MATH 332 - ADVANCED CALCULUS
- Double integrals over rectangular and
general regions
Question Bank - Set 4
Liberty University
Question 1
Question
Evaluate the double integral RRR(x2+y2)dx dy, where Ris the region in the
first quadrant bounded by the curves y= 0,y=x, and x= 1.
Solution
To evaluate the double integral over the region R, we must first set up the
bounds of integration for xand y.
Step 1: Determine the bounds for xand y. Since Ris in the first quadrant
bounded by y= 0,y=x, and x= 1, the bounds for xare from 0 to 1, and the
bounds for yare from 0 to x.
Step 2: Set up the double integral. The double integral becomes:
ZZR
(x2+y2)dx dy =Z1
0Zx
0
(x2+y2)dy dx
Step 3: Integrate with respect to y. Integrating (x2+y2)with respect to y
gives:
Zx
0
(x2+y2)dy =x2y+y3
3
x
0
=x3+x3
3=4
3x3
Step 4: Integrate with respect to x. Now integrate 4
3x3with respect to x
over the bounds 0 to 1:
Z1
0
4
3x3dx =4
3·x4
4
1
0
=1
3
Step 5: Final answer. Therefore, the value of the double integral RRR(x2+
y2)dx dy over the region Ris 1
3.
Question 2
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral over R, we first need to determine the limits
of integration for xand y. Let’s start by finding the points of intersection of
the two curves y=x2and y= 2x. Setting the two equations equal gives us
x2= 2x, which simplifies to x(x−2) = 0. This gives us x= 0 and x= 2 as the
points of intersection.
Step 1: Determine the limits of integration for xand y.
Since y=x2and y= 2xbound the region R, the limits of integration for x
will be 0≤x≤2. For y, the limits will be x2≤y≤2x.
Step 2: Set up the double integral with the given region and integrand.
Therefore, the double integral becomes:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2
=ex2+(2x)2−ex2+x4
Step 4: Substitute the result back into the double integral.
Z2
0
(ex2+(2x)2−ex2+x4)dx
Step 5: Evaluate the integral with respect to x.
Z2
0
(ex2+4x2−ex2+x4)dx =Z2
0
(e5x2−ex2+x4)dx
Step 6: Integrate e5x2and ex2+x4with respect to xthen evaluate the
integral using the limits of integration.
The final answer depends heavily on the form of the integrand. The exact
numerical computation may be challenging.
Therefore, the value of the double integral over the region Ris given by the
above expression.
2
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y= 2x, we need to set up the integral in terms of dx dy
and find the limits of integration for xand y.
Step 1: Find the limits of integration for xand yThe curves y=x2
and y= 2xintersect when x2= 2x, which gives x2−2x= 0. This equation can
be factored as x(x−2) = 0, leading to solutions x= 0 and x= 2.
So, the limits of integration for xare 0≤x≤2.
For each xin this interval, yranges from x2to 2x. Therefore, the limits of
integration for yare x2≤y≤2x.
Step 2: Set up and evaluate the double integral The double integral
becomes:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Now we integrate:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2=x2(2x) + (2x)3
3−x2(x2)−x3
3
= 2x3+8x3
3−x4−x3
3= 10x3−x4
Now integrate with respect to x:
Z2
0
10x3−x4dx ="10x4
4−x5
5#2
0
=10(2)4
4−(2)5
5=160
4−32
5= 40−32
5=208
5
Therefore, the value of the double integral is 208
5.
Question 4
Question
Evaluate the double integral ZZ
R
ex2−y2dx dy over the region Rbounded by the
curves y=x2,y= 1, and x= 0.
3
Solution
To evaluate the double integral over region R, we first need to determine the
limits of integration.
Step 1: Find the intersection points of the curves.
The intersection points are found by setting the equations of the curves equal
to each other:
For y=x2and y= 1:
x2= 1
x=±1
So the intersection points are (1,1) and (−1,1).
Step 2: Determine the limits of integration.
Since yranges from x2to 1, and xranges from 0 to 1, the limits of integration
are:
0≤x≤1
x2≤y≤1
Step 3: Evaluate the double integral.
Now we can evaluate the integral:
ZZ
R
ex2−y2dx dy =Z1
0Z1
x2
ex2−y2dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
ex2−y2dy =−ex2−1+ex2−x4
Step 5: Integrate with respect to x.
Z1
0
(−ex2−1+ex2−x4)dx =√e−e+√e(−1) −1
2
Therefore, the value of the double integral over region Ris √e−e+√e(−1)−1
2.
Question 5
Question
Evaluate the double integral RRR(x2+y)dA over the region Rbounded by the
curves y=x2,y= 4, and x= 0.
4
Solution
To evaluate the double integral over the given region R, we need to express the
region Rin terms of the limits of integration.
Step 1: Determine the limits of integration for xand y.
The curves y=x2and y= 4 bound the region in the ydirection, while
x= 0 bounds the region in the xdirection.
Since y=x2and y= 4 intersect at x2= 4, we have x=±2as the limits for
x. For y, the limits are from y=x2to y= 4.
Therefore, the limits of integration are −2≤x≤2and x2≤y≤4.
Step 2: Set up the double integral.
The double integral can be expressed as:
ZZR
(x2+y)dA =Z2
−2Z4
x2
(x2+y)dy dx
Step 3: Integrate with respect to yfirst.
Z2
−2Z4
x2
(x2+y)dy dx =Z2
−2x2y+y2
24
x2
dx
=Z2
−2x2(4) + 16
2−x2(x2)−x4
2dx
=Z2
−24x2+ 8 −x4−x4
2dx
=Z2
−2−3
2x4+ 4x2+ 8dx
Step 4: Evaluate the integral with respect to x.
Z2
−2−3
2x4+ 4x2+ 8dx =−3
10x5+4
3x3+ 8x2
−2
=−48
10 +32
3+ 16−48
10 −32
3−16
=64
3+ 32
Thus, the value of the double integral RRR(x2+y)dA over the region Ris
64
3+ 32.
Question 6
Question
Evaluate the double integral RRR(3x2y−2xy2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by y=x2and y= 2x. To find the limits of integration for
x, we set the two equations equal to each other:
x2= 2x=⇒x2−2x= 0 =⇒x(x−2) = 0
This gives us x= 0 and x= 2 as the bounds for x. Next, we find the bounds
for yby substituting x2and 2xinto the integrand:
x2≤y≤2x
Therefore, x2≤y≤2xare the bounds for y.
Step 2: The double integral becomes:
Z2
0Z2x
x2
(3x2y−2xy2)dydx
Step 3: Integrate with respect to yfirst:
Z2
03
2x2y2−2
3xy3
2x
x2dx
Step 4: Evaluate the integral after substituting the limits of integration:
Z2
03
2x2(2x)2−2
3x(2x)3−3
2x2(x2)2−2
3x(x2)3dx
Step 5: Simplify and continue integrating:
Z2
08x4−16
3x4−3x4−2
3x4dx
Step 6: Further simplify and integrate to find the final answer.
Question 7
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 2x, and y= 4.
Solution
Step 1: Determine the bounds of integration for xand y. To find the bounds
of integration for x, we need to determine the x-values at which the curves
intersect. Setting the curves equal to each other, we find:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
6
This gives us x= 0 and x= 2. Since we are given that y= 4 is the upper
bound, the bounds for xare 0≤x≤2.
Next, to find the bounds of integration for y, we need to determine the
y-values corresponding to the curves. Substituting y=x2into y= 2xgives:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
This gives us x= 0 and x= 2, which correspond to y= 0 and y= 4 respectively.
Therefore, the bounds for yare 0≤y≤4.
Step 2: Set up and evaluate the double integral. The double integral can be
set up as follows:
ZZR
x2y dA =Z2
0Z4
0
x2y dy dx
Now, we integrate with respect to yfirst:
Z4
0
x2y dy =x2y2
2
4
0
=x2·8 = 8x2
Next, we integrate the resulting expression with respect to x:
Z2
0
8x2dx =8
3x3
2
0
=8
3(23−0) = 8
3·8 = 64
3
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
3.
Question 8
Question
Calculate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, we first need to determine the limits of integration
by finding the intersection points of the two curves.
Step 1: Find the intersection points of the curves. Setting x2= 2x,
we get x2−2x= 0. Factoring, we obtain x(x−2) = 0, which gives x= 0 and
x= 2 as the intersection points.
Step 2: Determine the limits of integration. Since the region D
is bounded by the curves y=x2and y= 2x, the limits of integration are
0≤x≤2and x2≤y≤2x.
7
Step 3: Evaluate the double integral.
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
=Z2
0x2y+1
3y32x
x2
dx
=Z2
0x2(2x) + 1
3(2x)3−x2(x2)−1
3(x2)3dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=1
2x4+4
3x4−1
5x5−1
21x72
0
=1
2(16) + 4
3(16) −1
5(32) −1
21(128)
=8
2+64
3−32
5−128
21
= 4 + 64
3−32
5−128
21
=252
15 +320
15 −96
15 −360
15
=116
15 .
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 116
15 .
Question 9
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration by finding the intersection points of the two curves y=x2and y= 2x.
Step 1: Find the intersection points.
Setting x2= 2x, we get x2−2x= 0, which can be factored as x(x−2) = 0.
This gives us x= 0 and x= 2 as the intersection points. Thus, the region Dis
bounded by the curves y=x2and y= 2x, where 0≤x≤2.
8
Step 2: Set up the double integral.
The double integral can be written as:
ZZD
exy dA =Z2
0Z2x
x2
exy dy dx
Step 3: Integrate with respect to y.
Integrating with respect to yfirst, we get:
Z2x
x2
exy dy =exy
x2x
x2
=e2x2
x−ex3
x
Step 4: Integrate with respect to x.
Now we integrate with respect to x:
Z2
0 e2x2
x−ex3
x!dx
This integral may not have a simple closed-form solution and may require
numerical methods to approximate the value.
Therefore, the value of the double integral RRDexy dA over the region D
bounded by y=x2and y= 2xin the first quadrant is given by the last
integral, which may be approximated using numerical methods.
