Find${\int }_S 1dS$, where S is the portion of the surface determined by$z=x^2-\sqrt{3}y$that lies above the region in the xy-plane bounded by the x-axis and the lines with equations$$\begin{gathered} y=2x, \\ x=3 \end{gathered}$$.

Consider the given question,

$$\begin{gathered} z=x^2-\sqrt{3}y\mathit{,} \\ y\mathit{=}\mathit{2}x\mathit{,} \\ x\mathit{=}\mathit{3} \end{gathered}$$

If a surface S is given by $z=f\left(x,y\right)$for role="math" localid="1650346623982" $f(x,y)\in D\subseteq {\mathrm{ℝ}}^2$, then the surface area of the smooth surface is given below,

${\int }_S 1dS={\iint }_D \sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}dA......\left(i\right)$

Here, the surface S is the portion of the following surface, $z=x^2-\sqrt{3}y$.

Now first find role="math" localid="1650346754503" $\frac{\mathrm{\partial }\mathrm{z}}{\mathrm{\partial }\mathrm{x}},\frac{\mathrm{\partial }\mathrm{z}}{\mathrm{\partial }\mathrm{y}}$. The first partial derivates of z are given below,

$\begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x^2-\sqrt{3}y\right)\\ =2x\\ \begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }y}=\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(x^2-\sqrt{3}y\right)\\ =-\sqrt{3}\end{array}\end{array}$

The region of integration D is the region in the xy-plane that is bounded by the x-axis and the lines with the given equations as shown in the following figure,

Viewed it as a y-simple region, then the region of integration will be, $D=\left\{\right(x,y)\mid 0\le x\le 3,0\le y\le 2x\}$.

Substitute the values in equation (i), with the region of integration, $D=\left\{\right(x,y)\mid 0\le x\le 3,0\le y\le 2x\}$.

Then, the area of the surface is given below,

role="math" localid="1650347153613" $\begin{array}{r}{\int }_S 1dS={\iint }_D \sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}dA\\ ={\int }_0^3 {\int }_0^{2x} \sqrt{(2x{)}^2+(-\sqrt{3}{)}^2+1}dydx\\ \begin{array}{r}={\int }_0^3 {\int }_0^{2x} \sqrt{4\left(x^2+1\right)}dydx\\ ={\int }_0^3 {\int }_0^{2x} 2\sqrt{x^2+1}dydx\end{array}\end{array}$

On solving the above equation,

$\begin{array}{r}\begin{array}{r}{\int }_S 1dS={\int }_0^3 2\sqrt{x^2+1}[y{]}_0^{2x}dx\\ ={\int }_0^3 4x\sqrt{x^2+1}dx\end{array}\\ \begin{array}{r}={\left[\frac{4}{3}{\left(x^2+1\right)}^{3/2}\right]}_0^3\\ =\frac{4}{3}{\left(3^2+1\right)}^{3/2}-\frac{4}{3}{\left(0^2+1\right)}^{3/2}\\ \begin{array}{r}=\frac{4}{3}\cdot 10\sqrt{10}-\frac{4}{3}\cdot 1\\ =\frac{4}{3}(10\sqrt{10}-1)\end{array}\end{array}\end{array}$