SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS
|||| 16.3
S Click here for solutions.
1– 6 |||| Evaluate the iterated integral.
y y x dx dy y
1
0
3.
2
3
2
x
0
5.
1 x
yy
0
1 x
1
6.
0
yy 0
2
0
1
4.
dy dx
y y sin x dy dx 0
y
1
0
17.
x y
dA , D is bounded by y
0, y
x, x
1
yy xy dA,
D is the first-quadrant part of the disk with center
D
2 x
x
3 y 2 dy dx
0, 0 and radius 1 18.
2 y
x1
0
yy e D
y y y dx dy
2.
y
x
1
16.
0
y ys x
1
Double Integrals over General Regions
A Click here for answers.
1.
❙❙❙❙
dy dx
1
yy y
2
x
dA, D is bounded by x
y 2, x
3
2 y 2
D
19.
yy ye dA, x
D
7–19 |||| Evaluate the double integral.
7.
yy
xy dA dA,,
D
D is the triangular region with vertices 0, 0, 2, 4, and 6, 0
x,, y 0 x 1, x 2 y s x x
yy x
9.
yy x
bounded by y x,, y 1 x 2
x
2 x
x y
3, 1
10.
yy
bounded by y
2 xy dA dA,,
x x,, y 0
x 2 y 2 and above the region x and x y 2
x
2
21. Under the paraboloid z
x and x
3 x 2
y
2
y
y
22. Bounded by the paraboloid z
D
D
20. Under the paraboloid z
2 y dA dA,,
D
D
20–26 |||| Find the volume of the given solid.
D
8.
1,
s x y 2 x
x
0, y
0, z
0, x
y
23. Bounded by the cylinder x
x sin x sin y y dA dA,,
y
0, z
0, x
2 y
2
x 2
2
and above the region
y
2
4 and the planes
1 2
9 and the planes x 2 in the first octant
z
0,
D
D 11.
x,, y 0 x
dA,, yy x1 dA
D
y
2, 0
x x,, y 1
x
y
24. Bounded by the planes y
cos y cos y
e, y 2
6 x
x y
4
yy 3 x
13.
yy y
y
x x,, y 6 2
x
4, sin sin x x
y
cos x cos x
dA,, dA
14.
yy x
0, z
0, y
x, and
6
1
xy
and above the triangle with
2
z
2
9 and the planes y
3 x,
0 in the first octant
integration. x,, y 0 x
y
1,
y x
1
y
27. 2
y
dA,, yy 3 xy dA D
27–30 |||| Sketch the region of integration and change the order of
dA,, D is bounded by y dA
2
x , y
2
x
29.
D is bounded by y by y
x,, y x
x
2
4 x
y y f x, y dy dx
28.
yy
30.
x
1
0
2
D
15.
0, z
D
D
3 z
26. Bounded by the cylinder y
dA,, dA
y
xy
vertices 1, 1, 4, 1, and 3, 2
D
D
2 y
25. Under the surface z
D
12.
0
1
4
f x, y dx dy
2
0
2 y
y 2
0
y y yy 4
0
sin x
0
2
y2
f x, y dy dx
f x, y dx dy
2
❙ ❙ ❙ ❙ SECTION 1 6 .3 DOUBLE INTEGRALS OVER GENERAL REGIONS
Answers Click here for exercises.
E
S
1 1. 6
1 2. 3
3.
16 1
5.
1 2
7.
1 12
9.
√
2 7
− (1 − cos1)
4. 6. 8.
19 42
−
28.
13 6
− − 4 l n 2 − 34 3
10.
1 6
√
√
√
2
12.
3 2−1− 3 3 π + 14−13 4 8
13.
3 4
14.
16 5
16.
1 2
18.
−
20.
6 35
22.
13 6
2613 40
17.
1 8
19.
e6
21.
144 35
23.
1 6
1 24. 4
2
− 9e − 4
11√ 5 − 27 +
55 8
27.
f (x, y) dxdy
1 1 0 y
30.
2
e − 2e + 1
f (x, y) dydx +
24 5
26. 3
√
1 x 0 0
29.
9 sin −1 23 2
25.
1 π/2 f (x, y) dxdy 0 sin−1 y
5 2
11.
15.
Click here for solutions.
2 2x 0 0
f (x, y) dydx
2 2 −x 1 0
f (x, y) dydx
❙❙❙❙
SECTION 1 6 .3 DOUBLE INTEGRALS OVER GENERAL REGIONS
3
Solutions E
1. 2. 3.
Click here for exercises. 1 y 0 0
1 0
1 2
1 y 0 0
1 0
2
2 y 0
1 0
2 3 √ 0 x
2 0
2
1 1+x 0 1−x
5/2
9 2
1 0
4 3
3
4
1 + 14 4
1
1
x−1
0
7.
