Example 6. A point is thrown into the region $G$, bounded by the ellipse $x^{2}+4 y^{2}=8$. What is the probability that it will fall into the region $g$, bounded by this ellipse and the parabola $x^{2}-4 y=0$? (In Fig. 1.7, the region $g$ is shaded).

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-030.jpg?height=531&width=848&top_left_y=990&top_left_x=148)

Fig. 1.7.
Solution. The area of the region $G$ bounded by the ellipse
$x^{2}+4 y^{2}=8$, or $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$, is calculated using the formula $S_{G}=\pi a b$ (see example 5). Since in this case $a=2 \sqrt{2}, b=\sqrt{2}$, then $S=4 \pi$.

To calculate the area of the region $g$, we will use the formula

$$
S=\int_{a}^{b}\left(y_{2}-y_{1}\right) d x
$$

Solving the system of equations $x^{2}+4 y^{2}=8, x^{2}=4 y$, we find $x_{1}=-2$, $x_{2}=2$ - the x-coordinates of the points of intersection of the given lines; hence, $a=-2, b=2$. We solve each equation for $y$:

$$
y_{1}=\frac{x^{2}}{4}, y_{2}=\frac{\sqrt{8-x^{2}}}{2}
$$

We find the area of the region $g$:

$$
S_{g}=\int_{-2}^{2}\left(\frac{\sqrt{8-x^{2}}}{2}-\frac{x^{2}}{4}\right) d x=\frac{1}{2} \int_{-2}^{2} \sqrt{8-x^{2}} d x-\frac{1}{4} \int_{-2}^{2} x^{2} d x
$$

We compute the second integral directly:

$$
S_{2}=\frac{1}{4} \int_{-2}^{2} x^{2} d x=\left.\frac{1}{4} \cdot \frac{x^{3}}{3}\right|_{-2} ^{2}=\frac{1}{12}(8+8)=\frac{4}{3}
$$

To compute the first integral, we use the substitution $x=2 \sqrt{2} \sin t$; then $d x=2 \sqrt{2} \cos t d t, \alpha=-\pi / 4, \beta=\pi / 4$.

$$
\begin{aligned}
& S_{1}=\frac{1}{2} \int_{-2}^{2} \sqrt{8-x^{2}} d x=\frac{1}{2} \int_{-\pi / 4}^{\pi / 4} \sqrt{8-8 \sin ^{2} t} \cdot 2 \sqrt{2} \cos t d t= \\
& =2 \int_{-\pi / 4}^{\pi / 4}(1+\cos 2 t) d t=\left.2\left(t+\frac{1}{2} \sin 2 t\right)\right|_{-\pi / 4} ^{\pi / 4}=\pi+2
\end{aligned}
$$

Thus,

$$
S_{g}=S_{1}-S_{2}=\pi+2-\frac{4}{3}=\pi+\frac{2}{3}
$$

According to formula (1.5.1), we find the desired probability

$$
P(A)=\frac{S_{g}}{S_{G}}=\frac{\pi+2 / 3}{4 \pi}=\frac{1}{4}+\frac{1}{6 \pi} \approx 0.303
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-032.jpg?height=348&width=360&top_left_y=162&top_left_x=39)

Fig. 1.8