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Curve $C$ is closed plane curve, $a$ and $b$ are the moments of inertia about $x$ and $y$ axes, $\hat n$ is the unit outward vector and $r = \left|x \hat i + y \hat j\right|$.

Here's what I have:

$$\nabla(r^4) = 4 r^2 \vec r$$

$$\hat n = \frac{1}{\sqrt{x^2 + y^2}}( x \hat i + y \hat i)$$

Plugging it all in and simplifying, we get:

$$4 \int_C (x^2 + y^2)\sqrt{x^2 + y^2} ds$$

I know that $a = \int_C y^2 ds$ and $b = \int_C x^2 ds$. The answer is $16(a + b)$. I'm stuck. How do I get the $16$ and get rid of the square root?

Vincenzo Tibullo
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Haresh
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  • $\hat n$ is what you write only for a circle centered in $O$. For a general curve you have $$\hat n = \frac{-\dot y\hat i+\dot x\hat j}{\sqrt{\dot x^2+\dot y^2}}$$ – Vincenzo Tibullo Jul 24 '14 at 09:04

1 Answers1

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Using the divergence theorem, I get $$ \int_C\nabla r^4\cdot\hat nds=\int_D \Delta r^4 dx\,dy=16\int_D r^2 dx \, dy=16(a'+b') $$ where $D$ is the domain whose boundary is $C$, and $a'=\int_D y^2 dx\,dy$, $b'=\int_D x^2 dx\,dy$.

Vincenzo Tibullo
  • 10,955