Q: Find the volume of the solid bounded above by the cone $z^2 = x^2 + y^2$, below by the xy plane, and on the sides by the cylinder $x^2 + y^2 = 6x$.
I can't figure out what I'm doing wrong in my setup. I keep getting $0$ for the volume value based on my setup. However, if you sketch the volume out, it seems like there should be a volume in 3d for this shape.
My Work
$\int{_{0}^{2\pi}\int{_{0}^{6cos\theta}\int{_{0}^{r}r dzdrd\theta}}}$
- Since the xy plane was a bound, I assumed you needed the top part of the cone so $z = \sqrt{x^2+y^2}$ or $z = r$ for the top z bound.
- Since $x^2 + y^2 = 6x$, polar conversion equations give $r^2 = 6rcos\theta$ or $r^2 – 6rcos\theta=0$ so $r=0$ and $r = 6cos\theta$ from that.
- $x^2 + y^2 = 6x$ is a full circle so I let $\theta = 0$ to $\theta = 2\pi$ bounds. There is a shift in the circle but I don't believe it affects the $\theta$ bounds.
Best Answer
The formula $$V=\int_?^?\int_?^?\int_?^? r\> dz\>dr\>d\theta$$ refers to cylindrical coordinates. In particular, $\theta$ is the polar angle centered at $(0,0)$. Even though $\theta$ is not the "natural" parameter for the circle $(x-3)^2+y^2=9$ you can use $\theta$ to describe this circle in terms of $r$ and $\theta$ as well. Looking at the figure below one immediately verifies that the circle has the polar representation $$r(\theta)=6\cos\theta\qquad\left(-{\pi\over2}\leq\theta\leq{\pi\over2}\right)\ .$$ The volume in question is then given by $$V=\int_{-\pi/2}^{\pi/2}\int_0^{6\cos\theta}\int_0^r r\> dz\>dr\>d\theta=96\ .$$