1.2. Compute the double integral

$$
\iint_{\Omega}(x+y) d x d y
$$

where $\Omega$ is the triangle formed by the lines:

a) $x=0 ; y=0 ; x+y=3$;

b) $x=3 ; y=0 ; y=x$.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-042.jpg?height=265&width=225&top_left_y=657&top_left_x=911)
## Solution.

a) The region of integration is shown in Fig. 1.19.

We compute the double integral over $\Omega$ by repeated integration:

$$
\begin{aligned}
& \iint_{\Omega}(x+y) d x d y=\int_{0}^{3} d x \int_{0}^{3-x}(x+y) d y=\left.\int_{0}^{3}\left(x y+\frac{y^{2}}{2}\right)\right|_{0} ^{3-x} d x= \\
& =\int_{0}^{3}\left(x(3-x)+\frac{(3-x)^{2}}{2}\right) d x=\int_{0}^{3}\left(\frac{9}{2}+\frac{x^{2}}{2}\right) d x= \\
& =\left.\left(\frac{9}{2} x-\frac{x^{3}}{6}\right)\right|_{0} ^{3}=\frac{27}{2}-\frac{27}{6}=9
\end{aligned}
$$

b) The region of integration $\Omega$ is shown in Fig. 1.20. We perform the necessary calculations:

$$
\begin{aligned}
& \iint_{\Omega}(x+y) d x d y=\int_{0}^{3} d x \int_{0}^{x}(x+y) d y= \\
= & \left.\int_{0}^{3}\left(x y+\frac{y^{2}}{2}\right)\right|_{0} ^{x} d x=\frac{3}{2} \int_{0}^{3} x^{2} d x=\left.\frac{x^{3}}{2}\right|_{0} ^{3}=\frac{27}{2} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-042.jpg?height=251&width=225&top_left_y=1454&top_left_x=911)

Fig. 1.20

We change the order of integration:

$$
\begin{gathered}
\iint_{\Omega}(x+y) d x d y=\int_{0}^{3} d y \int_{y}^{3}(x+y) d x= \\
=\left.\int_{0}^{3}\left(\frac{x^{2}}{2}+x y\right)\right|_{y} ^{3} d y=\int_{0}^{3}\left(\frac{9}{2}-\frac{y^{2}}{2}+3 y-y^{2}\right) d y= \\
=\int_{0}^{3}\left(\frac{9}{2}+3 y-\frac{3}{2} y^{2}\right) d y=\left.\left(\frac{9}{2} y+\frac{3}{2} y^{2}-\frac{y^{3}}{2}\right)\right|_{0} ^{3}=\frac{27}{2}+\frac{27}{2}-\frac{27}{2}=\frac{27}{2}
\end{gathered}
$$