MATH 332 - ADVANCED CALCULUS
- Double integrals over rectangular and
general regions
Question Bank - Set 7
Liberty University
Question 1
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves x= 0,x= 1,y= 0, and y= 1.
Solution
To evaluate the given double integral, we need to integrate over the region R
bounded by the curves x= 0,x= 1,y= 0, and y= 1.
Step 1: First, express the given double integral as an iterated integral.
ZZR
ex2+y2dA =Z1
0Z1
0
ex2+y2dx dy
Step 2: Integrate with respect to xfirst.
Z1
0
ex2+y2dx =Z1
0
ey2ex2dx
Now, let u=x2, so du = 2x dx.
Zey2ex2dx =1
2Zey2eudu =1
2ey2Zeudu =1
2ey2eu+C=1
2ex2+y2+C
So now we have:
Z1
0
ex2+y2dx =1
2ex2+y21
0
=1
2(e1+y2
−ey2)
Step 3: Integrate the previous result with respect to y.
Z1
01
2(e1+y2
−ey2)dy =1
2Z1
0
e1+y2dy −1
2Z1
0
ey2dy
Integrating, we get:
1
21
e−1e1+y2
1
0−1
2√πerf(1)
Step 4: Simplifying the expression gives the final answer:
=1
21
e−1(e2−1) −1
2√πerf(1)
Question 2
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 4,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y.
The region Ris bounded by the curves y=x2,y= 4,x= 0, and x= 2.
To determine the limits of integration, we need to find the intersection points
of these curves.
- The curves y=x2and y= 4 intersect when x2= 4. Solving for xgives us
x=±2, but since xmust be non-negative in this case, the intersection point is
at x= 2. - The curve x= 0 is the y-axis. - The limits for xare from 0 to 2. -
The limits for yare from x2to 4.
Step 2: Set up and evaluate the double integral.
The double integral is given as:
ZZR
(x+y)dA =Z2
0Z4
x2
(x+y)dy dx
Now, we integrate with respect to yfirst and then with respect to x:
2
Z2
0Z4
x2
(x+y)dy dx =Z2
0xy +y2
24
x2
dx
=Z2
04x+16
2−x3−x4
2dx
=Z2
0
(4x+ 8 −x3−x4
2)dx
=2x2+ 8x−x4
4−x5
10 2
0
= 8 + 16 −4−32
10
= 20 −4−16
5
= 16 −16
5
=64
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 64
5.
Question 3
Question
Evaluate the double integral RRD(x2+y2)dx dy over the region Dbounded by
the curves y=x,y= 4 −x2,x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration, we need to determine the points of intersection between the curves
y=xand y= 4 −x2. Setting x=y, we have y= 4 −y2, which simplifies to
y2+y−4 = 0. Solving this quadratic equation gives y= 1 or y=−4, but since
y=x, we discard the negative value. Thus, the limits of integration for yare
1≤y≤4−x2.
Next, we determine the limits of integration for x. Since the region is
bounded by x= 0, we have 0≤x≤2.
Step 2: Evaluate the double integral. We can now rewrite the double integral
as:
ZZD
(x2+y2)dx dy =Z2
0Z4−x2
1
(x2+y2)dy dx
3
Step 3: Integrate with respect to yfirst.
Z4−x2
1
(x2+y2)dy =x2y+y3
3
4−x2
1
Step 4: Substitute the limits of integration and simplify.
=x2(4 −x2) + (4 −x2)3
3−(x21 + 1
3) = 2x2+7x4
3−23x2
3+23
3
Step 5: Integrate the result with respect to x.
Z2
02x2+7x4
3−23x2
3+23
3dx =2
3x3+7
15x5−23
9x3+23
3x
2
0
Step 6: Substitute the limits of integration and calculate the final answer.
=23
3+7(25)
15 −23(23)
9+23(2)
3−0
3+0
15 −0
9+23(0)
3
=8
3+64
15 −64
3+46
3
=−160
15
=−32
3
Therefore, the value of the double integral RRD(x2+y2)dx dy over the region
Dis −32
3.
Question 4
Question
Evaluate the double integral RRRsin(x+y)dA, where Ris the region bounded
by the lines y= 0,y=x,y= 2x, and y= 2.
Solution
Step 1: Sketch the region Rdescribed by the given bounds for yin terms of x.
xyy =xy = 2xy = 2
Step 2: Set up the double integral using these bounds:
ZZR
sin(x+y)dA =Z1
0Z2x
x
sin(x+y)dydx
4
Step 3: Integrate with respect to yfirst:
Z2x
x
sin(x+y)dy = [−cos(x+y)]2x
x=−cos(3x) + cos(2x)
Step 4: Now integrate the result from Step 3 with respect to x:
Z1
0
(−cos(3x) + cos(2x)) dx =−1
3sin(3x) + 1
2sin(2x)1
0
=−1
3sin(3) + 1
2sin(2)−(0 + 0)
=−1
3sin(3) + 1
2sin(2)
Thus, the value of the given double integral is −1
3sin(3) + 1
2sin(2).
Question 5
Question
Calculate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x,y= 2x, and y= 1.
Solution
To find the double integral RRRxy2dA over the region R, we first need to de-
termine the bounds of integration by examining the limits along the xand y
axes.
Step 1: Determine the bounds of integration in terms of x:
Since y=xand y= 2xare the bounding curves, we can set up the following
inequalities:
x≤y≤2x
Solving these inequalities for x, we find that 0≤x≤1
2.
Step 2: Determine the bounds of integration in terms of y:
Next, we consider the bounding curve y= 1. Since this line intersects the
region at y= 1, the upper bound for ywill be 1.
Step 3: Set up the double integral:
Putting the bounds together, the double integral becomes:
Z1
2
0Z2x
x
xy2dy dx
Step 4: Integrate with respect to y:
Integrating the inner integral with respect to y, we get:
Z1
2
01
3xy32x
x
dx
5
=Z1
2
07
3x4dx
Step 5: Integrate with respect to x:
Integrating the remaining integral with respect to x, we have:
7
15x5
1
2
0
=7
15 ·1
25
=7
15 ·1
32
=7
480
Therefore, the value of the double integral RRRxy2dA over the region Ris
7
480 .
Question 6
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy dA, we first need to determine the limits
of integration for xand yover the region R.
Step 1: Determine the limits of integration for xThe region Ris
bounded by the curves y=x2and y= 2x. To find the limits of integration
for x, we need to find the x-values at which these two curves intersect. Setting
x2= 2x, we find the intersection points at x= 0 and x= 2.
Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for yFor each value of x
in the region R,yvaries between the curves y=x2(bottom curve) and y= 2x
(top curve).
So, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral The given double
integral is RRRxy dA.
Rewriting the integrand in terms of xand y, we have xy.
Therefore, the double integral becomes:
Z2
0Z2x
x2
xy dy dx
6
Now, we integrate with respect to yfirst, then with respect to x.
Z2x
x2
xy dy =x1
2y22x
x2
=x1
2(2x)2−1
2(x2)2
=x2x2−1
2x4
= 2x3−1
2x5
Now we integrate this with respect to x:
Z2
0
(2x3−1
2x5)dx =1
2x4−1
12x62
0
=1
2(2)4−1
12(2)6−1
2(0)4−1
12(0)6
= (8 −64
3)−0
=24
3−64
3
=−40
3
Therefore, the value of the double integral RRRxy dA over the region Ris
−40
3.
Question 7
Question
Evaluate the double integral RRDex2+y2dx dy, where Dis the region bounded
by the curve y=x2and the lines x= 0 and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the curves y=x2,x= 0, and x= 1. To find
the limits of integration for x, we need to find the x-values where the curves
intersect.
Setting y=x2, we have:
x2=x⇒x2−x= 0 ⇒x(x−1) = 0
7
This gives us x= 0 and x= 1 as the x-values where the curves intersect.
So, the limits of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:
For each xin the region D, the corresponding yvalues are between the
curves y= 0 and y=x2.
So, the limits of integration for yare 0≤y≤x2.
Therefore, the double integral becomes:
ZZD
ex2+y2dx dy =Z1
0Zx2
0
ex2+y2dy dx
Step 3: Evaluate the integral:
Zx2
0
ex2+y2dy =ex2Zx2
0
ey2dy
Now, we use the substitution u=y2⇒du = 2y dy:
ex2Zx2
0
ey2dy =ex2
2Zx4
0
eudu
=ex2
2[eu]x4
0=ex2
2(ex4
−1)
Now, we integrate this expression with respect to xfrom 0to 1:
Z1
0
ex2
2(ex4
−1) dx =1
2Z1
0
ex2+x4
−ex2dx
This integral is not easy to solve in terms of elementary functions.
Question 8
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Begin by determining the intersection points of the curves y=x2and
y= 2x: Let’s set x2= 2xand solve for x.
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
8
Step 2: Determine the limits of integration in terms of x: We will integrate
with respect to yfirst. The lower bound is given by the curve y=x2and the
upper bound is given by the line y= 2x. Therefore, for a fixed x,yranges from
x2to 2x.
Step 3: Set up the double integral: The given double integral can be ex-
pressed as:
Z2
0Z2x
x2
(x+y)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x+y)dy =x(y) + y2
2
2x
x2
=x(2x) + (2x)2
2−x(x2)−(x2)2
2
= 2x2+ 2x2−x3−x4
2
= 4x2−x3−x4
2
Step 5: Evaluate the outer integral with respect to x:
Z2
0
(4x2−x3−x4
2)dx =4x3
3−x4
4−x5
10
2
0
=32
3−4−32
25 =40
3−32
25 =320 −96
75 =224
75
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 224
75 .
Question 9
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x,y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. The region Ris defined
by the curves y=x,y= 2x,x= 0, and x= 1. We will first find the bounds
for xand y.
For the lower bound of x,x= 0. For the upper bound of x, we need to find
the intersection point of y=xand y= 2x:x= 0 and y= 0, and y= 2(0) = 0.
Thus, the bounds for xare from 0to 1.
9
For the lower bound of y,y=x, and for the upper bound of y,y= 2x.
Thus, the bounds for yare from xto 2x.
Step 2: Set up the double integral with the appropriate bounds. The given
double integral can be expressed as:
Z1
0Z2x
x
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x
ex2+y2dy =hex2+y2iy=2x
y=x=e5x2
−e2x2
Step 4: Evaluate the outer integral with respect to x.
Z1
0
(e5x2
−e2x2)dx =1
5e5x2
−1
2e2x21
0
=1
5e5−1
2e2−1
5−1
2
Step 5: Simplify the expression.
1
5e5−1
2e2−1
5−1
2=1
5e5−1
2e2+1
10
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e5−1
2e2+1
10 .
Question 10
Question
Evaluate the double integral RRR(6x2−2y)dA where Ris the region bounded
by the curves y=x3,y= 0,x= 1, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Ris bounded
by the following curves: 1. y=x32. y= 0 3. x= 1 4. x= 2
Thus, the limits of integration for xare 1≤x≤2, and the limits of inte-
gration for yare 0≤y≤x3.
