$C$ be curve of intersection of hemisphere $x^2+y^2+z^2=2ax$ and cylinder $x^2+y^2=2bx$ ; to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ $C$ be the curve of intersection of the hemisphere $x^2+y^2+z^2=2ax$ and the cylinder $x^2+y^2=2bx$ , where $0<b<a$ ; how to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ using Stoke's theorem ? I can't parametrize the curve $C$ nor able to get the surface $S$ . Please help . Thanks in advance 
 A: There are at least to ways you can parametrize curve $C$.
Method 1. Since the curve is at the intersection of the sphere and the cylinder, it belongs to the cylinder, which has radius $b$ and is shifted $b$ units along the $x$ axis, so you can write:
$$
x=b\cos\theta+b, \quad y =b\sin\theta, \quad 0\le \theta \le 2\pi,
$$
but since it also belongs to the sphere, you can write
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(b\sin\theta)^2-(b\cos\theta+b-a)^2},
$$
which gives you a first parametrization.
Method 2. The cylinder has polar equation $r(\theta)=2b\cos\theta$, so 
$$
x=r(\theta)\cos\theta=2b\cos^2\theta, \quad y =2b\cos\theta\sin\theta, \quad -\pi/2\le \theta \le \pi/2,
$$
and again 
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(2b\cos\theta\sin\theta)^2-(2b\cos^2\theta-a)^2}.
$$
Note the difference of variations of $\theta$ in the two parametrizations.
This being said, Stokes theorem is probably more appropriate for this question, so you are going to need to parametrize the surface. You could do it in spherical coordinates, but I think cartesian coordinates are the way to go here. Take $x$ and $y$ as parameters and you have:
$$
x=x, \quad y=y, \quad z=\sqrt{a^2-y^2-(x-a)^2},
$$ 
where $x$ and $y$ belong to the projection of the cylinder in the $xy$ plane, that is,
$$
(x,y)\in D:=\{(r,\theta)\;|\; -\pi/2\le \theta \le \pi/2,0\le r\le 2b\cos\theta  \}
$$
It follows by the Stokes theorem that your integral equals
$$
\iint_D \nabla\times\vec{F}(x,y)\cdot \vec{r_x}\times \vec{r_y}\; dA,
$$
where $\nabla\times\vec{F}(x,y)$ is the curl of your field expressed as a field depending only on $x$ and $y$. I am sure you can takes computations from there. 
