Answer:
[tex]=\frac{364\pi}{3}[/tex] 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 = [tex]\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}[/tex]
