Question

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

Short answer

The volume obtained for the planes over the vertices is 4

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

Calculation of the volume integral bounded by the curve

Calculate the volume between the planes z = 2x + 3y + 6 and z = 2x + 7y + 8, and over the square in the (x,y) plane with vertices (0,0) , (1,0) (0,1) (1,1)

$$\begin{gathered} \mathrm{l}={\int }_0^1\mathrm{dx}{\int }_0^1\mathrm{dy}{\int }_{2\mathrm{x}+3\mathrm{y}+6}^{2\mathrm{x}-7\mathrm{y}+8}\mathrm{dz} \\ ={\int }_0^1\mathrm{dx}{\int }_0^1\mathrm{dy}(4\mathrm{y}+2) \\ ={\overline{){\int }_0^1\mathrm{dx}(2{\mathrm{y}}^2+2\mathrm{y})}}_0^1 \end{gathered}$$

Simplification and solving further.

$$\begin{gathered} {\overline{){\int }_0^{1}dx(2y^2+2y)}}_0^1={\int }_0^{1}4dx \\ =4 \end{gathered}$$

Therefore, the volume is 4.