Areas, Volumes, and Arc Lengths 303
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 12.9-15.)
R 1
0
y
x
2
(0,8) C B (2,8)
Figure 12.9-15
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 12.9-16.)
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 12.9-16
- 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.
Area of RectII=f(6)Δx 2 =(3.61)(4)= 14. 44.
Area of RectIII= f(10)Δx 3 =(4.58)(4)= 18. 32.
Total Area= 8. 96 + 14. 44 + 18. 32 = 41. 72.
The area under the curve is approximately
41.72.
