Volume of Rotation Let $W_n$ be the region in an $xy$-plane that’s bounded by the $y$-axis and the parabola $x=y^2-n^2$.
Find the value of $n$ for which the volume obtained by rotating $W_n$ about the $y$-axis is precisely 16 times the volume obtained by rotating $W_n$ about the $x$-axis.

Lets start with the y-axis revolution.
We are integrating from $-n$ to $n$.
We get $2π\displaystyle \int_0^{n}(n^2-y^2)^2\,dy = \frac{16}{15}πn^5$
I put a $2$ in front for symmetry and ease in calculation.
Next, we do x-axis revolution. This one is a little bit tricky, we have to rewrite in terms of $x$.
$x=y^2-n^2$ turns into $y=\sqrt{x+n^2}$
No plus or minus because either one would produce the needed solid. Using shells would cover the volume of the solid TWICE.
We get $π\displaystyle \int_{-n^2}^{0}(x+n^2)\,dx = \frac{1}{2}πn^4$.
So we get $\displaystyle \frac{16}{15}πn^5=16 \times \frac{1}{2}πn^4$. 
That means $\frac{16}{15}πn^5=8πn^4$.
We get $\frac{16}{15}n=8$.
Therefore, $\boxed{n=\displaystyle \frac{15}{2}}$. 
