Whats the surface area of the manifold $0 \leq z, (x-1)^2 + y^2 \leq 1$? The surface is the intersection of the sphere $x^2 + y^2 + z^2 = 4$ and a cylinder centered at $(1,0,0)$. I'm just not sure how to parameterisize the surface and then calculate the volume.
Any help would be welcome.
It seems most straightforward if you parameterize in cylidrical coordinates. Then along the surface, $x^2+y^2+z^2=r^2+z^2=4$, so $z=\sqrt{4-r^2}$, and $$\vec r=\langle r\cos\theta,r\sin\theta,\sqrt{4-r^2}\rangle$$ $$d\vec r=\langle\cos\theta,\sin\theta,\frac{-r}{\sqrt{4-r^2}}\rangle dr+ \langle -r\sin\theta,r\cos\theta,0\rangle d\theta$$ So the vector areal element is $$\begin{align}d^2\vec A & =\langle\cos\theta,\sin\theta,\frac{-r}{\sqrt{4-r^2}}\rangle dr\times \langle -r\sin\theta,r\cos\theta,0\rangle d\theta\\ & =\pm\langle\frac{r^2}{\sqrt{4-r^2}}\cos\theta,\frac{r^2}{\sqrt{4-r^2}}\sin\theta,r\rangle dr\,d\theta\end{align}$$ And the scalar areal element is $$d^2A=||d^2\vec A||=\sqrt{\frac{r^4}{4-r^2}+r^2}dr\,d\theta=\frac{2r}{\sqrt{4-r^2}}dr\,d\theta$$ On the boundary of the area, $(x-1)^2+y^2=r^2-2r\cos\theta+1=1$, so $r=2\cos\theta$, and $-\frac{\pi}2\le\theta\le\frac{\pi}2$. So now we can compute the area $$\begin{align}A & =\int d^2A=\int_{-\frac{\pi}2}^{\frac{\pi}2}\int_0^{2\cos\theta}\frac{2r}{\sqrt{4-r^2}}dr\,d\theta\\ & =\int_{-\frac{\pi}2}^{\frac{\pi}2}\left[-2\sqrt{4-r^2}\right]_0^{2\cos\theta}\,d\theta =-2\int_{-\frac{\pi}2}^{\frac{\pi}2}\left(2|\sin\theta|-2\right)\,d\theta\\ & =-8\int_0^{\frac{\pi}2}\left(\sin\theta-1\right)\,d\theta =-8\left[-\cos\theta-\theta\right]_0^{\frac{\pi}2}\\ & =4\pi-8\end{align}$$ Now there is a little ambiguity here: Do you want the area of the cap or some more areas as well?