MATH 332 - ADVANCED CALCULUS
- Double integrals over rectangular and
general regions
Question Bank - Set 10
Liberty University
Question 1
Question
Evaluate the double integral RRR(x2+y2)dA, where Ris the region bounded
by the curves y=x2and y= 2 −x2.
Solution
Step 1: First, we need to find the limits of integration for xand y.
The curves y=x2and y= 2−x2intersect when x2= 2−x2, which simplifies
to 2x2= 2. Thus, x2= 1 and x=±1. So, the limits of integration for xare
−1≤x≤1.
Step 2: Next, we need to find the limits of integration for yin terms of x.
The curves bound Rbetween y=x2(bottom curve) and y= 2−x2(top curve).
Therefore, the limits of integration for yare x2≤y≤2−x2.
Step 3: Now, we can evaluate the double integral.
ZZR
(x2+y2)dA
=Z1
−1Z2−x2
x2
(x2+y2)dy dx
=Z1
−1x2y+1
3y32−x2
x2
dx
=Z1
−1x2(2 −x2) + 1
3(2 −x2)3−x2(x2)−1
3(x2)3dx
=Z1
−1
(2x2−x4+8
3−4x2+ 4x4−8
3−x4−1
3x6)dx
=Z1
−1
(−5x4+ 6x2−1
3x6)dx
=−5
5x5+6
3x3−1
21x71
−1
=−x5+ 2x3−1
21x71
−1
=−1+2−1
21 −(−1−2 + 1
21)
= 3 −2
21
=61
21.
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 61
21 .
Question 2
Question
Let Rbe the region bounded by the curves y=x2−1,y= 0,x= 2, and x= 3.
Calculate the double integral RRRy dA using polar coordinates.
Solution
Step 1: Sketch the region Rin the xy-plane. The region Ris bounded by the
curves y=x2−1,y= 0,x= 2, and x= 3. The curves y=x2−1and y= 0
intersect at x= 1 and x=−1, but we only consider the part of the parabola
in the first quadrant. Therefore, the region consists of the area between y= 0
and y=x2−1from x= 2 to x= 3.
2
Step 2: Determine the limits of integration for rand θin polar coordinates.
The region Rcorresponds to 2≤x≤3and 0≤y≤x2−1. Converting
the equations to polar coordinates gives 2≤rcos(θ)≤3and 0≤r2sin(θ)−
1≤r2cos2(θ)−1. Simplifying the inequalities, we get 2≤r≤3
cos(θ)and
0≤r2sin(θ)−1≤r2cos2(θ)−1.
Step 3: Set up the double integral in polar coordinates. The double integral
RRRy dA can be expressed as Rθ2
θ1Rr2(θ)
r1(θ)f(r, θ)r dr dθ, where f(r, θ) = r2sin(θ).
Step 4: Calculate the double integral. We have Rπ
3
0R
3
cos(θ)
2r2sin(θ)r dr dθ.
Integrating with respect to rfirst, we get Rπ
3
0hr4
4sin(θ)i3
cos(θ)
2dθ. This simplifies
to Rπ
3
081
4 cos4(θ)−24sin(θ)dθ.
Step 5: Finally, evaluate the integral. We compute Rπ
3
0
81
4 cos4(θ)−16 sin(θ)dθ.
This integral can be challenging to evaluate directly, so we recommend using
appropriate trigonometric identities to simplify the integrand before proceeding
with the calculation.
Question 3
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: To evaluate the double integral, we first need to determine the bounds
for xand y.
Step 2: The region Ris a square with sides of length 2 centered at the origin.
Therefore, the bounds for xand yare −1≤x≤1and −1≤y≤1.
Step 3: The given double integral can now be rewritten as R1
−1R1
−1ex2−y2dx dy.
Step 4: We can now evaluate the inner integral with respect to xfirst:
Z1
−1
ex2−y2dx =hex2−y2i1
−1=e1−y2−e1−y2
Step 5: Substituting the bounds and simplifying, we get:
Z1
−1
ex2−y2dx = 0
Step 6: Now, the double integral becomes R1
−10dy.
Step 7: Integrating this with respect to ygives:
Z1
−1
0dy = [0]1
−1= 0
3
Step 8: Therefore, the value of the double integral RRRex2−y2dA over the
given region Ris 0.
Question 4
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: Begin by sketching the region Rin the xy-plane. The region Ris a
rectangle with vertices at (1,1),(−1,1),(−1,−1), and (1,−1).
Step 2: Rewrite the given double integral as an iterated integral by setting
up the limits of integration. In this case, we integrate first with respect to y
and then with respect to x. So, we have:
ZZR
ex2−y2dA =Z1
−1Zx
−x
ex2−y2dy dx.
Step 3: Integrate with respect to y:
Z1
−1Zx
−x
ex2−y2dy dx =Z1
−1hyex2−y2ix
−xdx
=Z1
−1
(xex2−x2−(−xex2−(−x)2)) dx
=Z1
−1
(x−xe4x2)dx.
Step 4: Integrate the inner integral with respect to x:
Z1
−1
(x−xe4x2)dx =1
2x2−1
8e4x21
−1
=1
2−1
8e4−1
2−1
8e4
= 0.
Therefore, the value of the double integral RRRex2−y2dA over the given
rectangular region Ris 0.
4
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration by finding the points of intersection between the curves y=x2and
y= 2x.
Step 1: Find the points of intersection. Setting x2= 2x, we get
x2−2x= 0, which factorizes to give us x(x−2) = 0. Thus, x= 0 or x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the double integral. The given region Rcan be described
by 0≤x≤2and x2≤y≤2x. Therefore, the double integral can be rewritten
as:
Z2
0Z2x
x2
ex+ydy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Zex+ydy =ex+y
2x
x2
=e3x−ex2
Step 4: Evaluate the outer integral. Integrating with respect to x, we
get:
Z2
0
e3x−ex2dx =e6−1
3−e4−1
2
Therefore, the value of the given double integral is e6−1
3−e4−1
2.
Question 6
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, we express the region Rin terms of the limits of integration. The
region Ris bounded by the curves y=x2,y=x,x= 0, and x= 1. This means
that the limits of integration are 0≤x≤1and x2≤y≤x.
5
Step 2: Rewrite the double integral with the given limits of integration. The
double integral becomes:
ZZR
ex+ydA =Z1
0Zx
x2
ex+ydydx
Step 3: Integrate with respect to yfirst.
Zx
x2
ex+ydy =ex+yx
x2=e2x−ex2
Step 4: Substitute back into the double integral.
Z1
0
(e2x−ex2)dx
Step 5: Integrate with respect to x.
Z1
0
(e2x−ex2)dx =1
2e2x1
0−Z1
0
ex2dx =1
2e2−Z1
0
ex2dx
Step 6: Evaluate the remaining integral. The integral R1
0ex2dx cannot be
expressed in terms of elementary functions, so we leave it as is.
Therefore, the value of the given double integral is 1
2e2−R1
0ex2dx.
Question 7
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of integra-
tion for xand yby finding the points of intersection of the two curves y=x2
and y= 2x.
Step 1: Find the points of intersection. Setting the two curves equal
to each other gives:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies x= 0 or x= 2. Substituting these values back into either
equation gives the y-coordinates of the points of intersection: (0,0) and (2,4).
6
Step 2: Set up the double integral. The region Dis bounded by the
lines y=x2and y= 2x. Thus, the limits of integration for yare x2and 2x,
and for xthey are 0and 2. The double integral is then:
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.Integrating with respect to yfirst,
we get:
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
Step 4: Integrate with respect to x.Now, we integrate with respect to
x:
Z2
0
(2x3+8x3
3−x4−x6
3)dx
=2x4
4+8x4
3·4−x5
5−x7
3·7
2
0
=16
4+64
12 −32
5−128
21
= 4 + 16
3−32
5−128
21
=642
105
Step 5: Conclusion. Therefore, the value of the double integral RRD(x2+
y2)dA over the region Dbounded by the curves y=x2and y= 2xis 642
105 .
Question 8
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 4.
7
Solution
To evaluate the given double integral, we first need to determine the bounds of
integration by finding the points of intersection of the curves y=x2and y= 4.
This will help us determine the limits of integration for the inner integral.
Step 1: Find the points of intersection. Setting x2= 4, we find the
points of intersection to be when y= 4:
x2= 4
x=±2
So the points of intersection are (2,4) and (−2,4).
Step 2: Set up the integral. Given that Ris bounded by y=x2and
y= 4, the limits of integration are:
−2≤x≤2
x2≤y≤4
Thus, the double integral becomes:
Z2
−2Z4
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Z4
x2
ex2+y2dy =ex2+x4−ex2
Step 4: Evaluate the outer integral. Integrating the result of the inner
integral from −2to 2, we get:
Z2
−2ex2+16 −ex2dx =Z2
−2
ex2+16 dx −Z2
−2
ex2dx
We can solve these integrals using techniques of integration, such as substi-
tution or recognizing standard integral forms.
Therefore, the value of the given double integral can be found by evaluating
the above expression.
Question 9
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y= 2x, and x= 1.
8
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yover the region R.
Step 1: Determining the limits of integration for y:The region R
is bounded by the lines y=x,y= 2x, and x= 1. We start by finding the y
limits. Along the line y=x,yvaries from xto 2x. Along the line x= 1,y
varies from 1to 2. Thus, the ylimits are from xto 2x.
Step 2: Determining the limits of integration for x:Next, we deter-
mine the limits for x. Since xvaries from 1to the intersection point of the lines
y=xand y= 2x, we need to find this intersection point. Setting x= 2x, we
find that x= 0. Therefore, xvaries from 1to 0which is the region where the
lines y=xand y= 2xintersect.
Step 3: Setting up and evaluating the double integral: The given
double integral can now be expressed as:
ZZR
ex2−y2dA =Z1
0Z2x
x
ex2−y2dy dx
We can now evaluate the inner integral with respect to y:
Z2x
x
ex2−y2dy =−ex2−4x2+ex2−x2=−e−3x2+ex2
Finally, we evaluate the outer integral with respect to x:
Z1
0
(−e−3x2+ex2)dx ="−e−3x2
6+ex2
2#1
0
=1
2−1
6=1
3
Therefore, the value of the double integral RRRex2−y2dA over the region R
is 1
3.
Question 10
Question
Evaluate the double integral RRDex2−y2dx dy, where Dis the region bounded
by the curves y=x,y= 3x, and x= 1.
Solution
Step 1: Find the intersection points of the curves to determine the bounds
of integration. The intersection points are obtained by solving the system of
equations:
y=xand y= 3x
9
This gives x= 1 and y= 1. So, the bounds of integration are 1≤x≤1and
x≤y≤3x.
Step 2: Setup the double integral. The given double integral can be rewritten
as:
Z1
1Z3x
x
ex2−y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we get:
Z3x
x
ex2−y2dy =−1
2ex2−y2
3x
x
=−1
2(ex2−(3x)2−ex2−x2)
=−1
2(ex2−9x2−1)
=−1
2(e−8x2−1)
Step 4: Evaluate the outer integral.