Step 2: Set up the integral. The given double integral is
ZZR
(6x2−2y)dA
We can rewrite this as the iterated integral:
Z2
1Zx3
0
(6x2−2y)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Zx3
0
(6x2−2y)dy =6x2y−y2x3
0
= 6x2(x3)−(x3)2−0
= 6x5−x6
Step 4: Substitute the result of the inner integral back into the original
expression and evaluate the outer integral with respect to x.
Z2
1
(6x5−x6)dx =6
6x6−1
7x72
1
=x6−1
7x7
2
1
= (26−1
727)−(16−1
717)
= (64 −128
7)−(1 −1
7)
=256
7−135
7
=121
7
Therefore, the value of the double integral over region Ris 121
7.
Question 11
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 2x,x= 0, and
x= 1. Calculate the double integral RRRy dA.
Solution
To compute the double integral RRRy dA, we first need to determine the bounds
of integration for xand yin the region R.
Step 1: Determining the bounds for xand y
The region Ris bounded by the curves y=x2and y= 2x, and lines x= 0
and x= 1. Therefore, for y, the lower bound is x2, the upper bound is 2x. For
x, the lower bound is 0 and the upper bound is 1.
Step 2: Setting up the integral
The double integral RRRy dA can be expressed as:
ZZR
y dA =Z1
0Z2x
x2
y dy dx
11
Step 3: Integrating with respect to y
Integrating with respect to yfirst, we have:
Z2x
x2
y dy =1
2y22x
x2
=1
2(4x2−x4) = 2x2−1
2x4
Step 4: Integrating with respect to x
Now, we integrate 2x2−1
2x4with respect to xover the bounds 0 to 1:
Z1
0
2x2−1
2x4dx =2
3x3−1
10x51
0
=2
3−1
10 =17
30
Step 5: Final answer
Therefore, the value of RRRy dA over the region Ris 17
30 .
Question 12
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
Step 1: Determine the limits of integration by finding the points of intersection
of the two curves.
x2= 2x
x2−2x= 0
x(x−2) = 0
x= 0, x = 2
So, the limits of integration for xare from 0 to 2.
Step 2: To find the limits of integration for y, we set up the inequalities for
R:
x2≤y≤2x
0≤x2≤y≤2x
Since y=x2is the lower curve and y= 2xis the upper curve, the limits of
integration for yare from x2to 2x.
Step 3: Rewrite the given integral with the limits of integration:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
12
Step 4: Integrate with respect to yfirst, keeping xconstant:
Z2x
x2
ex2+y2dy =ex2Z2x
x2
ey2dy
=ex2Z2x
x2
ey2dy
=ex2hey2i2x
x2
=ex2(e(2x)2
−ex4)
=ex2(e4x2
−ex4)
Step 5: Integrate the result from step 4 with respect to x:
Z2
0
ex2(e4x2
−ex4)dx =Z2
0
ex2+4x2
−ex2+x4dx
=Z2
0
e5x2
−ex2+x4dx
Step 6: Integrate to find the final result:
Z2
0
e5x2
−ex2+x4dx =1
5e5x22
0−1
3ex4+x22
0
=1
5e20 −1
5−1
3e16 −1
3
=1
5e20 −1
5−1
3e16 +1
3
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e20 −1
5−1
3e16 +1
3.
Question 13
Question
Evaluate the double integral RRDex2+y2dx dy over the region Denclosed by the
curves y=x2,y= 2,x= 0, and x= 1.
Solution
To evaluate the double integral RRDex2+y2dx dy over the region D, we first need
to determine the limits of integration.
Step 1: Determine the limits of integration.
The region Dis enclosed by the curves y=x2,y= 2,x= 0, and x= 1.
This implies that our limits of integration are: - For y:x2≤y≤2- For x:
0≤x≤1
13
Thus, the double integral becomes:
Z1
0Z2
x2
ex2+y2dy dx
Step 2: Evaluate the inner integral with respect to y.
Z2
x2
ex2+y2dy
=hex2+y2i2
x2
=ex2+4 −e2x2
Step 3: Evaluate the outer integral with respect to x.
Z1
0ex2+4 −e2x2dx
=Zex2+4 dx −Ze2x2dx1
0
=1
2ex2+4 −1
2√2e2x21
0
=1
2e5−1
2√2e6−1
2e4+1
2√2
Therefore, the value of the double integral RRDex2+y2dx dy over the region
Dis 1
2e5−1
2√2e6−1
2e4+1
2√2.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
lines y=x,y= 2x, and y= 4.
Solution
Step 1: Sketch the region Rto visualize the bounds of integration. The region
Ris bounded by the lines y=x,y= 2x, and y= 4. It is a triangular region in
the first quadrant.
Step 2: Determine the limits of integration. To evaluate the double integral
RRRex+ydA, we integrate with respect to xand yover the region R.
For the limits of integration with respect to y, we need to find the y-values
where the lines y=x,y= 2x, and y= 4 intersect. Setting y=xand y= 2x
equal to each other gives x= 2x, which simplifies to x= 0. This means the
14
lines intersect at the point (0,0). Setting y=xand y= 4 equal to each other
gives x= 4. This means the lines intersect at the point (4,4).
Thus, the limits of integration for yare from xto 4.
For the limits of integration with respect to x, we must determine the range
of xvalues for which xis between the lines y=xand y= 2x. This will give us
the limits of integration for x.
The line y=xintersects y= 2xat (0,0). So, xranges from 0to y/2.
Therefore, the limits of integration are:
0≤x≤y/2,0≤y≤4
Step 3: Evaluate the double integral. Now we can evaluate the double
integral:
ZZR
ex+ydA
=Z4
0Zy
2
0
ex+ydx dy
=Z4
0ex+yy
2
0dy
=Z4
0
(ey
2+y−ey)dy
=Z4
0
e3y
2−eydy
=2
3e3y
2−ey4
0
=2
3e6−e4−2
3−1
=2
3e6−e4+1
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
2
3e6−e4+1
3.
Question 15
Question
Let Rbe the region in the first quadrant bounded by y=x2and y= 2x.
Evaluate the double integral RRR
1
ydA.
Solution
Step 1: To find the limits of integration for xand y, we need to determine the
intersection points of the curves y=x2and y= 2x. Setting the two equations
15
equal gives us:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies either x= 0 or x= 2.
Step 2: Since the region Ris in the first quadrant, we take 0≤x≤2and
x2≤y≤2xas our limits of integration.
Step 3: The double integral becomes:
Z2
0Z2x
x2
1
ydy dx
Step 4: Solving the inner integral first, we have:
Z2x
x2
1
ydy = ln(2x)−lnx2= ln 2x
x2= ln 2
x
Step 5: Now, the double integral becomes:
Z2
0
ln 2
xdx
Step 6: Evaluating the integral gives:
xln 2
x−22
0
=2 ln 2
2−2−0 ln 2
0−0
= 2 ln(1) −2 = 2 ·0−2 = −2
Question 16
Question
Evaluate the double integral RRR(x2−y2)dA, where Ris the region in the first
quadrant bounded by the lines y=x,y= 2x, and the coordinate axes.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the lines y=xand y= 2x.
16
• For x, the limits of integration are from 0 to the x-value where y= 2x
intersects the x-axis, which is x= 0.
• For y, the limits of integration are from y=xto y= 2x.
Step 2: Set up the double integral The double integral is given by:
ZZR
(x2−y2)dA =Za
0Z2x
x
(x2−y2)dydx
Step 3: Evaluate the inner integral with respect to y
Z2x
x
(x2−y2)dy
=x2y−y3
3
2x
x
= (2x3−(2x)3
3)−(x3−x3
3)
= (2x3−8x3
3)−(x3−x3
3)
=2x3
3−8x3
3+2x3
3
=−4x3
3
Step 4: Evaluate the outer integral with respect to x
Za
0−4x3
3dx
=−4
3Za
0
x3dx
=−4
3x4
4a
0
=−1
3a4
Therefore, the value of the double integral RRR(x2−y2)dA over the region
Ris −1
3a4.
Question 17
Question
Evaluate the double integral RRRex2
−y2dA, where Ris the region bounded by
the curves y=x2and y= 2x.
17
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Determine limits of integration for x:The region Ris bounded
by y=x2and y= 2x. Setting these two equations equal gives us x2= 2x,
which simplifies to x2−2x= 0, or x(x−2) = 0. So, x= 0 and x= 2 are the
solutions.
Hence, the limits of integration for xare from 0 to 2.
Step 2: Determine limits of integration for y:The region Ris bounded
by y=x2and y= 2x. Therefore, the limits of integration for yare from x2to
2x.
Step 3: Set up and evaluate the double integral: The given double
integral is RRRex2
−y2dA. So we have:
Z2
0Z2x
x2
ex2
−y2dydx
Now evaluate this double integral:
Z2
0−1
2ex2
−y22x
x2
dx
=Z2
0−1
2ex2
−(2x)2+1
2ex2
−(x2)dx
=Z2
0−1
2ex2
−4x2+1
2e0dx
=Z2
0−1
2e−3x2+1
2dx
We will need to use a numerical method or software to evaluate the integral
of e−3x2, as it does not have a standard anti-derivative. This will yield the final
numerical result.
Question 18
Question
Evaluate the double integral RRRxye−x2
−y2dA over the region Rbounded by
the curves x= 0,x= 1,y= 0, and y=√3.
18
Solution
Step 1: Rewrite the given integral in polar coordinates. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =r dr dθ. The region Rbecomes 0≤r≤1,
0≤θ≤π
3. Therefore, our integral becomes:
Zπ
3
0Z1
0
r2cos(θ) sin(θ)e−r2dr dθ
Step 2: Evaluate the inner integral.
Z1
0
r2cos(θ) sin(θ)e−r2dr
Let u=r2, then du = 2r dr. The integral now becomes:
Z1
0
1
2cos(θ) sin(θ)e−udu
Evaluating this integral gives:
−1
4cos(θ) sin(θ)e−u
1
0
=−1
4cos(θ) sin(θ)(e−1−1)
Step 3: Evaluate the outer integral.
Zπ
3
0−1
4cos(θ) sin(θ)(e−1−1) dθ
This integral can be simplified to:
−1
4(e−1−1) Zπ
3
0
cos(θ) sin(θ)dθ
Using the identity Rcos(θ) sin(θ)dθ =−1
2cos2(θ) + C, we have:
−1
4(e−1−1) −1
2cos2(θ)
π
3
0
Evaluating this further gives the final answer.
Question 19
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2and y= 2x.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for the given region R. This region Ris bounded by the curves y=x2
and y= 2x. Let’s find the points of intersection between these two curves by
setting them equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Now, fixing xwithin the bounds x= 0 and x= 2, the corresponding yvalues
can be found by plugging them into the equations y=x2and y= 2x.
Step 1: Find the limits of integration At x= 0:
y1=x2= 02= 0
y2= 2x= 2(0) = 0
At x= 2:
y1=x2= 22= 4
y2= 2x= 2(2) = 4
Therefore, the limits of integration are 0≤x≤2and 0≤y≤4.
