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!