Z1
1−1
2(e−8x2−1) dx =−1
2Z1
1
(e−8x2−1) dx
=−1
2 e−8x2
−16 −x!
1
1
=−1
21
−16 −1−1
−16 −1
=−1
2−1
16 +1
16
= 0
Therefore, the value of the double integral is 0.
Question 11
Question
Evaluate the double integral RRRx2y dA where Ris the region bounded by the
curve y=x2, the x-axis, and the lines x= 0 and x= 2.
Solution
We begin by sketching the region Rdefined by the curve y=x2, the x-axis,
and the lines x= 0 and x= 2.
10
xy2; [onlymarks, mark =∗, color =red]coordinates(0,0)(2,0);
y=x2
x= 2
x
y
The region Ris a triangular region in the first quadrant, bounded by x= 0,
x= 2, and y=x2.
To evaluate the double integral RRRx2y dA, we set up the integral in terms
of xand y:
ZZR
x2y dA =Z2
0Zx2
0
x2y dy dx
Step 1: Integrate with respect to y:
Zx2
0
x2y dy =x2y2
2x2
0
=x2·(x2)2
2=x6
2
Step 2: Integrate with respect to x:
Z2
0
x6
2dx =1
2Z2
0
x6dx =1
2x7
72
0
=27
2·7=64
7
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
7.
Question 12
Question
Evaluate the double integral RRRex+ydA over the region Rbounded by y=x2,
y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y.
• The region Ris bounded by y=x2,y= 2x,x= 0, and x= 1.
• We can express the limits of integration for xin terms of yas 0≤x≤√y
(from x= 0 to x=√y) and √y≤x≤y
2(from x=√yto x=y
2).
11
• The limits of integration for yare 0≤y≤1.
Step 2: Set up the double integral.
• The double integral can be expressed as R1
0R√y
0ex+ydx dy+R1
0Ry
2
√yex+ydx dy.
Step 3: Evaluate the inner integral with respect to xfor the first region.
• Simplify R√y
0ex+ydx to get eyR√y
0exdx.
• Integrate to get ey[ex]
√y
0=ey(e√y−1).
Step 4: Evaluate the inner integral with respect to xfor the second region.
• Simplify Ry
2
√yex+ydx to get eyRy
2
√yexdx.
• Integrate to get ey[ex]
y
2
√y=eyey
2−e√y.
Step 5: Evaluate the outer integral with respect to y.
• Simplify the expression and integrate to get R1
0ey(e√y−1) dy+R1
0eyey
2−e√ydy.
• Integrate to get (e−1) −2√e+5
2−2e
3.
Hence, the value of the double integral RRRex+ydA over the region Ris
−2
3e−2
3√e+7
2.
Question 13
Question
Evaluate the double integral RRRxy dA, where Ris the region in the first quad-
rant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to find the limits of integra-
tion for xand yby determining the region R.
Step 1: Determine the limits of integration for xand y
The curves y=x2and y= 2xintersect at the point where x2= 2x. Solving
this equation, we find x2−2x= 0, which gives x(x−2) = 0. This implies x= 0
or x= 2.
Since Ris in the first quadrant, we have 0≤x≤2. The lower curve is
y=x2while the upper curve is y= 2x.
Therefore, the limits of integration are:
0≤x≤2and x2≤y≤2x
12
Step 2: Evaluate the double integral
The given double integral is:
ZZR
xy dA =Z2
0Z2x
x2
xy dy dx
We first integrate with respect to y:
Z2x
x2
xy dy =1
2xy22x
x2
=1
2(2x)(4x)−1
2(x)(x2) = 4x2−1
2x3
Now we integrate the result with respect to x:
Z2
0
4x2−1
2x3dx =4
3x3−1
8x42
0
=32
3−4 = 20
3
Therefore, the value of the double integral RRRxy dA over the region Ris
20
3.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=xand y=√x.
Solution
Step 1: We first need to determine the limits of integration for xand y. The
region Ris defined by y=xand y=√xin the first quadrant, so the limits of
integration are:
0≤x≤1
x≤y≤√x
Step 2: Rewrite the given double integral in terms of xand y:
ZZR
ex+ydA =Z1
0Z√x
x
ex+ydy dx
Step 3: Integrate with respect to yfirst:
Z√x
x
ex+ydy =e2x−ex
Step 4: Substitute back the result from Step 3 into the double integral:
Z1
0
e2x−exdx
13
Step 5: Integrate with respect to x:
Z1
0
e2x−exdx =e2x
2−ex1
0
=e2
2−e+ 1
Therefore, the value of the double integral RRRex+ydA over the region Ris
e2
2−e+ 1.
Question 15
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 1.
Solution
Step 1: Determine the limits of integration for x. Since y=x2and y= 1 bound
the region R, set the equations equal to find the x-values where they intersect:
x2= 1
x=±1
Thus, the limits of integration for xare −1to 1.
Step 2: Determine the limits of integration for y. The region Ris bounded
below by the curve y=x2and above by the line y= 1. So, the limits of
integration for yare x2to 1.
Step 3: Rewrite the integral and evaluate. The double integral becomes:
Z1
−1Z1
x2
(x2+y)dy dx
Now, evaluate the integral:
Z1
−1Z1
x2
(x2+y)dy dx =Z1
−1x2y+y2
21
x2
dx
=Z1
−1x2·1 + 1
2−x4−x4
2dx
=Z1
−1x2−x4−3x4
2+1
2dx
=x3
3−x5
5−3x5
10 +x
21
−1
14
=1
3−1
5−3
10 +1
2−−1
3+1
5+3
10 −1
2
=10
30 −6
30 =4
30 =2
15
Therefore, the value of the double integral RRR(x2+y)dA over the region R
is 2
15 .
Question 16
Question
Evaluate the double integral RRR(x2+xy)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2and y= 2x. We need to find the x-values at
which these curves intersect. Setting x2= 2xgives us x2−2x= 0, which factors
as x(x−2) = 0. Thus, the points of intersection are x= 0 and x= 2. Therefore,
the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a given xwithin
the region R, the lower bound for yis x2and the upper bound is 2x. Thus, the
limits for yare x2≤y≤2x.
Step 3: Evaluate the double integral: We can now express the given
double integral as:
ZZR
(x2+xy)dA =Z2
0Z2x
x2
(x2+xy)dy dx
Now we compute the integral with respect to yfirst, then x:
Z2x
x2
(x2+xy)dy =x2y+1
2y2
2x
x2
=x2(2x) + 1
2(2x)2−x2(x2)−1
2(x2)2
= 2x3+ 2x2−x4−1
2x4
= 2x3+ 2x2−3
2x4
15
Now, we compute the integral with respect to xas follows:
Z2
0
(2x3+ 2x2−3
2x4)dx =1
2x4+2
3x3−3
8x5
2
0
=1
2(2)4+2
3(2)3−3
8(2)5−1
2(0)4+2
3(0)3−3
8(0)5
=8 + 16
3−24−0
=8
3
Therefore, the value of the double integral RRR(x2+xy)dA over the region
Ris 8
3.
Question 17
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by the curves y=x2and y= 2xin the first quadrant. To find the limits of
integration for x, we need to determine the x-values at which these two curves
intersect. Setting x2= 2xgives x2−2x=x(x−2) = 0, which implies x= 0
and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Find the limits of integration for y:For each xin the interval
[0,2],yvaries between the curves y=x2and y= 2x. So, the limits of integration
for yare x2≤y≤2x.
Now, the integral becomes:
Z2
0Z2x
x2
ex2+y2dy dx
16
Step 3: Evaluate the integral:
Z2
0Z2x
x2
ex2+y2dy dx
=Z2
0hex2+y2iy=2x
y=x2dx
=Z2
0ex2+(2x)2−ex2+x4dx
=Z2
0e5x2−ex2(1+x2)dx
=Z2
0
e5x2dx −Z2
0
ex2(1+x2)dx
=1
2√5e5x22
0−Z2
0
ex2+x4dx
=1
2√5e20 −1−1
2√πex2erf(x)2
0
=1
2√5e20 −1−1
2√π(e16erf(2) −erf(0))
=1
2√5e20 −1−1
2√π(e16erf(2) −0)
=1
2√5e20 −1−1
2√πe16erf(2)
Therefore, the value of the double integral RRRex2+y2dA over Ris 1
2√5e20 −1−
1
2√πe16erf(2).
Question 18
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the curves y=x2,y= 0,x= 1, and x= 2.
Solution
Step 1: We first sketch the region Rto determine the limits of integration. The
region Ris bounded by y=x2,y= 0,x= 1, and x= 2. It is the region
enclosed between the parabola y=x2and the y-axis, and between x= 1 and
x= 2.
Step 2: To set up the double integral, we express xand yin terms of the
bounds of integration. The bounds for xare 1≤x≤2, and the bounds for y
are 0≤y≤x2.
17
Step 3: Now, we can express the given double integral in terms of xand y.
ZZR
ex2−y2dA =Z2
1Zx2
0
ex2−y2dy dx
Step 4: Evaluate the inner integral with respect to y.
Zx2
0
ex2−y2dy =h−ex2−y2ix2
0=−(ex2−x4−ex2)
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x.
Z2
1−(ex2−x4−ex2)dx
Step 6: Integrate the terms in the outer integral.
=−Z2
1
ex2−x4dx +Z2
1
ex2dx
Step 7: We may need to use techniques like substitution or manipulation to
evaluate the integrals in Step 6. The final result of the double integral requires
careful algebraic manipulation and integration.
Question 19
Question
Evaluate the double integral R RR(3x−y)dA, where Ris the region bounded by
the curves y=x2,y= 8 −x2,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Rcan be
described as 0≤x≤2and x2≤y≤8−x2. Thus, the limits of integration are
0≤x≤2and x2≤y≤8−x2.
Step 2: Rewrite the integrand in terms of xand y. The integrand (3x−y)
remains unchanged.
Step 3: Set up the double integral. The double integral can be set up as
follows:
Z2
0Z8−x2
x2
(3x−y)dy dx
Step 4: Integrate with respect to y. Integrating (3x−y)with respect to y
gives:
Z2
0
[3xy −y2
2]8−x2
x2dx
18
Step 5: Evaluate the integral. Substitute the limits of integration and sim-
plify the expression:
Z2
0
[(24x−(8 −x2)2/2) −(3x(x2)−x4/2)] dx
Step 6: Continue simplifying. After simplifying the expression, integrate
over [0,2]:
Z2
0
(24x−4−4x2+x4−3x3+x4/2) dx
Step 7: Integrate with respect to x. Integrating the expression above yields
the final answer:
12x2−4
3x3−x4+1
3x5−3
4x4+1
8x52
0
Step 8: Calculate the final answer. By substituting the limits of integration
into the expression and calculating, we find the final answer for the double
integral to be 560
3or approximately 186.67.
Question 20
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2and y=x.