Step 2: Evaluate the double integral RRR(x+y)dA We can now write
the given double integral in terms of xand yusing the limits of integration we
found earlier:
ZZR
(x+y)dA =Z2
0Z4
0
(x+y)dy dx
Let’s evaluate the inner integral first:
Z4
0
(x+y)dy =xy +y2
24
0
= 4x+ 8
Now, the double integral becomes:
Z2
0
(4x+ 8) dx =2x2+ 8x2
0= 8 + 16 = 24
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 24.
Question 20
Question
Evaluate the double integral RRD(x+y)dA, where Dis the region bounded by
the curves y=x2,y= 2x,x= 1, and x= 2.
20
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand y.
Step 1: Determine the limits of integration for x. The region Dis bounded
by the curves y=x2,y= 2x,x= 1, and x= 2. We can find the points
of intersection of these curves by setting them equal to each other. Solving
x2= 2x, we get x2−2x=x(x−2) = 0, which gives x= 0 and x= 2.
However, we only consider the intersection points within the interval 1≤
x≤2, so x= 0 is not relevant to our limits of integration for x. Therefore, the
limits of integration for xare 1≤x≤2.
Step 2: Determine the limits of integration for y. The lower bound for y
is given by the curve y=x2, and the upper bound is given by the line y= 2x.
Thus, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral. Putting it all together,
the given double integral becomes:
ZZD
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
Now, we evaluate this double integral:
Z2
1Z2x
x2
(x+y)dy dx =Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2+(x2)2
2)dx
=Z2
13x2−3
2x3dx
=x3−3
8x42
1
= (23−3
8·24)−(13−3
8·14)
= (8 −6) −(1 −3
8)
= 2 −5
8=11
8
Therefore, the value of the double integral RRD(x+y)dA over the region D
is 11
8.
Question 21
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x.
21
Solution
Step 1: First, we need to find the intersection points of the two curves. To find
the intersection points, we set x2= 2xand solve for x:x2−2x= 0
x(x−2) = 0
x= 0 or x= 2 So the intersection points are (0,0) and (2,4).
Step 2: Next, we need to determine the bounds of integration for xand y.
The region Dis bounded by the curves y=x2and y= 2x. Thus, for x, we will
integrate from 0 to 2, and for y, we will integrate from x2to 2x.
Step 3: Rewrite the integral in terms of xand ybounds. The double integral
becomes: R2
0R2x
x2exy dy dx
Step 4: Integrate with respect to yfirst. R2x
x2exy dy =1
xexy2x
x2
=1
xe2x2−1
xex3
Step 5: Integrate the result with respect to x.R2
01
xe2x2−1
xex3dx
=he2x2−ex3i2
0
=e8−e8−e2+e2
= 0
Therefore, the value of the double integral RRDexy dA over the region Dis
0.
Question 22
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over R, we need to set up the integral using the
given bounds and then compute the integral.
Step 1: Determine the bounds for the integral in terms of xand y. The
region Ris bounded by the curves y=x2,y= 2x, and x= 0. To find the limits
of integration, we need to determine the intersection points of these curves.
Let’s first find the x-coordinates of the intersection points x1and x2:
Setting x2= 2x, we get x2−2x= 0, which factors as x(x−2) = 0. So,
x= 0 or x= 2.
Therefore, the bounds for xwill be 0≤x≤2.
Now for y, the limits are given by the corresponding equations y=x2and
y= 2x. Since y=x2is the lower curve and y= 2xis the upper curve in the
region R, the bounds for ywill be x2≤y≤2x.
Step 2: Set up and evaluate the double integral. The double integral can
22
be set up as follows:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydy dx
Now, let’s compute this double integral:
Z2
0Z2x
x2
ex+ydy dx =Z2
0
exZ2x
x2
eydydx
=Z2
0
ex[ey]2x
x2dx
=Z2
0
ex(e2x−ex2)dx
=Z2
0
e3x−ex+x2dx
="e3x
3−ex+x2
1 + x2#2
0
=e6
3−e5
5−1
5
Therefore, the value of the double integral RRRex+ydA over the region Ris
e6
3−e5
5−1
5.
Question 23
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand yin the region R.
Step 1: Determine the limits of integration for x:
The curves y=x2and y= 2xintersect at the points where x2= 2x:x2−2x=
0⇒x(x−2) = 0 ⇒x= 0,2.
Therefore, the limits of integration for xare 0to 2.
Step 2: Determine the limits of integration for y:
For each value of xin the region R,yranges from x2to 2x. Therefore, the
limits of integration for yare x2to 2x.
Step 3: Set up the double integral:
The given double integral can be written as:
ZZR
ex2+y2dA =Zx=2
x=0 Zy=2x
y=x2
ex2+y2dy dx
23
Step 4: Evaluate the inner integral:
Zy=2x
y=x2
ex2+y2dy =ex2Zy=2x
y=x2
ey2
−x2dy
Note that the integral Rey2
−x2dy cannot be expressed in terms of elementary
functions. Therefore, the exact evaluation requires special techniques involving
Gaussian integrals or error functions.
Thus, the final answer would involve these special functions.
Question 24
Question
Evaluate the double integral RRDxy dA, where Dis the region bounded by the
lines y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral RRDxy dA over the region D, we first need to
determine the bounds of integration.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the lines y=xand y= 2x. These two lines
intersect at the point (0,0).
The line y=xintersects the line x= 1 at (1,1), and the line y= 2x
intersects x= 1 at (1,2).
Thus, the limits of integration for xare from x= 0 to x= 1.
Step 2: Determine the limits of integration for y:
The region Dis bounded by the lines y=xand y= 2x. To express yin
terms of xfor this region, we solve the system of equations y=xand y= 2x
to find y= 2x.
Thus, the limits of integration for yare from y=xto y= 2x.
Step 3: Set up and evaluate the double integral:
The double integral over the region Dis given by:
ZZD
xy dA =Z1
0Z2x
x
xy dy dx
Now, we evaluate the double integral:
Z1
0Z2x
x
xy dy dx =Z1
01
2x(2x)2−1
2x(x)2dx
=Z1
02x3−1
2x3dx =Z1
0
3
2x3dx
24
=3
8x4
1
0
=3
8(1)4−3
8(0)4=3
8
Therefore, the value of the double integral RRDxy dA over the region Dis 3
8.
Question 25
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
circle x2+y2= 1.
Solution
To evaluate the double integral RRRx2y dA over the given region R, we will
convert the integral to polar coordinates.
Step 1: Convert to Polar Coordinates The region Rcan be described
in polar coordinates as 0≤r≤1and 0≤θ≤2π. Also, dA =rdrdθ.
Step 2: Rewrite the Integral Plugging in the polar coordinates, we
obtain:
ZZR
x2y dA =Z2π
0Z1
0
(rcos2θ)(rsin θ)r drdθ
Step 3: Compute the Integral Let’s first simplify the integrand:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
0Z1
0
(r3cos2θsin θ)drdθ
To evaluate this integral, we perform the following steps:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
01
4cos2θsin θ1
0
dθ
Z2π
0
1
4cos2θsin θ dθ =1
4Z2π
0
cos2θsin θ dθ
This integral can be evaluated using trigonometric identities or by using
integration by parts.
Therefore, the value of the double integral RRRx2y dA over the region Ris
π
4.
25
Step 3: Integrate the previous result with respect to y.
Z1
01
2(e1+y2
−ey2)dy =1
2Z1
0
e1+y2dy −1
2Z1
0
ey2dy
Integrating, we get:
1
21
e−1e1+y2
1
0−1
2√πerf(1)
Step 4: Simplifying the expression gives the final answer:
=1
21
e−1(e2−1) −1
2√πerf(1)
Question 2
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 4,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y.
The region Ris bounded by the curves y=x2,y= 4,x= 0, and x= 2.
To determine the limits of integration, we need to find the intersection points
of these curves.
- The curves y=x2and y= 4 intersect when x2= 4. Solving for xgives us
x=±2, but since xmust be non-negative in this case, the intersection point is
at x= 2. - The curve x= 0 is the y-axis. - The limits for xare from 0 to 2. -
The limits for yare from x2to 4.
Step 2: Set up and evaluate the double integral.
The double integral is given as:
ZZR
(x+y)dA =Z2
0Z4
x2
(x+y)dy dx
Now, we integrate with respect to yfirst and then with respect to x:
2
Z2
0Z4
x2
(x+y)dy dx =Z2
0xy +y2
24
x2
dx
=Z2
04x+16
2−x3−x4
2dx
=Z2
0
(4x+ 8 −x3−x4
2)dx
=2x2+ 8x−x4
4−x5
10 2
0
= 8 + 16 −4−32
10
= 20 −4−16
5
= 16 −16
5
=64
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 64
5.
Question 3
Question
Evaluate the double integral RRD(x2+y2)dx dy over the region Dbounded by
the curves y=x,y= 4 −x2,x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration, we need to determine the points of intersection between the curves
y=xand y= 4 −x2. Setting x=y, we have y= 4 −y2, which simplifies to
y2+y−4 = 0. Solving this quadratic equation gives y= 1 or y=−4, but since
y=x, we discard the negative value. Thus, the limits of integration for yare
1≤y≤4−x2.
Next, we determine the limits of integration for x. Since the region is
bounded by x= 0, we have 0≤x≤2.
Step 2: Evaluate the double integral. We can now rewrite the double integral
as:
ZZD
(x2+y2)dx dy =Z2
0Z4−x2
1
(x2+y2)dy dx
3
Step 3: Integrate with respect to yfirst.
Z4−x2
1
(x2+y2)dy =x2y+y3
3
4−x2
1
Step 4: Substitute the limits of integration and simplify.
=x2(4 −x2) + (4 −x2)3
3−(x21 + 1
3) = 2x2+7x4
3−23x2
3+23
3
Step 5: Integrate the result with respect to x.
Z2
02x2+7x4
3−23x2
3+23
3dx =2
3x3+7
15x5−23
9x3+23
3x
2
0
Step 6: Substitute the limits of integration and calculate the final answer.
=23
3+7(25)
15 −23(23)
9+23(2)
3−0
3+0
15 −0
9+23(0)
3
=8
3+64
15 −64
3+46
3
=−160
15
=−32
3
Therefore, the value of the double integral RRD(x2+y2)dx dy over the region
Dis −32
3.
Question 4
Question
Evaluate the double integral RRRsin(x+y)dA, where Ris the region bounded
by the lines y= 0,y=x,y= 2x, and y= 2.
Solution
Step 1: Sketch the region Rdescribed by the given bounds for yin terms of x.
xyy =xy = 2xy = 2
Step 2: Set up the double integral using these bounds:
ZZR
sin(x+y)dA =Z1
0Z2x
x
sin(x+y)dydx
4
Step 3: Integrate with respect to yfirst:
Z2x
x
sin(x+y)dy = [−cos(x+y)]2x
x=−cos(3x) + cos(2x)
Step 4: Now integrate the result from Step 3 with respect to x:
Z1
0
(−cos(3x) + cos(2x)) dx =−1
3sin(3x) + 1
2sin(2x)1
0
=−1
3sin(3) + 1
2sin(2)−(0 + 0)
=−1
3sin(3) + 1
2sin(2)
Thus, the value of the given double integral is −1
3sin(3) + 1
2sin(2).
Question 5
Question
Calculate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x,y= 2x, and y= 1.
