Example 4. Calculate the mass of a quarter of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ $=1, x \geqslant 0, y \geqslant 0$, if the density is $\rho=x y$.
Solution. The mass of the arc of the ellipse will be calculated using the formula $m=$ $=\int_{L} x y d l$. We will use the parametric equations of the ellipse: $x=a \cos t, y=b \sin t, t \in\left[0 ; \frac{\pi}{2}\right]$. We have:

$x^{\prime}=-a \sin t, y^{\prime}=b \cos t, d l=\sqrt{x^{\prime 2}+{y^{\prime}}^{2}} d t=\sqrt{a^{2} \sin ^{2} t+b^{2} \cos ^{2} t} d t=$ $=\sqrt{\frac{1}{2}\left[a^{2}(1-\cos 2 t)+b^{2}(1+\cos 2 t)\right]} d t=\frac{1}{\sqrt{2}} \sqrt{a^{2}+b^{2}+\left(b^{2}-a^{2}\right) \cos 2 t} d t$,

$$
x \cdot y=a b \sin t \cos t=\frac{1}{2} a b \sin 2 t
$$

Substitute into the integral and make the substitution $u=\cos 2 t, d u=-2 \sin 2 t d t$; in this case, if $t=0$, then $u=1$, and if $t=\frac{\pi}{2}$, then $u=-1$. Thus,

$$
\begin{gathered}
m=\int_{L} x y d l=\int_{0}^{\pi / 2} \frac{1}{2 \sqrt{2}} a b \sin 2 t \sqrt{a^{2}+b^{2}+\left(b^{2}-a^{2}\right) \cos 2 t} d t= \\
=-\frac{a b}{4 \sqrt{2}} \int_{1}^{-1} \sqrt{a^{2}+b^{2}+\left(b^{2}-a^{2}\right) u} d u= \\
=\left.\frac{a b}{4 \sqrt{2}} \cdot \frac{2}{3} \cdot \frac{1}{b^{2}-a^{2}}\left(a^{2}+b^{2}+\left(b^{2}-a^{2}\right) u\right)^{3 / 2}\right|_{-1} ^{1}= \\
=\frac{a b \cdot 2 \sqrt{2}}{6 \sqrt{2}\left(b^{2}-a^{2}\right)}\left(b^{3}-a^{3}\right)=\frac{a b\left(a^{2}+a b+b^{2}\right)}{3(a+b)}
\end{gathered}
$$