Solution
Step 1: To evaluate the double integral, we need to find the limits of integration
for both xand y.
Step 2: The region Rcan be described by y=x2and y=x, which intersect
at the points (0,0) and (1,1). Therefore, the limits of integration for xare
0≤x≤1.
Step 3: For a given xin the interval [0,1], the corresponding yvalues are
x≤y≤x2. Hence, the limits of integration for yare x≤y≤x2.
Step 4: Now, we can rewrite the double integral as follows:
ZZR
(3x2+ 2y)dA =Z1
0Zx2
x
(3x2+ 2y)dy dx
Step 5: Compute the inner integral with respect to y:
Zx2
x
(3x2+ 2y)dy =3x2y+y2x2
x= 3x4+x4−(3x3+x) = 4x4−3x3−x
19
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(4x4−3x3−x)dx =4
5x5−3
4x4−1
2x21
0
=4
5−3
4−1
2=3
20
Step 7: Therefore, the value of the double integral RRR(3x2+ 2y)dA over
the region Ris 3
20 .
Question 21
Question
Evaluate the double integral RRRex
y2dA over the region Rbounded by the lines
y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determining limits of integration for x:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. We find the
intersection points of these lines to determine the limits of integration for x.
The intersection of y=xand y= 2xgives x=y= 0,x=y= 0 is outside
of R.
The intersection of y=xand x= 1 gives x=y= 1.
So the limits of integration for xare from x= 1 to x= 1.
Step 2: Determining limits of integration for y:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. Hence, y
varies from y=xto y= 2x.
Now, we can rewrite the double integral with the limits of integration for x
and y:
ZZR
ex
y2dA =Z1
1Z2x
x
ex
y2dy dx
Step 3: Evaluating the double integral:
Let’s evaluate the inner integral with respect to yfirst:
Z2x
x
ex
y2dy =−1
2ex
y22x
x
=−1
2e2−e1
4
Now, we can substitute this back into the double integral:
Z1
1−1
2e2−e1
4dx
20
=−1
2e2−e1
4Z1
1
dx
=−1
2e2−e1
4(1 −1)
= 0
Therefore, the value of the double integral RRRex
y2dA over the region Ris
0.
Question 22
Question
Let Rbe the region bounded by the curves y=x2and y=x4. Evaluate the
double integral RRRxy dA.
Solution
Step 1: Find the limits of integration for xand y. The region Ris bounded by
the curves y=x2and y=x4. To find the limits of integration for x, we need
to solve for the x-coordinates of the points of intersection of y=x2and y=x4.
Setting x2=x4, we get x4−x2= 0 ⇒x2(x2−1) = 0. This gives us x= 0
and x=±1as the x-coordinates of the points of intersection.
Therefore, the limits of integration for xare −1≤x≤1. For each xin
this range, yvaries between x2and x4. So, the limits of integration for yare
x2≤y≤x4.
Step 2: Set up the double integral. The double integral RRRxy dA over the
region Rcan be expressed as:
ZZR
xy dA =Z1
−1Zx4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y. Integrating xy with
respect to ywhile treating xas a constant, we get:
Zxy dy =1
2xy2
x4
x2=1
2x(x8−x4) = 1
2x9−1
2x5
Step 4: Evaluate the outer integral with respect to x. Substitute the limits
of integration and integrate 1
2x9−1
2x5with respect to x:
Z1
−11
2x9−1
2x5dx =1
21
10 −1
6=−1
15
Step 5: Final Answer Thus, the value of the double integral RRRxy dA over
the region Ris −1
15 .
21
Question 23
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y= 2x2and y= 3x, and xranges from 0 to 1.
Solution
To evaluate the given double integral over the region R, we first need to deter-
mine the limits of integration for xand ybased on the given region R.
Step 1: Determine the limits of integration for x. The curves y= 2x2and
y= 3xintersect at the point where 2x2= 3x. Solving for x, we get:
2x2= 3x
2x2−3x= 0
x(2x−3) = 0
x= 0 or x=3
2
Since xranges from 0 to 1, we will integrate xfrom 0 to 3
2.
Step 2: Determine the limits of integration for y. The region Ris bounded
by the curves y= 2x2and y= 3x. This means yranges from 2x2to 3x.
Therefore, the limits of integration for yare from 2x2to 3x.
Step 3: Set up the double integral. The given double integral becomes:
Z3
2
0Z3x
2x2
ex+ydy dx
Step 4: Evaluate the double integral.
Z3
2
0ex+y3x
2x2dx
Z3
2
0e4x−e3xdx
Integrating with respect to x:
1
4e4x−1
3e3x3
2
0
=1
4e6−1
3e9−1
4−1
3
=1
4e6−1
3e9−1
4+1
3
=1
4e6−1
4−1
3e9+1
3
22
Question 24
Question
Evaluate the double integral RRR
x−y
x+ydA, where Ris the region bounded by the
curves y=x,y= 2x, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we need to determine the intersection points of the curves
y=x,y= 2x, and x= 1. Setting y=xand y= 2xequal to each other gives
us x= 0. Setting y= 2xand x= 1 equal to each other gives us x= 1.
Therefore, the limits of integration for xare from 0 to 1.
To find the limits of integration for y, we need to determine the bounds in
terms of y. From the given region bounds, y=xand y= 2x, the limits of yare
from y=xto y= 2x.
Step 2: Rewrite the integral using the limits of integration. The double
integral can now be rewritten as:
Z1
0Z2x
x
x−y
x+ydy dx
Step 3: Evaluate the inner integral with respect to y. Integrating with
respect to y, we get:
Z2x
x
x−y
x+ydy =xy −y2
22x
x
=x(2x)−(2x)2
2−x(x)+x2
2= 2x2−2x2−x2+x2
2=−3
2x2
Step 4: Evaluate the outer integral with respect to x. Integrating the result
from the inner integral over the limits of x, we get:
Z1
0−3
2x2dx =−3
2x3
31
0
=−1
2
Therefore, the value of the double integral is −1
2.
Question 25
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y=
4x−x2. Find the volume of the solid that lies under the surface z= 12−x2−y2
and above the region R.
23
Solution
Step 1: Find the bounds for the double integral over the region R. The region
Ris bounded by the curves y=x2and y= 4x−x2. To find the bounds for x,
we set the two functions equal to each other:
x2= 4x−x2
2x2−4x= 0
2x(x−2) = 0
This implies x= 0 and x= 2. Therefore, the bounds for xare 0≤x≤2.
To find the bounds for y, we subtract the functions: 4x−x2−x2= 4x−2x2.
This tells us that the bounds for yare x2≤y≤4x−x2.
Step 2: Set up the double integral. The volume under the surface z=
12 −x2−y2can be expressed as:
ZZ
R
(12 −x2−y2)dA
where dA =dx dy.
Step 3: Evaluate the double integral. Now, we evaluate the double integral
over the region R:
Z2
0Z4x−x2
x2
(12 −x2−y2)dy dx
=Z2
012y−x2y−1
3y34x−x2
x2
dx
=Z2
012(4x−x2)−x2(4x−x2)−1
3(4x−x2)3−12x2+x4−1
3x6dx
=Z2
048x−12x2−4x3+x4−4
3x2(4x−x2) + 1
3(4x−x2)3dx
After integrating and simplifying, we get the volume of the solid that lies
under the surface and above the region R.
24
ZZR
(x2+y2)dA
=Z1
−1Z2−x2
x2
(x2+y2)dy dx
=Z1
−1x2y+1
3y32−x2
x2
dx
=Z1
−1x2(2 −x2) + 1
3(2 −x2)3−x2(x2)−1
3(x2)3dx
=Z1
−1
(2x2−x4+8
3−4x2+ 4x4−8
3−x4−1
3x6)dx
=Z1
−1
(−5x4+ 6x2−1
3x6)dx
=−5
5x5+6
3x3−1
21x71
−1
=−x5+ 2x3−1
21x71
−1
=−1+2−1
21 −(−1−2 + 1
21)
= 3 −2
21
=61
21.
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 61
21 .
Question 2
Question
Let Rbe the region bounded by the curves y=x2−1,y= 0,x= 2, and x= 3.
Calculate the double integral RRRy dA using polar coordinates.
Solution
Step 1: Sketch the region Rin the xy-plane. The region Ris bounded by the
curves y=x2−1,y= 0,x= 2, and x= 3. The curves y=x2−1and y= 0
intersect at x= 1 and x=−1, but we only consider the part of the parabola
in the first quadrant. Therefore, the region consists of the area between y= 0
and y=x2−1from x= 2 to x= 3.
2
Step 2: Determine the limits of integration for rand θin polar coordinates.
The region Rcorresponds to 2≤x≤3and 0≤y≤x2−1. Converting
the equations to polar coordinates gives 2≤rcos(θ)≤3and 0≤r2sin(θ)−
1≤r2cos2(θ)−1. Simplifying the inequalities, we get 2≤r≤3
cos(θ)and
0≤r2sin(θ)−1≤r2cos2(θ)−1.
Step 3: Set up the double integral in polar coordinates. The double integral
RRRy dA can be expressed as Rθ2
θ1Rr2(θ)
r1(θ)f(r, θ)r dr dθ, where f(r, θ) = r2sin(θ).
Step 4: Calculate the double integral. We have Rπ
3
0R
3
cos(θ)
2r2sin(θ)r dr dθ.
Integrating with respect to rfirst, we get Rπ
3
0hr4
4sin(θ)i3
cos(θ)
2dθ. This simplifies
to Rπ
3
081
4 cos4(θ)−24sin(θ)dθ.
Step 5: Finally, evaluate the integral. We compute Rπ
3
0
81
4 cos4(θ)−16 sin(θ)dθ.
This integral can be challenging to evaluate directly, so we recommend using
appropriate trigonometric identities to simplify the integrand before proceeding
with the calculation.
Question 3
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: To evaluate the double integral, we first need to determine the bounds
for xand y.
Step 2: The region Ris a square with sides of length 2 centered at the origin.
Therefore, the bounds for xand yare −1≤x≤1and −1≤y≤1.
Step 3: The given double integral can now be rewritten as R1
−1R1
−1ex2−y2dx dy.
Step 4: We can now evaluate the inner integral with respect to xfirst:
Z1
−1
ex2−y2dx =hex2−y2i1
−1=e1−y2−e1−y2
Step 5: Substituting the bounds and simplifying, we get:
Z1
−1
ex2−y2dx = 0
Step 6: Now, the double integral becomes R1
−10dy.
Step 7: Integrating this with respect to ygives:
Z1
−1
0dy = [0]1
−1= 0
3
Step 8: Therefore, the value of the double integral RRRex2−y2dA over the
given region Ris 0.
Question 4
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: Begin by sketching the region Rin the xy-plane. The region Ris a
rectangle with vertices at (1,1),(−1,1),(−1,−1), and (1,−1).
Step 2: Rewrite the given double integral as an iterated integral by setting
up the limits of integration. In this case, we integrate first with respect to y
and then with respect to x. So, we have:
ZZR
ex2−y2dA =Z1
−1Zx
−x
ex2−y2dy dx.
