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Consider the sphere of unit radius centered at $(0,0,1)$,and the cone of equation $z^2=x^2+y^2.$ Find the volume above the cone and inside the sphere.

The equation of the sphere is $$1 \leq x^2+y^2+(z-1)^2=x^2+y^2+z^2-2z+1$$

Using spherical coordinates this gives $$1 \leq \rho ^2\sin^2\varphi \cos^2\theta + \rho^2\sin^2\varphi sin^2\theta + \rho^2 \cos^2\varphi -2\rho \sin\varphi \sin\theta +1 $$

$$=\rho^2 -2\rho\sin\varphi\sin\theta+1$$

The solution says I should come to $\rho \leq 2\cos\varphi$. I don't understand where they have got this from though, could someone explain?

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Assuming the following spherical coordinates \begin{align} x & = \rho\sin\varphi\cos\theta \\ y & = \rho\sin\varphi\sin\theta \\ z & = \rho\cos\varphi \\ \end{align} Your expression in spherical coordinates for sphere is incorrect. It should be \begin{align} x^2 + y^2 + z^2 -2z + 1 =\rho^2 -2\rho\cos\varphi + 1 & \leq 1 \\ \rho^2 -2\rho\cos\varphi & \leq 0 \\ \rho & \leq 2 \cos\varphi \end{align} Hence the result.

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