, we first need to find the limits of
integration for xand y.
Step 1: Determine the limits of integration for y. The region Ris bounded
by y=x2,y= 4, and x= 0. Setting y=x2and y= 4 equal to each other gives
x2= 4, so x=±2. Therefore, the limits of integration for yare from x2to 4.
Step 2: Determine the limits of integration for x. Since Ris bounded by
x= 0 and x=±2, the limits of integration for xare from 0to 2.
Step 3: Setup the double integral. The given double integral can be written
as:
ZZR
x2y dA =Z2
0Z4
x2
x2y dy dx
Step 4: Evaluate the double integral.
Z2
0Z4
x2
x2y dy dx =Z2
0x2y2
24
x2
dx
=Z2
08x2−x6
2dx =Z2
0
8x2−x6
2dx
=8x3
3−x7
14 2
0
=64
3−128
14 =256
21
Hence, the value of the double integral RRRx2y dA over the region Ris 256
21 .
Question 17
Question
Evaluate the double integral ZZ
R
ex2+y2dA over the region Rbounded by the
curves y=x,y= 4x, and x= 1.
Solution
To evaluate the given double integral, we need to find the limits of integration
for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by y=xand y= 4x. Their intersection gives x=y= 0 and x=y= 1. So,
the limits of integration for xare from 0 to 1.
Step 2: Find the limits of integration for y:The region Ris also
bounded by x= 1. From the equation x= 1, we see that the region Rends at
x= 1. Combining this with y=xand y= 4x, we can express the limits of y
as x≤y≤4x.
16
Step 3: Set up and evaluate the double integral: The double integral
can be written as:
ZZ
R
ex2+y2dA =Z1
0Z4x
x
ex2+y2dy dx
Now, we calculate:
Z1
0Z4x
x
ex2+y2dy dx =Z1
0hex2+y2iy=4x
y=xdx
=Z1
0hex2+16x2
−ex2+x2idx
=Z1
0he17x2
−e2x2idx
=1
17e17x2
−1
2e2x21
0
=1
17e17 −1
2e2−1
17 −1
2
=1
17e17 −1
2e2+15
34
Therefore, the value of the double integral is 1
17 e17 −1
2e2+15
34 .
Question 18
Question
Evaluate the double integral RRR(xcos y)dA, where Ris the region bounded by
the curves y= 0,y= 2π,x= 0, and x=π.
Solution
Step 1: Rewrite the double integral in terms of xand y:
ZZR
(xcos y)dA =Z2π
0Zπ
0
(xcos y)dx dy
Step 2: Integrate with respect to x:
Zπ
0
(xcos y)dx =1
2x2cos y
π
0
=1
2(π2−0) cos y=π2
2cos y
Step 3: Substitute the result of the previous step back into the double inte-
gral:
Z2π
0
π2
2cos y dy
17
Step 4: Integrate with respect to y:
Z2π
0
π2
2cos y dy =π2
2Z2π
0
cos y dy
Step 5: Apply integration property of cosine function:
π2
2Z2π
0
cos y dy =π2
2sin y
2π
0
=π2
2(sin 2π−sin 0)
Step 6: Simplify the expression:
π2
2(0 −0) = 0
Therefore, the value of the double integral RRR(xcos y)dA over the region R
is 0.
Question 19
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 4x−x2.
Solution
To evaluate the double integral over the region D, we first need to determine
the limits of integration.
Step 1: Find the intersection points of the two curves. The intersec-
tion points occur when x2= 4x−x2. Simplifying, we get 2x2−4x= 0, which
gives 2x(x−2) = 0 ⇒x= 0 or x= 2.
Step 2: Determine the limits of integration for xand yin the
double integral. Since the region Dis bounded by y=x2and y= 4x−x2,
the limits of integration for xare from 0to 2and for yare from x2to 4x−x2.
