Consider a uniform disk of mass m and radius R, and let's calculate the moment of inertia about an axis passing through a diameter of the disk.
One easy way is to use the perpendicular axis theorem: since the moment of inertia $I_z$ through an axis perpendicular to the plane of the disk and passing through its center of mass is $\frac{mR^2}{2}$, we have that $I_z=I_x+I_y$, where $I_x$ and $I_y$ are moments of inertia about two perpendicular axes that pass through the center of mass in the plane of the disk. Since the disk is symmetric, $I_x$ and $I_y$ are the same: $\frac{mR^2}{4}$.
However I would like to do the calculation directly with the integral $\int_{disk}r^2 dm$, using Cartesian coordinates.
$$\int_{-R}^{R} \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} x^2 \rho dy dx$$
$$=2 \rho \int_{-R}^{R} x^2 \sqrt{R^2-x^2} dx$$
Is this approach correct, and if so, how does one deal with this integral? Maple can't seem to solve it.
Note that using polar coordinates is more straightforward, but I am curious about why it isn't apparently easy using Cartesian coordinates.
In polar the integral is just $$\int_0^{2\pi} \int_0^R (r\cos{\theta})^2 \rho r dr d\theta = \frac{mR^2}{4}$$
Best Answer
$ \displaystyle I_y = \int_{-R}^{R} \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} x^2 ~ \rho ~ dy ~ dx ~ $ is the correct set up in cartesian coordinates. Considering symmetry about x and y axis, we can consider only first quadrant and rewrite the integral as
$ \displaystyle I_y = 4 \rho \int_0^{R} x^2\sqrt{R^2-x^2} ~ dx ~ $
Now we can use the substitution $x = R \cos\theta$
$dx = - R \sin \theta ~d\theta, \sqrt{R^2 - x^2} = R \sin\theta$
At $x = 0, \theta = \pi/2$ and at $x = R, \theta = 0$
That leads to the integral,
$ \displaystyle I_y = 4 \rho R^4 \int_0^{\pi/2} \sin^2\theta \cos^2\theta ~ d\theta$
Note that $ \sin^2\theta \cos^2\theta = \frac{1}{8} (1 - \cos 4\theta)$. Integral of $\cos 4\theta$ for $(0, \frac{\pi}{2})$ is zero.
That leads to,
$ \displaystyle I_y = \frac{\pi \rho R^4}{4} = \frac{mR^2}{4}$