Page 236 - Calculus Demystified

CHAPTER 8
Applications of the Integral
√
! √ " 1/ 2 223
3
2 − 2x
= 2 −
6
0
! √ "
0 3 2 2
= 2 − − −
6 6
√
2 2
= .
3
You Try It: Calculate the volume of the solid with base in the plane an equilateral
triangle of side 1, with base on the x-axis, and with vertical cross-section parallel
to the y-axis consisting of an equilateral triangle.

EXAMPLE 8.4
Calculate the volume inside a sphere of radius 1.

SOLUTION
It is convenient for us to think of the sphere as centered at the origin in the
x-y plane. Thus (Fig. 8.11) the slice at position x, −1 ≤ x ≤ 1, is a disk. Since
we are working with base the unit circle, we may calculate ( just as in Example
√ √
2
2
8.2) that the diameter of this disk is 2 1 − x . Thus the radius is 1 − x and
the area is
2
2
A(x) = π · 1 − x 2 = π · (1 − x ).

Fig. 8.11
In conclusion, the volume we seek is

1

2
V = π(1 − x )dx.
−1