So far what I've got for solving this problem is:
$$ \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \int_{\sqrt{x^2+y^2}}^{3} dz dx$$
And I think that's correct, but I'm not sure.
How do I figure out the limits of the outermost integral (which will be with respect to $x$)? I can't use $\sqrt{9-x^2}$ because the limits of the outermost integral must be constants. But that equation is the only information I have about $x$, isn't it? Do I just set the limits to be from $0$ to $1$?
Any help is appreciated.
Remember the mass is given by \begin{align*} M = \iiint_{D}m(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z \end{align*}
In the present case, $D = \{(x,y,z)\in\textbf{R}^{3} \mid \sqrt{x^{2}+y^{2}}\leq z \leq 3\}$ and $m(x,y,z) = z$. Thus we have to solve the following integral \begin{align*} \int_{-3}^{3}\int_{-\sqrt{9-x^{2}}}^{+\sqrt{9-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{3}z\mathrm{d}z\mathrm{d}y\mathrm{d}x \end{align*}
Then apply the the cylindrical change of variables. Can you take it from here?