Page 233 - Calculus Demystified

Applications of the Integral
CHAPTER 8
220
the height of the upper edge is 3 − x/2, and that number is also the radius of
the circular slice at position x (Fig. 8.6). Thus the area of that slice is
x 2

A(x) = π 3 − .
2

thickness x
_
3 x/2

Fig. 8.6

We find then that the volume we seek is

6 6 3 6
x 2 2(3 − x/2)
V = A(x) dx = π 3 − dx =−π = 18π.
0 0 2 3 0
You Try It: Any book of tables (see [CRC]) will tell you that the volume of a right
1
2
circular cone of base radius r and height h is πr h. This formula is consistent
3
with the result that we obtained in the last example for r = 3 and h = 6. Use the
technique of Example 8.1 to verify this more general formula.
EXAMPLE 8.2
A solid has base the unit disk in the x-y plane. The vertical cross section at
position x isan equilateral triangle. Calculate the volume.

SOLUTION
2
2
Examine Fig. 8.7. The unit circle has equation x +y = 1. For our purposes,
this is more conveniently written as

2
y =± 1 − x . (.)
Thus the endpoints of the base of the equilateral triangle at position x are the
√
2
points (x, ± 1 − x ). In other words, the base of this triangle is

2
b = 2 1 − x .
Examine Fig. 8.8. We see that an equilateral triangle of side b has height
√ √
2
3b/2. Thus the area of the triangle is 3b /4. In our case then, the equilateral