Solution
To find the double integral RRRxy2dA over the region R, we first need to de-
termine the bounds of integration by examining the limits along the xand y
axes.
Step 1: Determine the bounds of integration in terms of x:
Since y=xand y= 2xare the bounding curves, we can set up the following
inequalities:
x≤y≤2x
Solving these inequalities for x, we find that 0≤x≤1
2.
Step 2: Determine the bounds of integration in terms of y:
Next, we consider the bounding curve y= 1. Since this line intersects the
region at y= 1, the upper bound for ywill be 1.
Step 3: Set up the double integral:
Putting the bounds together, the double integral becomes:
Z1
2
0Z2x
x
xy2dy dx
Step 4: Integrate with respect to y:
Integrating the inner integral with respect to y, we get:
Z1
2
01
3xy32x
x
dx
5
=Z1
2
07
3x4dx
Step 5: Integrate with respect to x:
Integrating the remaining integral with respect to x, we have:
7
15x5
1
2
0
=7
15 ·1
25
=7
15 ·1
32
=7
480
Therefore, the value of the double integral RRRxy2dA over the region Ris
7
480 .
Question 6
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy dA, we first need to determine the limits
of integration for xand yover the region R.
Step 1: Determine the limits of integration for xThe region Ris
bounded by the curves y=x2and y= 2x. To find the limits of integration
for x, we need to find the x-values at which these two curves intersect. Setting
x2= 2x, we find the intersection points at x= 0 and x= 2.
Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for yFor each value of x
in the region R,yvaries between the curves y=x2(bottom curve) and y= 2x
(top curve).
So, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral The given double
integral is RRRxy dA.
Rewriting the integrand in terms of xand y, we have xy.
Therefore, the double integral becomes:
Z2
0Z2x
x2
xy dy dx
6
Now, we integrate with respect to yfirst, then with respect to x.
Z2x
x2
xy dy =x1
2y22x
x2
=x1
2(2x)2−1
2(x2)2
=x2x2−1
2x4
= 2x3−1
2x5
Now we integrate this with respect to x:
Z2
0
(2x3−1
2x5)dx =1
2x4−1
12x62
0
=1
2(2)4−1
12(2)6−1
2(0)4−1
12(0)6
= (8 −64
3)−0
=24
3−64
3
=−40
3
Therefore, the value of the double integral RRRxy dA over the region Ris
−40
3.
Question 7
Question
Evaluate the double integral RRDex2+y2dx dy, where Dis the region bounded
by the curve y=x2and the lines x= 0 and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the curves y=x2,x= 0, and x= 1. To find
the limits of integration for x, we need to find the x-values where the curves
intersect.
Setting y=x2, we have:
x2=x⇒x2−x= 0 ⇒x(x−1) = 0
7
This gives us x= 0 and x= 1 as the x-values where the curves intersect.
So, the limits of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:
For each xin the region D, the corresponding yvalues are between the
curves y= 0 and y=x2.
So, the limits of integration for yare 0≤y≤x2.
Therefore, the double integral becomes:
ZZD
ex2+y2dx dy =Z1
0Zx2
0
ex2+y2dy dx
Step 3: Evaluate the integral:
Zx2
0
ex2+y2dy =ex2Zx2
0
ey2dy
Now, we use the substitution u=y2⇒du = 2y dy:
ex2Zx2
0
ey2dy =ex2
2Zx4
0
eudu
=ex2
2[eu]x4
0=ex2
2(ex4
−1)
Now, we integrate this expression with respect to xfrom 0to 1:
Z1
0
ex2
2(ex4
−1) dx =1
2Z1
0
ex2+x4
−ex2dx
This integral is not easy to solve in terms of elementary functions.
Question 8
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Begin by determining the intersection points of the curves y=x2and
y= 2x: Let’s set x2= 2xand solve for x.
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
8
Step 2: Determine the limits of integration in terms of x: We will integrate
with respect to yfirst. The lower bound is given by the curve y=x2and the
upper bound is given by the line y= 2x. Therefore, for a fixed x,yranges from
x2to 2x.
Step 3: Set up the double integral: The given double integral can be ex-
pressed as:
Z2
0Z2x
x2
(x+y)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x+y)dy =x(y) + y2
2
2x
x2
=x(2x) + (2x)2
2−x(x2)−(x2)2
2
= 2x2+ 2x2−x3−x4
2
= 4x2−x3−x4
2
Step 5: Evaluate the outer integral with respect to x:
Z2
0
(4x2−x3−x4
2)dx =4x3
3−x4
4−x5
10
2
0
=32
3−4−32
25 =40
3−32
25 =320 −96
75 =224
75
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 224
75 .
Question 9
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x,y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. The region Ris defined
by the curves y=x,y= 2x,x= 0, and x= 1. We will first find the bounds
for xand y.
For the lower bound of x,x= 0. For the upper bound of x, we need to find
the intersection point of y=xand y= 2x:x= 0 and y= 0, and y= 2(0) = 0.
Thus, the bounds for xare from 0to 1.
9
For the lower bound of y,y=x, and for the upper bound of y,y= 2x.
Thus, the bounds for yare from xto 2x.
Step 2: Set up the double integral with the appropriate bounds. The given
double integral can be expressed as:
Z1
0Z2x
x
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x
ex2+y2dy =hex2+y2iy=2x
y=x=e5x2
−e2x2
Step 4: Evaluate the outer integral with respect to x.
Z1
0
(e5x2
−e2x2)dx =1
5e5x2
−1
2e2x21
0
=1
5e5−1
2e2−1
5−1
2
Step 5: Simplify the expression.
1
5e5−1
2e2−1
5−1
2=1
5e5−1
2e2+1
10
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e5−1
2e2+1
10 .
Question 10
Question
Evaluate the double integral RRR(6x2−2y)dA where Ris the region bounded
by the curves y=x3,y= 0,x= 1, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Ris bounded
by the following curves: 1. y=x32. y= 0 3. x= 1 4. x= 2
Thus, the limits of integration for xare 1≤x≤2, and the limits of inte-
gration for yare 0≤y≤x3.
Step 2: Set up the integral. The given double integral is
ZZR
(6x2−2y)dA
We can rewrite this as the iterated integral:
Z2
1Zx3
0
(6x2−2y)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Zx3
0
(6x2−2y)dy =6x2y−y2x3
0
= 6x2(x3)−(x3)2−0
= 6x5−x6
Step 4: Substitute the result of the inner integral back into the original
expression and evaluate the outer integral with respect to x.
Z2
1
(6x5−x6)dx =6
6x6−1
7x72
1
=x6−1
7x7
2
1
= (26−1
727)−(16−1
717)
= (64 −128
7)−(1 −1
7)
=256
7−135
7
=121
7
Therefore, the value of the double integral over region Ris 121
7.
Question 11
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 2x,x= 0, and
x= 1. Calculate the double integral RRRy dA.
Solution
To compute the double integral RRRy dA, we first need to determine the bounds
of integration for xand yin the region R.
Step 1: Determining the bounds for xand y
The region Ris bounded by the curves y=x2and y= 2x, and lines x= 0
and x= 1. Therefore, for y, the lower bound is x2, the upper bound is 2x. For
x, the lower bound is 0 and the upper bound is 1.
Step 2: Setting up the integral
The double integral RRRy dA can be expressed as:
ZZR
y dA =Z1
0Z2x
x2
y dy dx
11
Step 3: Integrating with respect to y
Integrating with respect to yfirst, we have:
Z2x
x2
y dy =1
2y22x
x2
=1
2(4x2−x4) = 2x2−1
2x4
Step 4: Integrating with respect to x
Now, we integrate 2x2−1
2x4with respect to xover the bounds 0 to 1:
Z1
0
2x2−1
2x4dx =2
3x3−1
10x51
0
=2
3−1
10 =17
30
Step 5: Final answer
Therefore, the value of RRRy dA over the region Ris 17
30 .
Question 12
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
Step 1: Determine the limits of integration by finding the points of intersection
of the two curves.
x2= 2x
x2−2x= 0
x(x−2) = 0
x= 0, x = 2
So, the limits of integration for xare from 0 to 2.
Step 2: To find the limits of integration for y, we set up the inequalities for
R:
x2≤y≤2x
0≤x2≤y≤2x
Since y=x2is the lower curve and y= 2xis the upper curve, the limits of
integration for yare from x2to 2x.
Step 3: Rewrite the given integral with the limits of integration:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
12
Step 4: Integrate with respect to yfirst, keeping xconstant:
Z2x
x2
ex2+y2dy =ex2Z2x
x2
ey2dy
=ex2Z2x
x2
ey2dy
=ex2hey2i2x
x2
=ex2(e(2x)2
−ex4)
=ex2(e4x2
−ex4)
Step 5: Integrate the result from step 4 with respect to x:
Z2
0
ex2(e4x2
−ex4)dx =Z2
0
ex2+4x2
−ex2+x4dx
=Z2
0
e5x2
−ex2+x4dx
Step 6: Integrate to find the final result:
Z2
0
e5x2
−ex2+x4dx =1
5e5x22
0−1
3ex4+x22
0
=1
5e20 −1
5−1
3e16 −1
3
=1
5e20 −1
5−1
3e16 +1
3
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e20 −1
5−1
3e16 +1
3.
Question 13
Question
Evaluate the double integral RRDex2+y2dx dy over the region Denclosed by the
curves y=x2,y= 2,x= 0, and x= 1.
Solution
To evaluate the double integral RRDex2+y2dx dy over the region D, we first need
to determine the limits of integration.
Step 1: Determine the limits of integration.
The region Dis enclosed by the curves y=x2,y= 2,x= 0, and x= 1.
This implies that our limits of integration are: - For y:x2≤y≤2- For x:
0≤x≤1
13
Thus, the double integral becomes:
Z1
0Z2
x2
ex2+y2dy dx
Step 2: Evaluate the inner integral with respect to y.
Z2
x2
ex2+y2dy
=hex2+y2i2
x2
=ex2+4 −e2x2
Step 3: Evaluate the outer integral with respect to x.
Z1
0ex2+4 −e2x2dx
=Zex2+4 dx −Ze2x2dx1
0
=1
2ex2+4 −1
2√2e2x21
0
=1
2e5−1
2√2e6−1
2e4+1
2√2
Therefore, the value of the double integral RRDex2+y2dx dy over the region
Dis 1
2e5−1
2√2e6−1
2e4+1
2√2.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
lines y=x,y= 2x, and y= 4.
Solution
Step 1: Sketch the region Rto visualize the bounds of integration. The region
Ris bounded by the lines y=x,y= 2x, and y= 4. It is a triangular region in
the first quadrant.
Step 2: Determine the limits of integration. To evaluate the double integral
RRRex+ydA, we integrate with respect to xand yover the region R.
For the limits of integration with respect to y, we need to find the y-values
where the lines y=x,y= 2x, and y= 4 intersect. Setting y=xand y= 2x
equal to each other gives x= 2x, which simplifies to x= 0. This means the
14
lines intersect at the point (0,0). Setting y=xand y= 4 equal to each other
gives x= 4. This means the lines intersect at the point (4,4).
Thus, the limits of integration for yare from xto 4.
