# Volume inside Sphere, outside Cylinder

1. Jul 17, 2005

### 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.

Last edited: Jul 17, 2005
2. Jul 17, 2005

### 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.

3. Jul 17, 2005

### HallsofIvy

Staff Emeritus
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 $$\frac{16\sqrt{2}}{3}\pi-2\pi= \frac{16\sqrt{2}-6}{3}\pi$$.

4. Jul 17, 2005

### mattmns

5. Jul 18, 2005

### HallsofIvy

Staff Emeritus
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.