0
$\begingroup$

Find the moment of inertia about the $z$-axis of a wire which lies along the circle $x^2 + y^2 = R^2$, with density $δ(x, y) = x^2$, where $R$ is any finite radius.

Here's what I have so far:

$Iz$ = $\int\int_S (x^2+y^2)δd\sigma $

$Iz$ = $\int\int_S (x^2+y^2)x^2|r_u \times r_v|dudv$

I'm not quite sure what to do after this, though. I don't understand how to find $r_u$ or $r_v$, or what I should be setting up as my surface of integration. Would anyone be able to point me in the right direction here?

1 Answers 1

1

It is not clear to me why you are doing a surface integral when there is no surface at all. On a curve we can move in two directions, pretty much just like on a wire. On a surface we can move in an infinite amount of directions.

You are interested in computing the line integral,

$$\oint_{x^2+y^2=R^2} (x^2+y^2)\delta ds$$

This is not hard, just parametrize with $x=R \cos (\theta)$ and $y=R \sin (\theta)$ with $\theta \in [0,2\pi]$.

  • 0
    I looked up several different versions of the Moment of Integration formula and they were all variants of the one I posted, I didn't actually realize it was a different case for 2 dimensions. How would I interpret $ds$ in this case? – 2017-01-26
  • 0
    Remember the formula for arc-length $ds$ is pretty similar here , $ds=\sqrt{dx^2+dy^2}=\sqrt{ (\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2} d\theta$. A small, infinitesimal , change in arc-length. @Mock – 2017-01-26