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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.

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    What is your work on the subject ? – 2017-02-19
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    @JeanMarie I don't even know how to start. Of course, I know how to calculate the gradient.... But, after that I'm stuck. – 2017-02-21

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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

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    [Mathjax tutorial here, if it's helpful](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) – 2017-06-04