√
1 x 0 x2
1 0
1 0
2
1 2
2
1 3
3
1 0
3
4
π/2 cos y 0 0
e 1 e 1
π/4 cos x π/6 sin x π/4 π/6
2
5/2
2
2 7
7/2
3
4
2 1 2 y=2−x 2 2 y=x
2
1 5
1 0
5
1 0
4
16 5
15.
√
1+ 3 2
√
√ √ 3 2−1− 3 π + 14−13 3 8
34 3
1
4 x 3xydydx = 1 x2 −4x+4
19 42
0
π/2 0
1 6
e 1
4 1
3 2
4 1 4 1
3 2
1 4
1 x 0 0
x+y
e
1 0
4
3
5
3
4
3 2
16.
2 y=x y=x2 −4x+4
3 2
xy dx = x − x (x − 2) dx = x − (x − 2) − 2 (x − 2) dx x − (x − 2) − 2 (x − 2) = 64 − = = e − e dy = e − e dydx = e − 2e + 1 = 3 2
2
1 2
π 2
4
1 0
2
1 6
6
409 20
2613 40
x
2x
4
5 4 1
1 2x 2
x 1 0
2
1 2
17.
1 2 y=cos x 2 y=sin x
2
1 2
2
π/4 π/6
1 4
=
2
1 12
2 3 1
1 2
2 2−x √ x
π/4 π/6
π/4 π/6
6 1 0
1 6
2
π/2 1 2 0
e y 4 1 y2
3
3
3
7 3
1 6
π 4
2 −x2 x2
2 2x 1+x 2 2
2
1 2
12.
1 0
2
5 2
1 2
3 1
1 0
11.
2 −x2 −1 x2 1
x + y dydx = 2 x + y dydx x y + y dx = 2 −2x + 2 dx =2 = 4 − x + x =
x x2
1 3
10.
3 4
√
5
3 2x 1 1+x
1 2 −x √ 0 x
1 2 1 2 0
1 3 2
13 6
1 2
1 0
3 1
9.
1 4 4
0
2
1 2
8.
−
x−1
1
0
3 2 2
14.
0
0
1 2
3
4 1 0
1 4
=
1 0
2
2
1
1 0
2
2
0
2 2 x=1+y x=−y
1 2
3
sin x dydx = x sin x dx − cos x = (1 − cos1) = 2y y dydx = dx x + 1 x + 1 (1 − x) 4 = − dx = − x − 3 + dx x + 1 x + 1 = − x − 3x + 4ln(x + 1) = − 4 l n 2 xy dy dx = xy dx x − x dx = x − x = = (x − 2y) dydx = xy − y dx (1 + x) − 3x − x dx = = (1 + x) − x − x = − x − 2xy dydx = x y − xy dx −2x + 7x − 4x − x dx = = − = − x + x − 2x − x x sin y dxdy = cos y sin y dy = = − cos y (1/x) dxdy = ln y − ln y dy 2 ln y dy = 2 [y ln y − y] = 2 = (3x + y) dydx = 3xy + y dx 3x (cos x − sin x) + cos x − sin x dx = (sin x + cos x) dx = 3x (sin x + cos x)| − 3 + sin 2x √ 2 − · + 3 0 + + 1 − =3 1 2
1 0
2
2 7
0
1 4
− 4 +
1 1+y 0 −y
y − xy dxdy = xy − x y dy −y + y + y dy = − y + y + y = =
√
3 1+x 1−x 3
1 0
2
2
2
1 4
1 0 2
2
1 x 0 0
13.
Click here for answers.
1 2
7/2
2 7
=
1 2 3 √ 2 x
2
9 2
4 3
6.
1 6
1 3
2
3
5.
1 2 2
xdxdy = x dy = y dy = y dxdy = y dy = x + y dydx = x y + y dx 3x + − x − x dx = = 16 1 − = x + x− x − x 2x − 3y dydx = 2xy − y dx 4x − (1 + x) + (1 − x) dx = = x − (1 + x) − (1 − x) 2 0
4.
A
√ 1− 3 2
1 4
π/4 π/6
√
3 2
√
1 1−x2 0 0
√ xy dy dx = x − xxy 1 dxx 1 0
1
=
0
2
1 2
0
1−x2
3
2
2
dx =
2
2
−
x4 4
1
= 0
1 8
4
❙ ❙ ❙ ❙ SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS
18.
22.
3 −2y 2 −1 y2 1
1
2
2
−1
1
2
2
−1 1
9 4 2
−1
9 5 10
2
3
2 2
1 2
2 2
2 2
1 2
9 2
1
9 2
24 5
−1
1 1 −x 0 0
2
1 0
2
1 3 3
1 0
2
3
1 3
4
1 12
3
1 3
23.
