Problem 13 from Apostol's Calculus 2 book, 2nd ed.:
A plane region $R$ is bounded by a piecewise smooth Jordan curve $C$. The moments of inertia of $R$ about the $x$- and $y$-axes are known to be $a$ and $b$, respectively. Compute the line integral
$$\oint_C \nabla(r^4)\cdot \mathbf{n} \,ds $$
in terms of $a$ and $b$. Here $r = ||x\mathbf{i} + y\mathbf{j} ||$, $\mathbf{n}$ denotes the unit outward normal of $C$, and $s$ denotes arc length. The curve is traversed counterclockwise.
Answer: $16(a+b)$
I suppose that this question uses Green's ou Stokes' theorem, since it's in the chapter of surface integrals, but I'm not sure.
I don't really have time to answer this properly (need to learn the syntax). So I'm just going to show you the concept you'll use to solve this problem.
First, the problem: *"A PLANE region R is bounded by a..."
Therefore, a normal to any vector in the plane $(x, y)$ is: $(y, -x)$
Solution: $\nabla(r^{4}) = F = (F_{1}, F_{2})$, and given a tangent $T$ that describes the tangent of the curve $C$, we have that: $n = (T_{2}, -T_{1})$
Trick: rewrite the function as to: $$\langle F, n \rangle = F_{1}\cdot T_{2} - F_{2} \cdot T_{1} = \langle G, T \rangle.$$
Therefore: $$G = (-F_{2}, F_{1}).$$
Now apply Stoke's theorem for $G$. The integral you'll be left with will be 16 times the sum of both moments if inertia.
After I learn how to use the syntax, I'll edit this :P. Cheers