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I have trouble setting up a triple integral to find volume bound by equations, such as:

I'm not sure how to figure how to find the boundaries for the $z$-axis. How would I go about setting up the triple integral ? Find the volume bounded by $z^2 = x^2 + y^2$ and $z = \sqrt{9-x^2 - y^2}$

Duchamp Gérard H. E.
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Johhnnyy Elvis
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  • The general procedure: First figure out what types of surfaces those two equations represent. Then draw a picture. Then, based on any symmetries present, decide which coordinate system would work best. Finally, try to find inequalities which describe the region. –  Oct 30 '16 at 21:15
  • Also note, you don't necessarily have to bound the entire region with one set of inequalities. You are allowed to break up the region into parts if that's helpful. –  Oct 30 '16 at 21:20

2 Answers2

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It is a solid of revolution (around the z-axis), then use the second Pappus's centroid theorem. I add some details :

  1. $z^2=x^2+y^2$ is a cone (just think that with $r=\sqrt{x^2+y^2}$ it has equation $z=\pm r$. This cone is generated by the lines $z=x,y=0$ and $z=-x,y=0$ (just draw them :) rotating around the $z$-axis
  2. $z = \sqrt{9-x^2 - y^2}=\sqrt{9-(x^2 + y^2)}=\sqrt{9-r^2}$ is generated by the curve $z=\sqrt{9-x^2},y=0$ rotating around the $z$-axis. This is a semi-circle.

Hope it helps. Do not hesitate to interact.

Duchamp Gérard H. E.
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$$ \int_{0}^{2\pi}\int_{0}^{\frac{3}{\sqrt{2}}}\int_{0}^{r} r dz drd\theta + \int_{0}^{2\pi}\int_{\frac{3}{\sqrt{2}}}^{3}\int_{0}^{\sqrt{9-r^2}} r dz drd\theta $$

Anonymous
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