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$. 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.

 A: 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
$$V=\int_{-\pi/2}^{\pi/2}\int_0^{6\cos\theta}\int_0^r r\> dz\>dr\>d\theta=96\ .$$

A: The shift in the circle does affect the $\theta$ bounds. You can't use $0\le \theta \le 2\pi$ if you want to do F(b) - F(a) on the $d\theta$ integral because that $\theta$ domain does not produce a continuous curve. This may not be apparent from just drawing the circle, but an analysis of the $(r, \theta)$ polar coordinates reveals how the circle is drawn out based on $\theta$ values.  


*

*$\theta = 0: r(0) = 6cos(0)=6 \rightarrow (r,\theta) = (6, 0); (x,y) = (6,0)$

*$\theta = \pi/3: r(\pi/3) = 6cos(\pi/3)=3 \rightarrow (r,\theta) = (3, \pi/3); (x,y) = (3, 6sin(\pi/3))=(3, 5.2)$

*$\theta = \pi/2: r(\pi/2) = 6cos(\pi/2)=0 \rightarrow (r,\theta) = (0, \pi/2); (x,y) = (0,0)$

*$\theta = 2\pi/3: r(2\pi/3) = 6cos(2\pi/3)=-3 \rightarrow (r,\theta) = (-3, 2\pi/3); (x,y) = (3, 6sin(2\pi/3))=(3, 5.2)$

*$\theta = \pi: r(\pi) = 6cos(\pi)=-6 \rightarrow (r,\theta) = (-6, \pi); (x,y) = (6,0)$

*$\theta = 4\pi/3: r(4\pi/3) = 6cos(4\pi/3)=-3 \rightarrow (r,\theta) = (3, 4\pi/3); (x,y) = (3, 6sin(4\pi/3))=(3, 5.2)$

*$\theta = 3\pi/2: r(3\pi/2) = 6cos(3\pi/2)=0 \rightarrow (r,\theta) = (0, 3\pi/2); (x,y) = (0,0)$

*$\theta = 5\pi/3: r(5\pi/3) = 6cos(5\pi/3)=3 \rightarrow (r,\theta) = (3, 5\pi/3); (x,y) = (3, 6sin(5\pi/3))=(3, -5.2)$

*$\theta = 2\pi: r(2\pi) = 6cos(2\pi)=6 \rightarrow (r,\theta) = (6, 2\pi); (x,y) = (6,0)$


As shown above, $0\le \theta \le 2\pi$ does not draw out the circle continuously. The curve jumps from the origin at $\pi/2$ back to $(6, 0)$ at $\pi$ and starts drawing out the bottom half of the circle clockwise back towards the origin. In addition, that domain overlaps certain portions of the circle. What you want is one continuous domain going counterclockwise for $\theta$ values that draws out the circle one time with positive r values. These $\theta$ domains would work based on that criteria:


*

*$-\pi/2 \le \theta \le \pi/2$

*$3\pi/2 \le \theta \le 5\pi/2$


Note the above $\theta$ angles all produce positive $r$ values. Even if a $\theta$ angle draws out the correct circle based on $(x,y)$ angles, if it gives negative $r$ values in polar coordinates, it will not produce the correct volume value. For example, $\pi \le \theta \le 2\pi$ doesn't work, even though it the $(x,y)$ points show $(6,0)$, then $(0,0)$, and then $(6,0)$ because those angles have negative $r$ values.
