MA 3972-MA-Book April 11, 2018 16:1
Areas and Volumes 309
Using the Disc Method:
V=π
∫ 1
− 1
(y^2 +1)^2 dy
=π
∫ 1
− 1
(y^4 + 2 y^2 +1)dy
=π
[
y^5
5
+
2 y^3
3
+y
] 1
− 1
=π
[(
15
5
+
2(1)^3
3
+ 1
)
−
(
(−1)^5
5
+
2(−1)^3
3
+(−1)
)]
=π
(
28
15
+
28
15
)
=
56 π
15
Note that you can use the symmetry of
the region and find the volume by
2 π
∫ 1
0
(y^2 +1)^2 dy.
- Volume of solid by revolvingR:
VR=
∫ 4
0
π(3x)^2 dx=π
∫ 4
0
9 x^2 dx
=π[3x^2 ]^40 = 192 π
Set
∫ 4
0
π(3x)^2 dx=
192 π
2
⇒ 3 a^3 π= 96 π
a^3 = 32
a=(32)^1 /^3 =2(2)^2 /^3
You can verify your result by evaluating
∫2(2)^2 /^3
0
π( 3 x)^2 dx.The result is 96π.
Part B Calculators are allowed.
- (See Figure 13.8-10.)
y
y = x^3 y = x 2
0 1 x
Figure 13.8-10
Step 1: Using the Washer Method:
Points of Intersection: Set
x^3 =x^2 ⇒x^3 −x^2 = 0 ⇒
x^2 (x−1)=0orx=1.
Outer radius=x^2 ;
Inner radius=x^3.
Step 2: V=π
∫ 1
0
((
x^2
) 2
−
(
x^3
) 2 )
dx
=π
∫ 1
0
(x^4 −x^6 )dx
Step 3: Enter
∫
(π(x∧ 4 −x∧6),x,0,1)
and obtain
2 π
35
.
- (See Figure 13.8-11.)
y
x
0
2
2
Figure 13.8-11
