# Triple integral issue

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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Duplicate or specific case of another one. See for generalized case: math.stackexchange.com/questions/150251/… – 007resu Dec 26 '12 at 11:07
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? – Muhammad Khalifa Dec 26 '12 at 11:17
@MuhammadKhalifaTranCer: No. It is not true. – Babak S. Dec 26 '12 at 11:18
@BabakSorouh:could you tell me how I can find a relation between r and theta ? – Muhammad Khalifa Dec 26 '12 at 11:26
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. – Babak S. Dec 26 '12 at 11:33

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 $z=0$.
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.