Calculate the work done by the force field $F(x,y,z)=(y^2,z^2,x^2)$ along the curve of intersection of the sphere $x^2+y^2+z^2=1$, the cylinder $x^2+y^2=x$, and the halfspace $z>0$. The path is traversed in a direction that appears clockwise when viewed from the high above the $xy$-plane.
How do I even find the path for starters... after that it is straight forward.
Subbing the cylinder equation in the sphere gives $x+z^2=1$. Is this a right way to go?
Notice that the intersection of the cylinder and the $xy$-plane is a circle.
This means that the "shadow" of the path (assuming sunlight directly above in the positive $z$-direction) is a circle.
The unit circle $x^2+y^2=1$ is traditionally parametrized in the counterclockwise direction with $x=\cos(t)$ and $y=\sin(t)$ and $0\leq t \leq 2\pi$. We will modify this parametrization to give the one you want.
First, complete the square on $x^2+y^2=x$ to find the center and radius:
$(x-1/2)^2+y^2 = (1/2)^2$
Setting $x-1/2=1/2\cos(t)$ and $y=1/2\sin(t)$ parametrizes this, except it's going in the counterclockwise direction.
Use $x-1/2=-1/2\cos(t)$ and $y=1/2\cos(t)$ to fix this.
Now we want the $z$-coordinate to trace a path along the sphere as we are rotating around the cylinder, so we use the equation of the sphere to help us.
$x^2+y^2+z^2 = 1$
$(1/2 - 1/2\cos(t))^2+(1/2\sin(t))^2+z^2=1$
You'll get some cancellation because $\cos^2(x)+\sin^2(x)=1$ and eventually solve:
$z=\sqrt{1/2\cos(t)-1/2}$.
That's it! The parametrization is:
$x=1/2 - 1/2\cos(t)$
$y=1/2\sin(t)$
$z=\sqrt{1/2\cos(t)-1/2}$
for $0\leq t \leq 2\pi$.