The region between the graphs of y=x^2 and y=6x is rotated around the line x=6.
The volume of the resulting solid is ?

Question

The region between the graphs of y=x^2 and y=6x is rotated around the line x=6.
The volume of the resulting solid is ?

anonymous · Answer

h(x) = f(x) - g(x) = x² - 6x 
Vol.Rotate = 2*Pi * Integral( h(x)² ) dx

callisto · Answer

1. Find their intersection points.
    =>  Put y=x^2  into  y=6x  => x(x-6)=0  => x=0 or x=6
2. $$ volume=2π \int_{0}^{6} (6-x)(6x) - (6-x)(x^2) dx$$
Can you do it from here?
(PS: I'm not sure if it is correct :S)

anonymous · Answer

Yup I got it from here, Thanks a lot!

callisto · Answer

Welcome :)
Note: This time, I still use shell method to do it..