Moment of inertia of an ellipse

[Karol]

Homework Statement

To calculate I, the moment of inertia of an ellipse of mass m.
The radius are a and b, according to the drawing.

Homework 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

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 $\arcsin$.
The answer should be:
$$I=m(a^2+b^2)/4$$

 

[TSny]

$$I=4\frac{m}{\pi a b} \left( \frac{b}{a}\int_{x=0}^{a}x^2\sqrt{a^2-x^2}dx+\frac{b^3}{3a^3}\int_{x=0}^{a} (a^2-x^2)^{3/2}dx \right)$$
And it gives complicated expressions which include $\arcsin$.

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.

 

[Karol]

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

 

[Rellek]

May I suggest you work in polar coordinates?

 

[TSny]

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

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.

 

[Karol]

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

 

[TSny]

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