Question

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).

Short answer

The volume obtained for the planes over the vertices is 5

Step-by-step solution

Definition of double integral and mass formula

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 $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Drawing the area bounded by the curve in the plane

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

Calculation of the volume under the curve

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 first integral bounded by the curve

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 second integral bounded by the curve

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

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

Total volume bounded by the curve

Calculate total volume integral.

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

Therefore, the value is 5