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I have to find the volume below the plane $z=3-2y$ and above the paraboloid $z=x^2+y^2$.

Integrating by $z$ first, it looks like the "arrow" I draw parallel to z-axis enters the region at $z=x^2+y^2$ and exits the region at $z=3-2y$. So $\int_{x^2+y^2}^{3-2y}1 dz$.

I'm confused how I am supposed to find the other two integrals. Any hints please?

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Set $x^2 + y^2 = 3 - 2y$ to find the intersection in the xy-plane. You can then get the limits for x by solving for it in the above equation, or you can complete the square to get $x^2 + (y + 1)^2 = 4$ and solve for y.

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