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

Question

Evaluate $\int_cydx+zdy+zdz$ if c is intersection of upper hemisphere $x^2+y^2+z^2=4$ $z \geq0$ and $x^2+y^2=2x$ oriented counter clockwise from xy plane

First, I made the following parametricisation: $$g(x,y) = \begin{bmatrix}x\\y\\ \sqrt{4-x^2-y^2}\end{bmatrix}$$

naturally, I set $\omega=ydx+zdy+xdz$. This is where I am getting stumped. I want to evaluate $\int_Dg^*dw$ but I'm not sure how to get dw?

I did something janky: $d\omega = dy\wedge dx +dz\wedge dy + dx\wedge dz$. First of all, that doesn't seem right. Secondly, if it were, what would $g^*\omega$ be??

Answer 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 \mid\sin\theta \mid $

So for the intersection curve below $y = 0$, we can use the parametrization $ \displaystyle \vec r(\theta) = (1 + \cos2\theta, \sin2\theta, - 2 \sin\theta), -\pi/2 \leq \theta \leq 0$ and above $y = 0$, we can use the parametrization $ \displaystyle \vec r(\theta) = (1 + \cos2\theta, \sin2\theta, 2 \sin\theta), 0 \leq \theta \leq \pi/2$

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

So, $\vec F(\vec r(\theta)) = (\sin2\theta, - 2 \sin\theta, - 2\sin\theta)$ for $- \pi/2 \leq \theta \leq 0~$ and $~\vec F(\vec r(\theta)) = (\sin2\theta, 2 \sin\theta, 2\sin\theta)$ for $0 \leq \theta \leq \pi/2$

Can you take it from here?

Answer 2

You're proceeding just fine, as long as you realize you're going to want to integrable $d\omega$ on the portion of the sphere inside that curve. I would parametrize by polar coordinates to make life easier, but you'll end up there eventually.

Your formula for $d\omega$ is correct. Now pull back using your rules for pullback. $g^*dx = dx$, $g^*dy = dy$, $g^*dz = -(\frac x{\sqrt{4-x^2-y^2}}\,dx + \frac y{\sqrt{4-x^2-y^2}}\,dy)$. You should find that $g^*d\omega = -(\frac x{\sqrt{4-x^2-y^2}}+\frac y{\sqrt{4-x^2-y^2}}+1)dx\wedge dy$. Switching to polar coordinates, this will be $$-\left(\frac{r(\cos\theta+\sin\theta)}{\sqrt{4-r^2}} + 1\right)r\,dr\wedge d\theta.$$ You do need to stop and think about orientations. Given the orientation specified on the curve, is the region interior to it on the sphere given the usual upward orientation or its negative? So, is your parametrization orientation-preserving or reversing? Perhaps the negative sign will go away.

Now, over what region in the $xy$-plane are you going to do the double integral, now that you've pulled back? Does thinking about symmetry simplify the calculation slightly?

REMARK: This does seem to be a yucky problem. It does now occur to me that it comes from my text (although different orientation). My intent was a direct calculation of the line integral, as in MathLover's answer.