Region under graph of y=1/x is bound by x=1 and x=3, rotated around y=-1. Write integral equal to volume.?

Question

Region under graph of y=1/x is bound by x=1 and x=3, rotated around y=-1. Write integral equal to volume.?

Answer 1

Use the disk method of integration.

Radius of each disk is: [1/x - (-1)] = (1/x + 1)

Area of each disk is: π (r²) = π(1/x + 1)²

Volume of each disk is: π (r²) dx = π(1/x + 1)² dx =
π (1/x² + 2/x + 1) dx

The last equation is equivalent to: π [x^(-2) + 2/x + 1] dx

Now just integrate:

Total volume is: ƒπ [x^(-2) + 2/x + 1] dx from x = 1 to x = 3.

ƒπ [x^(-2) + 2/x + 1] dx = πƒ[x^(-2) + 2/x + 1] dx =

π[ƒ[x^(-2) dx + ƒ(2/x) dx + ƒ(1) dx] =

π[ƒ[x^(-2) dx + 2ƒ(1/x) dx + ƒ(1) dx] =

π[-x^(-1) + 2 (ln x) + x evaluated from x = 1 to x = 3.

Integral = π{[-1/3 + 2 (ln 3) + 3] - [(-1/1) + 2 (ln 1) + 1]} =

π{[-1/3 + 2 (ln 3) + 3] - [(-1) + 2 (0) + 1]} =

π{[-1/3 + 2 (ln 3) + 3] - [(-1 + 1) + 0 ]} =

π{[-1/3 + 2 (ln 3) + 3] - 0} =

π[3 - 1/3 + 2 (ln 3)] =

π[8/3 + 2 (ln 3)] cubic units

Check the steps out carefully. I did this very early in the morning (2 a.m.).