MATH 332 - ADVANCED CALCULUS
- Double integrals over rectangular and
general regions
Question Bank - Set 5
Liberty University
Question 1
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x,y= 0, and x= 1.
Solution
To evaluate the double integral, we first need to set up the integral using the
limits of integration determined by the region R.
Step 1: Sketch the region Rbounded by y=x,y= 0, and x= 1.
Step 2: Determine the limits of integration for xand y: Since the region R
is bounded by y=x,y= 0, and x= 1, we have the following limits: xranges
from 0 to 1, and yranges from 0 to x.
Step 3: Set up the double integral: The given double integral becomes:
ZZR
ex+ydA =Z1
0Zx
0
ex+ydy dx
Step 4: Evaluate the inner integral with respect to y:
Zx
0
ex+ydy =exZx
0
eydy =ex[ey]x
0=ex(ex−1)
Step 5: Evaluate the outer integral with respect to x:
Z1
0
ex(ex−1) dx =Z1
0
e2xdx −Z1
0
exdx=e2−1
2−(e−1)
Therefore, the value of the double integral RRRex+ydA over the region Ris
e2
−3
2.
Question 2
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves x= 0,x= 2,y= 0, and y=x2.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region D, we need to
first determine the bounds of integration.
Step 1: Bound of y:The lower bound of yis given by y= 0, and the
upper bound by y=x2.
Step 2: Bound of x:The lower bound of xis x= 0, and the upper bound
is x= 2.
Therefore, the double integral can be written as:
ZZD
(x2+y2)dA =Z2
0Zx2
0
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y:
Zx2
0
(x2+y2)dy =x2y+y3
3
x2
0
=x4+(x2)3
3−0−0
=x4+x6
3
Step 4: Evaluate the outer integral with respect to x:
Z2
0x4+x6
3dx =x5
5+x7
21
2
0
=25
5+27
21 −0−0
=32
5+128
21
=672
105 +640
105
=1312
105
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 1312
105 .
2
Question 3
Question
Compute the double integral RRDx2y dA over the region Dbounded by the
curves y=x2,y= 4,x= 1, and x= 2.
Solution
We first need to determine the limits of integration for xand y.
Step 1: Determine the limits of integration for y: The region Dis bounded
by the curves y=x2and y= 4. This suggests the limits for yshould be from
x2to 4.
Step 2: Determine the limits of integration for x: The region Dis bounded
by x= 1 and x= 2. Thus, the limits for xshould be from 1to 2.
Step 3: Express the double integral as an iterated integral: The double
integral becomes:
Z2
1Z4
x2
x2y dy dx
Step 4: Integrate with respect to y:
Z2
1x2·1
2y24
x2
dx
Simplify this to get:
Z2
18x2−1
2x6dx
Step 5: Integrate with respect to x:
8
3x3−1
14x72
1
This simplifies to: 107
3
Therefore, the double integral RRDx2y dA over the region Dis 107
3.
Question 4
Question
Evaluate the double integral RR
R
(x2+y2)dA, where Ris the region bounded by
the curves y=x2and y= 2 −x2in the first quadrant.
3
Solution
Step 1: To solve the double integral, we first need to find the limits of integration
by determining the intersection points of the curves y=x2and y= 2 −x2.
Let’s set the two equations equal to each other and solve for x:
x2= 2 −x2
2x2= 2
x2= 1
x=±1
Since we are looking at the first quadrant, we consider x= 1 as the intersection
point.
Step 2: Next, we find the bounds for yin terms of xwithin the region R.
The lower bound is y=x2and the upper bound is y= 2 −x2. Therefore, the
limits of integration for yare from x2to 2−x2.
Step 3: Now, we can express the double integral as an iterated integral:
ZZ
R
(x2+y2)dA =Z1
−1Z2−x2
x2
(x2+y2)dy dx
Step 4: We integrate with respect to yfirst:
Z2−x2
x2
(x2+y2)dy =x2y+y3
3
2−x2
x2
=x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3
Step 5: Now, we integrate the above expression with respect to x:
Z1
−1x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3dx
This integral can be computed algebraically to find the final answer.
Question 5
Question
Evaluate the double integral RRD
x2
y3dA where Dis the region bounded by the
curves y=x2,y= 4,x= 0, and x= 2.
4
Solution
To evaluate the double integral, we need to express the region Din terms of the
limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The bounds for
yare from the curve y=x2to y= 4, and the bounds for xare from 0 to 2.
Therefore, the limits of integration are:
0≤x≤2
x2≤y≤4
Step 2: Set up the integral. The double integral can be set up as:
ZZD
x2
y3dA =Z2
0Z4
x2
x2
y3dydx
Step 3: Evaluate the inner integral first.
Z4
x2
x2
y3dy =x2Z4
x2
1
y3dy
=x2−1
2y24
x2
=x2−1
32 +1
2x4
=−3
32 +1
2x2
Step 4: Plug the result into the outer integral and evaluate it.
Z2
0−3
32 +1
2x2dx
=−3
32 Z2
0
dx +1
2Z2
0
1
x2dx
=−3
32 [x]2
0+1
2−1
x2
0
=−3
16 −1
4
=−7
16
Therefore, the value of the double integral RRD
x2
y3dA over the region Dis
−7
16 .
5
Question 6
Question
Evaluate the double integral RRDex+ydA, where Dis the region in the first
quadrant bounded by the curves y=x2and x= 1.
Solution
Step 1: We begin by sketching the region Dbounded by y=x2and x= 1 in
the first quadrant. The region Dis a triangle with vertices at (0,0),(1,1), and
(1,0).
Step 2: To evaluate the given double integral, we first express it in terms of
iterated integrals. Since Dis a triangle, the bounds of integration will be:
0≤x≤1and 0≤y≤x2.
Step 3: We can then express the double integral as an iterated integral as
follows:
Z1
0Zx2
0
ex+ydy dx.
Step 4: We now evaluate the inner integral:
Zx2
0
ex+ydy =exZx2
0
eydy =ex[ey]x2
0=ex(ex2−1).
Step 5: Substituting the result of the inner integral back into the iterated
integral, we have:
Z1
0
ex(ex2−1) dx.
Step 6: We now evaluate the outer integral:
Z1
0
ex(ex2−1) dx =Z1
0
ex+x2−exdx.
Step 7: We evaluate the integral using integration techniques to get the final
answer.
Therefore, the value of the double integral RRDex+ydA over the region D
bounded by y=x2and x= 1 is to be calculated by evaluating the final integral.
Question 7
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x3,y=x,x= 0, and x= 1.
6
Solution
To evaluate the double integral over the region R, we need to express the region
Rin terms of the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by y=x3,y=x,x= 0, and x= 1. This gives us the following
limits of integration:
0≤x≤1
x3≤y≤x
Step 2: Set up the double integral The given integral can be expressed
as:
ZZR
ey2dA =Z1
0Zx
x3
ey2dydx
Step 3: Integrate with respect to yIntegrating with respect to yyields:
Z1
0Zx
x3
ey2dydx =Z1
0hex2−ex6idx
Step 4: Integrate with respect to xIntegrating the remaining expression
with respect to xgives:
Z1
0hex2−ex6idx ="ex2
2−ex6
6#
1
0
=e
2−e
6=2e−e
6=e
6
Therefore, the value of the double integral RRRey2dA over the region Ris
e
6.
Question 8
Question
Evaluate the double integral RRRe2x+3ydA over the region Rbounded by the
curves y= 0,y= 3 −x,y=x−3, and y=x.
Solution
To evaluate the double integral over the region R, we first need to determine
the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y= 0
and y=xbound the region along the x-axis, so the limits of integration for x
are 0≤x≤3.
Step 2: Determine the limits of integration for y:To find the bounds
for y, we need to consider the curves y= 3 −xand y=x−3. These curves
intersect at x= 0 and x= 3.
7
For 0≤x≤1, the curve y= 3 −xis above y=x, so the lower bound for y
is x. For 1≤x≤3, the curve y=x−3is below y=x, so the lower bound for
yis x−3.
Therefore, the limits of integration for yare R3−x
x−3dy for 1≤x≤3, and
R3−x
xdy for 0≤x≤1.
Step 3: Set up and evaluate the double integral: We can now set up
the double integral as follows:
ZZR
e2x+3ydA =Z3
0Z3−x
x−3
e2x+3ydy dx +Z1
0Z3−x
x
e2x+3ydy dx
Now, we evaluate the double integral step by step:
Z3−x
x−3
e2x+3ydy =e2x+3y
33−x
x−3
=1
3e2x+9−3x−e2x+3(3−x)
=1
3e−x+9 −e3x
Z3−x
x
e2x+3ydy =e2x+3y
33−x
x
=1
3e2x+3(3−x)−e2x
=1
3e3x−e2x
Finally, we substitute these results back into the double integral and evaluate
to obtain the final answer.
Question 9
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y=x4.
Solution
To evaluate the double integral RRRxy dA over the region R, we need to first
identify the bounds of integration. Step 1: Find the intersection points of the
curves y=x2and y=x4. Setting x2=x4, we get x4−x2= 0. Factoring, we
have x2(x2−1) = 0. Thus, x= 0 or x= 1. So the intersection points are (0,0)
and (1,1).
Step 2: Determine the bounds of integration for x. The bounds for xare
0≤x≤1.
Step 3: Determine the bounds of integration for y. For each xin [0,1],y
ranges from x2to x4. Therefore, the bounds for yare x2≤y≤x4.
8
Step 4: Set up and evaluate the double integral.
ZZR
xy dA =Z1
0Zx4
x2
xy dy dx.
Z1
0 Zx4
x2
xy dy!dx =Z1
0x·y2
2x4
x2
dx
=Z1
0x·x8
2−x·x4
2dx
=Z1
0x9
2−x5
2dx
=x10
20 −x6
12 1
0
=1
20 −1
12
=3
60 −5
60
=−1
30.
Therefore, the value of the double integral RRRxy dA over the region Ris
−1
30 .
Question 10
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 2.
Solution
Step 1: First, let’s determine the limits of integration for xand y.
The region Ris bounded by the curves y=x2and y= 2. To find the
limits of integration for x, we need to determine the x-values where the curves
intersect. Setting x2= 2, we find that x=±√2.
Thus, the limits of integration for xare −√2≤x≤√2.
To find the limits of integration for y, we note that yvaries between the
curves y=x2and y= 2. Thus, the limits of integration for yare x2≤y≤2.
Step 2: Now, we can set up the double integral.
The double integral to be evaluated is:
ZZR
(x2+y)dA =Z√2
−√2Z2
x2
(x2+y)dy dx
9
Step 3: Evaluate the inner integral with respect to y.
