Question

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

Answer 1

Answer:

$=\frac{364\pi}{3}$ cubic units

Step-by-step 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 = $\pi \int\limits^a_b {(2x+1)^2} \, dx \\= \pi \int\limits^4_0 {(4x^2+4x+1)} \, dx \\=\pi [\frac{4x^3}{3} +2x^2+x]^5_0\\\\=\pi[\frac{4*4^3}{3}+2*4^2+4-0]\\=\frac{364\pi}{3}$