Rotational Volume of $y^2 = \frac{x^3}{2a-x}$ around asymptote $x = 2a$ My approach is to use a shell method over the range $[0, 2a]$. One cylinder will be $C = 2\pi xy \ dx$. Let us only work with the positive quadrant and multiply by two for ease:
https://www.desmos.com/calculator/pvuquz8orz
then the volume is: $$V = 2 \int_{0}^{2a} 2 \pi (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx$$
One question I have is that when $x \rightarrow 2a$ it goes to infinity, would this mean I need to treat the definite integral as:
$$V = 2 \lim_{b\to2a}\int_{0}^{b} 2 \pi (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx =  4 \pi \lim_{b\to2a}\int_{0}^{b} (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx$$
I am not sure how to evaluate this integral either. Is this setup at least conceptually correct?
 A: Actually the radius of the shell is $2a-x$ and the height of the shell is $2y$ (accounting for below the $x-$axis as well). So the volume of the shell will be
$$dV=2\pi(2a-x)(2y) \,dx=4\pi(2a-x)\sqrt{\frac{x^3}{2a-x}}\,dx=4\pi x\sqrt{2ax-x^2}\,dx$$
Now we get the volume of the solid as
$$v=\int_0^{2a}dV=4\pi\int_0^{2a}x\sqrt{2ax-x^2}\,dx$$
Hopefully you can take it from here.
A: You're on the right track. As long as $x>0$ then
$$ 4\pi\int_0^{2a} (2a-x)x \sqrt{\frac{x}{2a-x}}\:dx = 4\pi\int_0^{2a} x \sqrt{x(2a-x)}\:dx$$
Then use the substitution $x=2a\sin^2\theta$:
$$= 64\pi a^3 \int_0^{\frac{\pi}{2}} \sin^4\theta\cos^2\theta\:d\theta$$
Denote $I$ as just the integral part without the constants. Under the variable interchange $\theta\leftrightarrow \frac{\pi}{2} -\theta$ we get that
$$2I = \int_0^{\frac{\pi}{2}}\sin^4\theta\cos^2\theta+\cos^4\theta\sin^2\theta\:d\theta = \frac{1}{8}\int_0^{\frac{\pi}{2}}\sin^2 2\theta\:2d\theta$$
Using the fact that $\int_a^b \sin^2 t \:dt = \frac{b-a}{2}$ whenever $a$ and $b$ are integer multiples of $\frac{\pi}{2}$, we can simplify this down to
$$I = \frac{\pi}{32}$$
which means our final answer is
$$2\pi^2 a^3$$