Z2
x2
(x2+y)dy =x2y+y2
22
x2
=x2(2)+ 22
2−x2(x2)−(x2)2
2= 2x2+2−x4−x4
2
Step 4: Substitute the result back into the double integral.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx
Step 5: Integrate with respect to x.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx =2x3
3+ 2x−x5
5−x5
10 √2
−√2
Step 6: Finally, substitute the limits of integration and solve for the value
of the double integral.
2(√2)3
3+ 2(√2) −(√2)5
5−(√2)5
10 !− 2(−√2)3
3+ 2(−√2) −(−√2)5
5−(−√2)5
10 !
Question 11
Question
Evaluate the double integral RR
R
(x+y)dA, where Ris the region enclosed by
the curves y=x2,y= 2x,x= 1, and x= 2.
Solution
Step 1: To find the limits of integration for xand y, we need to first sketch
the region Rdefined by the given curves. The region Ris bounded by y=x2,
y= 2x,x= 1, and x= 2. The curves y=x2and y= 2xintersect at x= 1
and x= 2. Therefore, the limits of integration for xare 1≤x≤2. The curve
y=x2is below y= 2xin the region, so the limits of integration for yare
x2≤y≤2x.
Step 2: Set up and evaluate the double integral using the given limits of
10
integration:
ZZ
R
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
=Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2)−(x2)2
2dx
=Z2
12x2+ 2x2−x3−x4
2dx
=Z2
1
(4x2−x3−x4
2)dx
=4x3
3−x4
4−x5
10 2
1
=32
3−4−32
10 −1
3+1
4+1
10
=8
3−4−16
5−1
3+1
4+1
10
=−111
30
=−37
10
Therefore, the value of the double integral RR
R
(x+y)dA over the region R
is −37
10 .
Question 12
Question
Compute the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
To compute the double integral over the region R, we need to express the region
Rin terms of the bounds for the double integral.
Step 1: Determine the bounds of integration for xand y.The
curves y=x2and y= 2xintersect at (0,0) and (2,4). Therefore, the region R
is bounded by y= 2x,y=x2, and x= 0. This gives us the following bounds:
Lower bound for x: 0,Upper bound for x: 2
11
Lower bound for y:x2,Upper bound for y: 2x
Step 2: Set up and evaluate the double integral. The double integral
we need to compute is:
ZZR
(x2+y)dA =Z2
0Z2x
x2
(x2+y)dy dx
To evaluate this integral, we first integrate with respect to yand then with
respect to x.
Z2
0Z2x
x2
(x2+y)dy dx =Z2
0x2y+y2
22x
x2
dx
=Z2
0x2(2x) + (2x)2
2−x2(x2)−(x2)2
2dx
=Z2
02x3+ 2x2−x4−x4
2dx
=Z2
0hx3(2 −x) + x2(2 −x
2)idx
Further, we compute:
=1
4x4−1
5x52
0
=1
4(2)4−1
5(2)5−1
4(0)4−1
5(0)5
=16
4−32
5= 4 −32
5=8
5
Thus, the value of the double integral RRR(x2+y)dA over the region Ris 8
5.
Question 13
Question
Evaluate the double integral RRRex2+y2dx dy over the region Renclosed by the
curves y= 0,y=x,y= 2x, and y= 4x2.
Solution
To evaluate the double integral over the given region R, we need to express the
region in the form of bounds for xand y.
Step 1: Sketch the region R
Let’s sketch the region Rin the xy-plane to understand its boundaries. The
region Ris enclosed by the curves y= 0,y=x,y= 2x, and y= 4x2.
Step 2: Find the intersection points of the curves
We need to find the points where the curves intersect: - y= 0 intersects y=x
12
at the point (0,0). - y=xintersects y= 2xat the point (0,0). - y= 2x
intersects y= 4x2at the points (0,0) and (1
4,1
2).
Step 3: Set up the integral bounds
To set up the double integral, we need to express the region Rin terms of bounds
for xand y. The region Rcan be expressed as:
R={(x, y)|0≤x≤1
4,0≤y≤4x2,4x2≤y≤2x, and 2x≤y≤x}
Step 4: Evaluate the double integral
Now, we can evaluate the double integral:
ZZR
ex2+y2dx dy =Z1
4
0Z2x
4x2
ex2+y2dy dx
Step 5: Integration
=Z1
4
0hex2+y2i2x
4x2dx
=Z1
4
0
e5x2−ey2dx
Finally, compute the integral to get the numerical value.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y=x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y=x, we need to first determine the limits of integration
for xand y.
Step 1: Determine the limits of integration for xTo find the limits
of integration for x, we set the two curves equal to each other and solve for x:
x2=x
x2−x= 0
x(x−1) = 0
This gives us x= 0 and x= 1. Therefore, the limits of integration for xare
from 0 to 1.
13
Step 2: Determine the limits of integration for yThe curve y=x2is
below the curve y=xwithin the region R. Therefore, the limits of integration
for yare from x2to x.
Step 3: Set up and evaluate the double integral The double integral
can be set up as follows:
ZZR
(x2+y2)dA =Z1
0Zx
x2
(x2+y2)dydx
Now, we can evaluate the double integral:
Z1
0Zx
x2
(x2+y2)dydx =Z1
0x2y+y3
3x
x2
dx
=Z1
0x3−x5+x6
3−x6
3dx
=Z1
0
(x3−x5)dx
=x4
4−x6
61
0
=1
4−1
6
=1
4−1
6
=1
12
So, the value of the double integral RRR(x2+y2)dA over the region Ris 1
12 .
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Sketch the region Rin the xy-plane to visualize the limits of integration.
xy2; [fill =gray!50]coordinates(0,0)(1,0)(1,1)(0,1)(0,0);
R
14
Step 2: Determine the limits of integration for xand y. Since Ris bounded
by y=x2,y= 1,x= 0, and x= 1, we have:
0≤x≤1
x2≤y≤1
Step 3: Set up the double integral in terms of xand y. The double integral
to be evaluated is:
ZZR
(x2+y2)dA =Z1
0Z1
x2
(x2+y2)dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
(x2+y2)dy =x2y+y3
31
x2
=x2(1) + 1
3−x6−x6
3
=x2+1
3−x6−x6
3
Step 5: Integrate the result from Step 4 with respect to x.
Z1
0x2+1
3−x6−x6
3dx =x3
3+x
3−x7
7−x4
12 1
0
=1
3+1
3−1
7−1
12
=23
84
Therefore, the value of the double integral over Ris 23
84 .
Question 16
Question
Evaluate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy2dA, we need to determine the limits of
integration for xand yover the region R.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at the points where x2= 2x. Solving this equation gives
us x= 0 and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
15
Step 2: Determine the limits of integration for y:Given that y=x2
is the lower curve and y= 2xis the upper curve, the limits of integration for y
are x2≤y≤2x.
Step 3: Compute the double integral:
ZZR
xy2dA =Z2
0Z2x
x2
xy2dy dx
=Z2
01
3xy32x
x2
dx
=Z2
08x4−x6
3dx
=2x5−x7
32
0
=64 −128
3
=−64
3.
Therefore, the value of the double integral RRRxy2dA over the region Ris
−64
3.
Question 17
Question
Evaluate the double integral RRReydA over the region Rbounded by the curve
y=x2,x= 0, and x= 2.
Solution
To evaluate the double integral over the region R, we need to determine the
limits of integration for xand y.
Step 1: Determine the limits of integration for x:Since xvaries from
0 to 2 along the x-axis, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a fixed x
between 0 and 2, yvaries from the curve y=x2to the x-axis. Thus, the limits
for yare 0≤y≤x2.
Step 3: Set up and evaluate the integral: The integral for the given
function over the region Ris:
ZZR
eydA =Z2
0Zx2
0
eydy dx
16
Now, we can evaluate the double integral step by step:
Z2
0Zx2
0
eydy dx =Z2
0
[ey]x2
0dx
=Z2
0
(ex2−e0)dx
=Z2
0
ex2−1dx
The integral Rex2dx does not have a simple elementary form, so we stop at
this step as this requires advanced methods for evaluation.
Therefore, the given double integral over the region Rcannot be evaluated
in closed form using elementary methods.
Question 18
Question
Evaluate the double integral RRR(3x2y+ 2y)dA over the region Rbounded by
the curves y=x2and y=−2x.
Solution
Step 1: First, we need to determine the limits of integration for xand yby
finding the points where the curves y=x2and y=−2xintersect.
Step 2: Setting x2=−2x, we find the intersection points:
−x2−2x= 0
x(x+ 2) = 0
x= 0 or x=−2
Step 3: The region Ris bounded by y=x2and y=−2x, so the limits for
ywill be −2x≤y≤x2, and the limits for xwill be −2≤x≤0.
Step 4: Now we can evaluate the double integral:
ZZR
(3x2y+ 2y)dA =Z0
−2Zx2
−2x
(3x2y+ 2y)dy dx
Step 5: Integrating with respect to yfirst, we get:
Zx2
−2x
(3x2y+ 2y)dy =3x2y2
2+y2x2
−2x
=3x2(x4−4x2) + x4−4x2−2x2+ 4x
2
17
Step 6: Simplifying, we have:
Z0
−2
3x6−12x4+ 4x4−16x2−2x2+ 4x
2dx
=Z0
−2
3x6−8x4−18x2+ 4x
2dx
Step 7: Finally, integrating with respect to x, we get:
x7
7−8x5
5−6x3+ 2x2
0
−2
0
7−0
5−0+0−−128
7+256
5+ 48 −8
=128
7−256
5−48 + 8
=128 ·5−256 ·7−48 ·35 + 8 ·35
35
=640 −1792 −1680 + 280
35
=−3552
35
=−101.14
Therefore, the value of the double integral over the region Ris −101.14.
Question 19
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 4, and x= 1.
Evaluate the double integral RRR(3x2y−xy2)dA.
Solution
To evaluate the double integral over the region R, we first need to set up the
bounds of integration.
Step 1: The region Ris bounded by y=x2,y= 4, and x= 1. So, the
bounds for xwill be from 1to 2and the bounds for ywill be from x2to 4.
Step 2: We rewrite the integrand in the integral as 3x2y−xy2as 3x2y−
x(y2).
18
Step 3: We set up and evaluate the double integral:
ZZR
(3x2y−xy2)dA =Z2
1Z4
x2
(3x2y−xy2)dydx
=Z2
13
2x2y2−x
3y3y=4
y=x2
dx
=Z2
148
2x2−64
3x−3
2x6+x4dx
=Z2
124x2−64
3x−3
2x6+x4dx
=8x3−32
3x2−1
2x7+1
5x52
1
= (64 −128
3−128
2+32
5)−(8 −32
3−1
2+1
5)
=64
15.
