Under the surface z = y(x+2) , and over the area bounded by $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\sqrt{\mathbf{x}}$.

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

The integral splits into two parts. First, have to calculate the individual then adding both of them gives a total volume bounded by the region.

The area is bounded by the graph $\mathrm{y}=\mathrm{x},\mathrm{y}=\sqrt{\mathrm{x}}$and y = 1 in the plane.

The volume under the surface z = y ( x + 2 ), and over the area bounded by $x+y=0,y=1,y=\sqrt{x}$.

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

$$\begin{gathered} l_1={\int }_{x=1}^{0}dx{\int }_{y=-x}^{1}dy\left[{\int }_{z=0}^{y(x+2)}dz\right] \\ ={\int }_{-1}^{0}dx{\int }_{-x}^{1}dyy(x-2) \\ ={\int }_{-1}^{0}dx(x+2){\overline{)\left(\frac{1}{2}{y}^2\right)}}_{-x}^1 \\ =\frac{1}{2}{\int }_{-1}^{0}dx(x+2)(1-x^2) \\ =\frac{1}{2}{\overline{)\left[\frac{x^2}{2}-\frac{x^4}{4}+2x-\frac{2}{3}{x}^3\right]}}_{-1}^0 \\ =\frac{13}{24} \end{gathered}$$

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

$$\begin{gathered} l_2={\int }_{x=0}^{1}dx{\int }_{y=\sqrt{x}}^{1}dy\left[{\int }_{z=0}^{y(x+2)}dz\right] \\ ={\int }_0^{1}dx{\int }_{\sqrt{x}}^{1}ydy(x+2) \\ =\frac{1}{2}{\int }_0^1(x+2){\overline{)dx\left(y^2\right)}}_{\sqrt{x}}^1 \\ =\frac{1}{2}{\int }_0^1(x+2)(1-x)dx \\ =\frac{1}{2}{\int }_0^1(2-x^2-x)dx \\ =\frac{1}{2}{\overline{)\left[2x-\frac{x^3}{3}-\frac{x^2}{2}\right]}}_0^1 \\ =\frac{14}{24} \end{gathered}$$

Total volume integral,

$$\begin{gathered} l=l_1+l_2 \\ =\frac{13}{24}+\frac{14}{24} \\ =\frac{9}{8} \end{gathered}$$

Therefore, the value is$\frac{9}{8}$