Parametric representation of surface of cylinder internally tangent to a sphere In an exercise, we are given the cylinder, $x^2+y^2=ax$ and the sphere $x^2+y^2+z^2=a^2$, and are asked to calculate the surface area of the part of the cylinder that's inside the sphere. The recommendation in the exercise is to represent this surface parametrically as $x(\theta, z)=x$ (and then I guess we can just calculate the integral), but I can't think of a way to do this. Could anyone explain to me how we can find a parametric representation of this surface?
Thanks!
 A: I evaluate one fourth of all area. You know that $S=\iint_S dS$ where in $S$ is your surface cut by the sphere from the cylinder and $dS=zds$. Cause the cylinder is generated by a normal lines (See the definition), so we can evaluate $dS$ respect to $ds$ along the $xy-$curve $x^2+y^2=ax$. So $$dS=zds=\sqrt{a^2-x^2-y^2}ds$$ Now we use the polar coordinate to depict $ds$ on the $xy-$plane. $x^2+y^2=ax\rightarrow r=a\cos(\theta)$ because we know that $x^2+y^2=r^2$ and $x=ar\cos(\theta)$ are the well-known relations linking two coordinates to each other. Now, we have $$ds=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta=ad\theta, a>0$$Since I regard one fourth rather than all surface, it can be easily seen that your range for $\theta$ will be $0≤\theta≤\frac{\pi}{2}$. So the whole surface say $A$ is $$A=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-x^2-y^2}ds=$$ $$=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-r^2}ds=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2\cos^2(\theta)}a d\theta=4\int_{0}^{\frac{\pi}{2}}a^2\sin(\theta)d\theta=4a^2$$