Step 4: Therefore, the value of the double integral RRR(3x2y−xy2)dA over
the region Ris 64
15 .
Question 20
Question
Evaluate the double integral RRD(6x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, where x∈[0,2].
Solution
To evaluate the double integral, we will first set up the iterated integral and
then compute it step by step.
Step 1: Determine the limits of integration for xand y. The region Dis
bounded by the curves y=x2and y= 2x. To find the limits of integration for
x, we need to find the x-values where the two curves intersect. Setting x2= 2x,
we get x2−2x= 0, which factors to (x)(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration for y. For a given x, the yvalues
range from the lower curve x2to the upper curve 2x. Thus, the limits for yare
x2to 2x.
Step 3: Set up the iterated integral. The double integral can be set up as:
ZZD
(6x2+y2)dA =Z2
0Z2x
x2
(6x2+y2)dy dx
19
Step 4: Evaluate the inner integral.
Z2x
x2
(6x2+y2)dy =6x2y+1
3y32x
x2
= 6x2(2x−x2) + 1
3(4x3−x6)
= 12x3−6x4+4
3x3−1
3x6
= 16x3−6x4−1
3x6
Step 5: Evaluate the outer integral.
Z2
0
(16x3−6x4−1
3x6)dx =4x4−6
5x5−1
21x72
0
= 4(2)4−6
5(2)5−1
21(2)7
= 64 −192
5−128
21
=644
105
Therefore, the value of the double integral RRD(6x2+y2)dA over the region
Dis 644
105 .
Question 21
Question
Evaluate the double integral RRD
y
x2+y2dA, where Dis the region bounded by
the curves y=x2,y= 1, and x= 0.
Solution
Step 1: We begin by sketching the region D. The region is bounded by the
curves y=x2,y= 1, and x= 0, as shown below:
Step 2: Next, we set up the double integral using the given region D:
ZZD
y
x2+y2dA =Z1
0Z√y
0
y
x2+y2dx dy
Step 3: We evaluate the inner integral first with respect to x, treating yas
a constant:
Z√y
0
y
x2+y2dx =1
2lnx2+y2√y
0
=1
2ln(2y)
20
Step 4: Now, we integrate the result from Step 3 with respect to y:
Z1
0
1
2ln(2y)dy =1
2[yln(2y)−y]1
0=1
2(1 ln(2) −1−0) = 1
2(ln 2 −1)
Step 5: Therefore, the value of the double integral RRD
y
x2+y2dA over the
region Dis 1
2(ln 2 −1) .
Question 22
Question
Evaluate the double integral RRR(x2+y)dA where Ris the region in the first
quadrant bounded by the curves x=y3and x= 1.
Solution
To evaluate the double integral over the region R, we need to express the region
in terms of xand yand then set up the double integral accordingly.
Step 1: Determine the limits of integration The region Rin the first
quadrant is bounded by the curves x=y3and x= 1. We can express these
curves as y=3
√xand y= 1 respectively. Therefore, the limits of integration
for yare 0≤y≤1and 3
√x≤y≤1. For x, the limits are 0≤x≤1.
Step 2: Set up the double integral We can now set up the double
integral as follows:
ZZR
(x2+y)dA =Z1
0Z1
3
√x
(x2+y)dy dx
Step 3: Evaluate the inner integral Integrating with respect to y, we
get:
Z1
3
√x
(x2+y)dy =x2y+y2
21
3
√x
=x2(1 −3
√x) + 1
2−x
2
Step 4: Evaluate the outer integral Now, we integrate with respect to
x:
Z1
0x2(1 −3
√x) + 1
2−x
2dx
This integral can be computed by expanding and simplifying the integrand.
Step 5: Compute the final result After integrating the outer integral
and simplifying the expression, the final result can be obtained by substituting
the limits of integration.
21
Question 23
Question
Evaluate the double integral RR
R
(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 will first find the limits of
integration by determining the intersection points of the given curves.
Step 1: Find the intersection points of the curves.
Setting y=x2and y= 2xequal to each other, we have:
x2= 2x
x2−2x= 0
Factoring out an x, we get:
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
Step 2: Determine the limits of integration.
The region Rcan be described as 0≤x≤1and x2≤y≤2x.
Therefore, the limits of integration will be:
R1
0R2x
x2(x2+y2)dy dx
Step 3: Evaluate the double integral.
R1
0R2x
x2(x2+y2)dy dx
=R1
0hx2y+y3
3i2x
x2dx
=R1
02x3−x4+8x3
3−8x4
3dx
=R1
010x3
3−11x4
3dx
=h5x4
3−11x5
15 i1
0
=5
3−11
15
=15
9−11
15
=50
45
So, the value of the double integral over region Ris 10
9.
Question 24
Question
Evaluate the double integral RRD(x2+y2)dx dy, where Dis the region in the
first quadrant bounded by the curves y=x,y=x2, and x= 1.
22
Solution
Step 1: First, sketch the region Din the xy-plane to better understand it. The
region Dis bounded by the curves y=x,y=x2, and the vertical line x= 1.
Step 2: To find the limits of integration, we need to determine the intersec-
tion points of y=xand y=x2. Setting these two equations equal to each other
gives x=x2, which simplifies to x2−x= 0. Solving this quadratic equation
gives x= 0 and x= 1.
Step 3: The limits of integration for ywill be from the curve y=x2to the
curve y=x, and the limits of integration for xwill be from x= 0 to x= 1.
Step 4: We can now set up the double integral:
ZZD
(x2+y2)dx dy =Z1
0Zx
x2
(x2+y2)dy dx
Step 5: Evaluate the inner integral with respect to y:
Zx
x2
(x2+y2)dy =x2y+1
3y3
x
x2
=x3−x5+1
3x3−1
3x6
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(x3−x5+1
3x3−1
3x6)dx
Step 7: Integrate each term individually:
Z1
0
(x3−x5+1
3x3−1
3x6)dx =1
4−1
6+1
12 −1
21 =23
84
Step 8: Thus, the value of the double integral RRD(x2+y2)dx dy over the
region Dis 23
84 .
Question 25
Question
Evaluate the double integral RRRexy dx dy, where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
Step 1: To find the limits of integration, we first need to determine the points
of intersection between the curves y=x2and y= 2x. Setting these two equal
to each other gives us x2= 2x, which simplifies to x2−2x= 0. Factoring out
an xgives x(x−2) = 0, so the solutions are x= 0 and x= 2.
Step 2: Next, we need to determine which curve is on top in the region R.
To do this, we can find the y-values of the curves at the points of intersection.
23
At x= 0, we have y= 02= 0 and y= 2 ·0=0, so they coincide at (0,0). At
x= 2, we have y= 22= 4 and y= 2 ·2 = 4, so they coincide at (2,4).
Step 3: Since the curve y= 2xis above the curve y=x2in the region R,
the limits of integration are 0≤x≤2and x2≤y≤2x.
Step 4: Now we can evaluate the double integral:
ZZR
exy dx dy
=Z2
0Z2x
x2
exy dy dx
=Z2
0exy
x2x
x2
dx
=Z2
0 e2x2
x−ex3
x!dx
=he2x2−ex3i2
0
= (e8−e8)−(1 −1)
= 0.
Therefore, the value of the double integral RRRexy dx dy over the region R
is 0.
24
Question 2
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves x= 0,x= 2,y= 0, and y=x2.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region D, we need to
first determine the bounds of integration.
Step 1: Bound of y:The lower bound of yis given by y= 0, and the
upper bound by y=x2.
Step 2: Bound of x:The lower bound of xis x= 0, and the upper bound
is x= 2.
Therefore, the double integral can be written as:
ZZD
(x2+y2)dA =Z2
0Zx2
0
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y:
Zx2
0
(x2+y2)dy =x2y+y3
3
x2
0
=x4+(x2)3
3−0−0
=x4+x6
3
Step 4: Evaluate the outer integral with respect to x:
Z2
0x4+x6
3dx =x5
5+x7
21
2
0
=25
5+27
21 −0−0
=32
5+128
21
=672
105 +640
105
=1312
105
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 1312
105 .
2
Question 3
Question
Compute the double integral RRDx2y dA over the region Dbounded by the
curves y=x2,y= 4,x= 1, and x= 2.
Solution
We first need to determine the limits of integration for xand y.
Step 1: Determine the limits of integration for y: The region Dis bounded
by the curves y=x2and y= 4. This suggests the limits for yshould be from
x2to 4.
Step 2: Determine the limits of integration for x: The region Dis bounded
by x= 1 and x= 2. Thus, the limits for xshould be from 1to 2.
Step 3: Express the double integral as an iterated integral: The double
integral becomes:
Z2
1Z4
x2
x2y dy dx
Step 4: Integrate with respect to y:
Z2
1x2·1
2y24
x2
dx
Simplify this to get:
Z2
18x2−1
2x6dx
Step 5: Integrate with respect to x:
8
3x3−1
14x72
1
This simplifies to: 107
3
Therefore, the double integral RRDx2y dA over the region Dis 107
3.
Question 4
Question
Evaluate the double integral RR
R
(x2+y2)dA, where Ris the region bounded by
the curves y=x2and y= 2 −x2in the first quadrant.
3
Solution
Step 1: To solve the double integral, we first need to find the limits of integration
by determining the intersection points of the curves y=x2and y= 2 −x2.
Let’s set the two equations equal to each other and solve for x:
x2= 2 −x2
2x2= 2
x2= 1
x=±1
Since we are looking at the first quadrant, we consider x= 1 as the intersection
point.
Step 2: Next, we find the bounds for yin terms of xwithin the region R.
The lower bound is y=x2and the upper bound is y= 2 −x2. Therefore, the
limits of integration for yare from x2to 2−x2.
Step 3: Now, we can express the double integral as an iterated integral:
ZZ
R
(x2+y2)dA =Z1
−1Z2−x2
x2
(x2+y2)dy dx
Step 4: We integrate with respect to yfirst:
Z2−x2
x2
(x2+y2)dy =x2y+y3
3
2−x2
x2
=x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3
Step 5: Now, we integrate the above expression with respect to x:
Z1
−1x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3dx
This integral can be computed algebraically to find the final answer.
Question 5
Question
Evaluate the double integral RRD
x2
y3dA where Dis the region bounded by the
curves y=x2,y= 4,x= 0, and x= 2.
4
Solution
To evaluate the double integral, we need to express the region Din terms of the
limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The bounds for
yare from the curve y=x2to y= 4, and the bounds for xare from 0 to 2.
