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I need to find a volume of an object restricted with the following planes:

$$z=x^{2}+y^{2}$$ $$z=1$$

Ok, pretty easy, I'm going for the polar coordinates:

$0 < r < 1$
$0 < \phi < \frac{\pi}{2}$

This is clear and obvious, but when it comes to the z variable...well, I think it should be:

$x^{2}+y^{2} < z < 1$

But the notes say it's rather:

$r^{2} < z < 1$

Why is that so, where did the r came from?

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I can't see that. If $x^{2}+y^{2}<z<1$ is true, and r = 1 = ...but wait, r should be a variable here...no, I can't see that. –  khernik Aug 2 '13 at 20:36
    
$x^2+y^2=r^2$ (I think that is what Ataraxia meant) –  kaine Aug 2 '13 at 20:36
    
Oh, ok, I got it now, thanks! Actually, in polar coordinates $x^{2}+y^{2}$ becomes $r^{2}$ –  khernik Aug 2 '13 at 20:40
    
@khernik Yea, typo, sorry about that. kaine is correct. –  Ataraxia Aug 2 '13 at 20:41

1 Answer 1

The limits in Cylindrical coordinates is as follows:

$$\theta|_0^{2\pi}, r|_{0}^1,z|_{r^2}^1$$ Note that the plane $z=1$ intersect $z=r^2=x^2+y^2$ at $x^2+y^2=1$. And that's why we choose above limits for $r$.

enter image description here

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+1, dear Babak! Helpful graph, here. –  amWhy Aug 3 '13 at 1:05
    
Each time I see these I get more and more envious of these nice graphs...+1 –  DonAntonio Aug 3 '13 at 11:50
1  
@BabakS.: I agree with DA, beautiful pictures! Congrats on answering over 1000 questions (wow) +1 –  Amzoti Aug 3 '13 at 12:44

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