Find the moment of inertia of the elliptical area

Find the moment of inertia of the elliptical area $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with respect to the major axis ($a>b$).

Step 1: Integrate over $y$

\[I_x=2\int_{-a}^a\int_0^{b\sqrt{1-x^2/a^2}}y^2\,dy\,dx = \frac{4b^3}3\int_0^a\left(1-\frac{x^2}{a^2}\right)^{3/2}dx\]

Step 2: Substitute $x=a\sin\theta$

\[I_x=\frac{4ab^3}3\int_0^{\pi/2}\cos^4\theta\,d\theta\]

Step 3: Evaluate $\int_0^{\pi/2}\cos^4\theta\,d\theta$

Using $\cos^4\theta=\frac{3+4\cos2\theta+\cos4\theta}8$: \[\int_0^{\pi/2}\cos^4\theta\,d\theta=\frac{3\pi}{16}\]

Step 4: Combine

\[I_x=\frac{4ab^3}3\cdot\frac{3\pi}{16}=\frac{\pi ab^3}4\]

\[\boxed{I_x=\frac{\pi ab^3}4}\]

Step 5: Minor axis and special case

By the same method, $I_y=\dfrac{\pi a^3b}4$ about the minor axis. When $a=b=R$, both reduce to the circle's $\dfrac{\pi R^4}4$ (Problem 146).