Find the volume between the planes z = 2x + 3y +6 and z = 2x + 7y + 8, and over the triangle with vertices, (0,0) (3,0) and (2,1).

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

The area is bounded by region for the integration in the xy plane.

The volume between the planes z = 2x + 3y + 6 and z = 2x + 7y + 8, and over the triangle with vertices (0,0),(3,0) and (2,1).

Splits the integral into two parts then calculate the individual and adding both of them gives a total volume bounded by the region.

$l=l_1+l_2$

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

$$\begin{gathered} l_1={\int }_0^{2}dx{\int }_0^{x/2}dy{\int }_{2x+3y+6}^{2x+7y+8}dz \\ ={\overline{){\int }_0^2\mathrm{dx}{\int }_0^{\mathrm{x}/2}\mathrm{dy}\mathrm{z}}}_{2x+3y+6}^{2x+7y+8} \end{gathered}$$

Substitute the z limits and calculate further

$$\begin{gathered} {\overline{){\int }_0^2\mathrm{dx}{\int }_0^{\mathrm{x}/2}\mathrm{dy}\mathrm{z}}}_{2x+3y+6}^{2x+7y+8}={\int }_0^{2}dx{\int }_0^{x/2}dy\left\{(2x+7y+8)-(2x+3y+6)\right\} \\ ={\int }_0^{2}dx{\int }_0^{x/2}dy(4y+2) \\ ={\overline{){\int }_0^{2}dx(2y^2+2y)}}_0^{x/2} \end{gathered}$$

Substitute the limits and calculate further.

$$\begin{gathered} {\overline{){\int }_0^{2}dx(2y^2+2y)}}_0^{x/2}=2{\int }_0^{2}dx\left\{\left(\frac{x^2}{4}+\frac{x}{2}\right)-0\right\} \\ =2{\int }_0^{2}dx\left(\frac{x^2}{4}+\frac{x}{2}\right) \\ =2{\overline{)\left[\frac{x^3}{12}+\frac{x^2}{4}\right]}}_0^2 \end{gathered}$$

Substitute the limits x and calculate further.

$$\begin{gathered} 2{\overline{)\left[\frac{x^3}{12}+\frac{x^2}{4}\right]}}_0^2=2\left(\frac{8}{12}+1\right) \\ =\frac{10}{3} \end{gathered}$$

Calculation of the value of ${\mathrm{l}}_2$.

$$\begin{gathered} l_2={\int }_2^{3}dx{\int }_0^{-x+3}dy(4y+2) \\ ={\overline{){\int }_2^{3}dx(2y^2+2y)}}_0^{-x+3} \end{gathered}$$

Substitute the y limits and calculate further

$$\begin{gathered} {\overline{){\int }_2^{3}dx\left(2y^2+2y\right)}}_0^{-x+3}=2{\int }_2^{3}dx({(-x+3)}^2-x+3) \\ =2{\int }_2^3(x^2-7x+12)dx \\ ={\overline{)2\left[\frac{x^3}{3}-\frac{7}{2}{x}^2+12x\right]}}_2^3 \end{gathered}$$

Substitute the x limits and calculate further

$$\begin{gathered} {\overline{)2\left[\frac{x^3}{3}-\frac{7}{2}{x}^2+12x\right]}}_2^3=2\left\{\left(\frac{27}{3}-\frac{7}{2}\times 9+12\times 3\right)-\left(\frac{8}{3}-\frac{7}{2}\times 4+12\times 2\right)\right\} \\ =2\left\{\left(\frac{54-189+216}{6}\right)-\left(\frac{16-84+144}{6}\right)\right\} \\ =2\left\{\left(\frac{81}{6}\right)-\left(\frac{76}{6}\right)\right\} \\ =\frac{5}{3} \end{gathered}$$

Calculate total volume integral.

$$\begin{gathered} \mathrm{l}={\mathrm{l}}_1+{\mathrm{l}}_2 \\ =5 \end{gathered}$$

Therefore, the value is 5