Volume of a cissoid? A cissoid with formula $y^2(2a - x) = x^3$ where $a>0$
How do I show the the volume of the cissoid rotated about the asymptote is:
$$V = 2\pi \int_0^{2a} (3a - x)(2ax - x^2)^{1/2} dx$$
 A: Because of limitations on my ability to draw pictures, the solution is being presented in very much the wrong way.  Please draw a picture, or have a calculator or computer program do so for you. For the price,  Wolfram Alpha does a pretty good job of making a quick sketch.
The asymptote referred to in the problem is the vertical line with equation $x=2a$.  
Whoever derived the formula we are trying to get at may have calculated the volume in not the best way.  We attempt to reconstruct that not so good way, and then do the problem in a simpler way.
A "Slicing" Calculation: Our derivation will be highly informal, and would upset a purist. Imagine making a horizontal slice of width $dy$ at height $y$, and rotating this slice about the line $x=2a$. The slice meets the cissoid at $x$, where $y^2(2a-x)=x^3$. So the length of the slice is the horizontal distance from $x$ to the asymptote, that is, $2a-x$.
When we rotate the slice, we get a thin coin that has radius $2a- x$ and thickness $dy$, and therefore volume $\pi(2a-x)^2 dy$.
"Add up", $y=-\infty$ to $y=\infty$.  Our volume is
$$\int_{y=-\infty}^\infty \pi(2a-x)^2dy.$$
It is useful to take advantage of symmetry and observe that our volume is
$$\int_{y=0}^\infty 2\pi (2a-x)^2 dy.$$
Our integral mixes variables. Let's express everything in terms of $x$. The limits are easy, they go from $x=0$ to $x=2a$.
To deal with the $dy$, we could solve explicitly for $y$ in terms of $x$, then differentiate.  Or else we could use implicit differentiation. I will do something halfway in between.
We have
$$y^2= \frac{x^3}{2a-x}.$$
Then 
$$2y \frac{dy}{dx}=\frac{6ax^2-2x^3}{(2a-x)^2}.$$
(some routine differentiation details are omitted).
Now solve for $dy$, using the fact that $y=x^{3/2}/(2a-x)^{1/2}$.
We get
$$dy= \frac{(3a-x)x^{1/2}}{(2a-x)^{3/2}}.$$
Finally, the volume is
$$\int_0^{2a} 2\pi (3a-x) x^{1/2}(2a-x)^{1/2}dx$$
This  is essentially what was to be shown.
A Cylindrical Shell Calculation:  Again, the reasoning will be highly informal.  Imagine drawing a vertical line at $x$, and another vertical line a tiny bit further on, at $x+dx$.  We obtain a thin strip that meets our curve at $\pm y$, where $y^2(2a-x)=x^3$.
Rotate this thin strip about the line $x=2a$.  We get a cylindrical shell.  The cylinder has radius $2a-x$, and height $2y$, where $y$ is the positive solution of $y^2(2a-x)=x^3$.  Thus the volume of the cylindrical shell is "about" $2\pi(2a-x)(2y)dx$.  Note that 
$y=x^{3/2}/(2a-x)^{1/2}$.  
Now "add up" the volumes of these shells.  We conclude that the volume we are looking for is 
$$\int_0^{2a} 4\pi (2a-x)^{1/2}x^{3/2}dx.$$ 
Of course this is not what was to be shown, it is simpler-looking!  
By fooling around for a while, we could uglify the above formula and turn it into what was to be shown.  Or start from the ugly formula, and write $3a-x$ as 
$2(2a-x) + (x-a)$.  Break up the integral into two parts.  The second part, the one involving the term $(x-a)$, integrates to $0$ over our interval, by symmetry.
