Question

Find the volume of the solid generated by revolving the region bounded by the curves y = x^2 and y = 4x − x^2 about the line y = 6.

Answer 1

Solution:

Solve a system

y=x²

y=4x-x²

We obtain x=0,x=2

Then in a new coordinates y'=y-6 we have two functions:y'₁=x²-6, y'₂=4x-x²-6

V=$\pi$ $\int$ ²₀((y'₁)²-(y'₂)²)dx=...=

π$\int$ ²₀(8x³-40x²+48x)dx=$\pi$ (2x⁴-40x³/3+24x²)|²₀=$\pi$ (32-320/3+96)=64$\pi$/3