Moment of inertia of an ellipse

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
[tex]I=mr^2[/tex]
Ellipse:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \Rightarrow y=b\sqrt{1-\frac{x^2}{a^2}}[/tex]
Area of an ellipse: [itex]\pi[/itex]ab

3. The attempt at a solution
[tex]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[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
[tex]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)[/tex]
And it gives complicated expressions which include [itex]\arcsin[/itex].
The answer should be:
[tex]I=m(a^2+b^2)/4[/tex]

Attached Images
File Type: jpg Ellipse.jpg (9.2 KB)

http://ift.tt/1nNGj9m

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