Triple integrals and cylindrical coordinates with hyperboloid

Question

please excuse me if I'll say some dumb things but I'm not very informed in this topic.
If I have an hyperboloid $x^2+y^2-0.12z^2=9$ and I want to find the volume between $z=0$ and $z=15$, I can integrate it like this $ \int_{0}^{15}{AreaOfCircle(z)dz} $ where Area of Circle is $\pi(9+0.12z^2)$.
My problem is that I haven't really understood how to use triple integrals with cylindrical coordinates.
I know that the formula is $ \iiint{r*f(cylindrical coordinates)}drdzd\theta $, but in this case I've read that even if $f(cylindrical coordinates)=r^2-0.12z^2-9$ only the integration limits change, and in the triple integral I have to put only $r$ –> $\int_0^{2\pi}\int_0^{15}\int_0^{\sqrt{9+0.12z^2}}rdrdzd\theta$ instead of $\int_0^{2\pi}\int_0^{15}\int_0^{\sqrt{9+0.12z^2}}(r^2-0.12z^2-9)rdrdzd\theta$.
I don't understand why. Thanks.

Answer 1

The volume of a 3-dimensional region is described by $$\iiint_{E}\,dV.$$ The statement about switching to cylindrical coordinates says that: If $T$ is a 3-dimensional region, and $T'$ is the same region described in cylindrical coordinates then $$\iiint_{T}f(x,y,z)\,dV = \iiint_{T'} rf(z,r,\theta)\,dzdrd\theta$$ Therefore, since you have a volume: $f(x,y,z)=1$ and $$\iiint_{E}\,dV = \iiint_{E'}r\,dzdrd\theta$$ where $E'$ is $E$ but in cylindrical coordinates