For the limits of integration with respect to x, we must determine the range
of xvalues for which xis between the lines y=xand y= 2x. This will give us
the limits of integration for x.
The line y=xintersects y= 2xat (0,0). So, xranges from 0to y/2.
Therefore, the limits of integration are:
0≤x≤y/2,0≤y≤4
Step 3: Evaluate the double integral. Now we can evaluate the double
integral:
ZZR
ex+ydA
=Z4
0Zy
2
0
ex+ydx dy
=Z4
0ex+yy
2
0dy
=Z4
0
(ey
2+y−ey)dy
=Z4
0
e3y
2−eydy
=2
3e3y
2−ey4
0
=2
3e6−e4−2
3−1
=2
3e6−e4+1
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
2
3e6−e4+1
3.
Question 15
Question
Let Rbe the region in the first quadrant bounded by y=x2and y= 2x.
Evaluate the double integral RRR
1
ydA.
Solution
Step 1: To find the limits of integration for xand y, we need to determine the
intersection points of the curves y=x2and y= 2x. Setting the two equations
15
equal gives us:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies either x= 0 or x= 2.
Step 2: Since the region Ris in the first quadrant, we take 0≤x≤2and
x2≤y≤2xas our limits of integration.
Step 3: The double integral becomes:
Z2
0Z2x
x2
1
ydy dx
Step 4: Solving the inner integral first, we have:
Z2x
x2
1
ydy = ln(2x)−lnx2= ln 2x
x2= ln 2
x
Step 5: Now, the double integral becomes:
Z2
0
ln 2
xdx
Step 6: Evaluating the integral gives:
xln 2
x−22
0
=2 ln 2
2−2−0 ln 2
0−0
= 2 ln(1) −2 = 2 ·0−2 = −2
Question 16
Question
Evaluate the double integral RRR(x2−y2)dA, where Ris the region in the first
quadrant bounded by the lines y=x,y= 2x, and the coordinate axes.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the lines y=xand y= 2x.
16
• For x, the limits of integration are from 0 to the x-value where y= 2x
intersects the x-axis, which is x= 0.
• For y, the limits of integration are from y=xto y= 2x.
Step 2: Set up the double integral The double integral is given by:
ZZR
(x2−y2)dA =Za
0Z2x
x
(x2−y2)dydx
Step 3: Evaluate the inner integral with respect to y
Z2x
x
(x2−y2)dy
=x2y−y3
3
2x
x
= (2x3−(2x)3
3)−(x3−x3
3)
= (2x3−8x3
3)−(x3−x3
3)
=2x3
3−8x3
3+2x3
3
=−4x3
3
Step 4: Evaluate the outer integral with respect to x
Za
0−4x3
3dx
=−4
3Za
0
x3dx
=−4
3x4
4a
0
=−1
3a4
Therefore, the value of the double integral RRR(x2−y2)dA over the region
Ris −1
3a4.
Question 17
Question
Evaluate the double integral RRRex2
−y2dA, where Ris the region bounded by
the curves y=x2and y= 2x.
17
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Determine limits of integration for x:The region Ris bounded
by y=x2and y= 2x. Setting these two equations equal gives us x2= 2x,
which simplifies to x2−2x= 0, or x(x−2) = 0. So, x= 0 and x= 2 are the
solutions.
Hence, the limits of integration for xare from 0 to 2.
Step 2: Determine limits of integration for y:The region Ris bounded
by y=x2and y= 2x. Therefore, the limits of integration for yare from x2to
2x.
Step 3: Set up and evaluate the double integral: The given double
integral is RRRex2
−y2dA. So we have:
Z2
0Z2x
x2
ex2
−y2dydx
Now evaluate this double integral:
Z2
0−1
2ex2
−y22x
x2
dx
=Z2
0−1
2ex2
−(2x)2+1
2ex2
−(x2)dx
=Z2
0−1
2ex2
−4x2+1
2e0dx
=Z2
0−1
2e−3x2+1
2dx
We will need to use a numerical method or software to evaluate the integral
of e−3x2, as it does not have a standard anti-derivative. This will yield the final
numerical result.
Question 18
Question
Evaluate the double integral RRRxye−x2
−y2dA over the region Rbounded by
the curves x= 0,x= 1,y= 0, and y=√3.
18
Solution
Step 1: Rewrite the given integral in polar coordinates. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =r dr dθ. The region Rbecomes 0≤r≤1,
0≤θ≤π
3. Therefore, our integral becomes:
Zπ
3
0Z1
0
r2cos(θ) sin(θ)e−r2dr dθ
Step 2: Evaluate the inner integral.
Z1
0
r2cos(θ) sin(θ)e−r2dr
Let u=r2, then du = 2r dr. The integral now becomes:
Z1
0
1
2cos(θ) sin(θ)e−udu
Evaluating this integral gives:
−1
4cos(θ) sin(θ)e−u
1
0
=−1
4cos(θ) sin(θ)(e−1−1)
Step 3: Evaluate the outer integral.
Zπ
3
0−1
4cos(θ) sin(θ)(e−1−1) dθ
This integral can be simplified to:
−1
4(e−1−1) Zπ
3
0
cos(θ) sin(θ)dθ
Using the identity Rcos(θ) sin(θ)dθ =−1
2cos2(θ) + C, we have:
−1
4(e−1−1) −1
2cos2(θ)
π
3
0
Evaluating this further gives the final answer.
Question 19
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2and y= 2x.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for the given region R. This region Ris bounded by the curves y=x2
and y= 2x. Let’s find the points of intersection between these two curves by
setting them equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Now, fixing xwithin the bounds x= 0 and x= 2, the corresponding yvalues
can be found by plugging them into the equations y=x2and y= 2x.
Step 1: Find the limits of integration At x= 0:
y1=x2= 02= 0
y2= 2x= 2(0) = 0
At x= 2:
y1=x2= 22= 4
y2= 2x= 2(2) = 4
Therefore, the limits of integration are 0≤x≤2and 0≤y≤4.
Step 2: Evaluate the double integral RRR(x+y)dA We can now write
the given double integral in terms of xand yusing the limits of integration we
found earlier:
ZZR
(x+y)dA =Z2
0Z4
0
(x+y)dy dx
Let’s evaluate the inner integral first:
Z4
0
(x+y)dy =xy +y2
24
0
= 4x+ 8
Now, the double integral becomes:
Z2
0
(4x+ 8) dx =2x2+ 8x2
0= 8 + 16 = 24
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 24.
Question 20
Question
Evaluate the double integral RRD(x+y)dA, where Dis the region bounded by
the curves y=x2,y= 2x,x= 1, and x= 2.
20
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand y.
Step 1: Determine the limits of integration for x. The region Dis bounded
by the curves y=x2,y= 2x,x= 1, and x= 2. We can find the points
of intersection of these curves by setting them equal to each other. Solving
x2= 2x, we get x2−2x=x(x−2) = 0, which gives x= 0 and x= 2.
However, we only consider the intersection points within the interval 1≤
x≤2, so x= 0 is not relevant to our limits of integration for x. Therefore, the
limits of integration for xare 1≤x≤2.
Step 2: Determine the limits of integration for y. The lower bound for y
is given by the curve y=x2, and the upper bound is given by the line y= 2x.
Thus, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral. Putting it all together,
the given double integral becomes:
ZZD
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
Now, we evaluate this double integral:
Z2
1Z2x
x2
(x+y)dy dx =Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2+(x2)2
2)dx
=Z2
13x2−3
2x3dx
=x3−3
8x42
1
= (23−3
8·24)−(13−3
8·14)
= (8 −6) −(1 −3
8)
= 2 −5
8=11
8
Therefore, the value of the double integral RRD(x+y)dA over the region D
is 11
8.
Question 21
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x.
21
Solution
Step 1: First, we need to find the intersection points of the two curves. To find
the intersection points, we set x2= 2xand solve for x:x2−2x= 0
x(x−2) = 0
x= 0 or x= 2 So the intersection points are (0,0) and (2,4).
Step 2: Next, we need to determine the bounds of integration for xand y.
The region Dis bounded by the curves y=x2and y= 2x. Thus, for x, we will
integrate from 0 to 2, and for y, we will integrate from x2to 2x.
Step 3: Rewrite the integral in terms of xand ybounds. The double integral
becomes: R2
0R2x
x2exy dy dx
Step 4: Integrate with respect to yfirst. R2x
x2exy dy =1
xexy2x
x2
=1
xe2x2−1
xex3
Step 5: Integrate the result with respect to x.R2
01
xe2x2−1
xex3dx
=he2x2−ex3i2
0
=e8−e8−e2+e2
= 0
Therefore, the value of the double integral RRDexy dA over the region Dis
0.
Question 22
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over R, we need to set up the integral using the
given bounds and then compute the integral.
Step 1: Determine the bounds for the integral in terms of xand y. The
region Ris bounded by the curves y=x2,y= 2x, and x= 0. To find the limits
of integration, we need to determine the intersection points of these curves.
Let’s first find the x-coordinates of the intersection points x1and x2:
Setting x2= 2x, we get x2−2x= 0, which factors as x(x−2) = 0. So,
x= 0 or x= 2.
Therefore, the bounds for xwill be 0≤x≤2.
Now for y, the limits are given by the corresponding equations y=x2and
y= 2x. Since y=x2is the lower curve and y= 2xis the upper curve in the
region R, the bounds for ywill be x2≤y≤2x.
Step 2: Set up and evaluate the double integral. The double integral can
22
be set up as follows:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydy dx
Now, let’s compute this double integral:
Z2
0Z2x
x2
ex+ydy dx =Z2
0
exZ2x
x2
eydydx
=Z2
0
ex[ey]2x
x2dx
=Z2
0
ex(e2x−ex2)dx
=Z2
0
e3x−ex+x2dx
="e3x
3−ex+x2
1 + x2#2
0
=e6
3−e5
5−1
5
Therefore, the value of the double integral RRRex+ydA over the region Ris
e6
3−e5
5−1
5.
Question 23
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand yin the region R.
Step 1: Determine the limits of integration for x:
The curves y=x2and y= 2xintersect at the points where x2= 2x:x2−2x=
0⇒x(x−2) = 0 ⇒x= 0,2.
Therefore, the limits of integration for xare 0to 2.
Step 2: Determine the limits of integration for y:
For each value of xin the region R,yranges from x2to 2x. Therefore, the
limits of integration for yare x2to 2x.
Step 3: Set up the double integral:
The given double integral can be written as:
ZZR
ex2+y2dA =Zx=2
x=0 Zy=2x
y=x2
ex2+y2dy dx
23
Step 4: Evaluate the inner integral:
Zy=2x
y=x2
ex2+y2dy =ex2Zy=2x
y=x2
ey2
−x2dy
Note that the integral Rey2
−x2dy cannot be expressed in terms of elementary
functions. Therefore, the exact evaluation requires special techniques involving
Gaussian integrals or error functions.
Thus, the final answer would involve these special functions.
Question 24
Question
Evaluate the double integral RRDxy dA, where Dis the region bounded by the
lines y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral RRDxy dA over the region D, we first need to
determine the bounds of integration.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the lines y=xand y= 2x. These two lines
intersect at the point (0,0).
