Question

How do you find the volume when the region bounded by y = x+3, y = 0, x = -3 and x = 3 is revolved around the x-axis?

Answer 1

`72pi`

Explanation:

Use the disk method: `A=piintr^2`

`r` is `y=x+3` because it's bound by `y=0`

`piint_-3^3(x+3)^2dx`

`piint_-3^3(x^2+6x+9)dx`

`pi((3^3/3+3(3^2)+9(3))-((-3)^3/3+3(-3)^2+9(-3)))=72pi`