MA 3972-MA-Book April 11, 2018 16:1
Areas and Volumes 311
- (See Figure 13.8-14.)
8
y
B (2, 8)
R 2
0 A (2, 0) x
Figure 13.8-14
Using the Washer Method:
Outer radius:x=2;
Inner radius:x=y^1 /^3.
V=π
∫ 8
0
(
22 −
(
y^1 /^3
) 2 )
dy
Using your calculator, you obtain
V=
64 π
5
.
- (See Figure 13.8-15.)
R 1
0
y
x
2
(0, 8) C B (2, 8)
Figure 13.8-15
Step 1: Using the Disc Method:
Radius =(8−x^3 ).
V=π
∫ 2
0
(
8 −x^3
) 2
dx
Step 2: Enter
∫ (
π∗
(
8 −x^3
) 2
,
x,0,2
)
and obtain
576 π
7
.
- (See Figure 13.8-16.)
Using the Disc Method:
Radius = 2 −x=
(
2 −y^1 /^3
)
.
V=π
∫ 8
0
(
2 −y^1 /^3
) 2
dy
Using your calculator, you obtain
V=
16 π
5
.
y
0 x
B (2, 8)
A (2, 0)
R 2
Figure 13.8-16
- Area =
∑^3
i= 1
f(xi)Δxi.
xi=midpoint of theith interval.
Length ofΔxi=
12 − 0
3
=4.
Area of RectI=f(2)Δx 1 =(2.24)(4)= 8. 96.
