Areas, Volumes, and Arc Lengths 299
- (See Figure 12.9-5.)
A=
∫ 4
0
(√
x−(−x)
)
dx
=
∫ 4
0
(
x^1 /^2 +x
)
dx=
[
2 x^3 /^2
3
+
x^2
2
] 4
0
=
(
2(4)^3 /^2
3
+
42
2
)
− 0
=
16
3
+ 8 =
40
3
04
y = –x y =
x = 4
y
x
√x
Figure 12.9-5
- (See Figure 12.9-6.)
Intersection points: 4=y^2 ⇒y=±2.
A=
∫ 2
− 2
(
4 −y^2
)
dy=
[
4 y−
y^3
3
] 2
− 2
=
(
4(2)−
23
3
)
−
(
4(−2)−
(−2)^3
3
)
=
(
8 −
8
3
)
−
(
− 8 +
8
3
)
=
16
3
+
16
3
=
32
3
y
x
0
2
–2
x = y^2
x = 4
Figure 12.9-6
You can use the symmetry of the region
and obtain the area= 2
∫ 2
0
(4−y^2 )dy.An
alternative method is to find the area by
setting up an integral with respect to the
x-axis and expressingx=y^2 asy=
√
x
andy=−
√
x.
- (See Figure 12.9-7.)
A=
∫π
π/ 2
sin
(
x
2
)
dx
Letu=
x
2
anddu=
dx
2
or 2du=dx.
0 x
y
ππ 2 2 π
x
2
2
x =π x = π
f(x) = sin( (
Figure 12.9-7
