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October 9th, 2017, 03:00 PM   #1
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3D polar coordinate revisit..

Find the volume below z = r, above the x-y plane, and inside $\displaystyle r = cos \theta $

Answer: 4/9

Would anyone show me how to solve this with double/triple integral with polar coordinates?
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October 9th, 2017, 06:16 PM   #2
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It's pretty much just what it says!

First, saying "below z= r" and "above the xy plane" means that z goes from 0 to r. Of course, the r and $\displaystyle \theta$ integration is just that for the "inside $\displaystyle r= cos(\theta)$" which means r goes from 0 to $\displaystyle cos(\theta)$.

$\displaystyle \int_0^{2\pi}\int_0^{cos(\theta)}\int_0^r dz (rdr d\theta)$
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October 9th, 2017, 07:00 PM   #3
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Why do you think theta limits be between 0 and $\displaystyle 2 \pi $?

Should the theta limits be between $\displaystyle - \pi / 2 $ and $\displaystyle \pi / 2 $??


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It's pretty much just what it says!

First, saying "below z= r" and "above the xy plane" means that z goes from 0 to r. Of course, the r and $\displaystyle \theta$ integration is just that for the "inside $\displaystyle r= cos(\theta)$" which means r goes from 0 to $\displaystyle cos(\theta)$.

$\displaystyle \int_0^{2\pi}\int_0^{cos(\theta)}\int_0^r dz (rdr d\theta)$
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October 9th, 2017, 07:26 PM   #4
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Originally Posted by zollen View Post
Why do you think theta limits be between 0 and $\displaystyle 2 \pi $?

Should the theta limits be between $\displaystyle - \pi / 2 $ and $\displaystyle \pi / 2 $??
yes, they should. The circle is to the right of the $y$ axis
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