Volume inside sphere, outside cylinder

[fishingspree2]

Find the volume inside the sphere [tex]x^{2}+y^{2}+z^{z}=16[/tex] and outside the cylinder [tex]x^{2}+y^{2}=4[/tex]. 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 [tex]x^{2}+y^{2}=4[/tex] and the two parts of the sphere are given by [tex]z=\pm\sqrt{16-x^{2}-y^{2}}[/tex]

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

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

What is wrong with my reasoning?
Thank you very much

 

[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?)

 

[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

 

[tiny-tim]

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

Sorry, no idea what you mean … is that the length of something?

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

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.

 

[arildno]

Find the volume inside the sphere [tex]x^{2}+y^{2}+z^{z}=16[/tex] and outside the cylinder [tex]x^{2}+y^{2}=4[/tex]. 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 [tex]x^{2}+y^{2}=4[/tex] and the two parts of the sphere are given by [tex]z=\pm\sqrt{16-x^{2}-y^{2}}[/tex]

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

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

What is wrong with my reasoning?
Thank you very much

Hi, fishingspree!

It might be easier solving this by:

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

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.