October 9th, 2017, 02:00 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2  3D polar coordinate revisit..
Find the volume below z = r, above the xy 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, 05:16 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,192 Thanks: 871 
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, 06:00 PM  #3  
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2 
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:
 
October 9th, 2017, 06:26 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,976 Thanks: 1026  

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coordinate, polar, revisit 
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