Volume inside Sphere, outside Cylinder

[mattmns]

Just wondering if I did this correct.



Find the volume of the region that lies inside the sphere x^2 + y^2 + z^2 = 2 and outside the cylinder x^2 + y^2 = 1
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Using cylindrical coordinates, and symmetry, I got:

I went up the z-axis, hitting z = 0 first, then exiting at z = \sqrt{2-r^2}

So, the projection is two circles, one with r=1 and the other r=\sqrt{2}

2 \int_{0}^{2\pi}\int_{1}^{\sqrt{2}}\int_{0}^{\sqrt{2-r^2}} r dz dr d\theta

Which is then

4\pi \int_{1}^{\sqrt{2}} r \sqrt{2-r^2} dr

Letu = 2-r^2<br /> =&gt; du = -2rdr <br /> =&gt; \frac{-du}{2} = rdr

Then I got:

-2\pi \int_{r=1}^{r=\sqrt{2}} \sqrt{u}du = \left[ -2\pi \frac{2}{3} u^\frac{3}{2} \right]_{r=1}^{r=\sqrt{2}}

\left[ \frac{-4\pi}{3}(2-r^2)^\frac{3}{2} \right]_{1}^{\sqrt{2}}

Which equals \frac{4\pi}{3}

Look ok? Thanks.

 

[OlderDan]

Looks good. Could also have been done by the methods of "concentric cylindrical shells" or "washers" ususually used to introduce volumes of revolution, but multiple integration gets you to the same place.

 

[HallsofIvy]

It's not really necessary to use calculus at all! The cylinder is given by x²+ y²= 1. That will intersect the sphere when x²+ y²+ z²= 1+ z²= 2: that is, when z= -1 and 1 so the cylinder has height 2. The volume of a sphere with radius \sqrt{2} is \frac{4}{3}\pi (\sqrt{2})^3= \frac{16\sqrt{2}}{3}\pi. The volume of a cylinder with radius 1 and height 2 is \pi (1)^2(2)= 2\pi. The volume of the region between them is \frac{16\sqrt{2}}{3}\pi-2\pi= \frac{16\sqrt{2}-6}{3}\pi.

 

[mattmns]

Well it may not be necessary to use calculus, but this is the way it is expected to be done for the class I am taking.

Also, I think you forgot the two areas (the top and bottom). Shown here:

http://img307.imageshack.us/img307/9403/areaforgot9ew.gif

 

[HallsofIvy]

Of course, the formula I gave was for a cylinder with flat tops. I suspect the problem means the part of the infinite cylinder contained within the sphere- that is, yes, you have to subtract off the two top and bottom volumes. You will need to use calculus to find those. The disk method ought to work.