The line y=xintersects the line x= 1 at (1,1), and the line y= 2x
intersects x= 1 at (1,2).
Thus, the limits of integration for xare from x= 0 to x= 1.
Step 2: Determine the limits of integration for y:
The region Dis bounded by the lines y=xand y= 2x. To express yin
terms of xfor this region, we solve the system of equations y=xand y= 2x
to find y= 2x.
Thus, the limits of integration for yare from y=xto y= 2x.
Step 3: Set up and evaluate the double integral:
The double integral over the region Dis given by:
ZZD
xy dA =Z1
0Z2x
x
xy dy dx
Now, we evaluate the double integral:
Z1
0Z2x
x
xy dy dx =Z1
01
2x(2x)2−1
2x(x)2dx
=Z1
02x3−1
2x3dx =Z1
0
3
2x3dx
24
=3
8x4
1
0
=3
8(1)4−3
8(0)4=3
8
Therefore, the value of the double integral RRDxy dA over the region Dis 3
8.
Question 25
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
circle x2+y2= 1.
Solution
To evaluate the double integral RRRx2y dA over the given region R, we will
convert the integral to polar coordinates.
Step 1: Convert to Polar Coordinates The region Rcan be described
in polar coordinates as 0≤r≤1and 0≤θ≤2π. Also, dA =rdrdθ.
Step 2: Rewrite the Integral Plugging in the polar coordinates, we
obtain:
ZZR
x2y dA =Z2π
0Z1
0
(rcos2θ)(rsin θ)r drdθ
Step 3: Compute the Integral Let’s first simplify the integrand:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
0Z1
0
(r3cos2θsin θ)drdθ
To evaluate this integral, we perform the following steps:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
01
4cos2θsin θ1
0
dθ
Z2π
0
1
4cos2θsin θ dθ =1
4Z2π
0
cos2θsin θ dθ
This integral can be evaluated using trigonometric identities or by using
integration by parts.
Therefore, the value of the double integral RRRx2y dA over the region Ris
π
4.
25
Step 3: Integrate the previous result with respect to y.
Z1
01
2(e1+y2
−ey2)dy =1
2Z1
0
e1+y2dy −1
2Z1
0
ey2dy
Integrating, we get:
1
21
e−1e1+y2
1
0−1
2√πerf(1)
Step 4: Simplifying the expression gives the final answer:
=1
21
e−1(e2−1) −1
2√πerf(1)
Question 2
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 4,x= 0, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y.
The region Ris bounded by the curves y=x2,y= 4,x= 0, and x= 2.
To determine the limits of integration, we need to find the intersection points
of these curves.
- The curves y=x2and y= 4 intersect when x2= 4. Solving for xgives us
x=±2, but since xmust be non-negative in this case, the intersection point is
at x= 2. - The curve x= 0 is the y-axis. - The limits for xare from 0 to 2. -
The limits for yare from x2to 4.
Step 2: Set up and evaluate the double integral.
The double integral is given as:
ZZR
(x+y)dA =Z2
0Z4
x2
(x+y)dy dx
Now, we integrate with respect to yfirst and then with respect to x:
2
Z2
0Z4
x2
(x+y)dy dx =Z2
0xy +y2
24
x2
dx
=Z2
04x+16
2−x3−x4
2dx
=Z2
0
(4x+ 8 −x3−x4
2)dx
=2x2+ 8x−x4
4−x5
10 2
0
= 8 + 16 −4−32
10
= 20 −4−16
5
= 16 −16
5
=64
5.
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 64
5.
Question 3
Question
Evaluate the double integral RRD(x2+y2)dx dy over the region Dbounded by
the curves y=x,y= 4 −x2,x= 0.
Solution
Step 1: Determine the limits of integration for xand y. To find the limits of
integration, we need to determine the points of intersection between the curves
y=xand y= 4 −x2. Setting x=y, we have y= 4 −y2, which simplifies to
y2+y−4 = 0. Solving this quadratic equation gives y= 1 or y=−4, but since
y=x, we discard the negative value. Thus, the limits of integration for yare
1≤y≤4−x2.
Next, we determine the limits of integration for x. Since the region is
bounded by x= 0, we have 0≤x≤2.
Step 2: Evaluate the double integral. We can now rewrite the double integral
as:
ZZD
(x2+y2)dx dy =Z2
0Z4−x2
1
(x2+y2)dy dx
3
Step 3: Integrate with respect to yfirst.
Z4−x2
1
(x2+y2)dy =x2y+y3
3
4−x2
1
Step 4: Substitute the limits of integration and simplify.
=x2(4 −x2) + (4 −x2)3
3−(x21 + 1
3) = 2x2+7x4
3−23x2
3+23
3
Step 5: Integrate the result with respect to x.
Z2
02x2+7x4
3−23x2
3+23
3dx =2
3x3+7
15x5−23
9x3+23
3x
2
0
Step 6: Substitute the limits of integration and calculate the final answer.
=23
3+7(25)
15 −23(23)
9+23(2)
3−0
3+0
15 −0
9+23(0)
3
=8
3+64
15 −64
3+46
3
=−160
15
=−32
3
Therefore, the value of the double integral RRD(x2+y2)dx dy over the region
Dis −32
3.
Question 4
Question
Evaluate the double integral RRRsin(x+y)dA, where Ris the region bounded
by the lines y= 0,y=x,y= 2x, and y= 2.
Solution
Step 1: Sketch the region Rdescribed by the given bounds for yin terms of x.
xyy =xy = 2xy = 2
Step 2: Set up the double integral using these bounds:
ZZR
sin(x+y)dA =Z1
0Z2x
x
sin(x+y)dydx
4
Step 3: Integrate with respect to yfirst:
Z2x
x
sin(x+y)dy = [−cos(x+y)]2x
x=−cos(3x) + cos(2x)
Step 4: Now integrate the result from Step 3 with respect to x:
Z1
0
(−cos(3x) + cos(2x)) dx =−1
3sin(3x) + 1
2sin(2x)1
0
=−1
3sin(3) + 1
2sin(2)−(0 + 0)
=−1
3sin(3) + 1
2sin(2)
Thus, the value of the given double integral is −1
3sin(3) + 1
2sin(2).
Question 5
Question
Calculate the double integral RRRxy2dA, where Ris the region bounded by the
curves y=x,y= 2x, and y= 1.
Solution
To find the double integral RRRxy2dA over the region R, we first need to de-
termine the bounds of integration by examining the limits along the xand y
axes.
Step 1: Determine the bounds of integration in terms of x:
Since y=xand y= 2xare the bounding curves, we can set up the following
inequalities:
x≤y≤2x
Solving these inequalities for x, we find that 0≤x≤1
2.
Step 2: Determine the bounds of integration in terms of y:
Next, we consider the bounding curve y= 1. Since this line intersects the
region at y= 1, the upper bound for ywill be 1.
Step 3: Set up the double integral:
Putting the bounds together, the double integral becomes:
Z1
2
0Z2x
x
xy2dy dx
Step 4: Integrate with respect to y:
Integrating the inner integral with respect to y, we get:
Z1
2
01
3xy32x
x
dx
5
=Z1
2
07
3x4dx
Step 5: Integrate with respect to x:
Integrating the remaining integral with respect to x, we have:
7
15x5
1
2
0
=7
15 ·1
25
=7
15 ·1
32
=7
480
Therefore, the value of the double integral RRRxy2dA over the region Ris
7
480 .
Question 6
Question
Evaluate the double integral RRRxy dA, where Ris the region bounded by the
curves y=x2and y= 2x.
Solution
To evaluate the double integral RRRxy dA, we first need to determine the limits
of integration for xand yover the region R.
Step 1: Determine the limits of integration for xThe region Ris
bounded by the curves y=x2and y= 2x. To find the limits of integration
for x, we need to find the x-values at which these two curves intersect. Setting
x2= 2x, we find the intersection points at x= 0 and x= 2.
Therefore, the limits of integration for xare 0≤x≤2.
Step 2: Determine the limits of integration for yFor each value of x
in the region R,yvaries between the curves y=x2(bottom curve) and y= 2x
(top curve).
So, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral The given double
integral is RRRxy dA.
Rewriting the integrand in terms of xand y, we have xy.
Therefore, the double integral becomes:
Z2
0Z2x
x2
xy dy dx
6
Now, we integrate with respect to yfirst, then with respect to x.
Z2x
x2
xy dy =x1
2y22x
x2
=x1
2(2x)2−1
2(x2)2
=x2x2−1
2x4
= 2x3−1
2x5
Now we integrate this with respect to x:
Z2
0
(2x3−1
2x5)dx =1
2x4−1
12x62
0
=1
2(2)4−1
12(2)6−1
2(0)4−1
12(0)6
= (8 −64
3)−0
=24
3−64
3
=−40
3
Therefore, the value of the double integral RRRxy dA over the region Ris
−40
3.
Question 7
Question
Evaluate the double integral RRDex2+y2dx dy, where Dis the region bounded
by the curve y=x2and the lines x= 0 and x= 1.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the curves y=x2,x= 0, and x= 1. To find
the limits of integration for x, we need to find the x-values where the curves
intersect.
Setting y=x2, we have:
x2=x⇒x2−x= 0 ⇒x(x−1) = 0
7
This gives us x= 0 and x= 1 as the x-values where the curves intersect.
So, the limits of integration for xare 0≤x≤1.
Step 2: Determine the limits of integration for y:
For each xin the region D, the corresponding yvalues are between the
curves y= 0 and y=x2.
So, the limits of integration for yare 0≤y≤x2.
Therefore, the double integral becomes:
ZZD
ex2+y2dx dy =Z1
0Zx2
0
ex2+y2dy dx
Step 3: Evaluate the integral:
Zx2
0
ex2+y2dy =ex2Zx2
0
ey2dy
Now, we use the substitution u=y2⇒du = 2y dy:
ex2Zx2
0
ey2dy =ex2
2Zx4
0
eudu
=ex2
2[eu]x4
0=ex2
2(ex4
−1)
Now, we integrate this expression with respect to xfrom 0to 1:
Z1
0
ex2
2(ex4
−1) dx =1
2Z1
0
ex2+x4
−ex2dx
This integral is not easy to solve in terms of elementary functions.
Question 8
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2,y= 2x, and x= 0.
Solution
Step 1: Begin by determining the intersection points of the curves y=x2and
y= 2x: Let’s set x2= 2xand solve for x.
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the intersection points.
8
Step 2: Determine the limits of integration in terms of x: We will integrate
with respect to yfirst. The lower bound is given by the curve y=x2and the
upper bound is given by the line y= 2x. Therefore, for a fixed x,yranges from
x2to 2x.
Step 3: Set up the double integral: The given double integral can be ex-
pressed as:
Z2
0Z2x
x2
(x+y)dy dx
Step 4: Evaluate the inner integral with respect to y:
Z2x
x2
(x+y)dy =x(y) + y2
2
2x
x2
=x(2x) + (2x)2
2−x(x2)−(x2)2
2
= 2x2+ 2x2−x3−x4
2
= 4x2−x3−x4
2
Step 5: Evaluate the outer integral with respect to x:
Z2
0
(4x2−x3−x4
2)dx =4x3
3−x4
4−x5
10
2
0
=32
3−4−32
25 =40
3−32
25 =320 −96
75 =224
75
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 224
75 .
