Volume inside sphere, outside cylinder

1. May 13, 2010

fishingspree2

Find the volume inside the sphere $$x^{2}+y^{2}+z^{z}=16$$ and outside the cylinder $$x^{2}+y^{2}=4$$. Use polar coordinates.

The sphere's center lies at the origin. The region of integration is the base of the cylinder, the radius 2 xy disk $$x^{2}+y^{2}=4$$ and the two parts of the sphere are given by $$z=\pm\sqrt{16-x^{2}-y^{2}}$$

Volume of sphere is 4 pi r^3 over 3 = 4 pi 4^3 over 3

Therefore:
Volume inside sphere but outside cylinder = $$\frac{4\pi 4^{3}}{3}-2\int_{0}^{2\pi}\int_{0}^{2}\left(\sqrt{16-r^{2}}\right)rdrd\theta$$

What is wrong with my reasoning?
Thank you very much

Last edited: May 13, 2010
2. May 13, 2010

tiny-tim

Hi fishingspree2!

(have a pi: π and a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box )

What is your √(16 - r2) supposed to be?

(and why are you finding the "missing" volume? isn't it easier just to find the volume given?)

3. May 13, 2010

fishingspree2

the √(16 - r2) is the top half of the sphere in polar coordinates

I thought about finding the volume directly but I don't know how to setup the integral.

thank you

4. May 13, 2010

tiny-tim

Sorry, no idea what you mean … is that the length of something?
Divide the volume into horizontal slices of height dz (each slice will be a "washer" with inner radius 2), find the volume of each slice, and integrate.

5. May 13, 2010

arildno

Hi, fishingspree!

It might be easier solving this by:

1. Determine at what z-level the cylinder intersects with the sphere:
$$x^{2}+y^{2}+z^{2}-16=0=x^{2}+y^{2}-4\to{z}^{2}=12\to{z}=\pm\sqrt{12}$$

2. Determine the volume INSIDE this region, using spherical coordinates, for example, for the volume of the two caps (along with the easy calculation of the volume of the cylinder).

3. Subtract this volume from the sphere's volume.