I was attempting the question:
Let $F(x,y,z) = (y, -x, 2z^2+x^2)$ and $S$ be the part of the sphere $x^2+y^2+z^2 = 25$ that lies below the plane $z = 4$. Evaluate the expression $\iint \operatorname{curl} F \cdot n \mathrm{d}\sigma$, where $n$ is the unit outward normal of $S$.
I applied Stokes theorem, since the given surface is piecewise smooth orientable & has a piecewise smooth boundary $C$ given by $C : x^2+y^2 = 9$, $z = 4$.
I then parametrized the curve $C$ as $C : r(t) = (3\cos t,3\sin t, 4)$; $0 \leq t \leq 2\pi$ and proceeded to compute the line integral of $F(r(t))$ over $C$, which evaluates to $-18\pi$.
But the answer given to the problem is $18\pi$, which has the opposite sign of the above answer. The explanation given in the exercise is that the boundary $C$ should be parametrized as $(3\sin t, 3\cos t, 4)$ and that the orientation of $C$ is clockwise when viewed from above.
Can anyone clarify which answer is correct, and why? Is the parametrization $(3\cos t, 3\sin t, 4)$ incorrect?
Best Answer
I realize images are discouraged in general; but I did not see another good way to do this.
I've drawn the partial spheres translated away from the origin since it was easier to draw and still illustrates the idea.
The first shows an example where the surface is an "upper" section of a sphere, that is above the cut by a plane $z=c$ and the second is representative of your problem.
In the first, the usual orientation convention for Stokes' theorem means that for the outward normal and the traversal of the boundary to follow a "right-hand" rule, then we require a counter-clockwise orientation of the boundary.
But in the second, for a consistent orientation, it needs to be clockwise. One way to see this is to just imagine flipping the first picture to the second (flip the partial sphere but keep $xyz$-space fixed), and the boundary circle reverses orientation relative to the coordinate axes - that is the key point.
This should explain the answer; if anyone has a different (or more rigorous) explanation please share it!