Originally Posted by

**Scott H** Volumes of revolution are found by integration just as two-dimensional areas are, only our cross-sections are discs or hollow cylinders (called 'shells') instead of rectangles.

We can find the answer to the first problem by rewriting the function as

$\displaystyle x=y^{1/4}$

and using the Shell Method. To do this, we note that the radius of each shell is $\displaystyle 3-y$ and the height $\displaystyle 2y^{1/4}$. The volume of each cross section will therefore be

$\displaystyle 2y^{1/4}\cdot 2\pi(3-y)\,dy.$

Integrating from $\displaystyle y=0$ to $\displaystyle y=1$, we obtain

$\displaystyle 4\pi\int_0^1(3y^{1/4}-y^{5/4})\,dy.$

(Actually, the correct way to rewrite the function is $\displaystyle x=\pm y^{1/4}$, but we accounted for both sides in our integral.)

For the second problem, the volume is infinite. Are you sure it's written correctly?