My Math Forum 3D polar coordinate revisit..

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 October 9th, 2017, 03:00 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 163 Thanks: 2 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?
 October 9th, 2017, 06:16 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 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 $\displaystyle cos(\theta)$. $\displaystyle \int_0^{2\pi}\int_0^{cos(\theta)}\int_0^r dz (rdr d\theta)$ Thanks from romsek and zollen
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 $\displaystyle \pi / 2$??

Quote:
 Originally Posted by Country Boy 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 $\displaystyle cos(\theta)$. $\displaystyle \int_0^{2\pi}\int_0^{cos(\theta)}\int_0^r dz (rdr d\theta)$

October 9th, 2017, 07:26 PM   #4
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Quote:
 Originally Posted by zollen 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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