Step 3: Integrate with respect to y:
Z1
−1Zx
−x
ex2−y2dy dx =Z1
−1hyex2−y2ix
−xdx
=Z1
−1
(xex2−x2−(−xex2−(−x)2)) dx
=Z1
−1
(x−xe4x2)dx.
Step 4: Integrate the inner integral with respect to x:
Z1
−1
(x−xe4x2)dx =1
2x2−1
8e4x21
−1
=1
2−1
8e4−1
2−1
8e4
= 0.
Therefore, the value of the double integral RRRex2−y2dA over the given
rectangular region Ris 0.
4
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration by finding the points of intersection between the curves y=x2and
y= 2x.
Step 1: Find the points of intersection. Setting x2= 2x, we get
x2−2x= 0, which factorizes to give us x(x−2) = 0. Thus, x= 0 or x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the double integral. The given region Rcan be described
by 0≤x≤2and x2≤y≤2x. Therefore, the double integral can be rewritten
as:
Z2
0Z2x
x2
ex+ydy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Zex+ydy =ex+y
2x
x2
=e3x−ex2
Step 4: Evaluate the outer integral. Integrating with respect to x, we
get:
Z2
0
e3x−ex2dx =e6−1
3−e4−1
2
Therefore, the value of the given double integral is e6−1
3−e4−1
2.
Question 6
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, we express the region Rin terms of the limits of integration. The
region Ris bounded by the curves y=x2,y=x,x= 0, and x= 1. This means
that the limits of integration are 0≤x≤1and x2≤y≤x.
5
Step 2: Rewrite the double integral with the given limits of integration. The
double integral becomes:
ZZR
ex+ydA =Z1
0Zx
x2
ex+ydydx
Step 3: Integrate with respect to yfirst.
Zx
x2
ex+ydy =ex+yx
x2=e2x−ex2
Step 4: Substitute back into the double integral.
Z1
0
(e2x−ex2)dx
Step 5: Integrate with respect to x.
Z1
0
(e2x−ex2)dx =1
2e2x1
0−Z1
0
ex2dx =1
2e2−Z1
0
ex2dx
Step 6: Evaluate the remaining integral. The integral R1
0ex2dx cannot be
expressed in terms of elementary functions, so we leave it as is.
Therefore, the value of the given double integral is 1
2e2−R1
0ex2dx.
Question 7
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of integra-
tion for xand yby finding the points of intersection of the two curves y=x2
and y= 2x.
Step 1: Find the points of intersection. Setting the two curves equal
to each other gives:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies x= 0 or x= 2. Substituting these values back into either
equation gives the y-coordinates of the points of intersection: (0,0) and (2,4).
6
Step 2: Set up the double integral. The region Dis bounded by the
lines y=x2and y= 2x. Thus, the limits of integration for yare x2and 2x,
and for xthey are 0and 2. The double integral is then:
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.Integrating with respect to yfirst,
we get:
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
Step 4: Integrate with respect to x.Now, we integrate with respect to
x:
Z2
0
(2x3+8x3
3−x4−x6
3)dx
=2x4
4+8x4
3·4−x5
5−x7
3·7
2
0
=16
4+64
12 −32
5−128
21
= 4 + 16
3−32
5−128
21
=642
105
Step 5: Conclusion. Therefore, the value of the double integral RRD(x2+
y2)dA over the region Dbounded by the curves y=x2and y= 2xis 642
105 .
Question 8
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 4.
7
Solution
To evaluate the given double integral, we first need to determine the bounds of
integration by finding the points of intersection of the curves y=x2and y= 4.
This will help us determine the limits of integration for the inner integral.
Step 1: Find the points of intersection. Setting x2= 4, we find the
points of intersection to be when y= 4:
x2= 4
x=±2
So the points of intersection are (2,4) and (−2,4).
Step 2: Set up the integral. Given that Ris bounded by y=x2and
y= 4, the limits of integration are:
−2≤x≤2
x2≤y≤4
Thus, the double integral becomes:
Z2
−2Z4
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Z4
x2
ex2+y2dy =ex2+x4−ex2
Step 4: Evaluate the outer integral. Integrating the result of the inner
integral from −2to 2, we get:
Z2
−2ex2+16 −ex2dx =Z2
−2
ex2+16 dx −Z2
−2
ex2dx
We can solve these integrals using techniques of integration, such as substi-
tution or recognizing standard integral forms.
Therefore, the value of the given double integral can be found by evaluating
the above expression.
Question 9
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y= 2x, and x= 1.
8
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yover the region R.
Step 1: Determining the limits of integration for y:The region R
is bounded by the lines y=x,y= 2x, and x= 1. We start by finding the y
limits. Along the line y=x,yvaries from xto 2x. Along the line x= 1,y
varies from 1to 2. Thus, the ylimits are from xto 2x.
Step 2: Determining the limits of integration for x:Next, we deter-
mine the limits for x. Since xvaries from 1to the intersection point of the lines
y=xand y= 2x, we need to find this intersection point. Setting x= 2x, we
find that x= 0. Therefore, xvaries from 1to 0which is the region where the
lines y=xand y= 2xintersect.
Step 3: Setting up and evaluating the double integral: The given
double integral can now be expressed as:
ZZR
ex2−y2dA =Z1
0Z2x
x
ex2−y2dy dx
We can now evaluate the inner integral with respect to y:
Z2x
x
ex2−y2dy =−ex2−4x2+ex2−x2=−e−3x2+ex2
Finally, we evaluate the outer integral with respect to x:
Z1
0
(−e−3x2+ex2)dx ="−e−3x2
6+ex2
2#1
0
=1
2−1
6=1
3
Therefore, the value of the double integral RRRex2−y2dA over the region R
is 1
3.
Question 10
Question
Evaluate the double integral RRDex2−y2dx dy, where Dis the region bounded
by the curves y=x,y= 3x, and x= 1.
Solution
Step 1: Find the intersection points of the curves to determine the bounds
of integration. The intersection points are obtained by solving the system of
equations:
y=xand y= 3x
9
This gives x= 1 and y= 1. So, the bounds of integration are 1≤x≤1and
x≤y≤3x.
Step 2: Setup the double integral. The given double integral can be rewritten
as:
Z1
1Z3x
x
ex2−y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we get:
Z3x
x
ex2−y2dy =−1
2ex2−y2
3x
x
=−1
2(ex2−(3x)2−ex2−x2)
=−1
2(ex2−9x2−1)
=−1
2(e−8x2−1)
Step 4: Evaluate the outer integral.
Z1
1−1
2(e−8x2−1) dx =−1
2Z1
1
(e−8x2−1) dx
=−1
2 e−8x2
−16 −x!
1
1
=−1
21
−16 −1−1
−16 −1
=−1
2−1
16 +1
16
= 0
Therefore, the value of the double integral is 0.
Question 11
Question
Evaluate the double integral RRRx2y dA where Ris the region bounded by the
curve y=x2, the x-axis, and the lines x= 0 and x= 2.
Solution
We begin by sketching the region Rdefined by the curve y=x2, the x-axis,
and the lines x= 0 and x= 2.
10
xy2; [onlymarks, mark =∗, color =red]coordinates(0,0)(2,0);
y=x2
x= 2
x
y
The region Ris a triangular region in the first quadrant, bounded by x= 0,
x= 2, and y=x2.
To evaluate the double integral RRRx2y dA, we set up the integral in terms
of xand y:
ZZR
x2y dA =Z2
0Zx2
0
x2y dy dx
Step 1: Integrate with respect to y:
Zx2
0
x2y dy =x2y2
2x2
0
=x2·(x2)2
2=x6
2
Step 2: Integrate with respect to x:
Z2
0
x6
2dx =1
2Z2
0
x6dx =1
2x7
72
0
=27
2·7=64
7
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
7.
Question 12
Question
Evaluate the double integral RRRex+ydA over the region Rbounded by y=x2,
y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y.
• The region Ris bounded by y=x2,y= 2x,x= 0, and x= 1.
• We can express the limits of integration for xin terms of yas 0≤x≤√y
(from x= 0 to x=√y) and √y≤x≤y
2(from x=√yto x=y
2).
11
• The limits of integration for yare 0≤y≤1.
Step 2: Set up the double integral.
• The double integral can be expressed as R1
0R√y
0ex+ydx dy+R1
0Ry
2
√yex+ydx dy.
Step 3: Evaluate the inner integral with respect to xfor the first region.
• Simplify R√y
0ex+ydx to get eyR√y
0exdx.
• Integrate to get ey[ex]
√y
0=ey(e√y−1).
Step 4: Evaluate the inner integral with respect to xfor the second region.
• Simplify Ry
2
√yex+ydx to get eyRy
2
√yexdx.
• Integrate to get ey[ex]
y
2
√y=eyey
2−e√y.
Step 5: Evaluate the outer integral with respect to y.
• Simplify the expression and integrate to get R1
0ey(e√y−1) dy+R1
0eyey
2−e√ydy.
• Integrate to get (e−1) −2√e+5
2−2e
3.
Hence, the value of the double integral RRRex+ydA over the region Ris
−2
3e−2
3√e+7
2.
Question 13
Question
Evaluate the double integral RRRxy dA, where Ris the region in the first quad-
rant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to find the limits of integra-
tion for xand yby determining the region R.
Step 1: Determine the limits of integration for xand y
The curves y=x2and y= 2xintersect at the point where x2= 2x. Solving
this equation, we find x2−2x= 0, which gives x(x−2) = 0. This implies x= 0
or x= 2.
Since Ris in the first quadrant, we have 0≤x≤2. The lower curve is
y=x2while the upper curve is y= 2x.
Therefore, the limits of integration are:
0≤x≤2and x2≤y≤2x
12
Step 2: Evaluate the double integral
The given double integral is:
ZZR
xy dA =Z2
0Z2x
x2
xy dy dx
We first integrate with respect to y:
Z2x
x2
xy dy =1
2xy22x
x2
=1
2(2x)(4x)−1
2(x)(x2) = 4x2−1
2x3
Now we integrate the result with respect to x:
Z2
0
4x2−1
2x3dx =4
3x3−1
8x42
0
=32
3−4 = 20
3
Therefore, the value of the double integral RRRxy dA over the region Ris
20
3.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=xand y=√x.
Solution
Step 1: We first need to determine the limits of integration for xand y. The
region Ris defined by y=xand y=√xin the first quadrant, so the limits of
integration are:
0≤x≤1
x≤y≤√x
Step 2: Rewrite the given double integral in terms of xand y:
ZZR
ex+ydA =Z1
0Z√x
x
ex+ydy dx
Step 3: Integrate with respect to yfirst:
Z√x
x
ex+ydy =e2x−ex
Step 4: Substitute back the result from Step 3 into the double integral:
Z1
0
e2x−exdx
13
Step 5: Integrate with respect to x:
Z1
0
e2x−exdx =e2x
2−ex1
0
=e2
2−e+ 1
Therefore, the value of the double integral RRRex+ydA over the region Ris
e2
2−e+ 1.
