I have 2 equations : The sphere $x^2+y^2+z^2 =R^2$ and the cylinder (filled on the inside) $\left(x − \dfrac{R}{2}\right)^2 + y^2 \le \left(\dfrac{R}{2}\right)^2$
I need to get the area of the intersection so I think I first need a parametrization to do the surface integral. How should I proceed?
We need only refer to the portion of this surface lying in the first octant; by symmetry the area of the entire surface in the question is four times the area of the surface in the first octant.
We can parametrize the portion of a sphere of radius $R$ lying in the first octant by $$\vec{r}(\phi,\theta)=\Big(R\sin(\phi)\cos(\theta),R\sin(\phi)\sin(\theta),R\cos(\phi)\Big): \phi,\theta\in[0,\pi/2]$$ Now notice $$(x-R/2)^2+y^2\leq R^2/4 \iff x^2+y^2\leq Rx$$ Replacing $x$ with $R\sin(\phi)\cos(\theta)$ and $y$ with $R\sin(\phi)\sin(\theta)$ gives us the useful condition $$0\leq\sin(\phi)\leq\cos(\theta)$$ Using the fact that $\theta,\phi\in[0,\pi/2]$ we see $\arcsin(\cos(\theta))=\pi/2-\theta$ and so $$0\leq \sin(\phi) \leq \cos(\theta) \iff 0\leq \phi \leq \theta-\pi/2$$ In other words, $$\vec{r}(\phi,\theta):\phi\in[0,\pi/2-\theta],\theta\in[0,\pi/2]$$ is a parametrization of the portion of this surface lying in the first octant. The surface area is now seen to be $$S=4\int_{0}^{\pi/2}\int_0^{\pi/2-\theta}||\vec{r}_\phi \times \vec{r}_{\theta}||d\phi d\theta$$ You can see this by clicking on THIS cool graphic.