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When I try Find the volume of the region $R$ lying below the plane $z = 3-2y$ and above the paraboloid $z = x^2 + y^2$

Solving the 2 equations together yields the cylinder $x^2 + (y+1)^2 = 4 $ How do I get the volume then???

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  • $\begingroup$ Duplicate or specific case of another one. See for generalized case: math.stackexchange.com/questions/150251/… $\endgroup$ – 007resu Dec 26 '12 at 11:07
  • $\begingroup$ what i don't understand is that when using polar coordinate to integrate my book deals with the region (which is the circle: x^2 + (y+1)^2 = 4) as if its center was the origin so theta: 0 --> 2PI , r: 0 --> 2 is that true? $\endgroup$ – Muhammad Khalifa Dec 26 '12 at 11:17
  • $\begingroup$ @MuhammadKhalifaTranCer: No. It is not true. $\endgroup$ – mrs Dec 26 '12 at 11:18
  • $\begingroup$ @BabakSorouh:could you tell me how I can find a relation between r and theta ? $\endgroup$ – Muhammad Khalifa Dec 26 '12 at 11:26
  • $\begingroup$ The problem is that the intersection area is not a circle with centered at (0,0). its center is indeed $(0,-1)$ with radius 2 as you noted above. $\endgroup$ – mrs Dec 26 '12 at 11:33
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First of all, I draw a plot for $x^2+(y+1)^2=4$ or $r^2+2r\sin(\theta)=3$ which is our integration area on plane $z=0$.

enter image description here

You see that $r$ varies from $r=3$ to $r=-\sin(\theta)+\sqrt{\sin(\theta)^2+3}$ and $\theta$ from $0$ to $pi/2$. As the volume is symmetric so you should double the result.

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  • $\begingroup$ Wow! I like the picture (and the answer!) + $\endgroup$ – Namaste Mar 1 '13 at 0:59
  • $\begingroup$ @amWhy: I my self made it! :-) $\endgroup$ – mrs Mar 1 '13 at 6:32

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