Question 15
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 1.
Solution
Step 1: Determine the limits of integration for x. Since y=x2and y= 1 bound
the region R, set the equations equal to find the x-values where they intersect:
x2= 1
x=±1
Thus, the limits of integration for xare −1to 1.
Step 2: Determine the limits of integration for y. The region Ris bounded
below by the curve y=x2and above by the line y= 1. So, the limits of
integration for yare x2to 1.
Step 3: Rewrite the integral and evaluate. The double integral becomes:
Z1
−1Z1
x2
(x2+y)dy dx
Now, evaluate the integral:
Z1
−1Z1
x2
(x2+y)dy dx =Z1
−1x2y+y2
21
x2
dx
=Z1
−1x2·1 + 1
2−x4−x4
2dx
=Z1
−1x2−x4−3x4
2+1
2dx
=x3
3−x5
5−3x5
10 +x
21
−1
14
=1
3−1
5−3
10 +1
2−−1
3+1
5+3
10 −1
2
=10
30 −6
30 =4
30 =2
15
Therefore, the value of the double integral RRR(x2+y)dA over the region R
is 2
15 .
Question 16
Question
Evaluate the double integral RRR(x2+xy)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2and y= 2x. We need to find the x-values at
which these curves intersect. Setting x2= 2xgives us x2−2x= 0, which factors
as x(x−2) = 0. Thus, the points of intersection are x= 0 and x= 2. Therefore,
the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a given xwithin
the region R, the lower bound for yis x2and the upper bound is 2x. Thus, the
limits for yare x2≤y≤2x.
Step 3: Evaluate the double integral: We can now express the given
double integral as:
ZZR
(x2+xy)dA =Z2
0Z2x
x2
(x2+xy)dy dx
Now we compute the integral with respect to yfirst, then x:
Z2x
x2
(x2+xy)dy =x2y+1
2y2
2x
x2
=x2(2x) + 1
2(2x)2−x2(x2)−1
2(x2)2
= 2x3+ 2x2−x4−1
2x4
= 2x3+ 2x2−3
2x4
15
Now, we compute the integral with respect to xas follows:
Z2
0
(2x3+ 2x2−3
2x4)dx =1
2x4+2
3x3−3
8x5
2
0
=1
2(2)4+2
3(2)3−3
8(2)5−1
2(0)4+2
3(0)3−3
8(0)5
=8 + 16
3−24−0
=8
3
Therefore, the value of the double integral RRR(x2+xy)dA over the region
Ris 8
3.
Question 17
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by the curves y=x2and y= 2xin the first quadrant. To find the limits of
integration for x, we need to determine the x-values at which these two curves
intersect. Setting x2= 2xgives x2−2x=x(x−2) = 0, which implies x= 0
and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Find the limits of integration for y:For each xin the interval
[0,2],yvaries between the curves y=x2and y= 2x. So, the limits of integration
for yare x2≤y≤2x.
Now, the integral becomes:
Z2
0Z2x
x2
ex2+y2dy dx
16
Step 3: Evaluate the integral:
Z2
0Z2x
x2
ex2+y2dy dx
=Z2
0hex2+y2iy=2x
y=x2dx
=Z2
0ex2+(2x)2−ex2+x4dx
=Z2
0e5x2−ex2(1+x2)dx
=Z2
0
e5x2dx −Z2
0
ex2(1+x2)dx
=1
2√5e5x22
0−Z2
0
ex2+x4dx
=1
2√5e20 −1−1
2√πex2erf(x)2
0
=1
2√5e20 −1−1
2√π(e16erf(2) −erf(0))
=1
2√5e20 −1−1
2√π(e16erf(2) −0)
=1
2√5e20 −1−1
2√πe16erf(2)
Therefore, the value of the double integral RRRex2+y2dA over Ris 1
2√5e20 −1−
1
2√πe16erf(2).
Question 18
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the curves y=x2,y= 0,x= 1, and x= 2.
Solution
Step 1: We first sketch the region Rto determine the limits of integration. The
region Ris bounded by y=x2,y= 0,x= 1, and x= 2. It is the region
enclosed between the parabola y=x2and the y-axis, and between x= 1 and
x= 2.
Step 2: To set up the double integral, we express xand yin terms of the
bounds of integration. The bounds for xare 1≤x≤2, and the bounds for y
are 0≤y≤x2.
17
Step 3: Now, we can express the given double integral in terms of xand y.
ZZR
ex2−y2dA =Z2
1Zx2
0
ex2−y2dy dx
Step 4: Evaluate the inner integral with respect to y.
Zx2
0
ex2−y2dy =h−ex2−y2ix2
0=−(ex2−x4−ex2)
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x.
Z2
1−(ex2−x4−ex2)dx
Step 6: Integrate the terms in the outer integral.
=−Z2
1
ex2−x4dx +Z2
1
ex2dx
Step 7: We may need to use techniques like substitution or manipulation to
evaluate the integrals in Step 6. The final result of the double integral requires
careful algebraic manipulation and integration.
Question 19
Question
Evaluate the double integral R RR(3x−y)dA, where Ris the region bounded by
the curves y=x2,y= 8 −x2,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Rcan be
described as 0≤x≤2and x2≤y≤8−x2. Thus, the limits of integration are
0≤x≤2and x2≤y≤8−x2.
Step 2: Rewrite the integrand in terms of xand y. The integrand (3x−y)
remains unchanged.
Step 3: Set up the double integral. The double integral can be set up as
follows:
Z2
0Z8−x2
x2
(3x−y)dy dx
Step 4: Integrate with respect to y. Integrating (3x−y)with respect to y
gives:
Z2
0
[3xy −y2
2]8−x2
x2dx
18
Step 5: Evaluate the integral. Substitute the limits of integration and sim-
plify the expression:
Z2
0
[(24x−(8 −x2)2/2) −(3x(x2)−x4/2)] dx
Step 6: Continue simplifying. After simplifying the expression, integrate
over [0,2]:
Z2
0
(24x−4−4x2+x4−3x3+x4/2) dx
Step 7: Integrate with respect to x. Integrating the expression above yields
the final answer:
12x2−4
3x3−x4+1
3x5−3
4x4+1
8x52
0
Step 8: Calculate the final answer. By substituting the limits of integration
into the expression and calculating, we find the final answer for the double
integral to be 560
3or approximately 186.67.
Question 20
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2and y=x.
Solution
Step 1: To evaluate the double integral, we need to find the limits of integration
for both xand y.
Step 2: The region Rcan be described by y=x2and y=x, which intersect
at the points (0,0) and (1,1). Therefore, the limits of integration for xare
0≤x≤1.
Step 3: For a given xin the interval [0,1], the corresponding yvalues are
x≤y≤x2. Hence, the limits of integration for yare x≤y≤x2.
Step 4: Now, we can rewrite the double integral as follows:
ZZR
(3x2+ 2y)dA =Z1
0Zx2
x
(3x2+ 2y)dy dx
Step 5: Compute the inner integral with respect to y:
Zx2
x
(3x2+ 2y)dy =3x2y+y2x2
x= 3x4+x4−(3x3+x) = 4x4−3x3−x
19
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(4x4−3x3−x)dx =4
5x5−3
4x4−1
2x21
0
=4
5−3
4−1
2=3
20
Step 7: Therefore, the value of the double integral RRR(3x2+ 2y)dA over
the region Ris 3
20 .
Question 21
Question
Evaluate the double integral RRRex
y2dA over the region Rbounded by the lines
y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determining limits of integration for x:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. We find the
intersection points of these lines to determine the limits of integration for x.
The intersection of y=xand y= 2xgives x=y= 0,x=y= 0 is outside
of R.
The intersection of y=xand x= 1 gives x=y= 1.
So the limits of integration for xare from x= 1 to x= 1.
Step 2: Determining limits of integration for y:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. Hence, y
varies from y=xto y= 2x.
Now, we can rewrite the double integral with the limits of integration for x
and y:
ZZR
ex
y2dA =Z1
1Z2x
x
ex
y2dy dx
Step 3: Evaluating the double integral:
Let’s evaluate the inner integral with respect to yfirst:
Z2x
x
ex
y2dy =−1
2ex
y22x
x
=−1
2e2−e1
4
Now, we can substitute this back into the double integral:
Z1
1−1
2e2−e1
4dx
20
=−1
2e2−e1
4Z1
1
dx
=−1
2e2−e1
4(1 −1)
= 0
Therefore, the value of the double integral RRRex
y2dA over the region Ris
0.
Question 22
Question
Let Rbe the region bounded by the curves y=x2and y=x4. Evaluate the
double integral RRRxy dA.
Solution
Step 1: Find the limits of integration for xand y. The region Ris bounded by
the curves y=x2and y=x4. To find the limits of integration for x, we need
to solve for the x-coordinates of the points of intersection of y=x2and y=x4.
Setting x2=x4, we get x4−x2= 0 ⇒x2(x2−1) = 0. This gives us x= 0
and x=±1as the x-coordinates of the points of intersection.
Therefore, the limits of integration for xare −1≤x≤1. For each xin
this range, yvaries between x2and x4. So, the limits of integration for yare
x2≤y≤x4.
Step 2: Set up the double integral. The double integral RRRxy dA over the
region Rcan be expressed as:
ZZR
xy dA =Z1
−1Zx4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y. Integrating xy with
respect to ywhile treating xas a constant, we get:
Zxy dy =1
2xy2
x4
x2=1
2x(x8−x4) = 1
2x9−1
2x5
Step 4: Evaluate the outer integral with respect to x. Substitute the limits
of integration and integrate 1
2x9−1
2x5with respect to x:
Z1
−11
2x9−1
2x5dx =1
21
10 −1
6=−1
15
Step 5: Final Answer Thus, the value of the double integral RRRxy dA over
the region Ris −1
15 .
21
Question 23
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y= 2x2and y= 3x, and xranges from 0 to 1.
Solution
To evaluate the given double integral over the region R, we first need to deter-
mine the limits of integration for xand ybased on the given region R.
Step 1: Determine the limits of integration for x. The curves y= 2x2and
y= 3xintersect at the point where 2x2= 3x. Solving for x, we get:
2x2= 3x
2x2−3x= 0
x(2x−3) = 0
x= 0 or x=3
2
Since xranges from 0 to 1, we will integrate xfrom 0 to 3
2.
Step 2: Determine the limits of integration for y. The region Ris bounded
by the curves y= 2x2and y= 3x. This means yranges from 2x2to 3x.
Therefore, the limits of integration for yare from 2x2to 3x.
Step 3: Set up the double integral. The given double integral becomes:
Z3
2
0Z3x
2x2
ex+ydy dx
Step 4: Evaluate the double integral.
