What would be the bounds for the triple integral over the surface of a cone, $$x^2+y^2 \leq z^2 $$ where $$0 \leq z \leq a$$
My first guess is to use polar coordinates for $x$ and $y$: $$x=rcos(\theta)$$ $$y=rsin(\theta)$$
I am unsure about what to use for $z$, and then I am unsure what the bounds of integration are for $x$ and $y$ when I go to set up my triple integral.
On
$$\int_0^a\int_0^{z}\int_0^{2\pi}r\mathrm d\theta\mathrm dr\mathrm dz=\frac{1}{3}\pi a^3$$
One way to see this is consider a section of the cone at certain heigh z. Its area is $\pi z^2$. Then you simply have to integrate it between $0$ and $a$. The two inner integrals give you the area, the third, the volume.
We have $r^2\le z^2$, thus $0\le r\le z$. On the other hand $0\le \theta\le 2\pi$, therefore $$\int_{0}^{2\pi}\int_{0}^{a}\int_{0}^{z}r\,\mathrm{d}r\mathrm{d}z\mathrm{d}\theta=\frac 13 a^3 \pi$$