Example 3.28. Compute the double integral $\iint_{D}(x+2 y) d x d y$, where the region of integration is bounded by the parabolas $y=x-x^{2}$, $y=1-x^{2}$, and the $O y$ axis (Fig. 48).

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-177.jpg?height=450&width=534&top_left_y=175&top_left_x=441)

Fig. 48
Solution. The parabolas intersect at point $A(1 ; 0)$. The region of integration $D$ is regular in the direction of the $O y$ axis and is defined by the inequalities

$$
\begin{aligned}
0 & \leq x \leq 1 \\
x-x^{2} & \leq y \leq 1-x^{2}
\end{aligned}
$$

Therefore,

$$
\iint_{D}(x+2 y) d x d y=\int_{0}^{1} d x \int_{x-x^{2}}^{1-x^{2}}(x+2 y) d y
$$

As a result,

$$
\begin{gathered}
\iint_{D}(x+2 y) d x d y=\left.\int_{0}^{1}\left(x y+y^{2}\right)\right|_{x-x^{2}} ^{1-x^{2}} d x= \\
\left.=\int_{0}^{1} x\left(1-x^{2}\right)+\left(1-x^{2}\right)^{2}-x\left(x-x^{2}\right)-\left(x-x^{2}\right)^{2}\right] d x= \\
=\int_{0}^{1}\left(2 x^{3}-4 x^{2}+x+1\right) d x=\frac{2}{3} .
\end{gathered}
$$