Z3
2
0ex+y3x
2x2dx
Z3
2
0e4x−e3xdx
Integrating with respect to x:
1
4e4x−1
3e3x3
2
0
=1
4e6−1
3e9−1
4−1
3
=1
4e6−1
3e9−1
4+1
3
=1
4e6−1
4−1
3e9+1
3
22
Question 24
Question
Evaluate the double integral RRR
x−y
x+ydA, where Ris the region bounded by the
curves y=x,y= 2x, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we need to determine the intersection points of the curves
y=x,y= 2x, and x= 1. Setting y=xand y= 2xequal to each other gives
us x= 0. Setting y= 2xand x= 1 equal to each other gives us x= 1.
Therefore, the limits of integration for xare from 0 to 1.
To find the limits of integration for y, we need to determine the bounds in
terms of y. From the given region bounds, y=xand y= 2x, the limits of yare
from y=xto y= 2x.
Step 2: Rewrite the integral using the limits of integration. The double
integral can now be rewritten as:
Z1
0Z2x
x
x−y
x+ydy dx
Step 3: Evaluate the inner integral with respect to y. Integrating with
respect to y, we get:
Z2x
x
x−y
x+ydy =xy −y2
22x
x
=x(2x)−(2x)2
2−x(x)+x2
2= 2x2−2x2−x2+x2
2=−3
2x2
Step 4: Evaluate the outer integral with respect to x. Integrating the result
from the inner integral over the limits of x, we get:
Z1
0−3
2x2dx =−3
2x3
31
0
=−1
2
Therefore, the value of the double integral is −1
2.
Question 25
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y=
4x−x2. Find the volume of the solid that lies under the surface z= 12−x2−y2
and above the region R.
23
Solution
Step 1: Find the bounds for the double integral over the region R. The region
Ris bounded by the curves y=x2and y= 4x−x2. To find the bounds for x,
we set the two functions equal to each other:
x2= 4x−x2
2x2−4x= 0
2x(x−2) = 0
This implies x= 0 and x= 2. Therefore, the bounds for xare 0≤x≤2.
To find the bounds for y, we subtract the functions: 4x−x2−x2= 4x−2x2.
This tells us that the bounds for yare x2≤y≤4x−x2.
Step 2: Set up the double integral. The volume under the surface z=
12 −x2−y2can be expressed as:
ZZ
R
(12 −x2−y2)dA
where dA =dx dy.
Step 3: Evaluate the double integral. Now, we evaluate the double integral
over the region R:
Z2
0Z4x−x2
x2
(12 −x2−y2)dy dx
=Z2
012y−x2y−1
3y34x−x2
x2
dx
=Z2
012(4x−x2)−x2(4x−x2)−1
3(4x−x2)3−12x2+x4−1
3x6dx
=Z2
048x−12x2−4x3+x4−4
3x2(4x−x2) + 1
3(4x−x2)3dx
After integrating and simplifying, we get the volume of the solid that lies
under the surface and above the region R.
24
ZZR
(x2+y2)dA
=Z1
−1Z2−x2
x2
(x2+y2)dy dx
=Z1
−1x2y+1
3y32−x2
x2
dx
=Z1
−1x2(2 −x2) + 1
3(2 −x2)3−x2(x2)−1
3(x2)3dx
=Z1
−1
(2x2−x4+8
3−4x2+ 4x4−8
3−x4−1
3x6)dx
=Z1
−1
(−5x4+ 6x2−1
3x6)dx
=−5
5x5+6
3x3−1
21x71
−1
=−x5+ 2x3−1
21x71
−1
=−1+2−1
21 −(−1−2 + 1
21)
= 3 −2
21
=61
21.
Therefore, the value of the double integral RRR(x2+y2)dA over the region
Ris 61
21 .
Question 2
Question
Let Rbe the region bounded by the curves y=x2−1,y= 0,x= 2, and x= 3.
Calculate the double integral RRRy dA using polar coordinates.
Solution
Step 1: Sketch the region Rin the xy-plane. The region Ris bounded by the
curves y=x2−1,y= 0,x= 2, and x= 3. The curves y=x2−1and y= 0
intersect at x= 1 and x=−1, but we only consider the part of the parabola
in the first quadrant. Therefore, the region consists of the area between y= 0
and y=x2−1from x= 2 to x= 3.
2
Step 2: Determine the limits of integration for rand θin polar coordinates.
The region Rcorresponds to 2≤x≤3and 0≤y≤x2−1. Converting
the equations to polar coordinates gives 2≤rcos(θ)≤3and 0≤r2sin(θ)−
1≤r2cos2(θ)−1. Simplifying the inequalities, we get 2≤r≤3
cos(θ)and
0≤r2sin(θ)−1≤r2cos2(θ)−1.
Step 3: Set up the double integral in polar coordinates. The double integral
RRRy dA can be expressed as Rθ2
θ1Rr2(θ)
r1(θ)f(r, θ)r dr dθ, where f(r, θ) = r2sin(θ).
Step 4: Calculate the double integral. We have Rπ
3
0R
3
cos(θ)
2r2sin(θ)r dr dθ.
Integrating with respect to rfirst, we get Rπ
3
0hr4
4sin(θ)i3
cos(θ)
2dθ. This simplifies
to Rπ
3
081
4 cos4(θ)−24sin(θ)dθ.
Step 5: Finally, evaluate the integral. We compute Rπ
3
0
81
4 cos4(θ)−16 sin(θ)dθ.
This integral can be challenging to evaluate directly, so we recommend using
appropriate trigonometric identities to simplify the integrand before proceeding
with the calculation.
Question 3
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: To evaluate the double integral, we first need to determine the bounds
for xand y.
Step 2: The region Ris a square with sides of length 2 centered at the origin.
Therefore, the bounds for xand yare −1≤x≤1and −1≤y≤1.
Step 3: The given double integral can now be rewritten as R1
−1R1
−1ex2−y2dx dy.
Step 4: We can now evaluate the inner integral with respect to xfirst:
Z1
−1
ex2−y2dx =hex2−y2i1
−1=e1−y2−e1−y2
Step 5: Substituting the bounds and simplifying, we get:
Z1
−1
ex2−y2dx = 0
Step 6: Now, the double integral becomes R1
−10dy.
Step 7: Integrating this with respect to ygives:
Z1
−1
0dy = [0]1
−1= 0
3
Step 8: Therefore, the value of the double integral RRRex2−y2dA over the
given region Ris 0.
Question 4
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the lines y=x,y=−x,y= 1, and y=−1.
Solution
Step 1: Begin by sketching the region Rin the xy-plane. The region Ris a
rectangle with vertices at (1,1),(−1,1),(−1,−1), and (1,−1).
Step 2: Rewrite the given double integral as an iterated integral by setting
up the limits of integration. In this case, we integrate first with respect to y
and then with respect to x. So, we have:
ZZR
ex2−y2dA =Z1
−1Zx
−x
ex2−y2dy dx.
Step 3: Integrate with respect to y:
Z1
−1Zx
−x
ex2−y2dy dx =Z1
−1hyex2−y2ix
−xdx
=Z1
−1
(xex2−x2−(−xex2−(−x)2)) dx
=Z1
−1
(x−xe4x2)dx.
Step 4: Integrate the inner integral with respect to x:
Z1
−1
(x−xe4x2)dx =1
2x2−1
8e4x21
−1
=1
2−1
8e4−1
2−1
8e4
= 0.
Therefore, the value of the double integral RRRex2−y2dA over the given
rectangular region Ris 0.
4
Question 5
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration by finding the points of intersection between the curves y=x2and
y= 2x.
Step 1: Find the points of intersection. Setting x2= 2x, we get
x2−2x= 0, which factorizes to give us x(x−2) = 0. Thus, x= 0 or x= 2.
Therefore, the points of intersection are (0,0) and (2,4).
Step 2: Set up the double integral. The given region Rcan be described
by 0≤x≤2and x2≤y≤2x. Therefore, the double integral can be rewritten
as:
Z2
0Z2x
x2
ex+ydy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Zex+ydy =ex+y
2x
x2
=e3x−ex2
Step 4: Evaluate the outer integral. Integrating with respect to x, we
get:
Z2
0
e3x−ex2dx =e6−1
3−e4−1
2
Therefore, the value of the given double integral is e6−1
3−e4−1
2.
Question 6
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y=x2,y=x,x= 0, and x= 1.
Solution
Step 1: First, we express the region Rin terms of the limits of integration. The
region Ris bounded by the curves y=x2,y=x,x= 0, and x= 1. This means
that the limits of integration are 0≤x≤1and x2≤y≤x.
5
Step 2: Rewrite the double integral with the given limits of integration. The
double integral becomes:
ZZR
ex+ydA =Z1
0Zx
x2
ex+ydydx
Step 3: Integrate with respect to yfirst.
Zx
x2
ex+ydy =ex+yx
x2=e2x−ex2
Step 4: Substitute back into the double integral.
Z1
0
(e2x−ex2)dx
Step 5: Integrate with respect to x.
Z1
0
(e2x−ex2)dx =1
2e2x1
0−Z1
0
ex2dx =1
2e2−Z1
0
ex2dx
Step 6: Evaluate the remaining integral. The integral R1
0ex2dx cannot be
expressed in terms of elementary functions, so we leave it as is.
Therefore, the value of the given double integral is 1
2e2−R1
0ex2dx.
Question 7
Question
Evaluate the double integral RRD(x2+y2)dA, where Dis the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of integra-
tion for xand yby finding the points of intersection of the two curves y=x2
and y= 2x.
Step 1: Find the points of intersection. Setting the two curves equal
to each other gives:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies x= 0 or x= 2. Substituting these values back into either
equation gives the y-coordinates of the points of intersection: (0,0) and (2,4).
6
Step 2: Set up the double integral. The region Dis bounded by the
lines y=x2and y= 2x. Thus, the limits of integration for yare x2and 2x,
and for xthey are 0and 2. The double integral is then:
ZZD
(x2+y2)dA =Z2
0Z2x
x2
(x2+y2)dy dx
Step 3: Integrate with respect to y.Integrating with respect to yfirst,
we get:
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
Step 4: Integrate with respect to x.Now, we integrate with respect to
x:
Z2
0
(2x3+8x3
3−x4−x6
3)dx
=2x4
4+8x4
3·4−x5
5−x7
3·7
2
0
=16
4+64
12 −32
5−128
21
= 4 + 16
3−32
5−128
21
=642
105
Step 5: Conclusion. Therefore, the value of the double integral RRD(x2+
y2)dA over the region Dbounded by the curves y=x2and y= 2xis 642
105 .
Question 8
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 4.
7
Solution
To evaluate the given double integral, we first need to determine the bounds of
integration by finding the points of intersection of the curves y=x2and y= 4.
This will help us determine the limits of integration for the inner integral.
Step 1: Find the points of intersection. Setting x2= 4, we find the
points of intersection to be when y= 4:
x2= 4
x=±2
So the points of intersection are (2,4) and (−2,4).