Therefore, the limits of integration are:
0≤x≤2
x2≤y≤4
Step 2: Set up the integral. The double integral can be set up as:
ZZD
x2
y3dA =Z2
0Z4
x2
x2
y3dydx
Step 3: Evaluate the inner integral first.
Z4
x2
x2
y3dy =x2Z4
x2
1
y3dy
=x2−1
2y24
x2
=x2−1
32 +1
2x4
=−3
32 +1
2x2
Step 4: Plug the result into the outer integral and evaluate it.
Z2
0−3
32 +1
2x2dx
=−3
32 Z2
0
dx +1
2Z2
0
1
x2dx
=−3
32 [x]2
0+1
2−1
x2
0
=−3
16 −1
4
=−7
16
Therefore, the value of the double integral RRD
x2
y3dA over the region Dis
−7
16 .
5
Question 6
Question
Evaluate the double integral RRDex+ydA, where Dis the region in the first
quadrant bounded by the curves y=x2and x= 1.
Solution
Step 1: We begin by sketching the region Dbounded by y=x2and x= 1 in
the first quadrant. The region Dis a triangle with vertices at (0,0),(1,1), and
(1,0).
Step 2: To evaluate the given double integral, we first express it in terms of
iterated integrals. Since Dis a triangle, the bounds of integration will be:
0≤x≤1and 0≤y≤x2.
Step 3: We can then express the double integral as an iterated integral as
follows:
Z1
0Zx2
0
ex+ydy dx.
Step 4: We now evaluate the inner integral:
Zx2
0
ex+ydy =exZx2
0
eydy =ex[ey]x2
0=ex(ex2−1).
Step 5: Substituting the result of the inner integral back into the iterated
integral, we have:
Z1
0
ex(ex2−1) dx.
Step 6: We now evaluate the outer integral:
Z1
0
ex(ex2−1) dx =Z1
0
ex+x2−exdx.
Step 7: We evaluate the integral using integration techniques to get the final
answer.
Therefore, the value of the double integral RRDex+ydA over the region D
bounded by y=x2and x= 1 is to be calculated by evaluating the final integral.
Question 7
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x3,y=x,x= 0, and x= 1.
6
Solution
To evaluate the double integral over the region R, we need to express the region
Rin terms of the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by y=x3,y=x,x= 0, and x= 1. This gives us the following
limits of integration:
0≤x≤1
x3≤y≤x
Step 2: Set up the double integral The given integral can be expressed
as:
ZZR
ey2dA =Z1
0Zx
x3
ey2dydx
Step 3: Integrate with respect to yIntegrating with respect to yyields:
Z1
0Zx
x3
ey2dydx =Z1
0hex2−ex6idx
Step 4: Integrate with respect to xIntegrating the remaining expression
with respect to xgives:
Z1
0hex2−ex6idx ="ex2
2−ex6
6#
1
0
=e
2−e
6=2e−e
6=e
6
Therefore, the value of the double integral RRRey2dA over the region Ris
e
6.
Question 8
Question
Evaluate the double integral RRRe2x+3ydA over the region Rbounded by the
curves y= 0,y= 3 −x,y=x−3, and y=x.
Solution
To evaluate the double integral over the region R, we first need to determine
the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y= 0
and y=xbound the region along the x-axis, so the limits of integration for x
are 0≤x≤3.
Step 2: Determine the limits of integration for y:To find the bounds
for y, we need to consider the curves y= 3 −xand y=x−3. These curves
intersect at x= 0 and x= 3.
7
For 0≤x≤1, the curve y= 3 −xis above y=x, so the lower bound for y
is x. For 1≤x≤3, the curve y=x−3is below y=x, so the lower bound for
yis x−3.
Therefore, the limits of integration for yare R3−x
x−3dy for 1≤x≤3, and
R3−x
xdy for 0≤x≤1.
Step 3: Set up and evaluate the double integral: We can now set up
the double integral as follows:
ZZR
e2x+3ydA =Z3
0Z3−x
x−3
e2x+3ydy dx +Z1
0Z3−x
x
e2x+3ydy dx
Now, we evaluate the double integral step by step:
Z3−x
x−3
e2x+3ydy =e2x+3y
33−x
x−3
=1
3e2x+9−3x−e2x+3(3−x)
=1
3e−x+9 −e3x
Z3−x
x
e2x+3ydy =e2x+3y
33−x
x
=1
3e2x+3(3−x)−e2x
=1
3e3x−e2x
Finally, we substitute these results back into the double integral and evaluate
to obtain the final answer.
Question 9
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y=x4.
Solution
To evaluate the double integral RRRxy dA over the region R, we need to first
identify the bounds of integration. Step 1: Find the intersection points of the
curves y=x2and y=x4. Setting x2=x4, we get x4−x2= 0. Factoring, we
have x2(x2−1) = 0. Thus, x= 0 or x= 1. So the intersection points are (0,0)
and (1,1).
Step 2: Determine the bounds of integration for x. The bounds for xare
0≤x≤1.
Step 3: Determine the bounds of integration for y. For each xin [0,1],y
ranges from x2to x4. Therefore, the bounds for yare x2≤y≤x4.
8
Step 4: Set up and evaluate the double integral.
ZZR
xy dA =Z1
0Zx4
x2
xy dy dx.
Z1
0 Zx4
x2
xy dy!dx =Z1
0x·y2
2x4
x2
dx
=Z1
0x·x8
2−x·x4
2dx
=Z1
0x9
2−x5
2dx
=x10
20 −x6
12 1
0
=1
20 −1
12
=3
60 −5
60
=−1
30.
Therefore, the value of the double integral RRRxy dA over the region Ris
−1
30 .
Question 10
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 2.
Solution
Step 1: First, let’s determine the limits of integration for xand y.
The region Ris bounded by the curves y=x2and y= 2. To find the
limits of integration for x, we need to determine the x-values where the curves
intersect. Setting x2= 2, we find that x=±√2.
Thus, the limits of integration for xare −√2≤x≤√2.
To find the limits of integration for y, we note that yvaries between the
curves y=x2and y= 2. Thus, the limits of integration for yare x2≤y≤2.
Step 2: Now, we can set up the double integral.
The double integral to be evaluated is:
ZZR
(x2+y)dA =Z√2
−√2Z2
x2
(x2+y)dy dx
9
Step 3: Evaluate the inner integral with respect to y.
Z2
x2
(x2+y)dy =x2y+y2
22
x2
=x2(2)+ 22
2−x2(x2)−(x2)2
2= 2x2+2−x4−x4
2
Step 4: Substitute the result back into the double integral.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx
Step 5: Integrate with respect to x.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx =2x3
3+ 2x−x5
5−x5
10 √2
−√2
Step 6: Finally, substitute the limits of integration and solve for the value
of the double integral.
2(√2)3
3+ 2(√2) −(√2)5
5−(√2)5
10 !− 2(−√2)3
3+ 2(−√2) −(−√2)5
5−(−√2)5
10 !
Question 11
Question
Evaluate the double integral RR
R
(x+y)dA, where Ris the region enclosed by
the curves y=x2,y= 2x,x= 1, and x= 2.
Solution
Step 1: To find the limits of integration for xand y, we need to first sketch
the region Rdefined by the given curves. The region Ris bounded by y=x2,
y= 2x,x= 1, and x= 2. The curves y=x2and y= 2xintersect at x= 1
and x= 2. Therefore, the limits of integration for xare 1≤x≤2. The curve
y=x2is below y= 2xin the region, so the limits of integration for yare
x2≤y≤2x.
Step 2: Set up and evaluate the double integral using the given limits of
10
integration:
ZZ
R
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
=Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2)−(x2)2
2dx
=Z2
12x2+ 2x2−x3−x4
2dx
=Z2
1
(4x2−x3−x4
2)dx
=4x3
3−x4
4−x5
10 2
1
=32
3−4−32
10 −1
3+1
4+1
10
=8
3−4−16
5−1
3+1
4+1
10
=−111
30
=−37
10
Therefore, the value of the double integral RR
R
(x+y)dA over the region R
is −37
10 .
Question 12
Question
Compute the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
To compute the double integral over the region R, we need to express the region
Rin terms of the bounds for the double integral.
Step 1: Determine the bounds of integration for xand y.The
curves y=x2and y= 2xintersect at (0,0) and (2,4). Therefore, the region R
is bounded by y= 2x,y=x2, and x= 0. This gives us the following bounds:
Lower bound for x: 0,Upper bound for x: 2
11
Lower bound for y:x2,Upper bound for y: 2x
Step 2: Set up and evaluate the double integral. The double integral
we need to compute is:
ZZR
(x2+y)dA =Z2
0Z2x
x2
(x2+y)dy dx
To evaluate this integral, we first integrate with respect to yand then with
respect to x.
Z2
0Z2x
x2
(x2+y)dy dx =Z2
0x2y+y2
22x
x2
dx
=Z2
0x2(2x) + (2x)2
2−x2(x2)−(x2)2
2dx
=Z2
02x3+ 2x2−x4−x4
2dx
=Z2
0hx3(2 −x) + x2(2 −x
2)idx
Further, we compute:
=1
4x4−1
5x52
0
=1
4(2)4−1
5(2)5−1
4(0)4−1
5(0)5
=16
4−32
5= 4 −32
5=8
5
Thus, the value of the double integral RRR(x2+y)dA over the region Ris 8
5.
Question 13
Question
Evaluate the double integral RRRex2+y2dx dy over the region Renclosed by the
curves y= 0,y=x,y= 2x, and y= 4x2.
Solution
To evaluate the double integral over the given region R, we need to express the
region in the form of bounds for xand y.
Step 1: Sketch the region R
Let’s sketch the region Rin the xy-plane to understand its boundaries. The
region Ris enclosed by the curves y= 0,y=x,y= 2x, and y= 4x2.
Step 2: Find the intersection points of the curves
We need to find the points where the curves intersect: - y= 0 intersects y=x
12
at the point (0,0). - y=xintersects y= 2xat the point (0,0). - y= 2x
intersects y= 4x2at the points (0,0) and (1
4,1
2).
Step 3: Set up the integral bounds
To set up the double integral, we need to express the region Rin terms of bounds
for xand y. The region Rcan be expressed as:
R={(x, y)|0≤x≤1
4,0≤y≤4x2,4x2≤y≤2x, and 2x≤y≤x}
Step 4: Evaluate the double integral
Now, we can evaluate the double integral:
ZZR
ex2+y2dx dy =Z1
4
0Z2x
4x2
ex2+y2dy dx
Step 5: Integration
=Z1
4
0hex2+y2i2x
4x2dx
=Z1
4
0
e5x2−ey2dx
Finally, compute the integral to get the numerical value.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y=x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y=x, we need to first determine the limits of integration
for xand y.
