Question

Above the square with vertices at, (0,0), (2,0),(0,2) and (2,2) and under the plane z = 8-x+y.

Short answer

The required solution is 32.

Step-by-step solution

Definition of double integral

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

Calculation of the volume under the curve

The volume above the square with vertices at (0,0),(2,0),(0,2) and (2,2), and under the plane z=8-x+y.

First, integrate over z:

$$\begin{gathered} |={\int }_{\mathrm{x}=0}^2{\int }_{\mathrm{y}=0}^2\left[{\int }_{\mathrm{z}=0}^{8-\mathrm{x}+\mathrm{y}}\mathrm{dz}\right]\mathrm{dxdy} \\ ={\int }_0^2\mathrm{dy}{\int }_0^2\mathrm{dx}\left(8-\mathrm{x}+\mathrm{y}\right) \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over x:

$$\begin{gathered} |={\int }_0^{2}dy{\overline{)\left(8x-\frac{1}{2}{x}^2+yx\right)}}_0^2 \\ ={\int }_0^{2}dy(16-2+2y) \end{gathered}$$

Further calculation of the volume of the bounded region

Now, integrate over y:

$$\begin{gathered} |={\int }_0^{2}dy(16-2+2y) \\ ={\overline{)(14y+y^2)}}_0^2 \\ =32 \end{gathered}$$

Therefore, the value is 32