# Moment of inertia

1. Apr 2, 2009

### Nyasha

Consider a thin plate of constant density which occupies the region in the first quadrant inside the curve:

$x^2+4y^2=4$

Find moment of inertia about line x=-3

Attempt to solution:

$y=\frac{\sqrt{4-x^2}}{2}$

$I(x=-3)=\frac{1\rho}{2}\int_0^2(x+3)^2\sqrt{4-x^2}$

$\text{Is there any easier way of integrating this thing without having to expand} (x+3)^2 \text{and then multiply it with}\sqrt{4-x^2}$

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Last edited: Apr 2, 2009
2. Apr 2, 2009

### Dr.D

In your solution for y(x), where did the 1/2 come from?

I don't see any options except to do it piece by piece.

3. Apr 2, 2009

### Nyasha

$$y=\sqrt{\frac{4-x^2}{4}}$$

4. Apr 2, 2009

### Dr.D

There is a conflict in your posted information. On the figure, it agrees with what you have used
y(x) = (1/2) * sqrt(4-x^2)
I see that now.
In your original post, you also wrote
LaTeX Code: x^2+y^2=4