Step 2: Set up the integral. Given that Ris bounded by y=x2and
y= 4, the limits of integration are:
−2≤x≤2
x2≤y≤4
Thus, the double integral becomes:
Z2
−2Z4
x2
ex2+y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we
get:
Z4
x2
ex2+y2dy =ex2+x4−ex2
Step 4: Evaluate the outer integral. Integrating the result of the inner
integral from −2to 2, we get:
Z2
−2ex2+16 −ex2dx =Z2
−2
ex2+16 dx −Z2
−2
ex2dx
We can solve these integrals using techniques of integration, such as substi-
tution or recognizing standard integral forms.
Therefore, the value of the given double integral can be found by evaluating
the above expression.
Question 9
Question
Evaluate the double integral RRRex2−y2dA over the region Rbounded by the
lines y=x,y= 2x, and x= 1.
8
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yover the region R.
Step 1: Determining the limits of integration for y:The region R
is bounded by the lines y=x,y= 2x, and x= 1. We start by finding the y
limits. Along the line y=x,yvaries from xto 2x. Along the line x= 1,y
varies from 1to 2. Thus, the ylimits are from xto 2x.
Step 2: Determining the limits of integration for x:Next, we deter-
mine the limits for x. Since xvaries from 1to the intersection point of the lines
y=xand y= 2x, we need to find this intersection point. Setting x= 2x, we
find that x= 0. Therefore, xvaries from 1to 0which is the region where the
lines y=xand y= 2xintersect.
Step 3: Setting up and evaluating the double integral: The given
double integral can now be expressed as:
ZZR
ex2−y2dA =Z1
0Z2x
x
ex2−y2dy dx
We can now evaluate the inner integral with respect to y:
Z2x
x
ex2−y2dy =−ex2−4x2+ex2−x2=−e−3x2+ex2
Finally, we evaluate the outer integral with respect to x:
Z1
0
(−e−3x2+ex2)dx ="−e−3x2
6+ex2
2#1
0
=1
2−1
6=1
3
Therefore, the value of the double integral RRRex2−y2dA over the region R
is 1
3.
Question 10
Question
Evaluate the double integral RRDex2−y2dx dy, where Dis the region bounded
by the curves y=x,y= 3x, and x= 1.
Solution
Step 1: Find the intersection points of the curves to determine the bounds
of integration. The intersection points are obtained by solving the system of
equations:
y=xand y= 3x
9
This gives x= 1 and y= 1. So, the bounds of integration are 1≤x≤1and
x≤y≤3x.
Step 2: Setup the double integral. The given double integral can be rewritten
as:
Z1
1Z3x
x
ex2−y2dy dx
Step 3: Evaluate the inner integral. Integrating with respect to y, we get:
Z3x
x
ex2−y2dy =−1
2ex2−y2
3x
x
=−1
2(ex2−(3x)2−ex2−x2)
=−1
2(ex2−9x2−1)
=−1
2(e−8x2−1)
Step 4: Evaluate the outer integral.
Z1
1−1
2(e−8x2−1) dx =−1
2Z1
1
(e−8x2−1) dx
=−1
2 e−8x2
−16 −x!
1
1
=−1
21
−16 −1−1
−16 −1
=−1
2−1
16 +1
16
= 0
Therefore, the value of the double integral is 0.
Question 11
Question
Evaluate the double integral RRRx2y dA where Ris the region bounded by the
curve y=x2, the x-axis, and the lines x= 0 and x= 2.
Solution
We begin by sketching the region Rdefined by the curve y=x2, the x-axis,
and the lines x= 0 and x= 2.
10
xy2; [onlymarks, mark =∗, color =red]coordinates(0,0)(2,0);
y=x2
x= 2
x
y
The region Ris a triangular region in the first quadrant, bounded by x= 0,
x= 2, and y=x2.
To evaluate the double integral RRRx2y dA, we set up the integral in terms
of xand y:
ZZR
x2y dA =Z2
0Zx2
0
x2y dy dx
Step 1: Integrate with respect to y:
Zx2
0
x2y dy =x2y2
2x2
0
=x2·(x2)2
2=x6
2
Step 2: Integrate with respect to x:
Z2
0
x6
2dx =1
2Z2
0
x6dx =1
2x7
72
0
=27
2·7=64
7
Therefore, the value of the double integral RRRx2y dA over the region Ris
64
7.
Question 12
Question
Evaluate the double integral RRRex+ydA over the region Rbounded by y=x2,
y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y.
• The region Ris bounded by y=x2,y= 2x,x= 0, and x= 1.
• We can express the limits of integration for xin terms of yas 0≤x≤√y
(from x= 0 to x=√y) and √y≤x≤y
2(from x=√yto x=y
2).
11
• The limits of integration for yare 0≤y≤1.
Step 2: Set up the double integral.
• The double integral can be expressed as R1
0R√y
0ex+ydx dy+R1
0Ry
2
√yex+ydx dy.
Step 3: Evaluate the inner integral with respect to xfor the first region.
• Simplify R√y
0ex+ydx to get eyR√y
0exdx.
• Integrate to get ey[ex]
√y
0=ey(e√y−1).
Step 4: Evaluate the inner integral with respect to xfor the second region.
• Simplify Ry
2
√yex+ydx to get eyRy
2
√yexdx.
• Integrate to get ey[ex]
y
2
√y=eyey
2−e√y.
Step 5: Evaluate the outer integral with respect to y.
• Simplify the expression and integrate to get R1
0ey(e√y−1) dy+R1
0eyey
2−e√ydy.
• Integrate to get (e−1) −2√e+5
2−2e
3.
Hence, the value of the double integral RRRex+ydA over the region Ris
−2
3e−2
3√e+7
2.
Question 13
Question
Evaluate the double integral RRRxy dA, where Ris the region in the first quad-
rant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to find the limits of integra-
tion for xand yby determining the region R.
Step 1: Determine the limits of integration for xand y
The curves y=x2and y= 2xintersect at the point where x2= 2x. Solving
this equation, we find x2−2x= 0, which gives x(x−2) = 0. This implies x= 0
or x= 2.
Since Ris in the first quadrant, we have 0≤x≤2. The lower curve is
y=x2while the upper curve is y= 2x.
Therefore, the limits of integration are:
0≤x≤2and x2≤y≤2x
12
Step 2: Evaluate the double integral
The given double integral is:
ZZR
xy dA =Z2
0Z2x
x2
xy dy dx
We first integrate with respect to y:
Z2x
x2
xy dy =1
2xy22x
x2
=1
2(2x)(4x)−1
2(x)(x2) = 4x2−1
2x3
Now we integrate the result with respect to x:
Z2
0
4x2−1
2x3dx =4
3x3−1
8x42
0
=32
3−4 = 20
3
Therefore, the value of the double integral RRRxy dA over the region Ris
20
3.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region in the first
quadrant bounded by the curves y=xand y=√x.
Solution
Step 1: We first need to determine the limits of integration for xand y. The
region Ris defined by y=xand y=√xin the first quadrant, so the limits of
integration are:
0≤x≤1
x≤y≤√x
Step 2: Rewrite the given double integral in terms of xand y:
ZZR
ex+ydA =Z1
0Z√x
x
ex+ydy dx
Step 3: Integrate with respect to yfirst:
Z√x
x
ex+ydy =e2x−ex
Step 4: Substitute back the result from Step 3 into the double integral:
Z1
0
e2x−exdx
13
Step 5: Integrate with respect to x:
Z1
0
e2x−exdx =e2x
2−ex1
0
=e2
2−e+ 1
Therefore, the value of the double integral RRRex+ydA over the region Ris
e2
2−e+ 1.
Question 15
Question
Evaluate the double integral RRR(x2+y)dA, where Ris the region bounded by
the curves y=x2and y= 1.
Solution
Step 1: Determine the limits of integration for x. Since y=x2and y= 1 bound
the region R, set the equations equal to find the x-values where they intersect:
x2= 1
x=±1
Thus, the limits of integration for xare −1to 1.
Step 2: Determine the limits of integration for y. The region Ris bounded
below by the curve y=x2and above by the line y= 1. So, the limits of
integration for yare x2to 1.
Step 3: Rewrite the integral and evaluate. The double integral becomes:
Z1
−1Z1
x2
(x2+y)dy dx
Now, evaluate the integral:
Z1
−1Z1
x2
(x2+y)dy dx =Z1
−1x2y+y2
21
x2
dx
=Z1
−1x2·1 + 1
2−x4−x4
2dx
=Z1
−1x2−x4−3x4
2+1
2dx
=x3
3−x5
5−3x5
10 +x
21
−1
14
=1
3−1
5−3
10 +1
2−−1
3+1
5+3
10 −1
2
=10
30 −6
30 =4
30 =2
15
Therefore, the value of the double integral RRR(x2+y)dA over the region R
is 2
15 .
Question 16
Question
Evaluate the double integral RRR(x2+xy)dA, where Ris the region bounded
by the curves y=x2and y= 2x.
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand yby sketching the region R.
Step 1: Determine the limits of integration for x:The region Ris
bounded by the curves y=x2and y= 2x. We need to find the x-values at
which these curves intersect. Setting x2= 2xgives us x2−2x= 0, which factors
as x(x−2) = 0. Thus, the points of intersection are x= 0 and x= 2. Therefore,
the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for y:For a given xwithin
the region R, the lower bound for yis x2and the upper bound is 2x. Thus, the
limits for yare x2≤y≤2x.
Step 3: Evaluate the double integral: We can now express the given
double integral as:
ZZR
(x2+xy)dA =Z2
0Z2x
x2
(x2+xy)dy dx
Now we compute the integral with respect to yfirst, then x:
Z2x
x2
(x2+xy)dy =x2y+1
2y2
2x
x2
=x2(2x) + 1
2(2x)2−x2(x2)−1
2(x2)2
= 2x3+ 2x2−x4−1
2x4
= 2x3+ 2x2−3
2x4
15
Now, we compute the integral with respect to xas follows:
Z2
0
(2x3+ 2x2−3
2x4)dx =1
2x4+2
3x3−3
8x5
2
0
=1
2(2)4+2
3(2)3−3
8(2)5−1
2(0)4+2
3(0)3−3
8(0)5
=8 + 16
3−24−0
=8
3
Therefore, the value of the double integral RRR(x2+xy)dA over the region
Ris 8
3.
Question 17
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Find the limits of integration for x:The region Ris bounded
by the curves y=x2and y= 2xin the first quadrant. To find the limits of
integration for x, we need to determine the x-values at which these two curves
intersect. Setting x2= 2xgives x2−2x=x(x−2) = 0, which implies x= 0
and x= 2. Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Find the limits of integration for y:For each xin the interval
[0,2],yvaries between the curves y=x2and y= 2x. So, the limits of integration
for yare x2≤y≤2x.