Question 10
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves x= 0,y= 0, and x2+y2= 1.
Solution
Step 1: Set up the integral in polar coordinates. In polar coordinates, x=
rcos(θ)and y=rsin(θ). The area element dA is r dr dθ. Therefore, the integral
becomes:
ZZR
ex2+y2dA =Zπ
2
0Z1
0
er2·r dr dθ
Step 2: Integrate with respect to r. This gives us:
Z1
0
er2·r dr =1
2(e−1)
Step 3: Substitute the result of the r-integral back into the expression:
Zπ
2
0
1
2(e−1) dθ
9
Step 4: Integrate with respect to θ:
1
2(e−1) Zπ
2
0
dθ =1
2(e−1) hπ
2i=π
4(e−1)
Therefore, the value of the double integral over the region Ris π
4(e−1).
Question 11
Question
Evaluate the double integral RRD(x2+ 3y)dA where Dis the region bounded
by the lines y=x,y=x2, and x= 0.
Solution
Step 1: Draw the region Ddescribed by the given bounds y=x,y=x2, and
x= 0 in the xy-plane.
Step 2: To set up the double integral, we need to determine the limits of
integration for xand y. Since Dis bounded by y=xand y=x2, the limits of
integration for ywill be x2≤y≤x. The bounds for xwill be 0≤x≤1.
Step 3: Rewrite the double integral using the determined limits of integra-
tion:
ZZD
(x2+ 3y)dA =Z1
0Zx
x2
(x2+ 3y)dy dx
Step 4: Evaluate the inner integral with respect to yfirst:
Zx
x2
(x2+ 3y)dy =x2y+3
2y2
x
x2
=x3−x4+3
2x2−3
2x4
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z1
0
(x3−x4+3
2x2−3
2x4)dx =1
2
Therefore, the value of the double integral is 1
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y= 4,x= 0, and x= 1.
10
Solution
Step 1: We will first find the limits of integration for the double integral by
considering the given region R. The region Ris bounded by the curves y=x2,
y= 4,x= 0, and x= 1. Thus, the limits of integration are:
0≤x≤1and x2≤y≤4
Step 2: Rewrite the given double integral with the limits of integration:
ZZR
xy dA =Z1
0Z4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y, treating xas a constant:
Z4
x2
xy dy =xy2
24
x2
=x16
2−x4
2=x(8 −x4
2) = 8x−x5
2
Step 4: Substitute the result of the inner integral back into the double
integral and evaluate the outer integral with respect to x:
Z1
0
(8x−x5
2)dx =4x2−x6
21
0
= (4 −1
2)−(0 −0) = 7
2
Step 5: Therefore, the value of the double integral RRRxy dA over the region
Ris 7
2.
Question 13
Question
Let Rbe the region bounded by the curves y=x2and y= 2xin the first
quadrant. Calculate the double integral RRR(2x−y)dA where dA =dxdy.
Solution
Step 1: Determine the limits of integration for xand y. Since Ris bounded by
y=x2and y= 2x, we need to find the x-values where these curves intersect.
Setting x2= 2x, we have x2−2x= 0, which gives us x(x−2) = 0. This implies
x= 0 and x= 2 are the points of intersection.
Therefore, 0≤x≤2and x2≤y≤2x.
Step 2: Set up the double integral. The double integral for the given function
over region Ris:
ZZR
(2x−y)dA =Z2
0Z2x
x2
(2x−y)dydx
11
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
(2x−y)dy = 2xy −y2
2
2x
x2
= 2x(2x)−(2x)2
2−[2x(x2)−(x2)2
2]
= 4x2−2x2−2x2+x2
2
= 0.5x2
Step 4: Evaluate the outer integral with respect to x.
Z2
0
0.5x2dx =1
2·x3
3
2
0
=1
2·23
3=8
6=4
3
Thus, the value of the double integral RRR(2x−y)dA over region Ris 4
3.
Question 14
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves x= 0,x= 1,y= 0, and y=x.
Solution
Step 1: To evaluate the double integral, we first need to set up the integral
bounds based on the region R.
Step 2: The region Ris a triangular region in the first quadrant bounded by
the lines x= 0,x= 1,y= 0, and y=x.
Step 3: Thus, the integral bounds for xwill go from 0 to 1, and for ywill
go from 0 to x.
Step 4: Therefore, the given double integral can be expressed as:
ZZR
ex2+y2dA =Z1
0Zx
0
ex2+y2dy dx
Step 5: To evaluate the inner integral with respect to y, we treat xas a
constant and integrate ex2+y2with respect to y.
Zx
0
ex2+y2dy =ex2Zx
0
ey2dy
12
Step 6: Since the antiderivative of ey2does not have a elementary represen-
tation, we know that the integral must be evaluated using other methods.
Step 7: Making a substitution u=y2,du = 2y dy, we get:
ex2Zx2
0
eudu
2=ex2
2Zx2
0
eudu
Step 8: The integral Reudu now can be evaluated and we obtain:
ex2
2[eu]x2
0=ex2
2(ex2−1)
Step 9: Substituting back the result of the inner integral into the original
double integral, we have:
Z1
0
ex2
2(ex2−1) dx
Step 10: Now we integrate this expression with respect to x, and simplify
the result to find the final answer.
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 4.
Solution
We can rewrite the given double integral as follows:
ZZR
(x2+y2)dA =ZZR
x2+y2dA
To evaluate this double integral over the given region R, we will first find
the limits of integration.
Step 1: Determine the limits of integration. The region Ris bounded
by y=x2and y= 4. To find the limits of integration for y, we set x2= 4 and
solve for x:
x2= 4 =⇒x=±2
So the limits of integration for yare from x2to 4, and for xthe limits are
from -2 to 2.
Step 2: Evaluate the double integral. Now we rewrite the given ex-
pression and evaluate the double integral:
ZZR
x2+y2dA =Z2
−2Z4
x2
(x2+y2)dy dx
13
=Z2
−2x2y+1
3y34
x2
dx
=Z2
−2x2(4) + 1
3(4)3−x2(x2)−1
3(x2)3dx
=Z2
−24x2+64
3−x4−1
3x6dx
=4x3
3+64x
3−x5
5−x7
21 2
−2
=32
3+128
3−32
5−128
21 −−32
3−128
3+32
5+128
21
=416
15 +496
35
=2496
105 =416
15
Therefore, the value of the double integral RRR(x2+y2)dA over the given
region Ris 416
15 .
Question 16
Question
Evaluate the double integral RRRexy dA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
Step 1: First, we need to determine the bounds of integration for xand y. The
region Ris bounded by the curves y=x2and y= 2x. Let’s find the points of
intersection by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the x-values where the curves intersect.
Step 2: The bounds of ywill go from x2to 2x, and the bounds of xwill go
from 0 to 2. Therefore, the double integral can be written as:
Z2
0Z2x
x2
exy dy dx
Step 3: To evaluate the inner integral with respect to y, we get:
Zexy dy =exy/x
2x
x2=e2x2/x −ex3/x
14
Step 4: Substituting the values of yand simplifying the integral, we get:
Z2
0
e2x2/x −ex3/x dx
Step 5: To solve this integral, we may need to use techniques like substitution
or integration by parts. After evaluating the integral, we get the final answer.
Step 6: Finally, calculate the definite integral and find the result of the
double integral over the region R.
Therefore, the value of the double integral RRRexy dA over the region R
bounded by y=x2and y= 2xin the first quadrant is obtained by evaluating
the definite integral found in Step 5.
Question 17
Question
Evaluate the double integral RRD(4x2−2y)dA, where Dis the region bounded
by the curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, sketch the region Dto determine the limits of integration. The
region Dis bounded by the curves y=x2,y=x,x= 0, and x= 1. This region
is a trapezoid with vertices at (0,0),(1,1),(1,0), and (0,0).
Step 2: To set up the double integral, we need to express the region Din
terms of xand ylimits. The integral can be written as:
ZZD
(4x2−2y)dA =Z1
0Zx
x2
(4x2−2y)dy dx
Step 3: Evaluate the inner integral with respect to yfirst:
Zx
x2
(4x2−2y)dy =4x2y−y2x
x2= 4x4−x2−4x2+x= 4x4−5x2
Step 4: Now, integrate the result from the previous step with respect to x:
Z1
0
(4x4−5x2)dx =4
5x5−5
3x31
0
=4
5−5
3=7
15
Step 5: Therefore, the value of the double integral RRD(4x2−2y)dA over
the region Dis 7
15 .
15
Question 18
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we first need to determine the points of intersection of the
curves. Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0, so x= 0 and
x= 2. Thus, the limits of integration for xare from 0 to 2. For y, the limits of
integration are from x2to 2x.
Step 2: Rewrite the double integral with the determined limits of integration.
The double integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2
0x2y+y3
3
2x
x2
dx
=Z2
0x2(2x) + (2x)3
3−x2(x2)−(x2)3
3dx
=Z2
02x3+8x3
3−x4−x6
3dx
Step 4: Evaluate the integral with respect to x.
=1
2x4+8
3x4−1
5x5−1
7x7
2
0
=1
2(2)4+8
3(2)4−1
5(2)5−1
7(2)7−0
=16
2+128
3−32
5−128
7
Therefore, the value of the double integral RRR(x2+y2)dA is 576
35 .
Question 19
Question
Evaluate the double integral RRRexy dA where Ris the region bounded by the
curves y=x2and y= 2x.
16
Solution
To evaluate the double integral, we need to express the region Ras a type I or
type II region and find the limits of integration.
Step 1: Determine the type of region
The region Ris bounded by the curves y=x2and y= 2x. This suggests that
it is a type I region.
Step 2: Find the limits of integration
To find the limits of integration, we need to determine the intersection points
of the curves y=x2and y= 2x. Setting the equations equal gives us x2= 2x,
which simplifies to x2−2x= 0. Factoring out an xgives x(x−2) = 0, so x= 0
or x= 2.
Thus, the limits of integration for xare 0≤x≤2, and for yare x2≤y≤2x.
