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How can I compute the volume of $x^2 + y^2 + z^2 = 4$ when it is cut with $x^2 + y^2 = 2y$?

The first one is a sphere and the second one is a cylinder. The general formula is

$$\int_{0}^{2\pi}\int_{0}^{\rho}\int_{0}^{h}r\mathop{dz}\mathop{dr}\mathop{d\theta} $$

I've drawn a picture but I still cannot get the bounds correct. Can someone help me?

EDIT: You can write $2y + z^2 = 4$, and plugging in $z = 0$, I believe a maximum occurs at $y = 2$

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