# 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

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

Let$$u = 2-r^2 => du = -2rdr => \frac{-du}{2} = rdr$$

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.

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#### OlderDan

Homework Helper
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

Homework Helper
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 [Broken]

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