Proof of Volume of a ball

  • Thread starter dhlee528
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  • #1
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Homework Statement



http://staff.washington.edu/dhlee528/003.JPG [Broken]

Homework Equations



x = r sin ( phi) cos ( theta)

y = r sin ( phi )sin (theta)

z = r cos ( phi )


The Attempt at a Solution



[tex]
vol=8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\int_0^r \rho^2 \sin(\phi)d\rho d\theta d\phi
[/tex]

[tex]
8 \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2} \sin(\phi)(\frac{\rho^3}{3}){|}_0^r d\theta d\phi
[/tex]

[tex]
\frac{4r^3 \pi}{3}\int_0^\frac{\pi}{2}sin(\phi)d\phi
[/tex]

[tex]
-\frac{4r^3\pi}{3}[0-1]=\frac{4\pi r^3}{3}
[/tex]



I think I got spherical coordinate right but don't know how to do for rectangular or spherical coordinate
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi dhlee528! Welcome to PF! :smile:

(have a theta: θ and a phi: φ and a pi: π :wink:)

For rectangular coordinates: obviouly the volume element is dxdydz, so decide which order you're going to integrate in … say keep z and y fixed, decide the limits on x; then keep z fixed, decide the limits on y.

What do you get? :smile:
 

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