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Find the volume formed by $x^2+y^2=9$, $z=0$, and $y=3z$.

I am having trouble with determining the limits of integration. $x^2+y^2=9$ describes a circle centered at the origin with a radius of $3$. It would also be helpful to know how to change the Cartesian limits to polar limits of integration.

I think that the $z$ value for the limit will be from $0$ to $\frac{y}{3}$ The $y$ limits of integration will be from $0$ to $3$. (since the radius is $3$). The $z$ limits of integration will be from $0$ to $2\pi$.

Integration is not the problem. It's just visualizing and labeling the limits of integration. Thank you.

Jihad
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user192989
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    It seems to be $\frac{1}{4}$ of the volume of cylinder $x^2+y^2 = 9$ and $z \in [0, 1]$ (since picture is pretty symmetric). So the answer will be $\frac{9}{4}\pi^2$. – Jihad Nov 16 '14 at 15:42
  • the problem for me is finding the limits of integration – user192989 Nov 16 '14 at 15:46
  • Just try to visualize it. Firstly infinite cylinder $x^2+y^2 = 9$ and then plane $y = 3z$. And then intersect them. – Jihad Nov 16 '14 at 15:48

2 Answers2

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Use cylindrical coordinates

$$\begin{align}x &= r\cos \theta \\y&= r\sin \theta\\ z&= z\end{align}$$

Where $0 \leq \theta \leq \pi ,\ 0\leq r\leq 3$ and $0 \leq z \leq \frac{1}{3}r \sin \theta$. Then

$$\int_{0}^{3}\int_{0}^{\pi}\int_{0}^{\frac{1}{3}r \sin \theta}f(r\cos\theta,r\sin \theta, z)\ r\ dz\ d\theta\ dr$$

Aaron Maroja
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There are two finite pieces. One of them has $y,z>0$, the other has $y,z<0$. They meet at the line $y=z=0$.
One piece has $0<\theta<\pi$,$0<r<3$ and $0<z<\frac13r\sin\theta$. Put them together into a cylindrical polar integral to find its volume. The negative piece will have the same volume.

Empy2
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