1.1. Compute the double integrals. Change the order of integration.

$$
\text { a) } \int_{0}^{2} d x \int_{0}^{3}\left(x^{2}+2 x y\right) d y
$$

b) $\int_{-2}^{0} d y \int_{0}^{y^{2}}(x+2 y) d x$
c) $\int_{0}^{5} d x \int_{0}^{5-x} \sqrt{4+x+y} d y$
Solution.

a) First, we compute the inner integral, treating $x$ as a constant:

$$
\int_{0}^{3}\left(x^{2}+2 x y\right) d y=\left.\left(x^{2} y+x y^{2}\right)\right|_{y=0} ^{y=3}=3 x^{2}+9 x .
$$

We substitute this function into the inner integral and compute it using the Newton-Leibniz formula:

$$
\int_{0}^{2}\left(3 x^{2}+9 x\right) d x=\left.\left(x^{3}+\frac{9}{2} x^{2}\right)\right|_{0} ^{2}=8+18=26
$$

Now, we change the order of integration and perform similar calculations:

$$
\begin{gathered}
\int_{0}^{2} d x \int_{0}^{3}\left(x^{2}+2 x y\right) d y=\int_{0}^{3} d y \int_{0}^{2}\left(x^{2}+2 x y\right) d x= \\
=\left.\int_{0}^{3}\left(\frac{x^{3}}{3}+x^{2} y\right)\right|_{x=0} ^{x=2} d y=\int_{0}^{3}\left(\frac{8}{3}+4 y\right) d y=\left.\left(\frac{8}{3} y+2 y^{2}\right)\right|_{0} ^{3}=8+18=26 .
\end{gathered}
$$

b) When computing the inner integral, the value of $y$ is considered constant:

$$
\begin{aligned}
& \int_{-2}^{0} d y \int_{0}^{y^{2}}(x+2 y) d x=\left.\int_{-2}^{0}\left(\frac{x^{2}}{2}+2 x y\right)\right|_{x=0} ^{x=y^{2}} d y= \\
= & \int_{-2}^{0}\left(\frac{y^{4}}{2}+2 y^{3}\right) d y=\left.\left(\frac{y^{5}}{10}+\frac{y^{4}}{2}\right)\right|_{-2} ^{0}=\frac{32}{10}-8=-\frac{24}{5} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-041.jpg?height=202&width=353&top_left_y=1046&top_left_x=162)

Fig. 1.17

To integrate in a different order, we first need to correctly set the limits. Figure 1.17 shows the region of integration $\Omega=\{(x, y)$ : $\left.0 \leqslant x \leqslant y^{2} ;-2 \leqslant y \leqslant 0\right\}$.

Lines parallel to the $O Y$ axis intersect the region $\Omega$ at points $y=-2$ and $y=-\sqrt{x}, 0 \leqslant x \leqslant 4$. Taking this into account, we get:

$$
\begin{aligned}
& \int_{0}^{4} d x \int_{-2}^{-\sqrt{x}}(x+2 y) d y=\left.\int_{0}^{4}\left(x y+y^{2}\right)\right|_{-2} ^{-\sqrt{x}} d x= \\
= & \int_{0}^{4}\left(3 x-x^{3 / 2}-4\right) d x=\left.\left(\frac{3}{2} x^{2}-\frac{2}{5} x^{5 / 2}-4 x\right)\right|_{0} ^{4}=24-\frac{64}{5}=-\frac{24}{5}
\end{aligned}
$$

c) Let's illustrate the region of integration $\Omega=\{(x, y): 0 \leqslant x \leqslant$ $\leqslant 5 ; 0 \leqslant y \leqslant 5-x\}$. This region is shown in Fig. 1.18.

In this example, the calculations are the same regardless of the order of integration:

$$
\begin{gathered}
\int_{0}^{5} d x \int_{0}^{5-x} \sqrt{4+x+y} d y=\left.\frac{2}{3} \int_{0}^{5}(4+x+y)^{3 / 2}\right|_{0} ^{5-x} d x=\underbrace{5}_{0} \\
=\frac{2}{3} \int_{0}^{5}\left(9^{3 / 2}-(4+x)^{3 / 2}\right) d x= \\
=\left.\frac{2}{3}\left(27 x-\frac{2}{5}(4+x)^{5 / 2}\right)\right|_{0} ^{5}=\frac{2}{3}\left(135-\frac{2}{5}\left(3^{5}-2^{5}\right)\right)=\frac{506}{3} .
\end{gathered}
$$