Step 3: Set up and evaluate the double integral. We have:
ZZD
(x2+y2)dA =Z2
0Z4x−x2
x2
(x2+y2)dydx
Now, substitute y=x2and y= 4x−x2into the integrand:
Z2
0Z4x−x2
x2
(x2+y2)dydx =Z2
0x2y+y3
34x−x2
x2
dx
Z2
0x2(4x−x2) + (4x−x2)3
3−x2(x2) + (x2)3
3dx
Now, simplify and evaluate the integral over the specified limits (from 0to
2) to get the final answer.
18
Question 20
Question
Evaluate the double integral RRR(x2+y2)dA over the region Renclosed by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over the given region R, we first need to deter-
mine the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at x= 0 and x= 1. Therefore, the limits of integration
for xare from 0 to 1.
Step 2: Determine the limits of integration for y:For each xbetween
0 and 1, the corresponding yvalues are between x2and 2x. Thus, the limits of
integration for yare from x2to 2x.
Step 3: Setting up the double integral: The given double integral can
be expressed as:
ZZR
(x2+y2)dA =Z1
0Z2x
x2
(x2+y2)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
Step 5: Simplify the expression:
= 2x3+8x3
3−x4−x6
3
Step 6: Evaluate the outer integral with respect to x:Now, we
integrate the result from Step 5 with respect to x:
Z1
0
(2x3+8x3
3−x4−x6
3)dx
Step 7: Simplify and compute the final answer:
=2x4
4+8x4
12 −x5
5−x7
21
1
0
=1
2+2
3−1
5−1
21
19
=105
42 +28
42 −10
42 −2
42
=121
42
Thus, the value of the double integral over the region Ris 121
42 .
Question 21
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 1,x= 1, and x= 2.
Solution
To solve this problem, we first need to determine the limits of integration for x
and y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 1,x= 1, and x= 2. Therefore, the
limits of integration for xare 1≤x≤2, and the limits of integration for yare
x2≤y≤1.
Step 2: Set up the double integral The double integral of ex+yover
region Rcan be set up as follows:
ZZR
ex+ydA =Z2
1Z1
x2
ex+ydy dx
Step 3: Integrate with respect to yIntegrating ex+ywith respect to y:
Zex+ydy =ex·Zeydy =ex·ey=ex+y
Step 4: Evaluate the inner integral Substitute the limits of integration
for yinto the integral:
Z1
x2
ex+ydy =ex+1 −ex+x2
Step 5: Evaluate the outer integral Integrate the inner integral with
respect to x:
Z2
1
(ex+1 −ex+x2)dx
Step 6: Evaluate the integral Now, evaluate the integral:
"ex+1 −ex2+x
1+2#2
1
=e3−e2−e4
3+e5
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
e3−e2−e4
3+e5
3.
20
Question 22
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y= 2x.
Calculate the double integral RRR(2y−x)dA where dA =dxdy.
Solution
Step 1: Determine the intersection points of the curves y=x2and y= 2x.
Setting x2= 2x, we get x2−2x= 0, which simplifies to x(x−2) = 0. So, x= 0
or x= 2.
Therefore, the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration for xand y. The region Ris
bounded by the curves y=x2and y= 2x. The limits of integration for yare
x2and 2x, and for xare from 0 to 2.
Step 3: Set up the integral. The given double integral becomes:
ZZR
(2y−x)dA =Z2
0Z2x
x2
(2y−x)dydx
Step 4: Evaluate the inner integral with respect to y.
Z2x
x2
(2y−x)dy =y2−xy2x
x2= [(2x)2−2x·2x]−[x2−x·x2]
= (4x2−4x)−(x2−x3) = 3x2−3x
Step 5: Evaluate the outer integral with respect to x.
Z2
0
3x2−3x dx =x3−3
2x22
0
= [23−3
2·22]−[0 −0]
= (8 −6) −0 = 2
Therefore, the value of the double integral over the region Ris 2.
Question 23
Question
Evaluate the double integral RRRe−x2
−y2dx dy, where Ris the region bounded
by the curves y=x2and y= 4.
21
Solution
To evaluate the double integral over R, we first need to determine the limits of
integration.