Step 3: Evaluate the integral
We can now set up and evaluate the double integral:
ZZR
exy dA =Z2
0Z2x
x2
exy dy dx
Step 4: Integrate with respect to yfirst
=Z2
0
[x·exy]2x
x2dx =Z2
0h2x·e2x2−x·ex3idx
Step 5: Integrate with respect to x
Now we integrate with respect to x:
=Z2x·e2x2dx −Zx·ex3dx2
0
=1
4e2x22
0−1
3ex32
0
=1
4e8−1
3e8−1
4+1
3
=1
12e8−1
4
Therefore, the value of the double integral RRRexy dA over the region Ris
1
12 e8−1
4.
Question 20
Question
Evaluate the double integral Z ZR
(x2+y)dA where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x.
17
Solution
To evaluate the double integral Z ZR
(x2+y)dA, where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x, we need to set up the integral
in terms of dx and dy using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The region Ris
bounded by the curves x= 0,y= 0,y=x2, and y= 2x. To find the limits of
integration, we need to consider the intersection points of these curves.
Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 and x= 2
are the x-values where y=x2and y= 2xintersect.
Therefore, the limits of integration for xare from x= 0 to x= 2, and the
limits of integration for yare from y= 0 to y= 2x.
Step 2: Set up the double integral. The double integral can be expressed
as:
Z2
0Z2x
0
(x2+y)dy dx
Step 3: Evaluate the inner integral first by integrating with respect to y.
Z2x
0
(x2+y)dy =x2y+y2
2
2x
0=x2(2x) + (2x)2
2−(0 + 0)
= 2x3+ 2x2
Step 4: Substitute back into the outer integral and evaluate.
Z2
0
(2x3+ 2x2)dx =2x4
4+2x3
3
2
0=2(2)4
4+2(2)3
3−0
=32
3
Therefore, the value of the double integral over the region Ris 32
3.
Question 21
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the double integral over the region R, we need to set up the integral
using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the limits of
18
integration for x, we set the curves y=x2and y= 2xequal to each other to
find the intersection points:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the intersection points. Since x= 1 is also a boundary
of the region, the limits of integration for xare from 0 to 1 and from 1 to 2.
The limits of integration for yare from x2to 2x.
Step 2: Set up the double integral The double integral RRR(x2+y2)dA
can be written as
Z1
0Z2x
x2
(x2+y2)dydx +Z2
1Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral Let’s first evaluate the inner
integral with respect to y:
Z(x2+y2)dy =x2y+1
3y3
2x
x2
=x2(2x) + 1
3(8x3−x6)−(0) = 2x3+8
3x3−1
3x6
=14
3x3−1
3x6
Now, we integrate this with respect to x:
Z1
014
3x3−1
3x6dx +Z2
114
3x3−1
3x6dx
Step 4: Final computation
14
12x4−1
24x71
0
+14
12x4−1
24x72
1
=14
12 −1
24−(0 −0) + 14
12 ×16 −1
24 ×128−14
12 −1
24
=139
12
Therefore, the value of the double integral is 139
12 .
Question 22
Question
Evaluate the double integral RR
R
1
x+ydA over the region Rbounded by the curves
y=x2,y= 2x,x= 0, and x= 1.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 1. We find the x-values
where the curves intersect by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:The region Ris
bounded by the curves y=x2and y= 2x. We find the y-values at which these
curves intersect by substituting y=x2into y= 2x:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for yare x2≤y≤2x.
Now, we can set up the double integral using the determined limits of inte-
gration.
Step 3: Set up the double integral: The integral to be evaluated is:
ZZ
R
1
x+ydA =Z1
0Z2x
x2
1
x+ydydx
Step 4: Evaluate the double integral:
Z1
0Z2x
x2
1
x+ydydx =Z1
0
[ln |x+y|]2x
x2dx
=Z1
0ln |3x| − ln |x+x2|dx
=Z1
0
(ln 3x−lnx2+x)dx
=1
2x2ln 3x−xlnx2+x−arctanx
21
0
=1
2ln 3 −ln 2 −π
4−1
2ln 2
Thus, the value of the double integral over the region Ris 1
2ln 3 −3
2ln 2 −π
4.
20
Question 23
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and yby determining the intersection points of the curves y=x2and y= 2x.
Step 1: Find the intersection points of the curves.
Setting x2= 2x, we have:
x2−2x= 0
x(x−2) = 0
So, x= 0 or x= 2. Substituting these values back into y=x2and y= 2x,
we find that the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration.
The limits of integration for yare x2and 2x, and for xare from 0to 2.
Therefore, the double integral can be expressed as:
Z2
0Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral.
We can now integrate with respect to yfirst, and then with respect to x:
Z2
0y3
3+xy
2x
x2
dx
=Z2
0(2x)3
3+ 2x(2x)−x3
3+x(x2)dx
=Z2
08x3
3+ 4x2−x3
3−x3dx
=Z2
0
7x3
3+ 3x2dx
=7x4
12 +x3
2
0
=7(2)4
12 + (2)3−0
=112
12 + 8
21
= 28 + 8
= 36
Therefore, the value of the double integral is 36 .
Question 24
Question
Evaluate the double integral RRRsin(xy)dA, where Ris the region bounded by
the curves y=x2and y=x.
Solution
Step 1: We first need to find the limits of integration for xand yby determining
the intersection points of the curves y=x2and y=x. Setting x2=x, we
have x2−x= 0, which gives x(x−1) = 0. This gives x= 0 and x= 1
as the intersection points. Thus, the limits of integration are 0≤x≤1and
x2≤y≤x.
Step 2: We can now set up the double integral:
ZZR
sin(xy)dA =Z1
0Zx
x2
sin(xy)dy dx
Step 3: We first integrate with respect to y:
Z1
0Zx
x2
sin(xy)dy dx =Z1
0−1
xcos(xy)x
x2
dx
Step 4: Continuing the integration, we get:
Z1
0−1
xcosx2+1
xcosx2dx =Z1
0
0dx = 0
Step 5: Therefore, the value of the double integral RRRsin(xy)dA over the
region Rbounded by y=x2and y=xis 0.
Question 25
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
22
Solution
Step 1: Find the limits of integration. The region Ris defined by the curves
y=x2and y= 2x. To find the limits of integration, we need to determine
the x-values at which these two curves intersect. Setting x2= 2x, we find the
intersection points:
x2−2x= 0
x(x−2) = 0
x= 0,2
So, Ris bounded by 0≤x≤2,x2≤y≤2x.
Step 2: Set up the double integral. The given double integral is
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
= 2x3+8x3
3−x4−x6
3
= 2x3+24x3
3−x4−x6
3
= 2x3+ 8x3−x4−x6
3= 10x3−x4−x6
3
Step 4: Integrate with respect to x.
Z2
0
(10x3−x4−x6
3)dx =10x4
4−x5
5−x7
21
2
0
=10(2)4
4−(2)5
5−(2)7
21 −(0)
= 40 −32
5−128
21
=840 −168 −640
21
=32
21
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 32
21 .
23
Question 2
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral over R, we first need to determine the limits
of integration for xand y. Let’s start by finding the points of intersection of
the two curves y=x2and y= 2x. Setting the two equations equal gives us
x2= 2x, which simplifies to x(x−2) = 0. This gives us x= 0 and x= 2 as the
points of intersection.
Step 1: Determine the limits of integration for xand y.
Since y=x2and y= 2xbound the region R, the limits of integration for x
will be 0≤x≤2. For y, the limits will be x2≤y≤2x.
Step 2: Set up the double integral with the given region and integrand.
Therefore, the double integral becomes:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2
=ex2+(2x)2−ex2+x4
Step 4: Substitute the result back into the double integral.
Z2
0
(ex2+(2x)2−ex2+x4)dx
Step 5: Evaluate the integral with respect to x.
Z2
0
(ex2+4x2−ex2+x4)dx =Z2
0
(e5x2−ex2+x4)dx
Step 6: Integrate e5x2and ex2+x4with respect to xthen evaluate the
integral using the limits of integration.
The final answer depends heavily on the form of the integrand. The exact
numerical computation may be challenging.
Therefore, the value of the double integral over the region Ris given by the
above expression.
2
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y= 2x, we need to set up the integral in terms of dx dy
and find the limits of integration for xand y.
Step 1: Find the limits of integration for xand yThe curves y=x2
and y= 2xintersect when x2= 2x, which gives x2−2x= 0. This equation can
be factored as x(x−2) = 0, leading to solutions x= 0 and x= 2.
So, the limits of integration for xare 0≤x≤2.
For each xin this interval, yranges from x2to 2x. Therefore, the limits of
integration for yare x2≤y≤2x.
Step 2: Set up and evaluate the double integral The double integral
becomes:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Now we integrate:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2=x2(2x) + (2x)3
3−x2(x2)−x3
3
= 2x3+8x3
3−x4−x3
3= 10x3−x4
Now integrate with respect to x:
Z2
0
10x3−x4dx ="10x4
4−x5
5#2
0
=10(2)4
4−(2)5
5=160
4−32
5= 40−32
5=208
5
Therefore, the value of the double integral is 208
5.
Question 4
Question
Evaluate the double integral ZZ
R
ex2−y2dx dy over the region Rbounded by the
curves y=x2,y= 1, and x= 0.
3
Solution
To evaluate the double integral over region R, we first need to determine the
limits of integration.
Step 1: Find the intersection points of the curves.
The intersection points are found by setting the equations of the curves equal
to each other:
For y=x2and y= 1:
x2= 1
x=±1
So the intersection points are (1,1) and (−1,1).
Step 2: Determine the limits of integration.
Since yranges from x2to 1, and xranges from 0 to 1, the limits of integration
are:
0≤x≤1
x2≤y≤1
Step 3: Evaluate the double integral.
Now we can evaluate the integral:
ZZ
R
ex2−y2dx dy =Z1
0Z1
x2
ex2−y2dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
ex2−y2dy =−ex2−1+ex2−x4
Step 5: Integrate with respect to x.
