My answers to the exercises

15.2 3, 8, 9, 13, 15, 17, 18, 21, 25, 27, 29
3–14 Calculate the iterated integral.
R4R2
3. 1 0 (6x2 y 2x) dy dx
R2
y=2
(6x2 y 2x) dy = [3x2 y 2 2xy]y=0 = 12x2 4x
R04
4
(12x2 4x) dx = [4x3 2x2 ]1 = (256 32) (4 2) =
1
222
R 3 R 5 ln y
8. 1 1
dy dx
xy
y=5
R 5 ln y
1 2
1 2
dy =
ln y
=
ln 5
1
xy
2x
2x
y=1
R3 1 2
3
ln 5 dx = 12 ln x ln2 5 1 = 21 ln 3 ln2 5
1
2x
R4R2 x y
9. 1 1
+
dy dx
y x
y=2
R2 x y
3
y2
= x ln 2 +
+
dy = x ln y +
1
y x
2x y=1
2x
4
2
R4
x
3 ln x
3
dx =
ln 2 +
= 15
x ln 2 +
ln 2 + 3 ln 2 =
2
1
2x
2
2 1
21
ln 2
R2 2 R
13. R0 0 r sin2 d dr
r sin2 d = 12 r
R02 1
2
r dr = 14 r2 0 =
0 2
15–22 Calculate the double integral.
RR
15. R sin (x y) dA,
R = f(x; y) : 0 x
=2; 0 y
=2g
R =2 R =2
The iterated integral is 0
sin (x y) dx dy.
0
R =2
=2
sin (x y) dx = [ cos (x y)]x=
= cos ( =2 y) +
x=0
0
cos ( y) = cos y sin y
R =2
(cos y sin y) dy = [sin y + cos y]0 =2 = 1 1 = 0. Obvi0
ous.
RR xy 2
17. R 2
dA,
R = f(x; y) : 0 x 1; 3 y 3g
x +1
R 1 R 3 xy 2
The iterated integral is 0 3 2
dy dx.
x +1
1
y=3
R3
xy 2
x
y3
18x
dy
=
= 2
3 2
2
x +1
x + 1 3 y= 3 x + 1
R 1 18x
1
dx = [9 ln (x2 + 1)]0 = 9 ln 2
0
x2 + 1
RR 1 + x2
18. R
dA,
R = f(x; y) : 0 x 1; 0
1 + y2
R 1 R 1 1 + x2
The iterated integral is 0 0
dx dy.
1 + y2
x=1
R 1 1 + x2
4=3
x + x3 =3
=
dx =
0
2
2
1+y
1+y
1 + y2
x=0
4
4 R1 1
dy = [arctan y]10 = =3
0
2
3
1+y
3
RR
xy
21. R ye
dA,
R = [0; 2] [0; 3]
R3R2
The iterated integral is 0 0 ye xy dx dy.
R2
x=2
ye xy dx = [ e xy ]x=0 = 1 e 2y
R03
3
(1 e 2y ) dy = y + 21 e 2y 0 = 3 + 12 e 6
0
y
1
2
=
1g
5
2
+ 21 e
6
25. Find the volume of the solid that lies under the plane 4x+6y 2z+
15 = 0 and above the rectangle R = f(x; y) : 1 x 2; 1 y 1g
The height of the plane above
R 2 the
R 1 point (x; y) is z = 2x+3y+15=2,
so the iterated integral is 1 1 (2x + 3y + 15=2) dy dx.
R1
y=1
(2x + 3y + 15=2) dy = [2xy + (3=2) y 2 + (15=2) y]y= 1 = (2x + 3=2 + 15=2)
1
( 2x + 3=2 15=2) = 4x + 15
R2
2
(4x + 15) dx = [2x2 + 15x] 1 = (8 + 30) (2 15) = 51
1
27. Find the volume of the solid lying under the elliptic paraboloid
x2 =4 + y 2 =9 + z = 1 and above the rectangle R = [ 1; 1] [ 2; 2]
The height of the surface above the
point
(x; y) is z = 1 x2 =4
R
R
1
2
y 2 =9, so the iterated integral is 1 2 (1 x2 =4 y 2 =9) dy dx.
Note that z is always positive for (x; y) in that region because
1=4 + 4=9 < 1.
R2
y=2
(1 x2 =4 y 2 =9) dy = [y x2 y=4 y 3 =27]y= 2 = (2 x2 =2 8=27)
2
( 2 + x2 =2 + 8=27) = 92=27 x2
R1
1
(92=27 x2 ) dx = [92x=27 x3 =3] 1 = (83=27) ( 83=27) =
1
166=27
29. Find the volume of the solid enclosed by the surface z = x sec2 y
2
and the planes x = 1, y = 0, y = =2, and z = 0, x = 0, x = 2,
y = 0, and y = =4.
That’s under the surface and above the rectangle [0; 2] [0; =4].
R 2 R =4
So the iterated integral is 0 0 x sec2 y dy dx.
R =4
R2
=4
x sec2 y dy = [x tan y]y=
= x and 0 x dx = 2.
y=0
0
15.3 1, 5, 7, 13, 17, 21
1–6 Evaluate the iterated integral.
R 4 R py
1. 0 0 xy 2 dx dy
p
R py 2
1 2 2 x= y
x
y
= 12 y 3
xy
dx
=
2
x=0
R04 1 3
4
y dy = 18 y 4 0 = 32.
0 2
R 1 R s2
5. 0 0 cos s3 dt ds
R s2
t=s2
cos s3 dt = [t cos s3 ]t=0 = s2 cos s2
0
R1 2
1
s cos s3 ds = 31 sin s3 0 = 13 sin 1.
0
7–10 Evaluate the double integral.
RR
7. D y 2 dA,
D = f(x; y) : 1 y 1,
y
R1 Ry
2
The iterated integral is 1 y 2 y dx dy.
Ry
y 2 dx = (2y + 2) y 2 = 2y 3 + 2y 2
R 1y 2 3
1
(2y + 2y 2 ) dy = 21 y 4 + 32 y 3 1 = 4=3.
1
2
x
yg
13–14 Express D as a region of type I and also as a region of type II.
Then evaluate the double integral in two ways.
RR
13. D x dA,
D is enclosed by the lines y = x, y = 0, x = 1
R1Rx
R1
As type I, the iterated integral is 0 0 x dy dx = 0 x2 dx =
1=3.
R1R1
As type II, the iterated integral is 0 y x dx dy.
R1
x=1
x dx = 21 x2 x=y = 12 (1 y 2 )
y
R1 1
1
(1 y 2 ) dy = 21 (y y 3 =3) 0 = 1=3.
0 2
17–22 Evaluate the double integral.
RR
17. D x cos y dA,
D is bounded by y = 0, y = x2 , x = 1
R 1 R x2
As type I, the iterated integral is 0 0 x cos y dy dx.
3
R x2
R01
0
RR
2
x cos y dy = [x sin y]xy=0 = x sin x2
1
1
x sin x2 dx =
cos x2 0 = 21 (1 cos 1)
2
21. D (2x y) dA,
D is bounded by the circle with center
the origin and radius 2
R 2 R p 4 x2
As type I, the iterated integral is 2 p4 x2 (2x y) dy dx.
p
p
R p 4 x2
4 x2
1 2 y= p
p
(2x
y)
dy
=
2xy
y
= 4x 4 x2
2
4 x2
4 x2
y=
i2
h
p
R2
4
2 3=2
2
(4 x )
= 0. Of course!
4x 4 x dx =
3
2
2
4

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