# Math Help - Moment of inertia.

1. ## Moment of inertia.

Find the moment of inertia with respect to the y axis of a thin sheet of constant density $\delta=1$ bounded by the curve $y=\frac{sin^2(x)}{x^2}$ and the interval $\pi \leq x \leq 2 \pi$.
I need to evaluate $I_y=\int_{\pi}^{2 \pi} \int_0^{\frac{sin^2(x)}{x^2}} y^2 \ dy \ dx=\frac{1}{3} \int_{\pi}^{2 \pi} \frac{sin^6 (x)}{x^6} dx$

As far as I know, this can't be integrated. This suggests that I have done something wrong with my initial integral.

Can anyone see what it is?

2. maybe not the way you want it done ...

$I = \int r^2 \, dm$

$r = x$

$dm = \delta \cdot y \, dx = \frac{\sin^2{x}}{x^2} \, dx$

$I = \int_{\pi}^{2\pi} x^2 \cdot \frac{\sin^2{x}}{x^2} \, dx$

$I = \frac{1}{2} \int_{\pi}^{2\pi} 1 - \cos(2x) \, dx$

$I = \frac{1}{2} \left[x - \frac{\sin(2x)}{2}\right]_{\pi}^{2\pi}
$

$I = \frac{\pi}{2}$

3. hey skeeter

when you write $dm = \delta \cdot y \, dx$ , does this come from $m = \delta \cdot V \mbox{ thus, in differential form } dm = \delta \, dV$ ?

4. actually, it's $dm = \delta \cdot dA$ since the region is planar, but you've got the right idea.

5. hi

Ok I see. Thanks!

$

I = \int r^2 \, dm
$
Is this an alternate definition for the moment of inertia?

In my book it has:

$I_x=\int y^2 \delta (x,y) dA$

$I_y=\int x^2 \delta (x,y) dA$

$I_0=\int (x^2+y^2) \delta (x,y) dA=I_x+I_y$

So I can see that what you've done is the last one, but haven't you used polar co-ordinates (ie. $x^2+y^2=r^2$)?