Question

$\mathbf{\int }\mathbf{\int }\mathbf{ydxdy}$over the triangle with vertices $\left(-1,0\right)\mathbf{,}\left(0,2\right)\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\left(2,0\right)$

Short answer

The required solution is S

Step-by-step solution

Definition of double integral

The double integral of$\mathbf{f}\left(x,y\right)$ over the areaA in the$\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\left(x,y\right)\mathbf{dxdy}\mathbf{.}$

Drawing the area bounded by the curve

The triangle with vertices (-1,0),(0,2), and (2,0).

Integration over the bounded curve

Now the total integral region can be divided into two parts and then adding them can be found in the final area.

$$\begin{gathered} \int {\int }_A\mathrm{ydydx}=\mathrm{l} \\ ={\mathrm{l}}_1+{\mathrm{l}}_2 \end{gathered}$$

The area of the left triangle

Calculation of the value, ${\mathrm{l}}_1$:

role="math" localid="1658895301367" $$\begin{gathered} {\mathrm{l}}_1={\int }_{\mathrm{x}=1}^0\left[{\int }_{\mathrm{y}=1}^{2+2\mathrm{x}}\mathrm{ydy}\right]\mathrm{dx} \\ ={\int }_{-1}^0\frac{1}{2}{\left(2+2\right)}^2\mathrm{dx} \\ ={\int }_0^1{\left(1+\mathrm{x}\right)}^2\mathrm{d}\left(1+\mathrm{x}\right) \\ =\frac{2}{3}{\left(1+\mathrm{x}\right)}^3\overline{){}_0^1} \\ =\frac{2}{3} \end{gathered}$$

The area of the right side triangle

Calculation of the value, ${\mathrm{l}}_2$:

$$\begin{gathered} {\mathrm{l}}_2={\int }_{\mathrm{x}=0}^2\left[{\int }_{\mathrm{y}=0}^{2-\mathrm{x}}\mathrm{ydy}\right]\mathrm{dx} \\ ={\int }_0^2\frac{1}{2}{\left(2+2\right)}^{2}d\left(2-x\right) \\ =\frac{1}{6}{\left(2+\mathrm{x}\right)}^3\overline{){}_0^2} \\ =\frac{4}{3} \end{gathered}$$

Total area bounded by the triangle

Hence:

$$\begin{gathered} \mathrm{l}={\mathrm{l}}_1+{\mathrm{l}}_2 \\ =2 \end{gathered}$$

Therefore, the value is 2.