TITLE: Trouble finding the volume of a solid $y^2=2x, x=2y$
QUESTION [2 upvotes]: I think I'm finally beginning to get the hang of these problems but I'm having trouble with this one:
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line (about the $y-axis$):
$$y^2=2x, x=2y$$
The first thing I did was draw the graph.  After drawing the graph I figured out that a slice would be that of a washer.  So I figured I need to make $y$ a function of $x$.
$$y^2=2x$$
$$\frac{y^2}{2}=x$$
And next I figured I needed to find points of intersection so I did the following:
$$y^2=2(2y)$$
$$y^2=4y$$
$$y=0, y=4$$  
I took those to be the bounds of my function.  So now I sought to find the area of the washer using the following formula:
$$A=\pi R^2-\pi r^2$$
$$A=\pi(R^2-r^2)$$
So plugging in my functions:
$$A=\pi((2y)^2-(\frac{y^2}{2})^2)$$
Simplifying this I get:
$$A=\pi (4y^2-\frac{y^4}{4})$$
But when I integrate using the following:
$$\pi \int_{0}^{4}4y^2-\frac{y^4}{4}dy$$
My anti-derivative is:
$$\pi [\frac{4y^3}{3}-\frac{y^5}{20}]|_{0}^{4}$$
But my answer is incorrect.  Do you all know what I'm doing wrong?

REPLY [3 votes]: CW answer to push it off unanswered queue:
Answer is $\frac{512\pi}{15}$. Everything you wrote before giving the final answer was correct. You did not show detail there, I assume it was a slip of arithmetic near the end. Substituting, I get $\pi(\frac{4^4}{3}−\frac{4^5}{20})$. The second term is $\frac{4^4}{5}$, now bring to a common denominator of $15$. 
(André Nicolas)
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