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**HallsofIvy** Actually, if ojones had not suggesed spherical coordinates, I would have been inclined to use cylindrical coordinates for the first problem and certainly for this one. In cylindrical coordinates, the equation of the sphere is $\displaystyle r^2+ z^2= 1$ and of the cone, z= r. Those intersect where $\displaystyle r^2+ r^2= 2r^2= 1$ so at $\displaystyle z= \frac{1}{\sqrt{2}$. At that z, the equation of the sphere becomes $\displaystyle r^2+ 1/2= 1$ or $\displaystyle r^2= 1/2$, $\displaystyle r= \frac{1}{\sqrt{2}}$.

The volume under the sphere, above the plane of intersection of the cone and sphere, is $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^{\frac{1}{\sqrt{2}}} (\sqrt{1- r^2}- \frac{1}{\sqrt{2}} r sin(\theta) drd\theta$.