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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.

Diagram

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!

2 Answers 2

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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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    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!2012-04-25
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\begin{align} V &= \int_0^2 2 \pi (2-x) (16-2x^3)dx\\ &= 57.6 \pi \end{align}