Question

Find the volume of the solid of revolution formed by rotating the bounded region about the x-axis.

f(x)=e-x, y=0, x=-2, x=1.

Answer 1

27.11514 pi

Explanation:

Given is a function exponential as

`f(x) = e^(-x)`

The region bounded by the above curve, y =0 , x=-2 x =1 is rotated about x axis.

The limits for x are -2 and 1

The volume when rotated through x axis is found by

`\pi\int\limits^b_a {f(x)^2} \, dx`

Here a = -2 and b =1

volume =
`\pi\int\limits^1_(-2) {(e^-x)^2} \, dx`

=
`\pi ((e^(-2x) )/(-2) )\\= (\pi)/(2) (-e^1+e^4)\\= 27.11514 \pi`