area of sphere in a cone help me to Find the area of the portion of the sphere of radius 1 (centered at the origin) that is in the cone 
 A: With $r$ constant, the differential element of area is $r^{2}\sin\theta~ d\theta~ d\phi$. The cone you have given is with $\theta=45^{o}$. Integrating as below
\begin{equation}
\int^{\pi/4}_{\theta=0}\int^{2\pi}_{\phi=0} r^{2}\sin\theta~ d\theta~ d\phi=2\pi r^{2}(1-\cos{\pi/4})
\end{equation}
A: The boundary surface, the sphere and the cone, intersect in a circle $(x^2+y^2 = \frac{1}{2})$, $z = \frac{1}{2}$.  $z = r^2 and z = 1-r^2$ Equate these two to get r. The center of the sphere is $(0,0,0)$, so the upper part of the surface is exactly one quarter of the sphere of radius $\frac{1}{\sqrt{2}}$ as you are only interested in the upper part of the cone.  Thus the area of the sphere = $4\pi{(\frac{1}{\sqrt{2}}})^{2}\frac{1}{4}= \frac{\pi}{2}$.
The surface area of the upper part of the cone with $0<=\theta<=\pi, 0<=\rho<=1$
we have $$S(cone) = \int_0^{\pi} \int_0^1 \frac{\rho}{\sqrt{2}} d\rho d\theta$$
$$ = \pi \frac {{\rho}^{2}}{2\sqrt{2}}|1,0 = \frac{\pi}{2\sqrt{2}} $$
Total area = $$\frac {\pi}{2}[1+\frac{1}{\sqrt{2}}]$$
