91Ó°ÊÓ

The function:

$x=9-y^2-z^2$ in the yz-plane where $xâ‰\yen 0$

The area of the smooth surface is given by,

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

The surface is given by $x=9-y^2-z^2$.

$$\begin{gathered} \frac{∂x}{∂y}=-2y \\ \frac{∂x}{∂z}=-2z \end{gathered}$$

The graph of paraboloid is:

So find the curve of intersection between $x=9-y^2-z^{2}andx=0$.

$$\begin{gathered} 9-y^2-z^2=0 \\ {y}^2+z^2=9 \end{gathered}$$

The paraboloid lies on the positive side of the yz plane, so it intersects the plane x=0 along the circle $y^2+z^2=9$.

The region of integration is shown as follows:

The region of integration in polar coordinates is:

$D=\left\{\right(r,\mathrm{θ}\left)\overline{)0≤r≤3,0≤\mathrm{θ}≤2\mathrm{π}}\right\}$

$$\begin{gathered} {∫}_{S}dS=∫{∫}_D\sqrt{{\left(\frac{∂x}{∂y}\right)}^2+{\left(\frac{∂x}{∂z}\right)}^2+1}dA \\ =∫{∫}_D\sqrt{{(-2y)}^2+{(-2z)}^2+1}dA \\ =∫{∫}_D\sqrt{4(y^2+z^2)+1}dA \end{gathered}$$

The area of the surface is given by,

$$\begin{gathered} {∫}_{S}dS=∫{∫}_D\sqrt{4(y^2+z^2)+1}dA \\ ={∫}_0^{2\mathrm{π}}\left[{∫}_0^3\sqrt{4r^2+1}rdr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}[\frac{1}{12}{(4r^2+1)}^{\frac{3}{2}}{]}_0^{3}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}[\frac{{\left(37\right)}^{{\displaystyle \frac{3}{2}}}}{12}-\frac{1}{12}]d\mathrm{θ} \\ =\frac{1}{12}[37\sqrt{37}-1]{∫}_0^{2\mathrm{π}}d\mathrm{θ} \\ =\frac{1}{12}[37\sqrt{37}-1]{\left[\mathrm{θ}\right]}_0^{2\mathrm{π}} \\ =\frac{2\mathrm{π}}{12}[37\sqrt{37}-1] \\ =\frac{\mathrm{π}}{6}[37\sqrt{37}-1] \end{gathered}$$