835. A homogeneous plate is bounded by two concentric ellipses with coinciding axes (elliptical ring). Find the moments of inertia of this plate relative to its axes.
Solution. Let the outer ellipse have semi-axes $a_{1}$ and $b_{1}$, and the inner ellipse have semi-axes $a_{2}$ and $b_{2}$. Suppose the axes of the rectangular coordinate system are directed along the symmetry axes of the plate. Then the equations of the ellipses will be

$$
\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1 \text { and } \frac{x^{2}}{a_{2}^{2}}+\frac{y^{2}}{b_{2}^{2}}=1,
$$

and the required moments of inertia will be $I_{x}$ and $I_{y}$. Applying formulas (3) and using the symmetry of the plate (D) relative to the coordinate axes, we get

$$
I_{x}=\iint_{D} y^{2} \delta d x d y=4 \delta\left(\iint_{D_{1}} y^{2} d x d y-\iint_{D_{2}} y^{2} d x d y\right)
$$

where $D_{1}$ and $D_{2}$ are the parts of the regions located in the first quadrant, bounded by the outer and inner ellipses, respectively.

We integrate first with respect to $y$, then with respect to $x$:

$$
\begin{aligned}
& I_{x}=4 \delta\left(\int_{0}^{a_{1}} d x \int_{0}^{t_{1}} y^{2} d y-\int_{0}^{a_{2}} d x \int_{0}^{y_{2}} y^{2} d y\right)=\frac{4}{3} \delta\left(\int_{0}^{a_{1}} y_{1}^{3} d x-\int_{0}^{a_{2}} y_{2}^{3} d x\right)= \\
& \quad=\frac{4}{3} \delta\left[\frac{b_{1}^{3}}{a_{1}^{3}} \int_{0}^{a_{1}}\left(a_{1}^{2}-x^{2}\right)^{\frac{3}{2}} d x-\frac{b_{2}^{3}}{a_{2}^{3}} \int_{0}^{a_{2}}\left(a_{2}^{2}-x^{2}\right)^{\frac{3}{2}} d x\right]
\end{aligned}
$$

To compute the integrals, we change the variable. In the first integral, use the formula $x=a_{1} \sin t$, in the second integral, use the formula $x=a_{2} \sin t$:

$$
\begin{gathered}
I_{x}=\frac{4}{3} \delta\left(a_{1} b_{1}^{3} \int_{0}^{\frac{\pi}{2}} \cos ^{4} t d t-a_{2} b_{2}^{3} \int_{0}^{\frac{\pi}{2}} \cos ^{4} t d t\right)= \\
=\frac{4}{3} \delta\left(a_{1} b_{1}^{3}-a_{2} b_{2}^{3}\right) \int_{0}^{\frac{\pi}{2}} \cos ^{4} t d t
\end{gathered}
$$

According to the solution of problem 505, the last integral is equal to $\frac{3 \pi}{16}$. Therefore,

$$
I_{x}=\frac{1}{4} \pi \delta\left(a_{1} b_{1}^{3}-a_{2} b_{2}^{3}\right)
$$

Similarly, applying the second of formulas (3), we find

$$
I_{y}=\frac{1}{4} \pi \delta\left(b_{1} a_{1}^{3}-b_{2} a_{2}^{3}\right)
$$

From this, when $a_{2}=b_{2}=0$, we obtain the moments of inertia of the complete ellipse:

$$
I_{x}=\frac{1}{4} \pi \delta a_{1} b_{1}^{3} ; \quad I_{y}=\frac{1}{4} \pi \delta b_{1} a_{1}^{3}
$$

When $b_{1}=a_{1}$ and $b_{2}=a_{2}$, we obtain the moment of inertia of a circular ring: $I=\frac{1}{4} \pi \delta\left(a_{1}^{4}-a_{2}^{4}\right)$, taken relative to any of its diameters.

For a circle of radius $a$, the moment of inertia relative to any diameter is $I=\frac{1}{4} \pi \delta a^{4}$.