Kristopher Rider Jackiewicz MAT 267 Spring 2021
Assignment Section 12.1 due 03/21/2020 at 10:00pm MST
1. (1 point) Consider the solid that lies above the square (in
the xy-plane) R= [0,2]×[0,2],
and below the elliptic paraboloid z=64 −x2−y2.
(A) Estimate the volume by dividing R into 4 equal squares
and choosing the sample points to lie in the lower left hand cor-
ners.
(B) Estimate the volume by dividing R into 4 equal squares
and choosing the sample points to lie in the upper right hand
corners.
(C) What is the average of the two answers from (A) and (B)?
Solution:
SOLUTION
(A)
f(x,y) = 64 −x2−y2,then
RRRf(x,y)dA ≈f(0,0)∆A+f(1,0)∆A+f(0,1)∆A+f(1,1)∆A
=64(1) + 63(1) + 63(1) + 62(1)
=252
;
(B)
RRRf(x,y)dA ≈f(1,1)∆A+f(2,1)∆A+f(1,2)∆A+f(2,2)∆A
=62(1) + 59(1) + 59(1) + 56(1)
=236
(C) A better approximation of the integral can be obtained by
taking the average of the values from part (A) and (B), that is,
RRRf(x,y)dA ≈252+236
2=244
Correct Answers:
•252
•236
•244
2. (1 point) Using geometry, calculate the volume of the solid
under z=p36 −x2−y2and over the circular disk x2+y2≤36.
Solution:
SOLUTION
The graph of z=p36 −x2−y2is the top half of the sphere
of radius 6 of equation x2+y2+z2=36. The volume of a sphere
of radius Ris given by V=4
3πR3.
Thus the volume of the solid under z=p36 −x2−y2and over
the circular disk x2+y2≤36 is given by
V=2
3π(6)3
Correct Answers:
•452.38934211693
3. (1 point) Evaluate the iterated integral R2
0R4
012x2y3dxdy
Solution:
SOLUTION:
Z2
0Z4
0
12x2y3dxdy =Z2
012x3
3y3x=4
x=0
dy
=Z2
0
1243
3y3dy
=256y4
4y=2
y=0
=25624
4
=1024
Correct Answers:
•1024
4. (1 point) Evaluate the iterated integral Z4
3Z5
4
(4x+
y)−2dydx
Solution:
SOLUTION:
1
Z4
3Z5
4
(4x+y)−2dydx =Z4
3−1
4x+yy=5
y=4
dx
=Z4
3−1
4x+5+1
4x+4dx
=−1
4ln(4x+5) + 1
4ln(4x+4)4
3
=1
4(−ln(21) + ln(17) + ln(20)−ln(16))
Correct Answers:
•0.00295861
5. (1 point) Evaluate the integral Zπ/3
0Z4
2
(ycosx+5)dydx.
Solution:
SOLUTION:
Zπ/3
0Z4
2
(ycosx+5)dydx =Zπ/3
0y2
2cosx+5yy=4
y=2
dx
=Zπ/3
042
2cosx+20 −22
2cosx+10dx
=Zπ/3
0
(6cosx+10)dx
= [6sinx+10x]π/3
0
=6sinπ
3+10π
3
=6 √3
2!+10
3π
Correct Answers:
•15.6681279346726
6. (1 point) Find Z6
3Z1
0
xyex+ydydx
Solution:
SOLUTION
Z6
3Z1
0
xyex+ydydx =R6
3R1
0xyexeydydx
=Z6
3
xexdx Z1
0
yeydy
Using integration by parts, we have, Rxexdx =ex(x−1). Thus
Z6
3Z1
0
xyex+ydydx = [ex(x−1)]6
3[ey(y−1)]1
0
=5e6−2e30e1+1e0
Correct Answers:
•1976.9728936173
7. (1 point) Find Z8
2Z10
9
(x+lny)dydx
Solution:
SOLUTION
Z8
2Z10
9
(x+lny)dydx =Z10
9Z8
2
(x+lny)dxdy
=Z1
9
0x2
2+xlnyx=8
x=2
dy
=Z1
9
082
2+8lny−22
2−2lnydy
=Z1
9
0[30 +6lny]dy
=Z1
9
030dy +6Z1
9
0lny dy
Using integration by parts in the second integral:
= [30y]1
90+6[ylny−y]1
90
=30 +6[10ln10 −10 −9ln9 +9]
=24 +60ln10 −54ln9
Correct Answers:
•43.5049784034869
8. (1 point) Calculate the double integral R RRxcos(2x+
y)dA where Ris the region: 0 ≤x≤1
3π,0≤y≤1
4π
Solution:
SOLUTION:
Z ZR
xcos(2x+y)dA =Z1
3π
0Z1
4π
0
xcos(2x+y)dydx
=Z1
3π
0
[xsin(2x+y)]y=1
4π
y=0dx
=Z1
3π
0xsin2x+1
4π−xsin(2x)dx
Using integration by parts separately for each term:
=hx
2cos(2x)−cos2x+1
4πi1
3π
0+Z1
3π
0
1
2cos2x+1
4π−cos(2x)dx
=1
6πcos2
3π−cos2
3π+1
4π+1
22sin2x+1
4π−sin(2x)
1
3π
0
=1
6πcos2
3π−cos11
12 π+1
4sin2
3π+1
4π−sin2
3π−sin1
4π
=1
6πcos2
3π−cos11
12 π+1
4sin11
12 π−sin2
3π−sin1
4π
≈ −0.0846201
Correct Answers:
•-0.0846201
9. (1 point) Consider the solid that lies above the square (in
the xy-plane) R= [0,2]×[0,2],
and below the elliptic paraboloid z=25 −x2−4y2.
