Evaluate $\int_cydx+zdy+zdz$ if c is intersection of upper hemisphere $x^2+y^2+z^2=4$ $z \geq0$ and $x^2+y^2=2x$

Solution 1:

You need to parametrize the curve which is intersection of the cylinder and the upper hemisphere. You have instead parametrized the spherical surface.

First note that the projection of the cylinder $x^2 + y^2 = 2x$ in xy-plane is a circle centered at $(1, 0)$, y-axis is tangent to it and it is formed in fourth and first quadrant.

Now in polar coordinates, $r^2 = 2 r \cos\theta \implies r = 2 \cos\theta$

That gives $x = 1 + \cos2\theta, y = \sin2\theta$

Also at the intersection, $2x + z^2 = 4 \implies z = \sqrt{4-2x} = 2 \sin\theta$

So for the intersection curve, we can use the parametrization $ \displaystyle \vec r(\theta) = (1 + \cos2\theta, \sin2\theta, 2 \sin\theta), -\pi/2 \leq \theta \leq \pi/2$

As per question, $\vec F = (y, z, z)$

$\vec F(\vec r(\theta)) = (\sin2\theta, 2 \sin\theta, 2\sin\theta)$

The line integral is,

$ \displaystyle \int_{-\pi/2}^{\pi/2} \vec F(\vec r(\theta)) \cdot \vec r'(\theta) ~ d\theta$

Can you take it from here?