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.
On
I am assuming that I can write $x=n^2-y^2$ and that the ranges of the variables are then given by $x\in[0,n^2],\,\,y\in[0,n]$. Now I'll invoke Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The volume is simply $V=2πRA$. Finally, I'll express $RA$ in the complex plane so that I can't compute it for the both the $x$ and $y$ components at the same time. Thus.
$$ \begin{align} RA &=\int_0^n\int_0^{n^2-y^2}(x+iy)dx\,dy\\ &=\int_0^n\left(\frac{x^2}{2}+ixy\right)\big|_0^{n^2-y^2}\,dy\\ &=\int_0^n\left(\frac{(n^2-y^2)^2}{2}+i(n^2-y^2)y\right)dy\\ &=\frac{1}{2}\left[n^5-\frac{2n^5}{3}+\frac{n^5}{5}\right]+i\left[\frac{n^4}{2}-\frac{n^4}{4}\right] \end{align} $$
Then setting
$$\frac{\Re\{RA\}}{\Im\{RA\}}=16=\frac{\frac{n}{2}\frac{8}{15}}{\frac{1}{4}}\Rightarrow n=15$$
I have verified this result numerically.
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, $\boxed{n=\displaystyle \frac{15}{2}}$.