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Consider the given question,

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

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

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

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

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

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

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, $D=\left\{\right(x,y)鈭�0鈮�x鈮�3,0鈮�y鈮�2x\}$.

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

Then, the area of the surface is given below,

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

On solving the above equation,

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