# Triple integrals and cylindrical coordinates with hyperboloid

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.

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
• Your volume is the triple integral, the function under the triple integral describes something more , usually a "density" in the case of a triple integral. Take some simpler examples: let $a>0$: then $\int_{0}^{a}\,dx=a$ (the length), $\int_{0}^{a}$ $\int_{0}^{a}\,dxdy=a^{2}$ (the area) and $\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}\,dxdydz = a^{3}$ (the volume Jul 18 '20 at 17:49
• Yes, or rather you have a $1$ under the triple integral. Jul 18 '20 at 18:02