# Moment of inertia of an ellipse

1. Apr 25, 2014

### Karol

1. The problem statement, all variables and given/known data
To calculate I, the moment of inertia of an ellipse of mass m.
The radius are a and b, according to the drawing.

2. Relevant equations
$$I=mr^2$$
Ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow y=b\sqrt{1-\frac{x^2}{a^2}}$$
Area of an ellipse: $\pi$ab

3. The attempt at a solution
$$I=4\frac{m}{\pi a b} \int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} x^2+y^2 dy$$
$$I=4\frac{m}{\pi a b} \left(\int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} x^2dy+\int_{x=0}^{a}dx \int_{y=0}^{b\sqrt{1-\frac{x^2}{a^2}}} y^2 dy \right)$$
$$I=4\frac{m}{\pi a b} \left(\int_{x=0}^{a} x^2 \left[y\right]_{0}^{b\sqrt{1-\frac{x^2}{a^2}}}dx+\int_{x=0}^{a}\frac{1}{3}\left[y^3\right]_{0}^{b\sqrt{1-\frac{x^2}{a^2}}}dx \right)$$
$$I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3}\int_{x=0}^{a}\left(1-\frac{x^2}{a^2}\right)^{3/2}dx\right)$$
$$I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3a^3}\int_{x=0}^{a} (a^2-x^2)^{3/2}dx \right)$$
And it gives complicated expressions which include $\arcsin$.
$$I=m(a^2+b^2)/4$$

#### Attached Files:

• ###### Ellipse.jpg
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2. Apr 25, 2014

### TSny

You result so far looks good. Those integrals are not too difficult to carry out. I don't see how you are getting arcsin expressions.

3. Apr 25, 2014

### Karol

From integrals tables:
$$\int x^2 \sqrt{a^2-x^2}dx=-\frac{x}{4}\sqrt{(a^2-x^2)^3}+\frac{a^2}{8}\left( x\sqrt{a^2-x^2}+a^2\arcsin \frac{x}{a}\right)$$

4. Apr 25, 2014

### Rellek

May I suggest you work in polar coordinates?

5. Apr 25, 2014

### TSny

OK. You can do it with just trig substitutions.

But you can use your expression from the tables. It will simplify very nicely when you substitute the limits.

6. Apr 25, 2014

### Karol

What trig substitution? if i make y=a*sinx it's complicated also

7. Apr 26, 2014

### TSny

Try $\; x = a \sin \theta$