Z1
0
(−ex2−1+ex2−x4)dx =√e−e+√e(−1) −1
2
Therefore, the value of the double integral over region Ris √e−e+√e(−1)−1
2.
Question 5
Question
Evaluate the double integral RRR(x2+y)dA over the region Rbounded by the
curves y=x2,y= 4, and x= 0.
4
Solution
To evaluate the double integral over the given region R, we need to express the
region Rin terms of the limits of integration.
Step 1: Determine the limits of integration for xand y.
The curves y=x2and y= 4 bound the region in the ydirection, while
x= 0 bounds the region in the xdirection.
Since y=x2and y= 4 intersect at x2= 4, we have x=±2as the limits for
x. For y, the limits are from y=x2to y= 4.
Therefore, the limits of integration are −2≤x≤2and x2≤y≤4.
Step 2: Set up the double integral.
The double integral can be expressed as:
ZZR
(x2+y)dA =Z2
−2Z4
x2
(x2+y)dy dx
Step 3: Integrate with respect to yfirst.
Z2
−2Z4
x2
(x2+y)dy dx =Z2
−2x2y+y2
24
x2
dx
=Z2
−2x2(4) + 16
2−x2(x2)−x4
2dx
=Z2
−24x2+ 8 −x4−x4
2dx
=Z2
−2−3
2x4+ 4x2+ 8dx
Step 4: Evaluate the integral with respect to x.
Z2
−2−3
2x4+ 4x2+ 8dx =−3
10x5+4
3x3+ 8x2
−2
=−48
10 +32
3+ 16−48
10 −32
3−16
=64
3+ 32
Thus, the value of the double integral RRR(x2+y)dA over the region Ris
64
3+ 32.
Question 6
Question
Evaluate the double integral RRR(3x2y−2xy2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by y=x2and y= 2x. To find the limits of integration for
x, we set the two equations equal to each other:
x2= 2x=⇒x2−2x= 0 =⇒x(x−2) = 0
This gives us x= 0 and x= 2 as the bounds for x. Next, we find the bounds
for yby substituting x2and 2xinto the integrand:
x2≤y≤2x
Therefore, x2≤y≤2xare the bounds for y.
Step 2: The double integral becomes:
Z2
0Z2x
x2
(3x2y−2xy2)dydx
Step 3: Integrate with respect to yfirst:
Z2
03
2x2y2−2
3xy3
2x
x2dx
Step 4: Evaluate the integral after substituting the limits of integration:
Z2
03
2x2(2x)2−2
3x(2x)3−3
2x2(x2)2−2
3x(x2)3dx
Step 5: Simplify and continue integrating:
Z2
08x4−16
3x4−3x4−2
3x4dx
Step 6: Further simplify and integrate to find the final answer.
Question 7
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 2x, and y= 4.
Solution
Step 1: Determine the bounds of integration for xand y. To find the bounds
of integration for x, we need to determine the x-values at which the curves
intersect. Setting the curves equal to each other, we find:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
6
This gives us x= 0 and x= 2. Since we are given that y= 4 is the upper
bound, the bounds for xare 0≤x≤2.
Next, to find the bounds of integration for y, we need to determine the
y-values corresponding to the curves. Substituting y=x2into y= 2xgives:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
This gives us x= 0 and x= 2, which correspond to y= 0 and y= 4 respectively.
Therefore, the bounds for yare 0≤y≤4.
Step 2: Set up and evaluate the double integral. The double integral can be
set up as follows:
ZZR
x2y dA =Z2
0Z4
0
x2y dy dx
Now, we integrate with respect to yfirst:
Z4
0
x2y dy =x2y2
2
4
0
=x2·8 = 8x2
Next, we integrate the resulting expression with respect to x:
Z2
0
8x2dx =8
3x3
2
0
=8
3(23−0) = 8
3·8 = 64
3
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
3.
Question 8
Question
Calculate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, we first need to determine the limits of integration
by finding the intersection points of the two curves.
Step 1: Find the intersection points of the curves. Setting x2= 2x,
we get x2−2x= 0. Factoring, we obtain x(x−2) = 0, which gives x= 0 and
x= 2 as the intersection points.
Step 2: Determine the limits of integration. Since the region D
is bounded by the curves y=x2and y= 2x, the limits of integration are
0≤x≤2and x2≤y≤2x.
7
Step 3: Evaluate the double integral.
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
=Z2
0x2y+1
3y32x
x2
dx
=Z2
0x2(2x) + 1
3(2x)3−x2(x2)−1
3(x2)3dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=1
2x4+4
3x4−1
5x5−1
21x72
0
=1
2(16) + 4
3(16) −1
5(32) −1
21(128)
=8
2+64
3−32
5−128
21
= 4 + 64
3−32
5−128
21
=252
15 +320
15 −96
15 −360
15
=116
15 .
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 116
15 .
Question 9
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration by finding the intersection points of the two curves y=x2and y= 2x.
Step 1: Find the intersection points.
Setting x2= 2x, we get x2−2x= 0, which can be factored as x(x−2) = 0.
This gives us x= 0 and x= 2 as the intersection points. Thus, the region Dis
bounded by the curves y=x2and y= 2x, where 0≤x≤2.
8
Step 2: Set up the double integral.
The double integral can be written as:
ZZD
exy dA =Z2
0Z2x
x2
exy dy dx
Step 3: Integrate with respect to y.
Integrating with respect to yfirst, we get:
Z2x
x2
exy dy =exy
x2x
x2
=e2x2
x−ex3
x
Step 4: Integrate with respect to x.
Now we integrate with respect to x:
Z2
0 e2x2
x−ex3
x!dx
This integral may not have a simple closed-form solution and may require
numerical methods to approximate the value.
Therefore, the value of the double integral RRDexy dA over the region D
bounded by y=x2and y= 2xin the first quadrant is given by the last
integral, which may be approximated using numerical methods.
Question 10
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves x= 0,y= 0, and x2+y2= 1.
Solution
Step 1: Set up the integral in polar coordinates. In polar coordinates, x=
rcos(θ)and y=rsin(θ). The area element dA is r dr dθ. Therefore, the integral
becomes:
ZZR
ex2+y2dA =Zπ
2
0Z1
0
er2·r dr dθ
Step 2: Integrate with respect to r. This gives us:
Z1
0
er2·r dr =1
2(e−1)
Step 3: Substitute the result of the r-integral back into the expression:
Zπ
2
0
1
2(e−1) dθ
9
Step 4: Integrate with respect to θ:
1
2(e−1) Zπ
2
0
dθ =1
2(e−1) hπ
2i=π
4(e−1)
Therefore, the value of the double integral over the region Ris π
4(e−1).
Question 11
Question
Evaluate the double integral RRD(x2+ 3y)dA where Dis the region bounded
by the lines y=x,y=x2, and x= 0.
Solution
Step 1: Draw the region Ddescribed by the given bounds y=x,y=x2, and
x= 0 in the xy-plane.
Step 2: To set up the double integral, we need to determine the limits of
integration for xand y. Since Dis bounded by y=xand y=x2, the limits of
integration for ywill be x2≤y≤x. The bounds for xwill be 0≤x≤1.
Step 3: Rewrite the double integral using the determined limits of integra-
tion:
ZZD
(x2+ 3y)dA =Z1
0Zx
x2
(x2+ 3y)dy dx
Step 4: Evaluate the inner integral with respect to yfirst:
Zx
x2
(x2+ 3y)dy =x2y+3
2y2
x
x2
=x3−x4+3
2x2−3
2x4
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z1
0
(x3−x4+3
2x2−3
2x4)dx =1
2
Therefore, the value of the double integral is 1
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y= 4,x= 0, and x= 1.
10
Solution
Step 1: We will first find the limits of integration for the double integral by
considering the given region R. The region Ris bounded by the curves y=x2,
y= 4,x= 0, and x= 1. Thus, the limits of integration are:
0≤x≤1and x2≤y≤4
Step 2: Rewrite the given double integral with the limits of integration:
ZZR
xy dA =Z1
0Z4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y, treating xas a constant:
Z4
x2
xy dy =xy2
24
x2
=x16
2−x4
2=x(8 −x4
2) = 8x−x5
2
Step 4: Substitute the result of the inner integral back into the double
integral and evaluate the outer integral with respect to x:
Z1
0
(8x−x5
2)dx =4x2−x6
21
0
= (4 −1
2)−(0 −0) = 7
2
Step 5: Therefore, the value of the double integral RRRxy dA over the region
Ris 7
2.
Question 13
Question
Let Rbe the region bounded by the curves y=x2and y= 2xin the first
quadrant. Calculate the double integral RRR(2x−y)dA where dA =dxdy.
Solution
Step 1: Determine the limits of integration for xand y. Since Ris bounded by
y=x2and y= 2x, we need to find the x-values where these curves intersect.
Setting x2= 2x, we have x2−2x= 0, which gives us x(x−2) = 0. This implies
x= 0 and x= 2 are the points of intersection.
Therefore, 0≤x≤2and x2≤y≤2x.
Step 2: Set up the double integral. The double integral for the given function
over region Ris:
ZZR
(2x−y)dA =Z2
0Z2x
x2
(2x−y)dydx
11
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
(2x−y)dy = 2xy −y2
2
2x
x2
= 2x(2x)−(2x)2
2−[2x(x2)−(x2)2
2]
= 4x2−2x2−2x2+x2
2
= 0.5x2
Step 4: Evaluate the outer integral with respect to x.
Z2
0
0.5x2dx =1
2·x3
3
2
0
=1
2·23
3=8
6=4
3
Thus, the value of the double integral RRR(2x−y)dA over region Ris 4
3.
Question 14
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves x= 0,x= 1,y= 0, and y=x.
Solution
Step 1: To evaluate the double integral, we first need to set up the integral
bounds based on the region R.
Step 2: The region Ris a triangular region in the first quadrant bounded by
the lines x= 0,x= 1,y= 0, and y=x.
Step 3: Thus, the integral bounds for xwill go from 0 to 1, and for ywill
go from 0 to x.