Question 9
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x,y= 2x,x= 0, and x= 1.
Solution
Step 1: Determine the limits of integration for xand y. The region Ris defined
by the curves y=x,y= 2x,x= 0, and x= 1. We will first find the bounds
for xand y.
For the lower bound of x,x= 0. For the upper bound of x, we need to find
the intersection point of y=xand y= 2x:x= 0 and y= 0, and y= 2(0) = 0.
Thus, the bounds for xare from 0to 1.
9
For the lower bound of y,y=x, and for the upper bound of y,y= 2x.
Thus, the bounds for yare from xto 2x.
Step 2: Set up the double integral with the appropriate bounds. The given
double integral can be expressed as:
Z1
0Z2x
x
ex2+y2dy dx
Step 3: Evaluate the inner integral with respect to y.
Z2x
x
ex2+y2dy =hex2+y2iy=2x
y=x=e5x2
−e2x2
Step 4: Evaluate the outer integral with respect to x.
Z1
0
(e5x2
−e2x2)dx =1
5e5x2
−1
2e2x21
0
=1
5e5−1
2e2−1
5−1
2
Step 5: Simplify the expression.
1
5e5−1
2e2−1
5−1
2=1
5e5−1
2e2+1
10
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e5−1
2e2+1
10 .
Question 10
Question
Evaluate the double integral RRR(6x2−2y)dA where Ris the region bounded
by the curves y=x3,y= 0,x= 1, and x= 2.
Solution
Step 1: Identify the limits of integration for xand y. The region Ris bounded
by the following curves: 1. y=x32. y= 0 3. x= 1 4. x= 2
Thus, the limits of integration for xare 1≤x≤2, and the limits of inte-
gration for yare 0≤y≤x3.
Step 2: Set up the integral. The given double integral is
ZZR
(6x2−2y)dA
We can rewrite this as the iterated integral:
Z2
1Zx3
0
(6x2−2y)dy dx
10
Step 3: Evaluate the inner integral with respect to y.
Zx3
0
(6x2−2y)dy =6x2y−y2x3
0
= 6x2(x3)−(x3)2−0
= 6x5−x6
Step 4: Substitute the result of the inner integral back into the original
expression and evaluate the outer integral with respect to x.
Z2
1
(6x5−x6)dx =6
6x6−1
7x72
1
=x6−1
7x7
2
1
= (26−1
727)−(16−1
717)
= (64 −128
7)−(1 −1
7)
=256
7−135
7
=121
7
Therefore, the value of the double integral over region Ris 121
7.
Question 11
Question
Let Rbe the region in the xy-plane bounded by y=x2,y= 2x,x= 0, and
x= 1. Calculate the double integral RRRy dA.
Solution
To compute the double integral RRRy dA, we first need to determine the bounds
of integration for xand yin the region R.
Step 1: Determining the bounds for xand y
The region Ris bounded by the curves y=x2and y= 2x, and lines x= 0
and x= 1. Therefore, for y, the lower bound is x2, the upper bound is 2x. For
x, the lower bound is 0 and the upper bound is 1.
Step 2: Setting up the integral
The double integral RRRy dA can be expressed as:
ZZR
y dA =Z1
0Z2x
x2
y dy dx
11
Step 3: Integrating with respect to y
Integrating with respect to yfirst, we have:
Z2x
x2
y dy =1
2y22x
x2
=1
2(4x2−x4) = 2x2−1
2x4
Step 4: Integrating with respect to x
Now, we integrate 2x2−1
2x4with respect to xover the bounds 0 to 1:
Z1
0
2x2−1
2x4dx =2
3x3−1
10x51
0
=2
3−1
10 =17
30
Step 5: Final answer
Therefore, the value of RRRy dA over the region Ris 17
30 .
Question 12
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region in the first
quadrant bounded by the curves y=x2and y= 2x.
Solution
Step 1: Determine the limits of integration by finding the points of intersection
of the two curves.
x2= 2x
x2−2x= 0
x(x−2) = 0
x= 0, x = 2
So, the limits of integration for xare from 0 to 2.
Step 2: To find the limits of integration for y, we set up the inequalities for
R:
x2≤y≤2x
0≤x2≤y≤2x
Since y=x2is the lower curve and y= 2xis the upper curve, the limits of
integration for yare from x2to 2x.
Step 3: Rewrite the given integral with the limits of integration:
ZZR
ex2+y2dA =Z2
0Z2x
x2
ex2+y2dy dx
12
Step 4: Integrate with respect to yfirst, keeping xconstant:
Z2x
x2
ex2+y2dy =ex2Z2x
x2
ey2dy
=ex2Z2x
x2
ey2dy
=ex2hey2i2x
x2
=ex2(e(2x)2
−ex4)
=ex2(e4x2
−ex4)
Step 5: Integrate the result from step 4 with respect to x:
Z2
0
ex2(e4x2
−ex4)dx =Z2
0
ex2+4x2
−ex2+x4dx
=Z2
0
e5x2
−ex2+x4dx
Step 6: Integrate to find the final result:
Z2
0
e5x2
−ex2+x4dx =1
5e5x22
0−1
3ex4+x22
0
=1
5e20 −1
5−1
3e16 −1
3
=1
5e20 −1
5−1
3e16 +1
3
Therefore, the value of the double integral RRRex2+y2dA over the region R
is 1
5e20 −1
5−1
3e16 +1
3.
Question 13
Question
Evaluate the double integral RRDex2+y2dx dy over the region Denclosed by the
curves y=x2,y= 2,x= 0, and x= 1.
Solution
To evaluate the double integral RRDex2+y2dx dy over the region D, we first need
to determine the limits of integration.
Step 1: Determine the limits of integration.
The region Dis enclosed by the curves y=x2,y= 2,x= 0, and x= 1.
This implies that our limits of integration are: - For y:x2≤y≤2- For x:
0≤x≤1
13
Thus, the double integral becomes:
Z1
0Z2
x2
ex2+y2dy dx
Step 2: Evaluate the inner integral with respect to y.
Z2
x2
ex2+y2dy
=hex2+y2i2
x2
=ex2+4 −e2x2
Step 3: Evaluate the outer integral with respect to x.
Z1
0ex2+4 −e2x2dx
=Zex2+4 dx −Ze2x2dx1
0
=1
2ex2+4 −1
2√2e2x21
0
=1
2e5−1
2√2e6−1
2e4+1
2√2
Therefore, the value of the double integral RRDex2+y2dx dy over the region
Dis 1
2e5−1
2√2e6−1
2e4+1
2√2.
Question 14
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
lines y=x,y= 2x, and y= 4.
Solution
Step 1: Sketch the region Rto visualize the bounds of integration. The region
Ris bounded by the lines y=x,y= 2x, and y= 4. It is a triangular region in
the first quadrant.
Step 2: Determine the limits of integration. To evaluate the double integral
RRRex+ydA, we integrate with respect to xand yover the region R.
For the limits of integration with respect to y, we need to find the y-values
where the lines y=x,y= 2x, and y= 4 intersect. Setting y=xand y= 2x
equal to each other gives x= 2x, which simplifies to x= 0. This means the
14
lines intersect at the point (0,0). Setting y=xand y= 4 equal to each other
gives x= 4. This means the lines intersect at the point (4,4).
Thus, the limits of integration for yare from xto 4.
For the limits of integration with respect to x, we must determine the range
of xvalues for which xis between the lines y=xand y= 2x. This will give us
the limits of integration for x.
The line y=xintersects y= 2xat (0,0). So, xranges from 0to y/2.
Therefore, the limits of integration are:
0≤x≤y/2,0≤y≤4
Step 3: Evaluate the double integral. Now we can evaluate the double
integral:
ZZR
ex+ydA
=Z4
0Zy
2
0
ex+ydx dy
=Z4
0ex+yy
2
0dy
=Z4
0
(ey
2+y−ey)dy
=Z4
0
e3y
2−eydy
=2
3e3y
2−ey4
0
=2
3e6−e4−2
3−1
=2
3e6−e4+1
3
Therefore, the value of the double integral RRRex+ydA over the region Ris
2
3e6−e4+1
3.
Question 15
Question
Let Rbe the region in the first quadrant bounded by y=x2and y= 2x.
Evaluate the double integral RRR
1
ydA.
Solution
Step 1: To find the limits of integration for xand y, we need to determine the
intersection points of the curves y=x2and y= 2x. Setting the two equations
15
equal gives us:
x2= 2x
x2−2x= 0
x(x−2) = 0
This implies either x= 0 or x= 2.
Step 2: Since the region Ris in the first quadrant, we take 0≤x≤2and
x2≤y≤2xas our limits of integration.
Step 3: The double integral becomes:
Z2
0Z2x
x2
1
ydy dx
Step 4: Solving the inner integral first, we have:
Z2x
x2
1
ydy = ln(2x)−lnx2= ln 2x
x2= ln 2
x
Step 5: Now, the double integral becomes:
Z2
0
ln 2
xdx
Step 6: Evaluating the integral gives:
xln 2
x−22
0
=2 ln 2
2−2−0 ln 2
0−0
= 2 ln(1) −2 = 2 ·0−2 = −2
Question 16
Question
Evaluate the double integral RRR(x2−y2)dA, where Ris the region in the first
quadrant bounded by the lines y=x,y= 2x, and the coordinate axes.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand y.
Step 1: Determine the limits of integration for xand yThe region
Ris bounded by the lines y=xand y= 2x.
16
• For x, the limits of integration are from 0 to the x-value where y= 2x
intersects the x-axis, which is x= 0.
• For y, the limits of integration are from y=xto y= 2x.
Step 2: Set up the double integral The double integral is given by:
ZZR
(x2−y2)dA =Za
0Z2x
x
(x2−y2)dydx
Step 3: Evaluate the inner integral with respect to y
Z2x
x
(x2−y2)dy
=x2y−y3
3
2x
x
= (2x3−(2x)3
3)−(x3−x3
3)
= (2x3−8x3
3)−(x3−x3
3)
=2x3
3−8x3
3+2x3
3
=−4x3
3
Step 4: Evaluate the outer integral with respect to x
Za
0−4x3
3dx
=−4
3Za
0
x3dx
=−4
3x4
4a
0
=−1
3a4
Therefore, the value of the double integral RRR(x2−y2)dA over the region
Ris −1
3a4.
Question 17
Question
Evaluate the double integral RRRex2
−y2dA, where Ris the region bounded by
the curves y=x2and y= 2x.
17
Solution
To evaluate the double integral, we need to find the limits of integration for x
and y.
Step 1: Determine limits of integration for x:The region Ris bounded
by y=x2and y= 2x. Setting these two equations equal gives us x2= 2x,
which simplifies to x2−2x= 0, or x(x−2) = 0. So, x= 0 and x= 2 are the
solutions.
Hence, the limits of integration for xare from 0 to 2.
Step 2: Determine limits of integration for y:The region Ris bounded
by y=x2and y= 2x. Therefore, the limits of integration for yare from x2to
2x.
