Question

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

f(x)=2x+1, y=0, x=0, x=4.

Answer 1

`=(364\pi)/(3)` cubic units

Explanation:

We are to find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

f(x)=2x+1, y=0, x=0, x=4.

The picture is given as shaded region.

This is rotated about x axis

Limits for x are already given as 0 and 4

f(x) is a straight line

The solid formed would be a cone

Volume =
`\pi \int\limits^a_b {(2x+1)^2} \, dx \\= \pi \int\limits^4_0 {(4x^2+4x+1)} \, dx \\=\pi [(4x^3)/(3) +2x^2+x]^5_0\\\\=\pi[(4*4^3)/(3)+2*4^2+4-0]\\=(364\pi)/(3)`