Calculating the work of the following vector field. Thanks in advance for reading my question. I am supposed to calculate $$\int_C F(r) \dot{} dr$$
of the following field: $F(x,y,z)=(y^2,z^2,x^2)$, and $C$ is the intersection of the sphere $\gamma_1: x^2+y^2+z^2=a^2$ with the cylinder $\gamma_2:x^2+y^2=ax$.
But i am having trouble in finding the intersection of $\gamma_1$ with $\gamma_2$. Any tips?
 A: It's easiest to start by finding a parameterization for the circle $x^2 + y^2 = ax$ in the $xy$-plane.  To do this, note that
$$
x^2 + y^2 = ax \implies x^2 - ax + \frac {a^2}{4} + y^2 = \frac{a^2}{4} \implies\\
(x - a/2)^2 + y^2 = (a/2)^2
$$
We can parametrize this in the usual way one parameterizes a circle:
$$
\tilde{\mathbf r}(t) = \frac a2( \cos(t), \sin(t) ) + (a/2,0) = \left(\frac a2(\cos(t) + 1), \frac a2 \sin(t) \right); \qquad t \in [0,2\pi]
$$
To get the formula for the intersection, note that the sphere can be written as 
$$
z = \pm \sqrt{a^2 - x^2 - y^2}
$$
The intersection of our cylinder with the upper hemisphere can be parameterized as
$$
\mathbf r(t) = (\tilde r_x, \tilde r_y, \sqrt{a^2 - \tilde r_x^2 + \tilde r_y^2}) \\
= \left(\frac a2(\cos(t) + 1), \frac a2 \sin(t), 
\sqrt{a^2 - a^2/2 - (a^2/2)\cos(t)}\right)
\\ 
=
\left(\frac a2(\cos(t) + 1), \frac a2 \sin(t), 
\frac{a}{\sqrt 2}\sqrt{1 - \cos(t)}\right); \qquad t \in [0,2\pi]
$$
We can separately parameterize the intersection with the lower hemisphere as
$$
\left(\frac a2(\cos(t) + 1), \frac a2 \sin(t), 
-\frac{a}{\sqrt 2}\sqrt{1 - \cos(t)}\right); \qquad t \in [0,2\pi]
$$ 
Interestingly, these parameterizations intersect at $t = \pi/2$.