Step 4: Therefore, the given double integral can be expressed as:
ZZR
ex2+y2dA =Z1
0Zx
0
ex2+y2dy dx
Step 5: To evaluate the inner integral with respect to y, we treat xas a
constant and integrate ex2+y2with respect to y.
Zx
0
ex2+y2dy =ex2Zx
0
ey2dy
12
Step 6: Since the antiderivative of ey2does not have a elementary represen-
tation, we know that the integral must be evaluated using other methods.
Step 7: Making a substitution u=y2,du = 2y dy, we get:
ex2Zx2
0
eudu
2=ex2
2Zx2
0
eudu
Step 8: The integral Reudu now can be evaluated and we obtain:
ex2
2[eu]x2
0=ex2
2(ex2−1)
Step 9: Substituting back the result of the inner integral into the original
double integral, we have:
Z1
0
ex2
2(ex2−1) dx
Step 10: Now we integrate this expression with respect to x, and simplify
the result to find the final answer.
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 4.
Solution
We can rewrite the given double integral as follows:
ZZR
(x2+y2)dA =ZZR
x2+y2dA
To evaluate this double integral over the given region R, we will first find
the limits of integration.
Step 1: Determine the limits of integration. The region Ris bounded
by y=x2and y= 4. To find the limits of integration for y, we set x2= 4 and
solve for x:
x2= 4 =⇒x=±2
So the limits of integration for yare from x2to 4, and for xthe limits are
from -2 to 2.
Step 2: Evaluate the double integral. Now we rewrite the given ex-
pression and evaluate the double integral:
ZZR
x2+y2dA =Z2
−2Z4
x2
(x2+y2)dy dx
13
=Z2
−2x2y+1
3y34
x2
dx
=Z2
−2x2(4) + 1
3(4)3−x2(x2)−1
3(x2)3dx
=Z2
−24x2+64
3−x4−1
3x6dx
=4x3
3+64x
3−x5
5−x7
21 2
−2
=32
3+128
3−32
5−128
21 −−32
3−128
3+32
5+128
21
=416
15 +496
35
=2496
105 =416
15
Therefore, the value of the double integral RRR(x2+y2)dA over the given
region Ris 416
15 .
Question 16
Question
Evaluate the double integral RRRexy dA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
Step 1: First, we need to determine the bounds of integration for xand y. The
region Ris bounded by the curves y=x2and y= 2x. Let’s find the points of
intersection by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the x-values where the curves intersect.
Step 2: The bounds of ywill go from x2to 2x, and the bounds of xwill go
from 0 to 2. Therefore, the double integral can be written as:
Z2
0Z2x
x2
exy dy dx
Step 3: To evaluate the inner integral with respect to y, we get:
Zexy dy =exy/x
2x
x2=e2x2/x −ex3/x
14
Step 4: Substituting the values of yand simplifying the integral, we get:
Z2
0
e2x2/x −ex3/x dx
Step 5: To solve this integral, we may need to use techniques like substitution
or integration by parts. After evaluating the integral, we get the final answer.
Step 6: Finally, calculate the definite integral and find the result of the
double integral over the region R.
Therefore, the value of the double integral RRRexy dA over the region R
bounded by y=x2and y= 2xin the first quadrant is obtained by evaluating
the definite integral found in Step 5.
Question 17
Question
Evaluate the double integral RRD(4x2−2y)dA, where Dis the region bounded
by the curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, sketch the region Dto determine the limits of integration. The
region Dis bounded by the curves y=x2,y=x,x= 0, and x= 1. This region
is a trapezoid with vertices at (0,0),(1,1),(1,0), and (0,0).
Step 2: To set up the double integral, we need to express the region Din
terms of xand ylimits. The integral can be written as:
ZZD
(4x2−2y)dA =Z1
0Zx
x2
(4x2−2y)dy dx
Step 3: Evaluate the inner integral with respect to yfirst:
Zx
x2
(4x2−2y)dy =4x2y−y2x
x2= 4x4−x2−4x2+x= 4x4−5x2
Step 4: Now, integrate the result from the previous step with respect to x:
Z1
0
(4x4−5x2)dx =4
5x5−5
3x31
0
=4
5−5
3=7
15
Step 5: Therefore, the value of the double integral RRD(4x2−2y)dA over
the region Dis 7
15 .
15
Question 18
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we first need to determine the points of intersection of the
curves. Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0, so x= 0 and
x= 2. Thus, the limits of integration for xare from 0 to 2. For y, the limits of
integration are from x2to 2x.
Step 2: Rewrite the double integral with the determined limits of integration.
The double integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2
0x2y+y3
3
2x
x2
dx
=Z2
0x2(2x) + (2x)3
3−x2(x2)−(x2)3
3dx
=Z2
02x3+8x3
3−x4−x6
3dx
Step 4: Evaluate the integral with respect to x.
=1
2x4+8
3x4−1
5x5−1
7x7
2
0
=1
2(2)4+8
3(2)4−1
5(2)5−1
7(2)7−0
=16
2+128
3−32
5−128
7
Therefore, the value of the double integral RRR(x2+y2)dA is 576
35 .
Question 19
Question
Evaluate the double integral RRRexy dA where Ris the region bounded by the
curves y=x2and y= 2x.
16
Solution
To evaluate the double integral, we need to express the region Ras a type I or
type II region and find the limits of integration.
Step 1: Determine the type of region
The region Ris bounded by the curves y=x2and y= 2x. This suggests that
it is a type I region.
Step 2: Find the limits of integration
To find the limits of integration, we need to determine the intersection points
of the curves y=x2and y= 2x. Setting the equations equal gives us x2= 2x,
which simplifies to x2−2x= 0. Factoring out an xgives x(x−2) = 0, so x= 0
or x= 2.
Thus, the limits of integration for xare 0≤x≤2, and for yare x2≤y≤2x.
Step 3: Evaluate the integral
We can now set up and evaluate the double integral:
ZZR
exy dA =Z2
0Z2x
x2
exy dy dx
Step 4: Integrate with respect to yfirst
=Z2
0
[x·exy]2x
x2dx =Z2
0h2x·e2x2−x·ex3idx
Step 5: Integrate with respect to x
Now we integrate with respect to x:
=Z2x·e2x2dx −Zx·ex3dx2
0
=1
4e2x22
0−1
3ex32
0
=1
4e8−1
3e8−1
4+1
3
=1
12e8−1
4
Therefore, the value of the double integral RRRexy dA over the region Ris
1
12 e8−1
4.
Question 20
Question
Evaluate the double integral Z ZR
(x2+y)dA where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x.
17
Solution
To evaluate the double integral Z ZR
(x2+y)dA, where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x, we need to set up the integral
in terms of dx and dy using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The region Ris
bounded by the curves x= 0,y= 0,y=x2, and y= 2x. To find the limits of
integration, we need to consider the intersection points of these curves.
Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 and x= 2
are the x-values where y=x2and y= 2xintersect.
Therefore, the limits of integration for xare from x= 0 to x= 2, and the
limits of integration for yare from y= 0 to y= 2x.
Step 2: Set up the double integral. The double integral can be expressed
as:
Z2
0Z2x
0
(x2+y)dy dx
Step 3: Evaluate the inner integral first by integrating with respect to y.
Z2x
0
(x2+y)dy =x2y+y2
2
2x
0=x2(2x) + (2x)2
2−(0 + 0)
= 2x3+ 2x2
Step 4: Substitute back into the outer integral and evaluate.
Z2
0
(2x3+ 2x2)dx =2x4
4+2x3
3
2
0=2(2)4
4+2(2)3
3−0
=32
3
Therefore, the value of the double integral over the region Ris 32
3.
Question 21
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the double integral over the region R, we need to set up the integral
using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the limits of
18
integration for x, we set the curves y=x2and y= 2xequal to each other to
find the intersection points:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the intersection points. Since x= 1 is also a boundary
of the region, the limits of integration for xare from 0 to 1 and from 1 to 2.
The limits of integration for yare from x2to 2x.
Step 2: Set up the double integral The double integral RRR(x2+y2)dA
can be written as
Z1
0Z2x
x2
(x2+y2)dydx +Z2
1Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral Let’s first evaluate the inner
integral with respect to y:
Z(x2+y2)dy =x2y+1
3y3
2x
x2
=x2(2x) + 1
3(8x3−x6)−(0) = 2x3+8
3x3−1
3x6
=14
3x3−1
3x6
Now, we integrate this with respect to x:
Z1
014
3x3−1
3x6dx +Z2
114
3x3−1
3x6dx
Step 4: Final computation
14
12x4−1
24x71
0
+14
12x4−1
24x72
1
=14
12 −1
24−(0 −0) + 14
12 ×16 −1
24 ×128−14
12 −1
24
=139
12
Therefore, the value of the double integral is 139
12 .
Question 22
Question
Evaluate the double integral RR
R
1
x+ydA over the region Rbounded by the curves
y=x2,y= 2x,x= 0, and x= 1.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 1. We find the x-values
where the curves intersect by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:The region Ris
bounded by the curves y=x2and y= 2x. We find the y-values at which these
curves intersect by substituting y=x2into y= 2x:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for yare x2≤y≤2x.
Now, we can set up the double integral using the determined limits of inte-
gration.
Step 3: Set up the double integral: The integral to be evaluated is:
ZZ
R
1
x+ydA =Z1
0Z2x
x2
1
x+ydydx
Step 4: Evaluate the double integral:
Z1
0Z2x
x2
1
x+ydydx =Z1
0
[ln |x+y|]2x
x2dx
=Z1
0ln |3x| − ln |x+x2|dx
=Z1
0
(ln 3x−lnx2+x)dx
=1
2x2ln 3x−xlnx2+x−arctanx
21
0
=1
2ln 3 −ln 2 −π
4−1
2ln 2
Thus, the value of the double integral over the region Ris 1
2ln 3 −3
2ln 2 −π
4.
20
Question 23
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and yby determining the intersection points of the curves y=x2and y= 2x.
Step 1: Find the intersection points of the curves.
