# Solids of Revolution - revolving $OAB$ around $x = 2$ (diagram included)

In my calculus one class, I was given a worksheet we could do for more practice (note: it is NOT homework). This particular problem, was a solids of revolution one. The goal is to find the volume of $OAB$ when it is rotated around $x = 2$. Below I have included a diagram, and further below is my work so far.

This seems to become nasty very quickly from what I notice.

$$V = \pi\int_0^{16}(16 - \sqrt[3]{\dfrac{y}{2}})^{2} - (\sqrt[3]{\dfrac{y}{2}})^{2} dy$$

In an earlier problem, I had done $\int_0^{16} (\sqrt[3]{\dfrac{y}{2}})^{2} dy$ and got $\frac{192\pi}{5}.$ So, I rewrote the integral as:

$$V = \pi\int_0^{16}(16 - \sqrt[3]{\dfrac{y}{2}})^{2} dy - \frac{192\pi}{5}$$

$$V = \pi\int_0^{16}256 -32\sqrt[3]{\dfrac{y}{2}}) + (\frac{y}{2})^{2/3} dy - \frac{192\pi}{5}$$

$$V = \pi\int_0^{16}256 -32(\frac{y}{2})^{2/3} + (\frac{y}{2})^{2/3} dy - \frac{192\pi}{5}$$

To easier see where the constant can be pulled out when integrating this, I broke it up into:

$$\pi[\int_0^{16} 256 dy - \int_0^{16} 32(\frac{y}{2})^{2/3} dy + \int_0^{16} (\frac{1}{2})^{2/3} y^{2/3} dy] - \frac {192\pi}{5}$$

Upon evaluating, I get an answer of approximately $9007$, which is wrong. Any guidance or help would be appreciated!

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It looks like you were supposed to square an exponent of 2/3 but got an exponent of 2/3 instead of 4/3 at one point.

The bigger deal is this: you are slicing horizontally (that's why you had to do the algebra to express $x$ in terms of $y$) so that your pieces look like washers. The inner radius, which is the distance from the vertical line to the curve, should be $(2 - (y/2)^{1/3})$ and the outer radius, which is the distance between the vertical lines $x = 0$ and $x = 2$, should just be 2. So the integrand (ignoring the $\pi$) should be $$(2)^2 - (2 - (y/2)^{1/3})^2$$

(note: comments below refer to a less detailed version of this from earlier.)

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Why $2$ - cube root rather than $16$? I was under the impression that if our lower and upper limits are in terms of $y$ ($0$ and $16$) that the outer radius would have to be $16$ - cube root. – Joe Apr 25 '12 at 4:02
So the point is that you're slicing this shape up horizontally, so that the pieces you get when you revolve it around this vertical line look like washers, with area pi times (outer radius squared minus inner radius squared). So the outer radius is the horizontal distance from the y axis (i.e. the line x = 0) to the line you're rotating about, which is 2, not 16. – user29743 Apr 25 '12 at 4:05
Ah, I'm sorry, there's another problem with the inner radius. Let me edit my answer. Do you see the other algebra error I mentioned? IT comes when you expand the square factor. – user29743 Apr 25 '12 at 4:06
Ah, I see it now. I was wrong from the beginning in setting up the integrand. I understand now (in fact, just got $\dfrac{288\pi}{5}$ with your integrand) which seems like a reasonable answer. Thank you! – Joe Apr 25 '12 at 4:17

\begin{align} V &= \int_0^2 2 \pi (2-x) (16-2x^3)dx\\ &= 57.6 \pi \end{align}

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