Step 1: Determine the limits of integration for xTo find the limits
of integration for x, we set the two curves equal to each other and solve for x:
x2=x
x2−x= 0
x(x−1) = 0
This gives us x= 0 and x= 1. Therefore, the limits of integration for xare
from 0 to 1.
13
Step 2: Determine the limits of integration for yThe curve y=x2is
below the curve y=xwithin the region R. Therefore, the limits of integration
for yare from x2to x.
Step 3: Set up and evaluate the double integral The double integral
can be set up as follows:
ZZR
(x2+y2)dA =Z1
0Zx
x2
(x2+y2)dydx
Now, we can evaluate the double integral:
Z1
0Zx
x2
(x2+y2)dydx =Z1
0x2y+y3
3x
x2
dx
=Z1
0x3−x5+x6
3−x6
3dx
=Z1
0
(x3−x5)dx
=x4
4−x6
61
0
=1
4−1
6
=1
4−1
6
=1
12
So, the value of the double integral RRR(x2+y2)dA over the region Ris 1
12 .
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Sketch the region Rin the xy-plane to visualize the limits of integration.
xy2; [fill =gray!50]coordinates(0,0)(1,0)(1,1)(0,1)(0,0);
R
14
Step 2: Determine the limits of integration for xand y. Since Ris bounded
by y=x2,y= 1,x= 0, and x= 1, we have:
0≤x≤1
x2≤y≤1
Step 3: Set up the double integral in terms of xand y. The double integral
to be evaluated is:
ZZR
(x2+y2)dA =Z1
0Z1
x2
(x2+y2)dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
(x2+y2)dy =x2y+y3
31
x2
=x2(1) + 1
3−x6−x6
3
=x2+1
3−x6−x6
3
Step 5: Integrate the result from Step 4 with respect to x.
Z1
0x2+1
3−x6−x6
3dx =x3
3+x
3−x7
7−x4
12 1
0
=1
3+1
3−1
7−1
12
=23
84
Therefore, the value of the double integral over Ris 23
84 .
Question 16
Question
Evaluate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy2dA, we need to determine the limits of
integration for xand yover the region R.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at the points where x2= 2x. Solving this equation gives
us x= 0 and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
15
Step 2: Determine the limits of integration for y:Given that y=x2
is the lower curve and y= 2xis the upper curve, the limits of integration for y
are x2≤y≤2x.
Step 3: Compute the double integral:
ZZR
xy2dA =Z2
0Z2x
x2
xy2dy dx
=Z2
01
3xy32x
x2
dx
=Z2
08x4−x6
3dx
=2x5−x7
32
0
=64 −128
3
=−64
3.
Therefore, the value of the double integral RRRxy2dA over the region Ris
−64
3.
Question 17
Question
Evaluate the double integral RRReydA over the region Rbounded by the curve
y=x2,x= 0, and x= 2.
Solution
To evaluate the double integral over the region R, we need to determine the
limits of integration for xand y.
Step 1: Determine the limits of integration for x:Since xvaries from
0 to 2 along the x-axis, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a fixed x
between 0 and 2, yvaries from the curve y=x2to the x-axis. Thus, the limits
for yare 0≤y≤x2.
Step 3: Set up and evaluate the integral: The integral for the given
function over the region Ris:
ZZR
eydA =Z2
0Zx2
0
eydy dx
16
Now, we can evaluate the double integral step by step:
Z2
0Zx2
0
eydy dx =Z2
0
[ey]x2
0dx
=Z2
0
(ex2−e0)dx
=Z2
0
ex2−1dx
The integral Rex2dx does not have a simple elementary form, so we stop at
this step as this requires advanced methods for evaluation.
Therefore, the given double integral over the region Rcannot be evaluated
in closed form using elementary methods.
Question 18
Question
Evaluate the double integral RRR(3x2y+ 2y)dA over the region Rbounded by
the curves y=x2and y=−2x.
Solution
Step 1: First, we need to determine the limits of integration for xand yby
finding the points where the curves y=x2and y=−2xintersect.
Step 2: Setting x2=−2x, we find the intersection points:
−x2−2x= 0
x(x+ 2) = 0
x= 0 or x=−2
Step 3: The region Ris bounded by y=x2and y=−2x, so the limits for
ywill be −2x≤y≤x2, and the limits for xwill be −2≤x≤0.
Step 4: Now we can evaluate the double integral:
ZZR
(3x2y+ 2y)dA =Z0
−2Zx2
−2x
(3x2y+ 2y)dy dx
Step 5: Integrating with respect to yfirst, we get:
Zx2
−2x
(3x2y+ 2y)dy =3x2y2
2+y2x2
−2x
=3x2(x4−4x2) + x4−4x2−2x2+ 4x
2
17
Step 6: Simplifying, we have:
Z0
−2
3x6−12x4+ 4x4−16x2−2x2+ 4x
2dx
=Z0
−2
3x6−8x4−18x2+ 4x
2dx
Step 7: Finally, integrating with respect to x, we get:
x7
7−8x5
5−6x3+ 2x2
0
−2
0
7−0
5−0+0−−128
7+256
5+ 48 −8
=128
7−256
5−48 + 8
=128 ·5−256 ·7−48 ·35 + 8 ·35
35
=640 −1792 −1680 + 280
35
=−3552
35
=−101.14
Therefore, the value of the double integral over the region Ris −101.14.
Question 19
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 4, and x= 1.
Evaluate the double integral RRR(3x2y−xy2)dA.
Solution
To evaluate the double integral over the region R, we first need to set up the
bounds of integration.
Step 1: The region Ris bounded by y=x2,y= 4, and x= 1. So, the
bounds for xwill be from 1to 2and the bounds for ywill be from x2to 4.
Step 2: We rewrite the integrand in the integral as 3x2y−xy2as 3x2y−
x(y2).
18
Step 3: We set up and evaluate the double integral:
ZZR
(3x2y−xy2)dA =Z2
1Z4
x2
(3x2y−xy2)dydx
=Z2
13
2x2y2−x
3y3y=4
y=x2
dx
=Z2
148
2x2−64
3x−3
2x6+x4dx
=Z2
124x2−64
3x−3
2x6+x4dx
=8x3−32
3x2−1
2x7+1
5x52
1
= (64 −128
3−128
2+32
5)−(8 −32
3−1
2+1
5)
=64
15.
Step 4: Therefore, the value of the double integral RRR(3x2y−xy2)dA over
the region Ris 64
15 .
Question 20
Question
Evaluate the double integral RRD(6x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, where x∈[0,2].
Solution
To evaluate the double integral, we will first set up the iterated integral and
then compute it step by step.
Step 1: Determine the limits of integration for xand y. The region Dis
bounded by the curves y=x2and y= 2x. To find the limits of integration for
x, we need to find the x-values where the two curves intersect. Setting x2= 2x,
we get x2−2x= 0, which factors to (x)(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration for y. For a given x, the yvalues
range from the lower curve x2to the upper curve 2x. Thus, the limits for yare
x2to 2x.
Step 3: Set up the iterated integral. The double integral can be set up as:
ZZD
(6x2+y2)dA =Z2
0Z2x
x2
(6x2+y2)dy dx
19
Step 4: Evaluate the inner integral.
Z2x
x2
(6x2+y2)dy =6x2y+1
3y32x
x2
= 6x2(2x−x2) + 1
3(4x3−x6)
= 12x3−6x4+4
3x3−1
3x6
= 16x3−6x4−1
3x6
Step 5: Evaluate the outer integral.
Z2
0
(16x3−6x4−1
3x6)dx =4x4−6
5x5−1
21x72
0
= 4(2)4−6
5(2)5−1
21(2)7
= 64 −192
5−128
21
=644
105
Therefore, the value of the double integral RRD(6x2+y2)dA over the region
Dis 644
105 .
Question 21
Question
Evaluate the double integral RRD
y
x2+y2dA, where Dis the region bounded by
the curves y=x2,y= 1, and x= 0.
Solution
Step 1: We begin by sketching the region D. The region is bounded by the
curves y=x2,y= 1, and x= 0, as shown below:
Step 2: Next, we set up the double integral using the given region D:
ZZD
y
x2+y2dA =Z1
0Z√y
0
y
x2+y2dx dy
Step 3: We evaluate the inner integral first with respect to x, treating yas
a constant:
Z√y
0
y
x2+y2dx =1
2lnx2+y2√y
0
=1
2ln(2y)
20
Step 4: Now, we integrate the result from Step 3 with respect to y:
Z1
0
1
2ln(2y)dy =1
2[yln(2y)−y]1
0=1
2(1 ln(2) −1−0) = 1
2(ln 2 −1)
Step 5: Therefore, the value of the double integral RRD
y
x2+y2dA over the
region Dis 1
2(ln 2 −1) .
Question 22
Question
Evaluate the double integral RRR(x2+y)dA where Ris the region in the first
quadrant bounded by the curves x=y3and x= 1.
Solution
To evaluate the double integral over the region R, we need to express the region
in terms of xand yand then set up the double integral accordingly.
Step 1: Determine the limits of integration The region Rin the first
quadrant is bounded by the curves x=y3and x= 1. We can express these
curves as y=3
√xand y= 1 respectively. Therefore, the limits of integration
for yare 0≤y≤1and 3
√x≤y≤1. For x, the limits are 0≤x≤1.
Step 2: Set up the double integral We can now set up the double
integral as follows:
ZZR
(x2+y)dA =Z1
0Z1
3
√x
(x2+y)dy dx
Step 3: Evaluate the inner integral Integrating with respect to y, we
get:
Z1
3
√x
(x2+y)dy =x2y+y2
21
3
√x
=x2(1 −3
√x) + 1
2−x
2
Step 4: Evaluate the outer integral Now, we integrate with respect to
x:
Z1
0x2(1 −3
√x) + 1
2−x
2dx
This integral can be computed by expanding and simplifying the integrand.
Step 5: Compute the final result After integrating the outer integral
and simplifying the expression, the final result can be obtained by substituting
the limits of integration.
21
Question 23
Question
Evaluate the double integral RR
R
(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 will first find the limits of
integration by determining the intersection points of the given curves.
Step 1: Find the intersection points of the curves.
Setting y=x2and y= 2xequal to each other, we have:
x2= 2x
x2−2x= 0
Factoring out an x, we get:
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
Step 2: Determine the limits of integration.
The region Rcan be described as 0≤x≤1and x2≤y≤2x.