Now, the integral becomes:
Z2
0Z2x
x2
ex2+y2dy dx
16
Step 3: Evaluate the integral:
Z2
0Z2x
x2
ex2+y2dy dx
=Z2
0hex2+y2iy=2x
y=x2dx
=Z2
0ex2+(2x)2−ex2+x4dx
=Z2
0e5x2−ex2(1+x2)dx
=Z2
0
e5x2dx −Z2
0
ex2(1+x2)dx
=1
2√5e5x22
0−Z2
0
ex2+x4dx
=1
2√5e20 −1−1
2√πex2erf(x)2
0
=1
2√5e20 −1−1
2√π(e16erf(2) −erf(0))
=1
2√5e20 −1−1
2√π(e16erf(2) −0)
=1
2√5e20 −1−1
2√πe16erf(2)
Therefore, the value of the double integral RRRex2+y2dA over Ris 1
2√5e20 −1−
1
2√πe16erf(2).
Question 18
Question
Evaluate the double integral RRRex2−y2dA, where Ris the region bounded by
the curves y=x2,y= 0,x= 1, and x= 2.
Solution
Step 1: We first sketch the region Rto determine the limits of integration. The
region Ris bounded by y=x2,y= 0,x= 1, and x= 2. It is the region
enclosed between the parabola y=x2and the y-axis, and between x= 1 and
x= 2.
Step 2: To set up the double integral, we express xand yin terms of the
bounds of integration. The bounds for xare 1≤x≤2, and the bounds for y
are 0≤y≤x2.
17
Step 3: Now, we can express the given double integral in terms of xand y.
ZZR
ex2−y2dA =Z2
1Zx2
0
ex2−y2dy dx
Step 4: Evaluate the inner integral with respect to y.
Zx2
0
ex2−y2dy =h−ex2−y2ix2
0=−(ex2−x4−ex2)
Step 5: Substitute the result back into the double integral and evaluate the
outer integral with respect to x.
Z2
1−(ex2−x4−ex2)dx
Step 6: Integrate the terms in the outer integral.
=−Z2
1
ex2−x4dx +Z2
1
ex2dx
Step 7: We may need to use techniques like substitution or manipulation to
evaluate the integrals in Step 6. The final result of the double integral requires
careful algebraic manipulation and integration.
Question 19
Question
Evaluate the double integral R RR(3x−y)dA, where Ris the region bounded by
the curves y=x2,y= 8 −x2,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Rcan be
described as 0≤x≤2and x2≤y≤8−x2. Thus, the limits of integration are
0≤x≤2and x2≤y≤8−x2.
Step 2: Rewrite the integrand in terms of xand y. The integrand (3x−y)
remains unchanged.
Step 3: Set up the double integral. The double integral can be set up as
follows:
Z2
0Z8−x2
x2
(3x−y)dy dx
Step 4: Integrate with respect to y. Integrating (3x−y)with respect to y
gives:
Z2
0
[3xy −y2
2]8−x2
x2dx
18
Step 5: Evaluate the integral. Substitute the limits of integration and sim-
plify the expression:
Z2
0
[(24x−(8 −x2)2/2) −(3x(x2)−x4/2)] dx
Step 6: Continue simplifying. After simplifying the expression, integrate
over [0,2]:
Z2
0
(24x−4−4x2+x4−3x3+x4/2) dx
Step 7: Integrate with respect to x. Integrating the expression above yields
the final answer:
12x2−4
3x3−x4+1
3x5−3
4x4+1
8x52
0
Step 8: Calculate the final answer. By substituting the limits of integration
into the expression and calculating, we find the final answer for the double
integral to be 560
3or approximately 186.67.
Question 20
Question
Evaluate the double integral RRR(3x2+ 2y)dA, where Ris the region bounded
by the curves y=x2and y=x.
Solution
Step 1: To evaluate the double integral, we need to find the limits of integration
for both xand y.
Step 2: The region Rcan be described by y=x2and y=x, which intersect
at the points (0,0) and (1,1). Therefore, the limits of integration for xare
0≤x≤1.
Step 3: For a given xin the interval [0,1], the corresponding yvalues are
x≤y≤x2. Hence, the limits of integration for yare x≤y≤x2.
Step 4: Now, we can rewrite the double integral as follows:
ZZR
(3x2+ 2y)dA =Z1
0Zx2
x
(3x2+ 2y)dy dx
Step 5: Compute the inner integral with respect to y:
Zx2
x
(3x2+ 2y)dy =3x2y+y2x2
x= 3x4+x4−(3x3+x) = 4x4−3x3−x
19
Step 6: Now, integrate the result from Step 5 with respect to x:
Z1
0
(4x4−3x3−x)dx =4
5x5−3
4x4−1
2x21
0
=4
5−3
4−1
2=3
20
Step 7: Therefore, the value of the double integral RRR(3x2+ 2y)dA over
the region Ris 3
20 .
Question 21
Question
Evaluate the double integral RRRex
y2dA over the region Rbounded by the lines
y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determining limits of integration for x:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. We find the
intersection points of these lines to determine the limits of integration for x.
The intersection of y=xand y= 2xgives x=y= 0,x=y= 0 is outside
of R.
The intersection of y=xand x= 1 gives x=y= 1.
So the limits of integration for xare from x= 1 to x= 1.
Step 2: Determining limits of integration for y:
The region Ris bounded by the lines y=x,y= 2x, and x= 1. Hence, y
varies from y=xto y= 2x.
Now, we can rewrite the double integral with the limits of integration for x
and y:
ZZR
ex
y2dA =Z1
1Z2x
x
ex
y2dy dx
Step 3: Evaluating the double integral:
Let’s evaluate the inner integral with respect to yfirst:
Z2x
x
ex
y2dy =−1
2ex
y22x
x
=−1
2e2−e1
4
Now, we can substitute this back into the double integral:
Z1
1−1
2e2−e1
4dx
20
=−1
2e2−e1
4Z1
1
dx
=−1
2e2−e1
4(1 −1)
= 0
Therefore, the value of the double integral RRRex
y2dA over the region Ris
0.
Question 22
Question
Let Rbe the region bounded by the curves y=x2and y=x4. Evaluate the
double integral RRRxy dA.
Solution
Step 1: Find the limits of integration for xand y. The region Ris bounded by
the curves y=x2and y=x4. To find the limits of integration for x, we need
to solve for the x-coordinates of the points of intersection of y=x2and y=x4.
Setting x2=x4, we get x4−x2= 0 ⇒x2(x2−1) = 0. This gives us x= 0
and x=±1as the x-coordinates of the points of intersection.
Therefore, the limits of integration for xare −1≤x≤1. For each xin
this range, yvaries between x2and x4. So, the limits of integration for yare
x2≤y≤x4.
Step 2: Set up the double integral. The double integral RRRxy dA over the
region Rcan be expressed as:
ZZR
xy dA =Z1
−1Zx4
x2
xy dydx
Step 3: Evaluate the inner integral with respect to y. Integrating xy with
respect to ywhile treating xas a constant, we get:
Zxy dy =1
2xy2
x4
x2=1
2x(x8−x4) = 1
2x9−1
2x5
Step 4: Evaluate the outer integral with respect to x. Substitute the limits
of integration and integrate 1
2x9−1
2x5with respect to x:
Z1
−11
2x9−1
2x5dx =1
21
10 −1
6=−1
15
Step 5: Final Answer Thus, the value of the double integral RRRxy dA over
the region Ris −1
15 .
21
Question 23
Question
Evaluate the double integral RRRex+ydA where Ris the region bounded by the
curves y= 2x2and y= 3x, and xranges from 0 to 1.
Solution
To evaluate the given double integral over the region R, we first need to deter-
mine the limits of integration for xand ybased on the given region R.
Step 1: Determine the limits of integration for x. The curves y= 2x2and
y= 3xintersect at the point where 2x2= 3x. Solving for x, we get:
2x2= 3x
2x2−3x= 0
x(2x−3) = 0
x= 0 or x=3
2
Since xranges from 0 to 1, we will integrate xfrom 0 to 3
2.
Step 2: Determine the limits of integration for y. The region Ris bounded
by the curves y= 2x2and y= 3x. This means yranges from 2x2to 3x.
Therefore, the limits of integration for yare from 2x2to 3x.
Step 3: Set up the double integral. The given double integral becomes:
Z3
2
0Z3x
2x2
ex+ydy dx
Step 4: Evaluate the double integral.
Z3
2
0ex+y3x
2x2dx
Z3
2
0e4x−e3xdx
Integrating with respect to x:
1
4e4x−1
3e3x3
2
0
=1
4e6−1
3e9−1
4−1
3
=1
4e6−1
3e9−1
4+1
3
=1
4e6−1
4−1
3e9+1
3
22
Question 24
Question
Evaluate the double integral RRR
x−y
x+ydA, where Ris the region bounded by the
curves y=x,y= 2x, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration for x, we need to determine the intersection points of the curves
y=x,y= 2x, and x= 1. Setting y=xand y= 2xequal to each other gives
us x= 0. Setting y= 2xand x= 1 equal to each other gives us x= 1.
Therefore, the limits of integration for xare from 0 to 1.
To find the limits of integration for y, we need to determine the bounds in
terms of y. From the given region bounds, y=xand y= 2x, the limits of yare
from y=xto y= 2x.
Step 2: Rewrite the integral using the limits of integration. The double
integral can now be rewritten as:
Z1
0Z2x
x
x−y
x+ydy dx
Step 3: Evaluate the inner integral with respect to y. Integrating with
respect to y, we get:
Z2x
x
x−y
x+ydy =xy −y2
22x
x
=x(2x)−(2x)2
2−x(x)+x2
2= 2x2−2x2−x2+x2
2=−3
2x2
Step 4: Evaluate the outer integral with respect to x. Integrating the result
from the inner integral over the limits of x, we get:
Z1
0−3
2x2dx =−3
2x3
31
0
=−1
2
Therefore, the value of the double integral is −1
2.
Question 25
Question
Let Rbe the region in the xy-plane bounded by the curves y=x2and y=
4x−x2. Find the volume of the solid that lies under the surface z= 12−x2−y2
and above the region R.
23
Solution
Step 1: Find the bounds for the double integral over the region R. The region
Ris bounded by the curves y=x2and y= 4x−x2. To find the bounds for x,
we set the two functions equal to each other:
x2= 4x−x2
2x2−4x= 0
2x(x−2) = 0
This implies x= 0 and x= 2. Therefore, the bounds for xare 0≤x≤2.
To find the bounds for y, we subtract the functions: 4x−x2−x2= 4x−2x2.
This tells us that the bounds for yare x2≤y≤4x−x2.
Step 2: Set up the double integral. The volume under the surface z=
12 −x2−y2can be expressed as:
ZZ
R
(12 −x2−y2)dA
where dA =dx dy.
Step 3: Evaluate the double integral. Now, we evaluate the double integral
over the region R:
Z2
0Z4x−x2
x2
(12 −x2−y2)dy dx
=Z2
012y−x2y−1
3y34x−x2
x2
dx
=Z2
012(4x−x2)−x2(4x−x2)−1
3(4x−x2)3−12x2+x4−1
3x6dx
=Z2
048x−12x2−4x3+x4−4
3x2(4x−x2) + 1
3(4x−x2)3dx
After integrating and simplifying, we get the volume of the solid that lies
under the surface and above the region R.
24