2
x + y + 4 dydx x y + y + 4y dx = x − x + (1 − x) + 4(1 − x) dx = = = x − x − (1 − x) − 2 1 − x
V =
2 2 x=3−2y x=y 2
1 2
y − x dxdy = xy − x dy 3 − 2y y − 3 − 2y − y y + y dy = − y + 9y − dy = = − = − y + 3y − y
1 4
y=1−x y=0 3
4
1
2
0
13 6
19.
√ 9 − x dydx = y√ 9 − x √ 9 − x − x√ 9 − x dx = √ 9 − x dx + −2x√ 9 − x dx = √ 9 − x = x 9 − x + sin (x/3) +
V = 4 6 −y 0 y/2
4 0
x
x x=6−y x=y/2
2 0 2 0
[ye ] dy ye dxdy = ye − ye dy = − 2e + −e = y −e 4 0
6−y
y/2
6−y
y/2
4
6−y
0
y/2
+ 4e
=
−12e
2
6
+ 3e + e
6
=
2
− 4 = e − 9e − 4
1 6
1 4
9 sin −1 23 + 56 2
√ 11
2 0
2
−1
9 2
2
dx
2
2
√ = 5+
0
1 2
2
2 y=1−x/2 y=0
2 0
2
1 2
4
(by parts separately for each term) 2
2 1 −x/2 0 0
9
5 − 27 +
√ 5 −
sin 2
2 2 3/2
1 6
−1
0
1 (27) 6 2
3
24.
20.
√
1 x 0 x2
2
√
y= x y=x2
1 0
2
1 3 3
1 0
5/2
4
1 3
5
2 15
2 7
7/2
1 5
3/2
1 3
5/2
6
1 21
7
1 0
18 105
21.
=
3/4 (3 −y)/3 1 3 0 y
(6 − 6x − 2y) dxdy (3 − y) x − x = dy (3 − y) − 2y + y dy = = − (3 − y) − y + y
V =
2
x + y dydx x y + y dx = x − x + x − x dx = = x − x + x − x =
V =
3/4 0
2 3
3/4 0
1 9
2 x=(3−y)/3 x=y
2
3
1 27
6 35
27 64
=
5 2 3
− −
9 + 15 16 64
2
+1 =
5 3 3/4 9 0 1 4
25.
2 y 0 y 2 −y
2
2
3x + y dxdy x + y x dy = 2y − y − 3y + 4y − 2y dy = = − y + y − y + y =
V =
2 0 2 0
3
x=y x=y 2 −y
2
3
1 7 7
6
1 6 2
5
4 5 5
4
4 2 0
3
144 35
2
5
1
2y
y
2
− (1 + xy) dxdy = x + − = 6 − 3y − y − 3y + 12y dy = 6y + y − y − y =
V =
1
1
2
3
1
2
9
2
2
3
8
3
4
2
3
2
55
1
8
x2 y 2 1
x=5−y x=2y −1
dy
SECTION 1 6 .3 DOUBLE INTEGRALS OVER GENERAL REGIONS
❙❙❙❙
5
29.
26.
3
y/ 3
0
0
3 1
9 − y dxdy = y 9 − y =3 = − 9 − y
V =
2
0
2
1
3/2
9
3
2
To reverse the order, we must break the region into two
dy
3
separate type I regions. Because the region of integration is
0
D = (x, y) y 2
| ≤ x ≤ 2 − y, 0 ≤ y ≤ 1 √ = {(x, y) | 0 ≤ y ≤ x, 0 ≤ x ≤ 1 } ∪ {0 ≤ y ≤ 2 − x, 1 ≤ x ≤ 2}
27.
we have 1
2
y
− 0
y
2
f (x, y) dA √ f (x, y) dydx + − =
f (x, y) dxdy = 0
D
x
1
0
2
2
1
0
x
Because the region of integration is 30.
D = (x, y) 0
{ | ≤ y ≤ x, 0 ≤ x ≤ 1} = {(x, y) | y ≤ x ≤ 1, 0 ≤ y ≤ 1 }
we have 1
x
0
0
f (x, y) dA f (x, y) dxdy =
f (x, y) dydx =
D
1
1
0
y
28.
Because the region of integration is
D = (x, y) y/2
{ | ≤ x ≤ 2, 0 ≤ y ≤ 4} = {(x, y) | 0 ≤ y ≤ 2x, 0 ≤ x ≤ 2 }
we have 4
2
0
Because the region of integration is π
| ≤ y ≤ sin x, 0 ≤ x ≤ = (x, y) | sin− y ≤ x ≤ , 0 ≤ y ≤ 1
D = (x, y) 0
2
π
1
2
we have π/ 2
sin
0
0
x
f (x, y) dA f (x, y) dxdy =
f (x, y) dydx =
D
1
π/ 2
0
sin−1
y
f (x, y) dA f (x, y) dydx =
f (x, y) dxdy = y/2
D
2
2 x
0
0
f (x, y) dydx
