Verify Stokes' theorem on a hemisphere using a line integral

Question

When solving this question, I wrote $r(t)$ as $\langle\cos t,0,\sin t\rangle$ then I differentiated it and got $dr=\langle-\sin t,0,\cos t\rangle\,dt$, then i substituted in the formula for line integrals. However, in the solution, $dr$ isn't like that? Also, why did we assume $y = 0$?

Answer 1

Simply put, Stokes'theorem tells us that if we are interested in calculating the flux through a certain orientable surface in Euclidean space, we can simply consider the flux on a contour around that surface to end up with the same value. The important part here is that we always keep the same orientation. We naturally assume anti-clockwise orientation belonging to the upward direction, this is often called a right-handed coordinate system.

The problem states that we need an orientation in the direction of the positive $y$-axis, this means that our curves will run in an anti-clockwise direction see: http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx

This is often called the natural direction. When we consider what closed curves we can form on our surface $C$, we realise these will be the great circles on the hemisphere, it is easiest to just take the one that intersects the $xz$ plane: This parametrisation is given by: $$r(t)=(\sin(t), 0, \cos(t)) $$ where $0\leq t \leq 2\pi$, so our begin point is $(0,0,1)$, we then move towards $(\frac{1}{2}\sqrt{2},0,\frac{1}{2}\sqrt{2}$) etc, before finally returning to $(0,0,1)$, your method runs in the other direction (and has a phase difference of $\frac{\pi}{2}$), let's see if this should make a difference. Call it:

$$s(t)=(\cos(t), 0, \sin(t)) $$ If we compute our line integrals with $r(t)$ and $r'(t)$: $$ \oint_{t=0} ^{2\pi} (0, \cos(t) , \sin(t) ) \cdot (\cos(t),0,-\sin(t))dt=$$ $$ \oint_{t=0} ^{2\pi} - \sin^2(t)dt=-\pi$$ And now via your method $s(t)$ and $s'(t)$: $$ \oint_{t=0} ^{2\pi} (0, \sin(t) , \cos(t) ) \cdot (-\sin(t),0,\cos(t))dt =\oint_{t=0} ^{2\pi} \cos^2(t)dt = \pi$$

So indeed the difference is just a minus sign because of reversed orientation. Your method actually calculates the flux ($\pi$) that enters the lower hemisphere and apparently (logically) this should leave ($- \pi$)the upper hemisphere.