Setting x2= 2x, we have:
x2−2x= 0
x(x−2) = 0
So, x= 0 or x= 2. Substituting these values back into y=x2and y= 2x,
we find that the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration.
The limits of integration for yare x2and 2x, and for xare from 0to 2.
Therefore, the double integral can be expressed as:
Z2
0Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral.
We can now integrate with respect to yfirst, and then with respect to x:
Z2
0y3
3+xy
2x
x2
dx
=Z2
0(2x)3
3+ 2x(2x)−x3
3+x(x2)dx
=Z2
08x3
3+ 4x2−x3
3−x3dx
=Z2
0
7x3
3+ 3x2dx
=7x4
12 +x3
2
0
=7(2)4
12 + (2)3−0
=112
12 + 8
21
= 28 + 8
= 36
Therefore, the value of the double integral is 36 .
Question 24
Question
Evaluate the double integral RRRsin(xy)dA, where Ris the region bounded by
the curves y=x2and y=x.
Solution
Step 1: We first need to find the limits of integration for xand yby determining
the intersection points of the curves y=x2and y=x. Setting x2=x, we
have x2−x= 0, which gives x(x−1) = 0. This gives x= 0 and x= 1
as the intersection points. Thus, the limits of integration are 0≤x≤1and
x2≤y≤x.
Step 2: We can now set up the double integral:
ZZR
sin(xy)dA =Z1
0Zx
x2
sin(xy)dy dx
Step 3: We first integrate with respect to y:
Z1
0Zx
x2
sin(xy)dy dx =Z1
0−1
xcos(xy)x
x2
dx
Step 4: Continuing the integration, we get:
Z1
0−1
xcosx2+1
xcosx2dx =Z1
0
0dx = 0
Step 5: Therefore, the value of the double integral RRRsin(xy)dA over the
region Rbounded by y=x2and y=xis 0.
Question 25
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
22
Solution
Step 1: Find the limits of integration. The region Ris defined by the curves
y=x2and y= 2x. To find the limits of integration, we need to determine
the x-values at which these two curves intersect. Setting x2= 2x, we find the
intersection points:
x2−2x= 0
x(x−2) = 0
x= 0,2
So, Ris bounded by 0≤x≤2,x2≤y≤2x.
Step 2: Set up the double integral. The given double integral is
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
= 2x3+8x3
3−x4−x6
3
= 2x3+24x3
3−x4−x6
3
= 2x3+ 8x3−x4−x6
3= 10x3−x4−x6
3
Step 4: Integrate with respect to x.
Z2
0
(10x3−x4−x6
3)dx =10x4
4−x5
5−x7
21
2
0
=10(2)4
4−(2)5
5−(2)7
21 −(0)
= 40 −32
5−128
21
=840 −168 −640
21
=32
21
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 32
21 .
23
Question 2
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral over R, we first need to determine the limits
of integration for xand y. Let’s start by finding the points of intersection of
the two curves y=x2and y= 2x. Setting the two equations equal gives us
x2= 2x, which simplifies to x(x−2) = 0. This gives us x= 0 and x= 2 as the
points of intersection.
Step 1: Determine the limits of integration for xand y.
Since y=x2and y= 2xbound the region R, the limits of integration for x
will be 0≤x≤2. For y, the limits will be x2≤y≤2x.
Step 2: Set up the double integral with the given region and integrand.
Therefore, the double integral becomes:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
ex2+y2dy =ex2+y2
2x
x2
=ex2+(2x)2−ex2+x4
Step 4: Substitute the result back into the double integral.
Z2
0
(ex2+(2x)2−ex2+x4)dx
Step 5: Evaluate the integral with respect to x.
Z2
0
(ex2+4x2−ex2+x4)dx =Z2
0
(e5x2−ex2+x4)dx
Step 6: Integrate e5x2and ex2+x4with respect to xthen evaluate the
integral using the limits of integration.
The final answer depends heavily on the form of the integrand. The exact
numerical computation may be challenging.
Therefore, the value of the double integral over the region Ris given by the
above expression.
2
Question 3
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y= 2x, we need to set up the integral in terms of dx dy
and find the limits of integration for xand y.
Step 1: Find the limits of integration for xand yThe curves y=x2
and y= 2xintersect when x2= 2x, which gives x2−2x= 0. This equation can
be factored as x(x−2) = 0, leading to solutions x= 0 and x= 2.
So, the limits of integration for xare 0≤x≤2.
For each xin this interval, yranges from x2to 2x. Therefore, the limits of
integration for yare x2≤y≤2x.
Step 2: Set up and evaluate the double integral The double integral
becomes:
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Now we integrate:
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2=x2(2x) + (2x)3
3−x2(x2)−x3
3
= 2x3+8x3
3−x4−x3
3= 10x3−x4
Now integrate with respect to x:
Z2
0
10x3−x4dx ="10x4
4−x5
5#2
0
=10(2)4
4−(2)5
5=160
4−32
5= 40−32
5=208
5
Therefore, the value of the double integral is 208
5.
Question 4
Question
Evaluate the double integral ZZ
R
ex2−y2dx dy over the region Rbounded by the
curves y=x2,y= 1, and x= 0.
3
Solution
To evaluate the double integral over region R, we first need to determine the
limits of integration.
Step 1: Find the intersection points of the curves.
The intersection points are found by setting the equations of the curves equal
to each other:
For y=x2and y= 1:
x2= 1
x=±1
So the intersection points are (1,1) and (−1,1).
Step 2: Determine the limits of integration.
Since yranges from x2to 1, and xranges from 0 to 1, the limits of integration
are:
0≤x≤1
x2≤y≤1
Step 3: Evaluate the double integral.
Now we can evaluate the integral:
ZZ
R
ex2−y2dx dy =Z1
0Z1
x2
ex2−y2dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
ex2−y2dy =−ex2−1+ex2−x4
Step 5: Integrate with respect to x.
Z1
0
(−ex2−1+ex2−x4)dx =√e−e+√e(−1) −1
2
Therefore, the value of the double integral over region Ris √e−e+√e(−1)−1
2.
Question 5
Question
Evaluate the double integral RRR(x2+y)dA over the region Rbounded by the
curves y=x2,y= 4, and x= 0.
4
Solution
To evaluate the double integral over the given region R, we need to express the
region Rin terms of the limits of integration.
Step 1: Determine the limits of integration for xand y.
The curves y=x2and y= 4 bound the region in the ydirection, while
x= 0 bounds the region in the xdirection.
Since y=x2and y= 4 intersect at x2= 4, we have x=±2as the limits for
x. For y, the limits are from y=x2to y= 4.
Therefore, the limits of integration are −2≤x≤2and x2≤y≤4.
Step 2: Set up the double integral.
The double integral can be expressed as:
ZZR
(x2+y)dA =Z2
−2Z4
x2
(x2+y)dy dx
Step 3: Integrate with respect to yfirst.
Z2
−2Z4
x2
(x2+y)dy dx =Z2
−2x2y+y2
24
x2
dx
=Z2
−2x2(4) + 16
2−x2(x2)−x4
2dx
=Z2
−24x2+ 8 −x4−x4
2dx
=Z2
−2−3
2x4+ 4x2+ 8dx
Step 4: Evaluate the integral with respect to x.
Z2
−2−3
2x4+ 4x2+ 8dx =−3
10x5+4
3x3+ 8x2
−2
=−48
10 +32
3+ 16−48
10 −32
3−16
=64
3+ 32
Thus, the value of the double integral RRR(x2+y)dA over the region Ris
64
3+ 32.
Question 6
Question
Evaluate the double integral RRR(3x2y−2xy2)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
5
Solution
Step 1: First, we need to determine the limits of integration for xand y. The
region Ris bounded by y=x2and y= 2x. To find the limits of integration for
x, we set the two equations equal to each other:
x2= 2x=⇒x2−2x= 0 =⇒x(x−2) = 0
This gives us x= 0 and x= 2 as the bounds for x. Next, we find the bounds
for yby substituting x2and 2xinto the integrand:
x2≤y≤2x
Therefore, x2≤y≤2xare the bounds for y.
Step 2: The double integral becomes:
Z2
0Z2x
x2
(3x2y−2xy2)dydx
Step 3: Integrate with respect to yfirst:
Z2
03
2x2y2−2
3xy3
2x
x2dx
Step 4: Evaluate the integral after substituting the limits of integration:
Z2
03
2x2(2x)2−2
3x(2x)3−3
2x2(x2)2−2
3x(x2)3dx
Step 5: Simplify and continue integrating:
Z2
08x4−16
3x4−3x4−2
3x4dx
Step 6: Further simplify and integrate to find the final answer.
Question 7
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
curves y=x2,y= 2x, and y= 4.
Solution
Step 1: Determine the bounds of integration for xand y. To find the bounds
of integration for x, we need to determine the x-values at which the curves
intersect. Setting the curves equal to each other, we find:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
6
This gives us x= 0 and x= 2. Since we are given that y= 4 is the upper
bound, the bounds for xare 0≤x≤2.
Next, to find the bounds of integration for y, we need to determine the
y-values corresponding to the curves. Substituting y=x2into y= 2xgives:
x2= 2x⇒x2−2x= 0 ⇒x(x−2) = 0
This gives us x= 0 and x= 2, which correspond to y= 0 and y= 4 respectively.
Therefore, the bounds for yare 0≤y≤4.
Step 2: Set up and evaluate the double integral. The double integral can be
set up as follows:
ZZR
x2y dA =Z2
0Z4
0
x2y dy dx
Now, we integrate with respect to yfirst:
Z4
0
x2y dy =x2y2
2
4
0
=x2·8 = 8x2
Next, we integrate the resulting expression with respect to x:
Z2
0
8x2dx =8
3x3
2
0
=8
3(23−0) = 8
3·8 = 64
3
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
3.
Question 8
Question
Calculate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, we first need to determine the limits of integration
by finding the intersection points of the two curves.
Step 1: Find the intersection points of the curves. Setting x2= 2x,
we get x2−2x= 0. Factoring, we obtain x(x−2) = 0, which gives x= 0 and
x= 2 as the intersection points.