Therefore, the limits of integration will be:
R1
0R2x
x2(x2+y2)dy dx
Step 3: Evaluate the double integral.
R1
0R2x
x2(x2+y2)dy dx
=R1
0hx2y+y3
3i2x
x2dx
=R1
02x3−x4+8x3
3−8x4
3dx
=R1
010x3
3−11x4
3dx
=h5x4
3−11x5
15 i1
0
=5
3−11
15
=15
9−11
15
=50
45
So, the value of the double integral over region Ris 10
9.
Question 24
Question
Evaluate the double integral RRD(x2+y2)dx dy, where Dis the region in the
first quadrant bounded by the curves y=x,y=x2, and x= 1.
22
Solution
Step 1: First, sketch the region Din the xy-plane to better understand it. The
region Dis bounded by the curves y=x,y=x2, and the vertical line x= 1.
Step 2: To find the limits of integration, we need to determine the intersec-
tion points of y=xand y=x2. Setting these two equations equal to each other
gives x=x2, which simplifies to x2−x= 0. Solving this quadratic equation
gives x= 0 and x= 1.
Step 3: The limits of integration for ywill be from the curve y=x2to the
curve y=x, and the limits of integration for xwill be from x= 0 to x= 1.
Step 4: We can now set up the double integral:
ZZD
(x2+y2)dx dy =Z1
0Zx
x2
(x2+y2)dy dx
Step 5: Evaluate the inner integral with respect to y:
Zx
x2
(x2+y2)dy =x2y+1
3y3
x
x2
=x3−x5+1
3x3−1
3x6
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(x3−x5+1
3x3−1
3x6)dx
Step 7: Integrate each term individually:
Z1
0
(x3−x5+1
3x3−1
3x6)dx =1
4−1
6+1
12 −1
21 =23
84
Step 8: Thus, the value of the double integral RRD(x2+y2)dx dy over the
region Dis 23
84 .
Question 25
Question
Evaluate the double integral RRRexy dx dy, where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
Step 1: To find the limits of integration, we first need to determine the points
of intersection between the curves y=x2and y= 2x. Setting these two equal
to each other gives us x2= 2x, which simplifies to x2−2x= 0. Factoring out
an xgives x(x−2) = 0, so the solutions are x= 0 and x= 2.
Step 2: Next, we need to determine which curve is on top in the region R.
To do this, we can find the y-values of the curves at the points of intersection.
23
At x= 0, we have y= 02= 0 and y= 2 ·0=0, so they coincide at (0,0). At
x= 2, we have y= 22= 4 and y= 2 ·2 = 4, so they coincide at (2,4).
Step 3: Since the curve y= 2xis above the curve y=x2in the region R,
the limits of integration are 0≤x≤2and x2≤y≤2x.
Step 4: Now we can evaluate the double integral:
ZZR
exy dx dy
=Z2
0Z2x
x2
exy dy dx
=Z2
0exy
x2x
x2
dx
=Z2
0 e2x2
x−ex3
x!dx
=he2x2−ex3i2
0
= (e8−e8)−(1 −1)
= 0.
Therefore, the value of the double integral RRRexy dx dy over the region R
is 0.
24
Question 2
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves x= 0,x= 2,y= 0, and y=x2.
Solution
To evaluate the double integral RRD(x2+y2)dA over the region D, we need to
first determine the bounds of integration.
Step 1: Bound of y:The lower bound of yis given by y= 0, and the
upper bound by y=x2.
Step 2: Bound of x:The lower bound of xis x= 0, and the upper bound
is x= 2.
Therefore, the double integral can be written as:
ZZD
(x2+y2)dA =Z2
0Zx2
0
(x2+y2)dy dx
Step 3: Evaluate the inner integral with respect to y:
Zx2
0
(x2+y2)dy =x2y+y3
3
x2
0
=x4+(x2)3
3−0−0
=x4+x6
3
Step 4: Evaluate the outer integral with respect to x:
Z2
0x4+x6
3dx =x5
5+x7
21
2
0
=25
5+27
21 −0−0
=32
5+128
21
=672
105 +640
105
=1312
105
Therefore, the value of the double integral RRD(x2+y2)dA over the region
Dis 1312
105 .
2
Question 3
Question
Compute the double integral RRDx2y dA over the region Dbounded by the
curves y=x2,y= 4,x= 1, and x= 2.
Solution
We first need to determine the limits of integration for xand y.
Step 1: Determine the limits of integration for y: The region Dis bounded
by the curves y=x2and y= 4. This suggests the limits for yshould be from
x2to 4.
Step 2: Determine the limits of integration for x: The region Dis bounded
by x= 1 and x= 2. Thus, the limits for xshould be from 1to 2.
Step 3: Express the double integral as an iterated integral: The double
integral becomes:
Z2
1Z4
x2
x2y dy dx
Step 4: Integrate with respect to y:
Z2
1x2·1
2y24
x2
dx
Simplify this to get:
Z2
18x2−1
2x6dx
Step 5: Integrate with respect to x:
8
3x3−1
14x72
1
This simplifies to: 107
3
Therefore, the double integral RRDx2y dA over the region Dis 107
3.
Question 4
Question
Evaluate the double integral RR
R
(x2+y2)dA, where Ris the region bounded by
the curves y=x2and y= 2 −x2in the first quadrant.
3
Solution
Step 1: To solve the double integral, we first need to find the limits of integration
by determining the intersection points of the curves y=x2and y= 2 −x2.
Let’s set the two equations equal to each other and solve for x:
x2= 2 −x2
2x2= 2
x2= 1
x=±1
Since we are looking at the first quadrant, we consider x= 1 as the intersection
point.
Step 2: Next, we find the bounds for yin terms of xwithin the region R.
The lower bound is y=x2and the upper bound is y= 2 −x2. Therefore, the
limits of integration for yare from x2to 2−x2.
Step 3: Now, we can express the double integral as an iterated integral:
ZZ
R
(x2+y2)dA =Z1
−1Z2−x2
x2
(x2+y2)dy dx
Step 4: We integrate with respect to yfirst:
Z2−x2
x2
(x2+y2)dy =x2y+y3
3
2−x2
x2
=x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3
Step 5: Now, we integrate the above expression with respect to x:
Z1
−1x2(2 −x2) + (2 −x2)3
3−x2(x2)−(x2)3
3dx
This integral can be computed algebraically to find the final answer.
Question 5
Question
Evaluate the double integral RRD
x2
y3dA where Dis the region bounded by the
curves y=x2,y= 4,x= 0, and x= 2.
4
Solution
To evaluate the double integral, we need to express the region Din terms of the
limits of integration for xand y.
Step 1: Determine the limits of integration for xand y. The bounds for
yare from the curve y=x2to y= 4, and the bounds for xare from 0 to 2.
Therefore, the limits of integration are:
0≤x≤2
x2≤y≤4
Step 2: Set up the integral. The double integral can be set up as:
ZZD
x2
y3dA =Z2
0Z4
x2
x2
y3dydx
Step 3: Evaluate the inner integral first.
Z4
x2
x2
y3dy =x2Z4
x2
1
y3dy
=x2−1
2y24
x2
=x2−1
32 +1
2x4
=−3
32 +1
2x2
Step 4: Plug the result into the outer integral and evaluate it.
Z2
0−3
32 +1
2x2dx
=−3
32 Z2
0
dx +1
2Z2
0
1
x2dx
=−3
32 [x]2
0+1
2−1
x2
0
=−3
16 −1
4
=−7
16
Therefore, the value of the double integral RRD
x2
y3dA over the region Dis
−7
16 .
5
Question 6
Question
Evaluate the double integral RRDex+ydA, where Dis the region in the first
quadrant bounded by the curves y=x2and x= 1.
Solution
Step 1: We begin by sketching the region Dbounded by y=x2and x= 1 in
the first quadrant. The region Dis a triangle with vertices at (0,0),(1,1), and
(1,0).
Step 2: To evaluate the given double integral, we first express it in terms of
iterated integrals. Since Dis a triangle, the bounds of integration will be:
0≤x≤1and 0≤y≤x2.
Step 3: We can then express the double integral as an iterated integral as
follows:
Z1
0Zx2
0
ex+ydy dx.
Step 4: We now evaluate the inner integral:
Zx2
0
ex+ydy =exZx2
0
eydy =ex[ey]x2
0=ex(ex2−1).
Step 5: Substituting the result of the inner integral back into the iterated
integral, we have:
Z1
0
ex(ex2−1) dx.
Step 6: We now evaluate the outer integral:
Z1
0
ex(ex2−1) dx =Z1
0
ex+x2−exdx.
Step 7: We evaluate the integral using integration techniques to get the final
answer.
Therefore, the value of the double integral RRDex+ydA over the region D
bounded by y=x2and x= 1 is to be calculated by evaluating the final integral.
Question 7
Question
Evaluate the double integral RRRey2dA, where Ris the region bounded by the
curves y=x3,y=x,x= 0, and x= 1.
6
Solution
To evaluate the double integral over the region R, we need to express the region
Rin terms of the limits of integration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by y=x3,y=x,x= 0, and x= 1. This gives us the following
limits of integration:
0≤x≤1
x3≤y≤x
Step 2: Set up the double integral The given integral can be expressed
as:
ZZR
ey2dA =Z1
0Zx
x3
ey2dydx
Step 3: Integrate with respect to yIntegrating with respect to yyields:
Z1
0Zx
x3
ey2dydx =Z1
0hex2−ex6idx
Step 4: Integrate with respect to xIntegrating the remaining expression
with respect to xgives:
Z1
0hex2−ex6idx ="ex2
2−ex6
6#
1
0
=e
2−e
6=2e−e
6=e
6
Therefore, the value of the double integral RRRey2dA over the region Ris
e
6.
Question 8
Question
Evaluate the double integral RRRe2x+3ydA over the region Rbounded by the
curves y= 0,y= 3 −x,y=x−3, and y=x.
Solution
To evaluate the double integral over the region R, we first need to determine
the limits of integration for xand y.
Step 1: Determine the limits of integration for x:The curves y= 0
and y=xbound the region along the x-axis, so the limits of integration for x
are 0≤x≤3.
Step 2: Determine the limits of integration for y:To find the bounds
for y, we need to consider the curves y= 3 −xand y=x−3. These curves
intersect at x= 0 and x= 3.
7
For 0≤x≤1, the curve y= 3 −xis above y=x, so the lower bound for y
is x. For 1≤x≤3, the curve y=x−3is below y=x, so the lower bound for
yis x−3.
Therefore, the limits of integration for yare R3−x
x−3dy for 1≤x≤3, and
R3−x
xdy for 0≤x≤1.
