Areas, Volumes, and Arc Lengths 285
x
y
y = –3
y = ex
0
–3
ln 2
Figure 12.4-11
Step 2. Determine the radius from a cross section.
r=y−(−3)=y+ 3 =ex+ 3
Step 3. Set up an integral.
V=π
∫ln2
0
(ex+ 3 )^2 dx
Step 4. Evaluate the integral.
Enter
∫ (
π(e∧(x)+ 3 )∧2, x, 0 ln (2)
)
and obtainπ
(
9ln2+
15
2
)
= 13. 7383 π.
The volume of the solid is approximately 13. 7383 π.
TIP • Remember: iff′is increasing, thenf′′>0 and the graph offis concave upward.
The Washer Method
The volume of a solid (with a hole in the middle) generated by revolving a region bounded
by 2 curves:
About thex-axis:
V=π
∫b
a
[
(f(x))^2 −(g(x))^2
]
dx; wheref(x)=outer radius andg(x)=inner radius.
About they-axis:
V=π
∫d
c
[
(p(y))^2 −(q(y))^2
]
dy; wherep(y)=outer radius andq(y)=inner radius.