Step 2: Determine the limits of integration. Since the region D
is bounded by the curves y=x2and y= 2x, the limits of integration are
0≤x≤2and x2≤y≤2x.
7
Step 3: Evaluate the double integral.
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
=Z2
0x2y+1
3y32x
x2
dx
=Z2
0x2(2x) + 1
3(2x)3−x2(x2)−1
3(x2)3dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=Z2
0
(2x3+8
3x3−x4−1
3x6)dx
=1
2x4+4
3x4−1
5x5−1
21x72
0
=1
2(16) + 4
3(16) −1
5(32) −1
21(128)
=8
2+64
3−32
5−128
21
= 4 + 64
3−32
5−128
21
=252
15 +320
15 −96
15 −360
15
=116
15 .
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 116
15 .
Question 9
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration by finding the intersection points of the two curves y=x2and y= 2x.
Step 1: Find the intersection points.
Setting x2= 2x, we get x2−2x= 0, which can be factored as x(x−2) = 0.
This gives us x= 0 and x= 2 as the intersection points. Thus, the region Dis
bounded by the curves y=x2and y= 2x, where 0≤x≤2.
8
Step 2: Set up the double integral.
The double integral can be written as:
ZZD
exy dA =Z2
0Z2x
x2
exy dy dx
Step 3: Integrate with respect to y.
Integrating with respect to yfirst, we get:
Z2x
x2
exy dy =exy
x2x
x2
=e2x2
x−ex3
x
Step 4: Integrate with respect to x.
Now we integrate with respect to x:
Z2
0 e2x2
x−ex3
x!dx
This integral may not have a simple closed-form solution and may require
numerical methods to approximate the value.
Therefore, the value of the double integral RRDexy dA over the region D
bounded by y=x2and y= 2xin the first quadrant is given by the last
integral, which may be approximated using numerical methods.
Question 10
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves x= 0,y= 0, and x2+y2= 1.
Solution
Step 1: Set up the integral in polar coordinates. In polar coordinates, x=
rcos(θ)and y=rsin(θ). The area element dA is r dr dθ. Therefore, the integral
becomes:
ZZR
ex2+y2dA =Zπ
2
0Z1
0
er2·r dr dθ
Step 2: Integrate with respect to r. This gives us:
Z1
0
er2·r dr =1
2(e−1)
Step 3: Substitute the result of the r-integral back into the expression:
Zπ
2
0
1
2(e−1) dθ
9
Step 4: Integrate with respect to θ:
1
2(e−1) Zπ
2
0
dθ =1
2(e−1) hπ
2i=π
4(e−1)
Therefore, the value of the double integral over the region Ris π
4(e−1).
Question 11
Question
Evaluate the double integral RRD(x2+ 3y)dA where Dis the region bounded
by the lines y=x,y=x2, and x= 0.
Solution
Step 1: Draw the region Ddescribed by the given bounds y=x,y=x2, and
x= 0 in the xy-plane.
Step 2: To set up the double integral, we need to determine the limits of
integration for xand y. Since Dis bounded by y=xand y=x2, the limits of
integration for ywill be x2≤y≤x. The bounds for xwill be 0≤x≤1.
Step 3: Rewrite the double integral using the determined limits of integra-
tion:
ZZD
(x2+ 3y)dA =Z1
0Zx
x2
(x2+ 3y)dy dx
Step 4: Evaluate the inner integral with respect to yfirst:
Zx
x2
(x2+ 3y)dy =x2y+3
2y2
x
x2
=x3−x4+3
2x2−3
2x4
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x:
Z1
0
(x3−x4+3
2x2−3
2x4)dx =1
2
Therefore, the value of the double integral is 1
2.
Question 12
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2,y= 4,x= 0, and x= 1.
10
Solution
Step 1: We will first find the limits of integration for the double integral by
considering the given region R. The region Ris bounded by the curves y=x2,
y= 4,x= 0, and x= 1. Thus, the limits of integration are:
0≤x≤1and x2≤y≤4
Step 2: Rewrite the given double integral with the limits of integration:
ZZR
xy dA =Z1
0Z4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y, treating xas a constant:
Z4
x2
xy dy =xy2
24
x2
=x16
2−x4
2=x(8 −x4
2) = 8x−x5
2
Step 4: Substitute the result of the inner integral back into the double
integral and evaluate the outer integral with respect to x:
Z1
0
(8x−x5
2)dx =4x2−x6
21
0
= (4 −1
2)−(0 −0) = 7
2
Step 5: Therefore, the value of the double integral RRRxy dA over the region
Ris 7
2.
Question 13
Question
Let Rbe the region bounded by the curves y=x2and y= 2xin the first
quadrant. Calculate the double integral RRR(2x−y)dA where dA =dxdy.
Solution
Step 1: Determine the limits of integration for xand y. Since Ris bounded by
y=x2and y= 2x, we need to find the x-values where these curves intersect.
Setting x2= 2x, we have x2−2x= 0, which gives us x(x−2) = 0. This implies
x= 0 and x= 2 are the points of intersection.
Therefore, 0≤x≤2and x2≤y≤2x.
Step 2: Set up the double integral. The double integral for the given function
over region Ris:
ZZR
(2x−y)dA =Z2
0Z2x
x2
(2x−y)dydx
11
Step 3: Evaluate the inner integral with respect to y.
Z2x
x2
(2x−y)dy = 2xy −y2
2
2x
x2
= 2x(2x)−(2x)2
2−[2x(x2)−(x2)2
2]
= 4x2−2x2−2x2+x2
2
= 0.5x2
Step 4: Evaluate the outer integral with respect to x.
Z2
0
0.5x2dx =1
2·x3
3
2
0
=1
2·23
3=8
6=4
3
Thus, the value of the double integral RRR(2x−y)dA over region Ris 4
3.
Question 14
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves x= 0,x= 1,y= 0, and y=x.
Solution
Step 1: To evaluate the double integral, we first need to set up the integral
bounds based on the region R.
Step 2: The region Ris a triangular region in the first quadrant bounded by
the lines x= 0,x= 1,y= 0, and y=x.
Step 3: Thus, the integral bounds for xwill go from 0 to 1, and for ywill
go from 0 to x.
Step 4: Therefore, the given double integral can be expressed as:
ZZR
ex2+y2dA =Z1
0Zx
0
ex2+y2dy dx
Step 5: To evaluate the inner integral with respect to y, we treat xas a
constant and integrate ex2+y2with respect to y.
Zx
0
ex2+y2dy =ex2Zx
0
ey2dy
12
Step 6: Since the antiderivative of ey2does not have a elementary represen-
tation, we know that the integral must be evaluated using other methods.
Step 7: Making a substitution u=y2,du = 2y dy, we get:
ex2Zx2
0
eudu
2=ex2
2Zx2
0
eudu
Step 8: The integral Reudu now can be evaluated and we obtain:
ex2
2[eu]x2
0=ex2
2(ex2−1)
Step 9: Substituting back the result of the inner integral into the original
double integral, we have:
Z1
0
ex2
2(ex2−1) dx
Step 10: Now we integrate this expression with respect to x, and simplify
the result to find the final answer.
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 4.
Solution
We can rewrite the given double integral as follows:
ZZR
(x2+y2)dA =ZZR
x2+y2dA
To evaluate this double integral over the given region R, we will first find
the limits of integration.
Step 1: Determine the limits of integration. The region Ris bounded
by y=x2and y= 4. To find the limits of integration for y, we set x2= 4 and
solve for x:
x2= 4 =⇒x=±2
So the limits of integration for yare from x2to 4, and for xthe limits are
from -2 to 2.
Step 2: Evaluate the double integral. Now we rewrite the given ex-
pression and evaluate the double integral:
ZZR
x2+y2dA =Z2
−2Z4
x2
(x2+y2)dy dx
13
=Z2
−2x2y+1
3y34
x2
dx
=Z2
−2x2(4) + 1
3(4)3−x2(x2)−1
3(x2)3dx
=Z2
−24x2+64
3−x4−1
3x6dx
=4x3
3+64x
3−x5
5−x7
21 2
−2
=32
3+128
3−32
5−128
21 −−32
3−128
3+32
5+128
21
=416
15 +496
35
=2496
105 =416
15
Therefore, the value of the double integral RRR(x2+y2)dA over the given
region Ris 416
15 .
Question 16
Question
Evaluate the double integral RRRexy dA, where Ris the region bounded by the
curves y=x2and y= 2x, in the first quadrant.
Solution
Step 1: First, we need to determine the bounds of integration for xand y. The
region Ris bounded by the curves y=x2and y= 2x. Let’s find the points of
intersection by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the x-values where the curves intersect.
Step 2: The bounds of ywill go from x2to 2x, and the bounds of xwill go
from 0 to 2. Therefore, the double integral can be written as:
Z2
0Z2x
x2
exy dy dx
Step 3: To evaluate the inner integral with respect to y, we get:
Zexy dy =exy/x
2x
x2=e2x2/x −ex3/x
14
Step 4: Substituting the values of yand simplifying the integral, we get:
Z2
0
e2x2/x −ex3/x dx
Step 5: To solve this integral, we may need to use techniques like substitution
or integration by parts. After evaluating the integral, we get the final answer.
Step 6: Finally, calculate the definite integral and find the result of the
double integral over the region R.
Therefore, the value of the double integral RRRexy dA over the region R
bounded by y=x2and y= 2xin the first quadrant is obtained by evaluating
the definite integral found in Step 5.
Question 17
Question
Evaluate the double integral RRD(4x2−2y)dA, where Dis the region bounded
by the curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, sketch the region Dto determine the limits of integration. The
region Dis bounded by the curves y=x2,y=x,x= 0, and x= 1. This region
is a trapezoid with vertices at (0,0),(1,1),(1,0), and (0,0).