Using iterated integrals, compute the exact value of the vol-
ume.
Solution:
SOLUTION
2
V=Z2
0Z2
025 −x2−4y2dxdy
=Z2
025x−x3
3−4y2xx=2
x=0
dy
=Z2
0142
3−8y2dy
=142
3y−8y3
32
0
=284
3−64
3
=220
3
Correct Answers:
•73.3333
10. (1 point) Find the average value of f(x,y) = 3ey√x+ey
over the rectangle R= [0,5]×[0,3].
Average value =
Solution:
SOLUTION:
The area of Ris A(R) = 5(3) = 15, so
fave =1
A(R)ZZR
f(x,y)dA
=1
15 Z5
0Z3
0
3ey√x+eydydx
=3
15 Z5
02
3(x+ey)3/2y=3
y=0
dx
=3
15 ·2
3Z5
0hx+e33/2−(x+1)3/2idx
=2
15 2
5x+e35/2−2
5(x+1)5/25
0
=4
75 5+e35/2−65/2−e
15
2+1
=4
75 5+e35/2−65/2−e
15
2+1
Correct Answers:
•67.0173266771719
11. (1 point)
The table below gives values of f(x,y), the number of mil-
ligrams of mosquito larvae per square meter in a swamp.
x=0x=5x=10
y=0 1 2 3
y=6 2 4 6
y=12 5 9 13
If xand yare in meters and Ris the rectangle 0 ≤x≤10,
0≤y≤12, estimate RRf(x,y)dA.
RRf(x,y)dA ≈
(Include units in your answer.)
Hint: Evaluate the Riemann sums using the lower left corners
and upper right corners. Then take the average of the two Rie-
mann sums.
Solution:
SOLUTION
We use four subrectangles to find an overestimate and under-
estimate of the integral:
Overestimate = (5)(6)(4+6+9+13) = 960,
and
Underestimate = (5)(6)(1+2+2+4) = 270.
A better estimate of the integral is the average of the two:
ZR
f(x,y)dA ≈960 +270
2=615 mg.
The units of the integral are milligrams, and the integral repre-
sents the total number of mg of mosquito larvae in this 10 meter
by 12 meter section of swamp.
Correct Answers:
•615 mg
12. (1 point)
Calculate a Riemann sum S3,3on the square R= [0,3]×[0,3]
for the function g(x,y) = f(x,y)−7.
The contour plot of f(x,y)is shown in Figure 4.
Choose sample points and use the plot to find the values of
f(x,y)at these points.
Use the values of f(x,y)to evaluate g(x,y)accordingly.
S3,3=
Solution:
Solution: Each subrectangle is a square of side length 1,
hence its area is ∆A=12=1.
We choose the sample points shown in the figure. The contour
plot shows the following values of fat the sample points:
f(P11) = 2f(P21) = 3f(P31) = 4
f(P12) = 3f(P22) = 4f(P32) = 7
f(P13) = 5f(P23) = 6f(P33) = 10
Evaluating g(x,y) = f(x,y)−7 at these points gives
g(P11) = −5g(P21) = −4g(P31) = −3
g(P12) = −4g(P22) = −3g(P32) = 0
g(P13) = −2g(P23) = −1g(P33) = 3
3
The Riemann sum S3,3is thus
S3,3=
3
∑
i=1
3
∑
j=1
g(Pi j)∆A=
3
∑
i=1
3
∑
j=1
g(Pi j)·1=−5−4−3−4−3+0−2−1+3=−19
Correct Answers:
•-19
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