91Ó°ÊÓ

Let a, b, and c be nonzero constants.

Consider the following surface:
The surface S is the portion of the plane with equation $ax+by+cz=k$, that lies above the rectangle$\left[\mathrm{Î\pm },\mathrm{β}\right]×\left[\mathrm{γ},\mathrm{δ}\right]$ in the xy-plane, where$a,b,c,k$ are nonzero constants.
The objective is to find the general formula for the area of this surface.
Formula for Finding the Area of a Surface$z=f(x,y)$ :
If a surface S is given by $z=f(x,y)$for $f(x,y)∈D⊆R^2$, then the surface area of the smooth surface is,

$$\begin{gathered} {∫}_{s}1dS={∫}_s\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA} \\ =∫{∫}_D\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA}...................\left(1\right) \end{gathered}$$

$z=\frac{1}{c}(z-ax-by)$

Now first find $\frac{∂z}{∂x}$and $\frac{∂z}{∂y}$ . the first partial derivatives of z are

$$\begin{gathered} \frac{∂z}{∂x}=\frac{∂}{∂x}\left(\frac{1}{c}\left(k-ax-by\right)\right) \\ =\frac{1}{c}.\frac{∂}{∂x}\left(k-ax-by\right) \\ =\frac{1}{c}.(-a) \\ =-\frac{a}{c} \end{gathered}$$

and,

$$\begin{gathered} \frac{∂z}{∂y}=\frac{∂}{∂y}\left(\frac{1}{c}\left(k-ax-by\right)\right) \\ =\frac{1}{c}.\frac{∂}{∂y}(k-ax-by) \\ =\frac{1}{c}.(-b) \\ =-\frac{b}{c} \end{gathered}$$

Then the value of dSwill be,

$$\begin{gathered} dS=\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA} \\ =\sqrt{{\left(-\frac{a}{c}\right)}^2+{\left(-\frac{b}{c}\right)}^2+1dA} \\ =\sqrt{\frac{a^2}{c^2}+\frac{b^2}{c^2}+1dA} \\ =\sqrt{\frac{a^2+b^2+c^2}{c^2}}dA \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}dA \end{gathered}$$

The surface S is the portion of the plane with equation $ax+by+cz=k$that lies above the rectangle $\left[\mathrm{Î\pm },\mathrm{β}\right]×\left[\mathrm{γ},\mathrm{δ}\right]$ in the xy-plane ,so the region of integration will be,

$D=\left\{\left(x,y\right)\left|\mathrm{Î\pm }≤x≤\mathrm{β},\mathrm{γ}≤y≤\mathrm{δ}\right|\right\}$

Then the area of the surface is,

localid="1650353315748" $$\begin{gathered} {∫}_{s}1dS={∫}_s\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}dA \\ ={∫}_s\frac{1}{c}\sqrt{a^2+b^2+c^2}dA \\ =∫{∫}_D\frac{1}{c}\sqrt{a^2+b^2+c^2}dA \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}∫{∫}_{D}dA \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}{∫}_{\mathrm{γ}}^{\mathrm{δ}}{∫}_{\mathrm{Î\pm }}^{\mathrm{β}}dxdy \end{gathered}$$

Simplify the last integral as follows:

localid="1650353897980" $$\begin{gathered} {∫}_{s}1dS=\frac{1}{c}\sqrt{a^2+b^2+c^2}{∫}_{\mathrm{γ}}^{\mathrm{δ}}{∫}_{\mathrm{Î\pm }}^{\mathrm{β}}dxdy \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}{∫}_{\mathrm{γ}}^{\mathrm{δ}}\left[{∫}_{\mathrm{Î\pm }}^{\mathrm{β}}dx\right]dy \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}{∫}_{\mathrm{γ}}^{\mathrm{δ}}{\left[x\right]}_{\mathrm{Î\pm }}^{\mathrm{β}}dy \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}{∫}_{\mathrm{γ}}^{\mathrm{δ}}(\mathrm{β}-\mathrm{Î\pm })dy \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}(\mathrm{β}-\mathrm{Î\pm }){\left[y\right]}_{\mathrm{γ}}^{\mathrm{δ}} \\ =\frac{1}{c}\sqrt{a^2+b^2+c^2}\left(\mathrm{β}-\mathrm{Î\pm }\right)\left(\mathrm{δ}-\mathrm{γ}\right) \\ =\frac{1}{c}\left(\mathrm{β}-\mathrm{Î\pm }\right)\left(\mathrm{δ}-\mathrm{γ}\right)\sqrt{a^2+b^2+c^2} \end{gathered}$$

Therefore,the general formula for the area of this surface is,localid="1650354129906" $=\frac{1}{c}\left(\mathrm{β}-\mathrm{Î\pm }\right)\left(\mathrm{δ}-\mathrm{γ}\right)\sqrt{a^2+b^2+c^2}$