Step 3: Set up and evaluate the double integral: We can now set up
the double integral as follows:
ZZR
e2x+3ydA =Z3
0Z3−x
x−3
e2x+3ydy dx +Z1
0Z3−x
x
e2x+3ydy dx
Now, we evaluate the double integral step by step:
Z3−x
x−3
e2x+3ydy =e2x+3y
33−x
x−3
=1
3e2x+9−3x−e2x+3(3−x)
=1
3e−x+9 −e3x
Z3−x
x
e2x+3ydy =e2x+3y
33−x
x
=1
3e2x+3(3−x)−e2x
=1
3e3x−e2x
Finally, we substitute these results back into the double integral and evaluate
to obtain the final answer.
Question 9
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y=x4.
Solution
To evaluate the double integral RRRxy dA over the region R, we need to first
identify the bounds of integration. Step 1: Find the intersection points of the
curves y=x2and y=x4. Setting x2=x4, we get x4−x2= 0. Factoring, we
have x2(x2−1) = 0. Thus, x= 0 or x= 1. So the intersection points are (0,0)
and (1,1).
Step 2: Determine the bounds of integration for x. The bounds for xare
0≤x≤1.
Step 3: Determine the bounds of integration for y. For each xin [0,1],y
ranges from x2to x4. Therefore, the bounds for yare x2≤y≤x4.
8
Step 4: Set up and evaluate the double integral.
ZZR
xy dA =Z1
0Zx4
x2
xy dy dx.
Z1
0 Zx4
x2
xy dy!dx =Z1
0x·y2
2x4
x2
dx
=Z1
0x·x8
2−x·x4
2dx
=Z1
0x9
2−x5
2dx
=x10
20 −x6
12 1
0
=1
20 −1
12
=3
60 −5
60
=−1
30.
Therefore, the value of the double integral RRRxy dA over the region Ris
−1
30 .
Question 10
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 2.
Solution
Step 1: First, let’s determine the limits of integration for xand y.
The region Ris bounded by the curves y=x2and y= 2. To find the
limits of integration for x, we need to determine the x-values where the curves
intersect. Setting x2= 2, we find that x=±√2.
Thus, the limits of integration for xare −√2≤x≤√2.
To find the limits of integration for y, we note that yvaries between the
curves y=x2and y= 2. Thus, the limits of integration for yare x2≤y≤2.
Step 2: Now, we can set up the double integral.
The double integral to be evaluated is:
ZZR
(x2+y)dA =Z√2
−√2Z2
x2
(x2+y)dy dx
9
Step 3: Evaluate the inner integral with respect to y.
Z2
x2
(x2+y)dy =x2y+y2
22
x2
=x2(2)+ 22
2−x2(x2)−(x2)2
2= 2x2+2−x4−x4
2
Step 4: Substitute the result back into the double integral.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx
Step 5: Integrate with respect to x.
Z√2
−√2
(2x2+ 2 −x4−x4
2)dx =2x3
3+ 2x−x5
5−x5
10 √2
−√2
Step 6: Finally, substitute the limits of integration and solve for the value
of the double integral.
2(√2)3
3+ 2(√2) −(√2)5
5−(√2)5
10 !− 2(−√2)3
3+ 2(−√2) −(−√2)5
5−(−√2)5
10 !
Question 11
Question
Evaluate the double integral RR
R
(x+y)dA, where Ris the region enclosed by
the curves y=x2,y= 2x,x= 1, and x= 2.
Solution
Step 1: To find the limits of integration for xand y, we need to first sketch
the region Rdefined by the given curves. The region Ris bounded by y=x2,
y= 2x,x= 1, and x= 2. The curves y=x2and y= 2xintersect at x= 1
and x= 2. Therefore, the limits of integration for xare 1≤x≤2. The curve
y=x2is below y= 2xin the region, so the limits of integration for yare
x2≤y≤2x.
Step 2: Set up and evaluate the double integral using the given limits of
10
integration:
ZZ
R
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
=Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2)−(x2)2
2dx
=Z2
12x2+ 2x2−x3−x4
2dx
=Z2
1
(4x2−x3−x4
2)dx
=4x3
3−x4
4−x5
10 2
1
=32
3−4−32
10 −1
3+1
4+1
10
=8
3−4−16
5−1
3+1
4+1
10
=−111
30
=−37
10
Therefore, the value of the double integral RR
R
(x+y)dA over the region R
is −37
10 .
Question 12
Question
Compute the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
To compute the double integral over the region R, we need to express the region
Rin terms of the bounds for the double integral.
Step 1: Determine the bounds of integration for xand y.The
curves y=x2and y= 2xintersect at (0,0) and (2,4). Therefore, the region R
is bounded by y= 2x,y=x2, and x= 0. This gives us the following bounds:
Lower bound for x: 0,Upper bound for x: 2
11
Lower bound for y:x2,Upper bound for y: 2x
Step 2: Set up and evaluate the double integral. The double integral
we need to compute is:
ZZR
(x2+y)dA =Z2
0Z2x
x2
(x2+y)dy dx
To evaluate this integral, we first integrate with respect to yand then with
respect to x.
Z2
0Z2x
x2
(x2+y)dy dx =Z2
0x2y+y2
22x
x2
dx
=Z2
0x2(2x) + (2x)2
2−x2(x2)−(x2)2
2dx
=Z2
02x3+ 2x2−x4−x4
2dx
=Z2
0hx3(2 −x) + x2(2 −x
2)idx
Further, we compute:
=1
4x4−1
5x52
0
=1
4(2)4−1
5(2)5−1
4(0)4−1
5(0)5
=16
4−32
5= 4 −32
5=8
5
Thus, the value of the double integral RRR(x2+y)dA over the region Ris 8
5.
Question 13
Question
Evaluate the double integral RRRex2+y2dx dy over the region Renclosed by the
curves y= 0,y=x,y= 2x, and y= 4x2.
Solution
To evaluate the double integral over the given region R, we need to express the
region in the form of bounds for xand y.
Step 1: Sketch the region R
Let’s sketch the region Rin the xy-plane to understand its boundaries. The
region Ris enclosed by the curves y= 0,y=x,y= 2x, and y= 4x2.
Step 2: Find the intersection points of the curves
We need to find the points where the curves intersect: - y= 0 intersects y=x
12
at the point (0,0). - y=xintersects y= 2xat the point (0,0). - y= 2x
intersects y= 4x2at the points (0,0) and (1
4,1
2).
Step 3: Set up the integral bounds
To set up the double integral, we need to express the region Rin terms of bounds
for xand y. The region Rcan be expressed as:
R={(x, y)|0≤x≤1
4,0≤y≤4x2,4x2≤y≤2x, and 2x≤y≤x}
Step 4: Evaluate the double integral
Now, we can evaluate the double integral:
ZZR
ex2+y2dx dy =Z1
4
0Z2x
4x2
ex2+y2dy dx
Step 5: Integration
=Z1
4
0hex2+y2i2x
4x2dx
=Z1
4
0
e5x2−ey2dx
Finally, compute the integral to get the numerical value.
Question 14
Question
Evaluate the double integral RRR(x2+y2)dA over the region Rbounded by the
curves y=x2and y=x.
Solution
To evaluate the double integral RRR(x2+y2)dA over the region Rbounded by
the curves y=x2and y=x, we need to first determine the limits of integration
for xand y.
Step 1: Determine the limits of integration for xTo find the limits
of integration for x, we set the two curves equal to each other and solve for x:
x2=x
x2−x= 0
x(x−1) = 0
This gives us x= 0 and x= 1. Therefore, the limits of integration for xare
from 0 to 1.
13
Step 2: Determine the limits of integration for yThe curve y=x2is
below the curve y=xwithin the region R. Therefore, the limits of integration
for yare from x2to x.
Step 3: Set up and evaluate the double integral The double integral
can be set up as follows:
ZZR
(x2+y2)dA =Z1
0Zx
x2
(x2+y2)dydx
Now, we can evaluate the double integral:
Z1
0Zx
x2
(x2+y2)dydx =Z1
0x2y+y3
3x
x2
dx
=Z1
0x3−x5+x6
3−x6
3dx
=Z1
0
(x3−x5)dx
=x4
4−x6
61
0
=1
4−1
6
=1
4−1
6
=1
12
So, the value of the double integral RRR(x2+y2)dA over the region Ris 1
12 .
Question 15
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2,y= 1,x= 0, and x= 1.
Solution
Step 1: Sketch the region Rin the xy-plane to visualize the limits of integration.
xy2; [fill =gray!50]coordinates(0,0)(1,0)(1,1)(0,1)(0,0);
R
14
Step 2: Determine the limits of integration for xand y. Since Ris bounded
by y=x2,y= 1,x= 0, and x= 1, we have:
0≤x≤1
x2≤y≤1
Step 3: Set up the double integral in terms of xand y. The double integral
to be evaluated is:
ZZR
(x2+y2)dA =Z1
0Z1
x2
(x2+y2)dy dx
Step 4: Integrate with respect to yfirst.
Z1
x2
(x2+y2)dy =x2y+y3
31
x2
=x2(1) + 1
3−x6−x6
3
=x2+1
3−x6−x6
3
Step 5: Integrate the result from Step 4 with respect to x.
Z1
0x2+1
3−x6−x6
3dx =x3
3+x
3−x7
7−x4
12 1
0
=1
3+1
3−1
7−1
12
=23
84
Therefore, the value of the double integral over Ris 23
84 .
Question 16
Question
Evaluate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy2dA, we need to determine the limits of
integration for xand yover the region R.
Step 1: Determine the limits of integration for x:The curves y=x2
and y= 2xintersect at the points where x2= 2x. Solving this equation gives
us x= 0 and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
15
Step 2: Determine the limits of integration for y:Given that y=x2
is the lower curve and y= 2xis the upper curve, the limits of integration for y
are x2≤y≤2x.
Step 3: Compute the double integral:
ZZR
xy2dA =Z2
0Z2x
x2
xy2dy dx
=Z2
01
3xy32x
x2
dx
=Z2
08x4−x6
3dx
=2x5−x7
32
0
=64 −128
3
=−64
3.
Therefore, the value of the double integral RRRxy2dA over the region Ris
−64
3.
Question 17
Question
Evaluate the double integral RRReydA over the region Rbounded by the curve
y=x2,x= 0, and x= 2.
Solution
To evaluate the double integral over the region R, we need to determine the
limits of integration for xand y.
Step 1: Determine the limits of integration for x:Since xvaries from
0 to 2 along the x-axis, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a fixed x
between 0 and 2, yvaries from the curve y=x2to the x-axis. Thus, the limits
for yare 0≤y≤x2.
