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I am trying to find the volume of the solid enclosed by three cylinders given by $x^2+y^2=1$, $x^2+z^2=1$, and $y^2+z^2=1$. I'm supposed to be using a triple integral, and I assume, cylindrical coordinates.

So far, I've figured out that I need to evaluate a triple integral of $dV$, which is equal to $rdzdrd\theta$. However, I am having trouble figuring out what bounds to use for $r, \theta, z$.

Any assistance or hints with this problem would be greatly appreciated!

NeverendingRae
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    Here's a good source on Steinmetz Solids to act as supplementary material to the answer below. When you're solving problems like this, graphing and drawing detailed pictures can be very helpful in visualizing the boundaries of the solid. – B. Standage Jul 26 '17 at 23:17
  • @B.Standage: interesting, I did not know such a solid had a name ! drawing is always helpful, but in 3D can be many times also misleading, so I personally found by experience that inequalities set are unavoidable – G Cab Jul 26 '17 at 23:28
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    I agree, sometimes it can get a little messy. If you're interested, here's another source that gives a few good drawings of this shape. It also adds a visual aid to the below solution as their methods are almost identical. – B. Standage Jul 26 '17 at 23:32
  • @B.Standage: thanks, it is helpful. – G Cab Jul 27 '17 at 11:09
  • This comment is to link this post as one of the (abstract) duplicates to the current choice of mother/target post, which merit is not in the content nor being the oldest but merely having an existing link. – Lee David Chung Lin Jan 22 '19 at 12:04

1 Answers1

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Take the cylinders as solid, so $x^2+y^2 \le 1$ and so on.

Take for symmetry the first octant, i.e. $0 \le x,y,z$.

Then change the coordinates to cylindrical to obtain $$ \left\{ \matrix{ 0 \le r\,(by\,def.) \hfill \cr 0 \le r\cos \theta ,r\sin \theta ,z \hfill \cr r^2 \left( {\cos ^2 \theta + \sin ^2 \theta } \right) \le 1 \hfill \cr z^2 + r^2 \cos ^2 \theta \le 1 \hfill \cr z^2 + r^2 \sin ^2 \theta \le 1 \hfill \cr} \right. $$

and simplify to $$ \left\{ \matrix{ 0 \le \theta \le \pi /2 \hfill \cr 0 \le r \le 1 \hfill \cr 0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta } \hfill \cr 0 \le z \le \sqrt {1 - r^2 \sin ^2 \theta } \hfill \cr} \right. $$

To solve for the last two bounds in $z$, again using symmetry, just reduce the angle to $\pi /4$, and integrate with these conditions $$ \left\{ \matrix{ 0 \le \theta \le \pi /4 \hfill \cr 0 \le r \le 1 \hfill \cr 0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta } \hfill \cr} \right. $$

Then finally you shall multiply by $16$.

G Cab
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  • Thank you! So, just to clarify, the integral is multiplied by 16 because we were looking at an 8th of the solid and then cut that in half by reducing the angle further? – NeverendingRae Jul 26 '17 at 23:30
  • @C.Thomson: Yes, exactly because that. – G Cab Jul 27 '17 at 10:47
  • @C.Thomson: sorry, I made an error in the last set: for $0 \le \theta \le \pi /4$ the "leading" inequality is that in $cos$, not in $sin$. I have amended my answer accordingly. – G Cab Jul 27 '17 at 10:54