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

  • $\begingroup$ Thanks. Sorry but I don't understand why my volume is f(x,y,z)=1 $\endgroup$ Jul 18 '20 at 17:39
  • $\begingroup$ 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 $\endgroup$ Jul 18 '20 at 17:49
  • $\begingroup$ Ok, so If I have to calculate a volume I have no function under the triple integral, right? $\endgroup$ Jul 18 '20 at 17:59
  • $\begingroup$ Yes, or rather you have a $1$ under the triple integral. $\endgroup$ Jul 18 '20 at 18:02
  • $\begingroup$ Ok, thanks! Now I've understood. $\endgroup$ Jul 18 '20 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.