Step 3: Set up and evaluate the integral: The integral for the given
function over the region Ris:
ZZR
eydA =Z2
0Zx2
0
eydy dx
16
Now, we can evaluate the double integral step by step:
Z2
0Zx2
0
eydy dx =Z2
0
[ey]x2
0dx
=Z2
0
(ex2−e0)dx
=Z2
0
ex2−1dx
The integral Rex2dx does not have a simple elementary form, so we stop at
this step as this requires advanced methods for evaluation.
Therefore, the given double integral over the region Rcannot be evaluated
in closed form using elementary methods.
Question 18
Question
Evaluate the double integral RRR(3x2y+ 2y)dA over the region Rbounded by
the curves y=x2and y=−2x.
Solution
Step 1: First, we need to determine the limits of integration for xand yby
finding the points where the curves y=x2and y=−2xintersect.
Step 2: Setting x2=−2x, we find the intersection points:
−x2−2x= 0
x(x+ 2) = 0
x= 0 or x=−2
Step 3: The region Ris bounded by y=x2and y=−2x, so the limits for
ywill be −2x≤y≤x2, and the limits for xwill be −2≤x≤0.
Step 4: Now we can evaluate the double integral:
ZZR
(3x2y+ 2y)dA =Z0
−2Zx2
−2x
(3x2y+ 2y)dy dx
Step 5: Integrating with respect to yfirst, we get:
Zx2
−2x
(3x2y+ 2y)dy =3x2y2
2+y2x2
−2x
=3x2(x4−4x2) + x4−4x2−2x2+ 4x
2
17
Step 6: Simplifying, we have:
Z0
−2
3x6−12x4+ 4x4−16x2−2x2+ 4x
2dx
=Z0
−2
3x6−8x4−18x2+ 4x
2dx
Step 7: Finally, integrating with respect to x, we get:
x7
7−8x5
5−6x3+ 2x2
0
−2
0
7−0
5−0+0−−128
7+256
5+ 48 −8
=128
7−256
5−48 + 8
=128 ·5−256 ·7−48 ·35 + 8 ·35
35
=640 −1792 −1680 + 280
35
=−3552
35
=−101.14
Therefore, the value of the double integral over the region Ris −101.14.
Question 19
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 4, and x= 1.
Evaluate the double integral RRR(3x2y−xy2)dA.
Solution
To evaluate the double integral over the region R, we first need to set up the
bounds of integration.
Step 1: The region Ris bounded by y=x2,y= 4, and x= 1. So, the
bounds for xwill be from 1to 2and the bounds for ywill be from x2to 4.
Step 2: We rewrite the integrand in the integral as 3x2y−xy2as 3x2y−
x(y2).
18
Step 3: We set up and evaluate the double integral:
ZZR
(3x2y−xy2)dA =Z2
1Z4
x2
(3x2y−xy2)dydx
=Z2
13
2x2y2−x
3y3y=4
y=x2
dx
=Z2
148
2x2−64
3x−3
2x6+x4dx
=Z2
124x2−64
3x−3
2x6+x4dx
=8x3−32
3x2−1
2x7+1
5x52
1
= (64 −128
3−128
2+32
5)−(8 −32
3−1
2+1
5)
=64
15.
Step 4: Therefore, the value of the double integral RRR(3x2y−xy2)dA over
the region Ris 64
15 .
Question 20
Question
Evaluate the double integral RRD(6x2+y2)dA over the region Dbounded by
the curves y=x2and y= 2x, where x∈[0,2].
Solution
To evaluate the double integral, we will first set up the iterated integral and
then compute it step by step.
Step 1: Determine the limits of integration for xand y. The region Dis
bounded by the curves y=x2and y= 2x. To find the limits of integration for
x, we need to find the x-values where the two curves intersect. Setting x2= 2x,
we get x2−2x= 0, which factors to (x)(x−2) = 0. So, x= 0 or x= 2.
Step 2: Determine the limits of integration for y. For a given x, the yvalues
range from the lower curve x2to the upper curve 2x. Thus, the limits for yare
x2to 2x.
Step 3: Set up the iterated integral. The double integral can be set up as:
ZZD
(6x2+y2)dA =Z2
0Z2x
x2
(6x2+y2)dy dx
19
Step 4: Evaluate the inner integral.
Z2x
x2
(6x2+y2)dy =6x2y+1
3y32x
x2
= 6x2(2x−x2) + 1
3(4x3−x6)
= 12x3−6x4+4
3x3−1
3x6
= 16x3−6x4−1
3x6
Step 5: Evaluate the outer integral.
Z2
0
(16x3−6x4−1
3x6)dx =4x4−6
5x5−1
21x72
0
= 4(2)4−6
5(2)5−1
21(2)7
= 64 −192
5−128
21
=644
105
Therefore, the value of the double integral RRD(6x2+y2)dA over the region
Dis 644
105 .
Question 21
Question
Evaluate the double integral RRD
y
x2+y2dA, where Dis the region bounded by
the curves y=x2,y= 1, and x= 0.
Solution
Step 1: We begin by sketching the region D. The region is bounded by the
curves y=x2,y= 1, and x= 0, as shown below:
Step 2: Next, we set up the double integral using the given region D:
ZZD
y
x2+y2dA =Z1
0Z√y
0
y
x2+y2dx dy
Step 3: We evaluate the inner integral first with respect to x, treating yas
a constant:
Z√y
0
y
x2+y2dx =1
2lnx2+y2√y
0
=1
2ln(2y)
20
Step 4: Now, we integrate the result from Step 3 with respect to y:
Z1
0
1
2ln(2y)dy =1
2[yln(2y)−y]1
0=1
2(1 ln(2) −1−0) = 1
2(ln 2 −1)
Step 5: Therefore, the value of the double integral RRD
y
x2+y2dA over the
region Dis 1
2(ln 2 −1) .
Question 22
Question
Evaluate the double integral RRR(x2+y)dA where Ris the region in the first
quadrant bounded by the curves x=y3and x= 1.
Solution
To evaluate the double integral over the region R, we need to express the region
in terms of xand yand then set up the double integral accordingly.
Step 1: Determine the limits of integration The region Rin the first
quadrant is bounded by the curves x=y3and x= 1. We can express these
curves as y=3
√xand y= 1 respectively. Therefore, the limits of integration
for yare 0≤y≤1and 3
√x≤y≤1. For x, the limits are 0≤x≤1.
Step 2: Set up the double integral We can now set up the double
integral as follows:
ZZR
(x2+y)dA =Z1
0Z1
3
√x
(x2+y)dy dx
Step 3: Evaluate the inner integral Integrating with respect to y, we
get:
Z1
3
√x
(x2+y)dy =x2y+y2
21
3
√x
=x2(1 −3
√x) + 1
2−x
2
Step 4: Evaluate the outer integral Now, we integrate with respect to
x:
Z1
0x2(1 −3
√x) + 1
2−x
2dx
This integral can be computed by expanding and simplifying the integrand.
Step 5: Compute the final result After integrating the outer integral
and simplifying the expression, the final result can be obtained by substituting
the limits of integration.
21
Question 23
Question
Evaluate the double integral RR
R
(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 will first find the limits of
integration by determining the intersection points of the given curves.
Step 1: Find the intersection points of the curves.
Setting y=x2and y= 2xequal to each other, we have:
x2= 2x
x2−2x= 0
Factoring out an x, we get:
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
Step 2: Determine the limits of integration.
The region Rcan be described as 0≤x≤1and x2≤y≤2x.
Therefore, the limits of integration will be:
R1
0R2x
x2(x2+y2)dy dx
Step 3: Evaluate the double integral.
R1
0R2x
x2(x2+y2)dy dx
=R1
0hx2y+y3
3i2x
x2dx
=R1
02x3−x4+8x3
3−8x4
3dx
=R1
010x3
3−11x4
3dx
=h5x4
3−11x5
15 i1
0
=5
3−11
15
=15
9−11
15
=50
45
So, the value of the double integral over region Ris 10
9.
Question 24
Question
Evaluate the double integral RRD(x2+y2)dx dy, where Dis the region in the
first quadrant bounded by the curves y=x,y=x2, and x= 1.
22
Solution
Step 1: First, sketch the region Din the xy-plane to better understand it. The
region Dis bounded by the curves y=x,y=x2, and the vertical line x= 1.
Step 2: To find the limits of integration, we need to determine the intersec-
tion points of y=xand y=x2. Setting these two equations equal to each other
gives x=x2, which simplifies to x2−x= 0. Solving this quadratic equation
gives x= 0 and x= 1.
Step 3: The limits of integration for ywill be from the curve y=x2to the
curve y=x, and the limits of integration for xwill be from x= 0 to x= 1.
Step 4: We can now set up the double integral:
ZZD
(x2+y2)dx dy =Z1
0Zx
x2
(x2+y2)dy dx
Step 5: Evaluate the inner integral with respect to y:
Zx
x2
(x2+y2)dy =x2y+1
3y3
x
x2
=x3−x5+1
3x3−1
3x6
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(x3−x5+1
3x3−1
3x6)dx
Step 7: Integrate each term individually:
Z1
0
(x3−x5+1
3x3−1
3x6)dx =1
4−1
6+1
12 −1
21 =23
84
Step 8: Thus, the value of the double integral RRD(x2+y2)dx dy over the
region Dis 23
84 .
Question 25
Question
Evaluate the double integral RRRexy dx dy, where Ris the region bounded by
the curves y=x2and y= 2x.
Solution
Step 1: To find the limits of integration, we first need to determine the points
of intersection between the curves y=x2and y= 2x. Setting these two equal
to each other gives us x2= 2x, which simplifies to x2−2x= 0. Factoring out
an xgives x(x−2) = 0, so the solutions are x= 0 and x= 2.
Step 2: Next, we need to determine which curve is on top in the region R.
To do this, we can find the y-values of the curves at the points of intersection.
23
At x= 0, we have y= 02= 0 and y= 2 ·0=0, so they coincide at (0,0). At
x= 2, we have y= 22= 4 and y= 2 ·2 = 4, so they coincide at (2,4).
Step 3: Since the curve y= 2xis above the curve y=x2in the region R,
the limits of integration are 0≤x≤2and x2≤y≤2x.
Step 4: Now we can evaluate the double integral:
ZZR
exy dx dy
=Z2
0Z2x
x2
exy dy dx
=Z2
0exy
x2x
x2
dx
=Z2
0 e2x2
x−ex3
x!dx
=he2x2−ex3i2
0
= (e8−e8)−(1 −1)
= 0.
Therefore, the value of the double integral RRRexy dx dy over the region R
is 0.
24