Step 1: Find the limits of integration for x:The curves y=x2and
y= 4 intersect when x2= 4, so x=±2. Therefore, the limits of integration for
xare −2and 2.
Step 2: Find the limits of integration for y:The curve y=x2intersects
y= 4 at y= 4. Therefore, the limits of integration for yare x2and 4.
Thus, the double integral becomes:
ZZR
e−x2
−y2dx dy =Z2
−2Z4
x2
e−x2
−y2dy dx
Step 3: Evaluate the integral: Start by evaluating the inner integral
with respect to y:
Z4
x2
e−x2
−y2dy =h−e−x2
−y2i4
x2
=−e−x2
−42+e−x2
−x2
=−e−x2
−16 +e−2x2
Now, integrate this result with respect to x:
Z2
−2−e−x2
−16 +e−2x2dx
=16e−x2
−16 −1
2e−2x22
−2
= 16e−4−16 −1
2e−8−16e−4−16 −1
2e−8
= 2e−20 −e−8
Therefore, the value of the double integral is 2e−20 −e−8.
Question 24
Question
Evaluate the double integral RRD(2x+y)dA, where Dis the region bounded by
the curves y=x2,y= 4,x=−2, and x= 2.
22
Solution
Step 1: Draw the region Ddescribed in the question. It is the region bounded
by the curves y=x2,y= 4,x=−2, and x= 2. This region is a rectangle with
a parabolic cutout along one side.
Step 2: Set up the double integral in terms of xand y:
ZZD
(2x+y)dA =Z2
−2Z4
x2
(2x+y)dy dx
Step 3: Integrate with respect to yfirst:
Z2
−2Z4
x2
(2x+y)dy dx =Z2
−22xy +1
2y24
x2
dx
Step 4: Substitute the limits of integration and simplify the expression:
=Z2
−28x+ 8 −2x3−x4dx
Step 5: Integrate with respect to x:
=4x2+ 8x−1
2x4−1
5x52
−2
Step 6: Evaluate the integral at the limits of integration:
= (16 + 16 −8−32
5)−(16 −16+8+32
5)
Step 7: Simplify the expression to find the final answer:
=192
5
Therefore, the value of the double integral RRD(2x+y)dA over the region
Dis 192
5.
Question 25
Question
Evaluate the double integral RRDex2
−y2dx dy, where Dis the region bounded
by y=x2and y= 2x−x2.
23
Solution
To evaluate the given double integral, we need to find the bounds of integration
for xand yby determining the region D.
Step 1: Determine the bounds of integration for xand y
First, find the points of intersection of the curves y=x2and y= 2x−x2:
Setting x2= 2x−x2, we get 2x−2x2= 0 =⇒x(1 −2x) = 0. Thus, x= 0
or x=1
2.
Now, by observing the graph of the curves, we see that y= 2x−x2is above
y=x2in the interval 0≤x≤1
2and below y=x2in the interval 1
2≤x≤1.
So, the bounds of integration for xare 0≤x≤1
2.
For y, the bounds are given by the curves:
• Lower bound: y=x2
• Upper bound: y= 2x−x2
Thus, the bounds of integration for yare x2≤y≤2x−x2.
Step 2: Evaluate the integral
The double integral can now be expressed as:
ZZD
ex2
−y2dx dy =Z1
2
0Z2x−x2
x2
ex2
−y2dy dx
Let’s compute this integral step by step.
Step 3: Integrate with respect to y
Z2x−x2
x2
ex2
−y2dy =h−ex2
−y2i2x−x2
x2
=−ex2
−(2x−x2)2+ex2
−x2
=−ex2
−4x2+4x4+e0
=−e4x4
−3x2+ 1
Step 4: Integrate with respect to x
Z1
2
0−e4x4
−3x2+ 1 dx =−Z1
2
0
e4x4
−3x2dx +Z1
2
0
1dx
This integral may not have a closed-form solution and would typically be
calculated by numerical methods.
Therefore, the double integral RRDex2
−y2dx dy over the region Dbounded
by y=x2and y= 2x−x2needs to be evaluated numerically.
24
