1.16. Find the center of gravity of a plate bounded by the lines $x=0, y=0, \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, y \geqslant 0, x \geqslant 0$, if the surface density $P=x y$.
Solution. The coordinates of the center of gravity of a plane figure $x_{C}, y_{C}$ are determined by the formulas:

$$
x_{C}=\frac{m_{y}}{M}, y_{C}=\frac{m_{x}}{M}
$$

where $m_{y}=\iint_{\Omega} p x d x d y$ - the static moment of the plate relative to the $O Y$ axis; $m_{x}=\iint_{\Omega} p y d x d y$ - the static moment of the plate relative to the $O X$ axis; $M=\iint_{\Omega} p d x d y$ - the mass of the plate.

In this problem,

$$
\Omega=\left\{(x, y): \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1, x \geqslant 0, \quad y \geqslant 0\right\}
$$

We will perform the necessary calculations:

$$
\begin{aligned}
& M=\iint_{\Omega} p d x d y=\iint_{\Omega} x y d x d y=\int_{0}^{a} x d x \int_{0}^{b \sqrt{1-\frac{x^{2}}{a^{2}}}} y d y= \\
& =\frac{b^{2}}{2} \int_{0}^{a} x\left(1-\frac{x^{2}}{a^{2}}\right) d x=\left.\frac{b^{2}}{2}\left(\frac{x^{2}}{2}-\frac{x^{4}}{4 a^{2}}\right)\right|_{0} ^{a}=\frac{a^{2} b^{2}}{8} \\
& m_{y}=\iint_{\Omega} p x d x d y=\iint_{\Omega} x^{2} y d x d y=\int_{0}^{a} x^{2} d x \int_{0}^{b \sqrt{1-\frac{x^{2}}{a^{2}}}} y d y= \\
& =\frac{b^{2}}{2} \int_{0}^{a} x^{2}\left(1-\frac{x^{2}}{a^{2}}\right) d x=\left.\frac{b^{2}}{2}\left(\frac{x^{3}}{3}-\frac{x^{5}}{5 a^{2}}\right)\right|_{0} ^{a}=\frac{a^{3} b^{2}}{15} \\
& m_{x}=\iint_{\Omega} p y d x d y=\iint_{\Omega} x y^{2} d x d y=\int_{0}^{b} y^{2} d y \int_{0}^{a \sqrt{1-x^{2}}} \\
& =\frac{a^{2}}{2} \int_{0}^{b} y^{2}\left(1-\frac{y^{2}}{b^{2}}\right) d y=\left.\frac{a^{2}}{2}\left(\frac{y^{3}}{3}-\frac{y^{5}}{5 b^{2}}\right)\right|_{0} ^{b}=\frac{a^{2} b^{3}}{15}
\end{aligned}
$$

From this, we determine the required coordinates of the center of gravity of the plate:

$$
x_{C}=\frac{m_{y}}{M}=\frac{8}{15} a ; \quad y_{c}=\frac{m_{x}}{M}=\frac{8}{15} b
$$