Volume of rotation $y=x^3$ ; lines $x=0$ ; $y=8$ ; about $x=2$?

Question

I'm having trouble with this problem, I'm not sure how to proceed. I know if we were to rotate about the $y$-axis ($x=0$), then I would have the following disk method.

$$\pi \int_0^8 y^{2/3}dy = \frac{(96\pi)}{5}$$

In my case, it becomes a washer method, but I'm unable to properly setup the formula to calculate.

Thanks for any help!

Answer 1

Using the cylindrical shell method:

$$ V=\int_0^22\pi rh\,dx$$

where $r=2-x$ and $h=8-x^3$.

Using the annulus or "disk/washer" method:

$$ V=\int_0^8 \pi(R^2-r^2)\,dy$$

where $R=2$ and $r=2-y^{\frac{1}{3}}$

Answer 2

In these types of problems, the best way to start is to make a sketch of the function. This looks almost like a parabola from $(0,0)$ to $(2,8)$ that you rotate around a vertical line at $x=2$. It's a series of disks, with smaller and smaller radius as $y$ increases, with the largest disk at $y=0$. If you look at the picture, observe the lines $x=0$, $y=x^3$, and $x=2$. It should be obvious that the radius that you care about is is $2-x=2-y^\frac{1}{3}$. You should now be able to finish the problem.