Step 2: To set up the double integral, we need to express the region Din
terms of xand ylimits. The integral can be written as:
ZZD
(4x2−2y)dA =Z1
0Zx
x2
(4x2−2y)dy dx
Step 3: Evaluate the inner integral with respect to yfirst:
Zx
x2
(4x2−2y)dy =4x2y−y2x
x2= 4x4−x2−4x2+x= 4x4−5x2
Step 4: Now, integrate the result from the previous step with respect to x:
Z1
0
(4x4−5x2)dx =4
5x5−5
3x31
0
=4
5−5
3=7
15
Step 5: Therefore, the value of the double integral RRD(4x2−2y)dA over
the region Dis 7
15 .
15
Question 18
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we first need to determine the points of intersection of the
curves. Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0, so x= 0 and
x= 2. Thus, the limits of integration for xare from 0 to 2. For y, the limits of
integration are from x2to 2x.
Step 2: Rewrite the double integral with the determined limits of integration.
The double integral becomes:
Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2
0x2y+y3
3
2x
x2
dx
=Z2
0x2(2x) + (2x)3
3−x2(x2)−(x2)3
3dx
=Z2
02x3+8x3
3−x4−x6
3dx
Step 4: Evaluate the integral with respect to x.
=1
2x4+8
3x4−1
5x5−1
7x7
2
0
=1
2(2)4+8
3(2)4−1
5(2)5−1
7(2)7−0
=16
2+128
3−32
5−128
7
Therefore, the value of the double integral RRR(x2+y2)dA is 576
35 .
Question 19
Question
Evaluate the double integral RRRexy dA where Ris the region bounded by the
curves y=x2and y= 2x.
16
Solution
To evaluate the double integral, we need to express the region Ras a type I or
type II region and find the limits of integration.
Step 1: Determine the type of region
The region Ris bounded by the curves y=x2and y= 2x. This suggests that
it is a type I region.
Step 2: Find the limits of integration
To find the limits of integration, we need to determine the intersection points
of the curves y=x2and y= 2x. Setting the equations equal gives us x2= 2x,
which simplifies to x2−2x= 0. Factoring out an xgives x(x−2) = 0, so x= 0
or x= 2.
Thus, the limits of integration for xare 0≤x≤2, and for yare x2≤y≤2x.
Step 3: Evaluate the integral
We can now set up and evaluate the double integral:
ZZR
exy dA =Z2
0Z2x
x2
exy dy dx
Step 4: Integrate with respect to yfirst
=Z2
0
[x·exy]2x
x2dx =Z2
0h2x·e2x2−x·ex3idx
Step 5: Integrate with respect to x
Now we integrate with respect to x:
=Z2x·e2x2dx −Zx·ex3dx2
0
=1
4e2x22
0−1
3ex32
0
=1
4e8−1
3e8−1
4+1
3
=1
12e8−1
4
Therefore, the value of the double integral RRRexy dA over the region Ris
1
12 e8−1
4.
Question 20
Question
Evaluate the double integral Z ZR
(x2+y)dA where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x.
17
Solution
To evaluate the double integral Z ZR
(x2+y)dA, where Ris the region bounded
by the curves x= 0,y= 0,y=x2, and y= 2x, we need to set up the integral
in terms of dx and dy using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The region Ris
bounded by the curves x= 0,y= 0,y=x2, and y= 2x. To find the limits of
integration, we need to consider the intersection points of these curves.
Setting x2= 2x, we get x2−2x= 0 ⇒x(x−2) = 0. So, x= 0 and x= 2
are the x-values where y=x2and y= 2xintersect.
Therefore, the limits of integration for xare from x= 0 to x= 2, and the
limits of integration for yare from y= 0 to y= 2x.
Step 2: Set up the double integral. The double integral can be expressed
as:
Z2
0Z2x
0
(x2+y)dy dx
Step 3: Evaluate the inner integral first by integrating with respect to y.
Z2x
0
(x2+y)dy =x2y+y2
2
2x
0=x2(2x) + (2x)2
2−(0 + 0)
= 2x3+ 2x2
Step 4: Substitute back into the outer integral and evaluate.
Z2
0
(2x3+ 2x2)dx =2x4
4+2x3
3
2
0=2(2)4
4+2(2)3
3−0
=32
3
Therefore, the value of the double integral over the region Ris 32
3.
Question 21
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2,y= 2x, and x= 1.
Solution
To evaluate the double integral over the region R, we need to set up the integral
using the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the curves y=x2,y= 2x, and x= 1. To find the limits of
18
integration for x, we set the curves y=x2and y= 2xequal to each other to
find the intersection points:
x2= 2x
x2−2x= 0
x(x−2) = 0
So, x= 0 and x= 2 are the intersection points. Since x= 1 is also a boundary
of the region, the limits of integration for xare from 0 to 1 and from 1 to 2.
The limits of integration for yare from x2to 2x.
Step 2: Set up the double integral The double integral RRR(x2+y2)dA
can be written as
Z1
0Z2x
x2
(x2+y2)dydx +Z2
1Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral Let’s first evaluate the inner
integral with respect to y:
Z(x2+y2)dy =x2y+1
3y3
2x
x2
=x2(2x) + 1
3(8x3−x6)−(0) = 2x3+8
3x3−1
3x6
=14
3x3−1
3x6
Now, we integrate this with respect to x:
Z1
014
3x3−1
3x6dx +Z2
114
3x3−1
3x6dx
Step 4: Final computation
14
12x4−1
24x71
0
+14
12x4−1
24x72
1
=14
12 −1
24−(0 −0) + 14
12 ×16 −1
24 ×128−14
12 −1
24
=139
12
Therefore, the value of the double integral is 139
12 .
Question 22
Question
Evaluate the double integral RR
R
1
x+ydA over the region Rbounded by the curves
y=x2,y= 2x,x= 0, and x= 1.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2,y= 2x,x= 0, and x= 1. We find the x-values
where the curves intersect by setting x2= 2x:
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:The region Ris
bounded by the curves y=x2and y= 2x. We find the y-values at which these
curves intersect by substituting y=x2into y= 2x:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives x= 0 and x= 2 as the intersection points for x. Therefore, the limits
of integration for yare x2≤y≤2x.
Now, we can set up the double integral using the determined limits of inte-
gration.
Step 3: Set up the double integral: The integral to be evaluated is:
ZZ
R
1
x+ydA =Z1
0Z2x
x2
1
x+ydydx
Step 4: Evaluate the double integral:
Z1
0Z2x
x2
1
x+ydydx =Z1
0
[ln |x+y|]2x
x2dx
=Z1
0ln |3x| − ln |x+x2|dx
=Z1
0
(ln 3x−lnx2+x)dx
=1
2x2ln 3x−xlnx2+x−arctanx
21
0
=1
2ln 3 −ln 2 −π
4−1
2ln 2
Thus, the value of the double integral over the region Ris 1
2ln 3 −3
2ln 2 −π
4.
20
Question 23
Question
Evaluate the double integral RRR(x2+y2)dA where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we need to find the limits of integration for x
and yby determining the intersection points of the curves y=x2and y= 2x.
Step 1: Find the intersection points of the curves.
Setting x2= 2x, we have:
x2−2x= 0
x(x−2) = 0
So, x= 0 or x= 2. Substituting these values back into y=x2and y= 2x,
we find that the intersection points are (0,0) and (2,4).
Step 2: Determine the limits of integration.
The limits of integration for yare x2and 2x, and for xare from 0to 2.
Therefore, the double integral can be expressed as:
Z2
0Z2x
x2
(x2+y2)dydx
Step 3: Evaluate the double integral.
We can now integrate with respect to yfirst, and then with respect to x:
Z2
0y3
3+xy
2x
x2
dx
=Z2
0(2x)3
3+ 2x(2x)−x3
3+x(x2)dx
=Z2
08x3
3+ 4x2−x3
3−x3dx
=Z2
0
7x3
3+ 3x2dx
=7x4
12 +x3
2
0
=7(2)4
12 + (2)3−0
=112
12 + 8
21
= 28 + 8
= 36
Therefore, the value of the double integral is 36 .
Question 24
Question
Evaluate the double integral RRRsin(xy)dA, where Ris the region bounded by
the curves y=x2and y=x.
Solution
Step 1: We first need to find the limits of integration for xand yby determining
the intersection points of the curves y=x2and y=x. Setting x2=x, we
have x2−x= 0, which gives x(x−1) = 0. This gives x= 0 and x= 1
as the intersection points. Thus, the limits of integration are 0≤x≤1and
x2≤y≤x.
Step 2: We can now set up the double integral:
ZZR
sin(xy)dA =Z1
0Zx
x2
sin(xy)dy dx
Step 3: We first integrate with respect to y:
Z1
0Zx
x2
sin(xy)dy dx =Z1
0−1
xcos(xy)x
x2
dx
Step 4: Continuing the integration, we get:
Z1
0−1
xcosx2+1
xcosx2dx =Z1
0
0dx = 0
Step 5: Therefore, the value of the double integral RRRsin(xy)dA over the
region Rbounded by y=x2and y=xis 0.
Question 25
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
22
Solution
Step 1: Find the limits of integration. The region Ris defined by the curves
y=x2and y= 2x. To find the limits of integration, we need to determine
the x-values at which these two curves intersect. Setting x2= 2x, we find the
intersection points:
x2−2x= 0
x(x−2) = 0
x= 0,2
So, Ris bounded by 0≤x≤2,x2≤y≤2x.
Step 2: Set up the double integral. The given double integral is
ZZR
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.
Z2x
x2
(x2+y2)dy =x2y+y3
3
2x
x2
=x2(2x) + (2x)3
3−x2(x2)−(x2)3
3
= 2x3+8x3
3−x4−x6
3
= 2x3+24x3
3−x4−x6
3
= 2x3+ 8x3−x4−x6
3= 10x3−x4−x6
3
Step 4: Integrate with respect to x.
Z2
0
(10x3−x4−x6
3)dx =10x4
4−x5
5−x7
21
2
0
=10(2)4
4−(2)5
5−(2)7
21 −(0)
= 40 −32
5−128
21
=840 −168 −640
21
=32
21
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 32
21 .
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