Example 6. Calculate the areas of figures bounded by the curves:

a) $\left\{\begin{array}{l}y=x \sqrt{9-x^{2}} \\ y=0(0 \leqslant x \leqslant 3)\end{array}\right.$

b) $\left\{\begin{array}{l}y=2 x-x^{2}+3 \\ y=x^{2}-4 x+3\end{array}\right.$
Solution. We will construct the corresponding regions (Fig. 3.8) and determine the appropriate limits of integration from them, omitting the systems of inequalities. Thus:

a) $S=\int_{D} \int d x d y=\int_{0}^{3} d x \int_{0}^{x \sqrt{9-x^{2}}} d y=$

$=\left.\int_{0}^{3} d x \cdot y\right|_{0} ^{x \sqrt{9-x^{2}}}=\int_{0}^{3} x \sqrt{9-x^{2}} d x=$

$=-\int_{0}^{3} \frac{1}{2}\left(9-x^{2}\right)^{1 / 2} d\left(9-x^{2}\right)=$

$=-\left.\frac{1}{3}\left(9-x^{2}\right)^{3 / 2}\right|_{0} ^{3}=9 ;$

b) $S=\iint_{D} d x d y=$

$$
=\int_{0}^{3}\left(2 x-x^{2}+3-x^{2}+4 x-3\right) d x=
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-052.jpg?height=569&width=431&top_left_y=1128&top_left_x=2037)

Fig. 3.9

$$
=\left.\left(-\frac{2 x^{3}}{3}+3 x^{2}\right)\right|_{0} ^{3}=9
$$

Answer. a) $S=9 ;$ b) $S=9$.