The volume of the solid obtained by revolving the region bounded by $y^2=x$ and $x=2{y}$ about the $y$-axis.

Question

I know the answer already, I just want to know how to solve it.

Answer 1

$$\pi\int\limits_0^2(4y^2-y^4)dy=\pi\left(\frac{32}{3}-\frac{32}{5}\right)=\frac{64\pi}{15}$$

Added: Look at both curves as functions of the form $\,x(y)\,$. Draw a diagram if necessary and find out where they meet.

The function $\,x=2y\,$ is above the function $\,x=y^2\,$ on $\,y\in [0,2]\,$ , i.e. $\,2y\geq y^2\,\,,\forall\,y\in [0,2]\,$ , so the volume of revolution is Pi times the integral on the given interval of the difference of squares of the given functions.