Step 3: Set up and evaluate the double integral: The given double
integral is RRRex2
−y2dA. So we have:
Z2
0Z2x
x2
ex2
−y2dydx
Now evaluate this double integral:
Z2
0−1
2ex2
−y22x
x2
dx
=Z2
0−1
2ex2
−(2x)2+1
2ex2
−(x2)dx
=Z2
0−1
2ex2
−4x2+1
2e0dx
=Z2
0−1
2e−3x2+1
2dx
We will need to use a numerical method or software to evaluate the integral
of e−3x2, as it does not have a standard anti-derivative. This will yield the final
numerical result.
Question 18
Question
Evaluate the double integral RRRxye−x2
−y2dA over the region Rbounded by
the curves x= 0,x= 1,y= 0, and y=√3.
18
Solution
Step 1: Rewrite the given integral in polar coordinates. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =r dr dθ. The region Rbecomes 0≤r≤1,
0≤θ≤π
3. Therefore, our integral becomes:
Zπ
3
0Z1
0
r2cos(θ) sin(θ)e−r2dr dθ
Step 2: Evaluate the inner integral.
Z1
0
r2cos(θ) sin(θ)e−r2dr
Let u=r2, then du = 2r dr. The integral now becomes:
Z1
0
1
2cos(θ) sin(θ)e−udu
Evaluating this integral gives:
−1
4cos(θ) sin(θ)e−u
1
0
=−1
4cos(θ) sin(θ)(e−1−1)
Step 3: Evaluate the outer integral.
Zπ
3
0−1
4cos(θ) sin(θ)(e−1−1) dθ
This integral can be simplified to:
−1
4(e−1−1) Zπ
3
0
cos(θ) sin(θ)dθ
Using the identity Rcos(θ) sin(θ)dθ =−1
2cos2(θ) + C, we have:
−1
4(e−1−1) −1
2cos2(θ)
π
3
0
Evaluating this further gives the final answer.
Question 19
Question
Evaluate the double integral RRR(x+y)dA, where Ris the region bounded by
the curves y=x2and y= 2x.
19
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for the given region R. This region Ris bounded by the curves y=x2
and y= 2x. Let’s find the points of intersection between these two curves by
setting them equal to each other:
x2= 2x
x2−2x= 0
x(x−2) = 0
This gives us x= 0 and x= 2 as the points of intersection.
Now, fixing xwithin the bounds x= 0 and x= 2, the corresponding yvalues
can be found by plugging them into the equations y=x2and y= 2x.
Step 1: Find the limits of integration At x= 0:
y1=x2= 02= 0
y2= 2x= 2(0) = 0
At x= 2:
y1=x2= 22= 4
y2= 2x= 2(2) = 4
Therefore, the limits of integration are 0≤x≤2and 0≤y≤4.
Step 2: Evaluate the double integral RRR(x+y)dA We can now write
the given double integral in terms of xand yusing the limits of integration we
found earlier:
ZZR
(x+y)dA =Z2
0Z4
0
(x+y)dy dx
Let’s evaluate the inner integral first:
Z4
0
(x+y)dy =xy +y2
24
0
= 4x+ 8
Now, the double integral becomes:
Z2
0
(4x+ 8) dx =2x2+ 8x2
0= 8 + 16 = 24
Therefore, the value of the double integral RRR(x+y)dA over the region R
is 24.
Question 20
Question
Evaluate the double integral RRD(x+y)dA, where Dis the region bounded by
the curves y=x2,y= 2x,x= 1, and x= 2.
20
Solution
To evaluate the given double integral, we first need to determine the limits of
integration for xand y.
Step 1: Determine the limits of integration for x. The region Dis bounded
by the curves y=x2,y= 2x,x= 1, and x= 2. We can find the points
of intersection of these curves by setting them equal to each other. Solving
x2= 2x, we get x2−2x=x(x−2) = 0, which gives x= 0 and x= 2.
However, we only consider the intersection points within the interval 1≤
x≤2, so x= 0 is not relevant to our limits of integration for x. Therefore, the
limits of integration for xare 1≤x≤2.
Step 2: Determine the limits of integration for y. The lower bound for y
is given by the curve y=x2, and the upper bound is given by the line y= 2x.
Thus, the limits of integration for yare x2≤y≤2x.
Step 3: Set up and evaluate the double integral. Putting it all together,
the given double integral becomes:
ZZD
(x+y)dA =Z2
1Z2x
x2
(x+y)dy dx
Now, we evaluate this double integral:
Z2
1Z2x
x2
(x+y)dy dx =Z2
1xy +y2
22x
x2
dx
=Z2
12x2+(2x)2
2−x(x2+(x2)2
2)dx
=Z2
13x2−3
2x3dx
=x3−3
8x42
1
= (23−3
8·24)−(13−3
8·14)
= (8 −6) −(1 −3
8)
= 2 −5
8=11
8
Therefore, the value of the double integral RRD(x+y)dA over the region D
is 11
8.
Question 21
Question
Evaluate the double integral RRDexy dA, where Dis the region bounded by the
curves y=x2and y= 2x.
21
Solution
Step 1: First, we need to find the intersection points of the two curves. To find
the intersection points, we set x2= 2xand solve for x:x2−2x= 0
x(x−2) = 0
x= 0 or x= 2 So the intersection points are (0,0) and (2,4).
Step 2: Next, we need to determine the bounds of integration for xand y.
The region Dis bounded by the curves y=x2and y= 2x. Thus, for x, we will
integrate from 0 to 2, and for y, we will integrate from x2to 2x.
Step 3: Rewrite the integral in terms of xand ybounds. The double integral
becomes: R2
0R2x
x2exy dy dx
Step 4: Integrate with respect to yfirst. R2x
x2exy dy =1
xexy2x
x2
=1
xe2x2−1
xex3
Step 5: Integrate the result with respect to x.R2
01
xe2x2−1
xex3dx
=he2x2−ex3i2
0
=e8−e8−e2+e2
= 0
Therefore, the value of the double integral RRDexy dA over the region Dis
0.
Question 22
Question
Evaluate the double integral RRRex+ydA, where Ris the region bounded by the
curves y=x2,y= 2x, and x= 0.
Solution
To evaluate the double integral over R, we need to set up the integral using the
given bounds and then compute the integral.
Step 1: Determine the bounds for the integral in terms of xand y. The
region Ris bounded by the curves y=x2,y= 2x, and x= 0. To find the limits
of integration, we need to determine the intersection points of these curves.
Let’s first find the x-coordinates of the intersection points x1and x2:
Setting x2= 2x, we get x2−2x= 0, which factors as x(x−2) = 0. So,
x= 0 or x= 2.
Therefore, the bounds for xwill be 0≤x≤2.
Now for y, the limits are given by the corresponding equations y=x2and
y= 2x. Since y=x2is the lower curve and y= 2xis the upper curve in the
region R, the bounds for ywill be x2≤y≤2x.
Step 2: Set up and evaluate the double integral. The double integral can
22
be set up as follows:
ZZR
ex+ydA =Z2
0Z2x
x2
ex+ydy dx
Now, let’s compute this double integral:
Z2
0Z2x
x2
ex+ydy dx =Z2
0
exZ2x
x2
eydydx
=Z2
0
ex[ey]2x
x2dx
=Z2
0
ex(e2x−ex2)dx
=Z2
0
e3x−ex+x2dx
="e3x
3−ex+x2
1 + x2#2
0
=e6
3−e5
5−1
5
Therefore, the value of the double integral RRRex+ydA over the region Ris
e6
3−e5
5−1
5.
Question 23
Question
Evaluate the double integral RRRex2+y2dA, where Ris the region bounded by
the curves y=x2and y= 2x, in the first quadrant.
Solution
To evaluate the double integral, we first need to determine the limits of inte-
gration for xand yin the region R.
Step 1: Determine the limits of integration for x:
The curves y=x2and y= 2xintersect at the points where x2= 2x:x2−2x=
0⇒x(x−2) = 0 ⇒x= 0,2.
Therefore, the limits of integration for xare 0to 2.
Step 2: Determine the limits of integration for y:
For each value of xin the region R,yranges from x2to 2x. Therefore, the
limits of integration for yare x2to 2x.
Step 3: Set up the double integral:
The given double integral can be written as:
ZZR
ex2+y2dA =Zx=2
x=0 Zy=2x
y=x2
ex2+y2dy dx
23
Step 4: Evaluate the inner integral:
Zy=2x
y=x2
ex2+y2dy =ex2Zy=2x
y=x2
ey2
−x2dy
Note that the integral Rey2
−x2dy cannot be expressed in terms of elementary
functions. Therefore, the exact evaluation requires special techniques involving
Gaussian integrals or error functions.
Thus, the final answer would involve these special functions.
Question 24
Question
Evaluate the double integral RRDxy dA, where Dis the region bounded by the
lines y=x,y= 2x, and x= 1.
Solution
To evaluate the double integral RRDxy dA over the region D, we first need to
determine the bounds of integration.
Step 1: Determine the limits of integration for x:
The region Dis bounded by the lines y=xand y= 2x. These two lines
intersect at the point (0,0).
The line y=xintersects the line x= 1 at (1,1), and the line y= 2x
intersects x= 1 at (1,2).
Thus, the limits of integration for xare from x= 0 to x= 1.
Step 2: Determine the limits of integration for y:
The region Dis bounded by the lines y=xand y= 2x. To express yin
terms of xfor this region, we solve the system of equations y=xand y= 2x
to find y= 2x.
Thus, the limits of integration for yare from y=xto y= 2x.
Step 3: Set up and evaluate the double integral:
The double integral over the region Dis given by:
ZZD
xy dA =Z1
0Z2x
x
xy dy dx
Now, we evaluate the double integral:
Z1
0Z2x
x
xy dy dx =Z1
01
2x(2x)2−1
2x(x)2dx
=Z1
02x3−1
2x3dx =Z1
0
3
2x3dx
24
=3
8x4
1
0
=3
8(1)4−3
8(0)4=3
8
Therefore, the value of the double integral RRDxy dA over the region Dis 3
8.
Question 25
Question
Evaluate the double integral RRRx2y dA, where Ris the region bounded by the
circle x2+y2= 1.
Solution
To evaluate the double integral RRRx2y dA over the given region R, we will
convert the integral to polar coordinates.
Step 1: Convert to Polar Coordinates The region Rcan be described
in polar coordinates as 0≤r≤1and 0≤θ≤2π. Also, dA =rdrdθ.
Step 2: Rewrite the Integral Plugging in the polar coordinates, we
obtain:
ZZR
x2y dA =Z2π
0Z1
0
(rcos2θ)(rsin θ)r drdθ
Step 3: Compute the Integral Let’s first simplify the integrand:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
0Z1
0
(r3cos2θsin θ)drdθ
To evaluate this integral, we perform the following steps:
Z2π
0Z1
0
(r3cos2θsin θ)drdθ =Z2π
01
4cos2θsin θ1
0
dθ
Z2π
0
1
4cos2θsin θ dθ =1
4Z2π
0
cos2θsin θ dθ
This integral can be evaluated using trigonometric identities or by using
integration by parts.
Therefore, the value of the double integral RRRx2y dA over the region Ris
π
4.
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