I am not sure if I am doing the right thing here in setting up the integrals using cylindrical coordinates. I need to set up the integral using cylindrical coordinates to solve for the volume such that the triple integral is $\iiint xyz \,dV$ and S is the region bounded by the cylinders $x^2+y^2=25$, $x^2+z^2=25$, and the first octant.

My attempt here is that I converted the cylinders to its cylindrical coordinates; hence, it become

$$ x^2+y^2=25 \implies r =5$$ $$ x^2+z^2 = 25 \implies r^2\cos^2(\theta) + z^2 = 25$$

If I consider the intersection $ y = \pm z$, then are the bounds for z here will be $0 \leq z \leq r \sin(\theta)$ and the bounds for r here will be $0 \leq r \leq 5$?

  • $\begingroup$ use symmetry with intersections $y=\pm z $ $\endgroup$
    – Narasimham
    May 24, 2021 at 0:17
  • $\begingroup$ @Aiden was my answer helpful? $\endgroup$
    – Math Lover
    Jun 16, 2021 at 6:27

1 Answer 1


For cylindrical coordinates, think this way. If you pick any of the two cylinders, the area of the cross section is bound by that cylinder and the height is bound by the other cylinder.

So if we pick the cylinder $x^2 + y^2 \leq 25$, we have

$x = r \cos \theta, y = r \sin\theta, 0 \leq r \leq 5, 0 \leq \theta \leq \frac{\pi}{2}$. Upper bound of $\theta$ is given we are limited in first octant.

Now, $z$ is bound by the other cylinder. So,
$0 \leq z \leq \sqrt{25 - x^2} = \sqrt{25 - r^2 \cos^2\theta}$

Finally the integral,

$ I = \ \displaystyle \frac{1}{2} \ \int_0^{\pi/2} \int_0^5 \int_0^\sqrt{25 - r^2 \cos^2\theta} \: z \ r^3 \sin 2\theta \ dz \ dr \ d\theta$

Once you integrate wrt $dz$, you get $z^2$ term and as you can see, the integral is straightforward.


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.