91Ó°ÊÓ

The portion of paraboloid have equation $z=9-x^2-y^2$that lies above cartesian plane.

It is given by $z=f\left(x,y\right)$

Surface Area of Smooth Surface is given by

$∫dS={âˆ\neg }_D\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}dA$

$z=9-x^2-y^2$

Solving, we get

$\frac{∂z}{∂x}=\frac{∂}{∂x}\left(9-x^2-y^2\right)$

$\frac{∂z}{∂x}=-2x$

Similarly

$\frac{∂z}{∂y}=\frac{∂}{∂y}\left(9-x^2-y^2\right)$

$\frac{∂z}{\widehat{∂}y}=-2y$

As paraboloid lies above cartesian plane, $z=0$

To determine curve of intersection between to surfaces, equate the $z$coordinates.

$9-x^2-y^2=0$

$⇒x^2+y^2=9$

Hence, the region of integration D is

$D=\left\{\right(r,\mathrm{θ})âˆ\pounds 0≤r≤3,0≤\mathrm{θ}≤2\mathrm{Ï€}\}$

Using values of partial derivative, we get

$∫dS={âˆ\neg }_D\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}dA$

$={âˆ\neg }_D\sqrt{4x^2+4y^2+1}dA$

$={âˆ\neg }_0\sqrt{4\left(x^2+y^2\right)+1}dA$

Changing to polar coordinates, the required area is:

$∫dS={âˆ\neg }_D\sqrt{4\left(x^2+y^2\right)+1}dA$

$={∫}_0^{2\mathrm{π}}\left[{∫}_0^3\sqrt{4r^2+1}dr\right]d\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{\left[\frac{1}{12}{\left(4r^2+1\right)}^{3z}\right]}_0^{3}d\mathrm{θ}$

$={∫}_0^7\left[\frac{1}{12}{\left(4(3{)}^2+1\right)}^{3/2}-\frac{1}{12}{\left(4(0{)}^2+1\right)}^{3/2}\right]d\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}\frac{1}{12}(37\sqrt{37}-1)d\mathrm{θ}$

$=\frac{1}{12}(37\sqrt{37}-1)[2\mathrm{Ï€}-0]$

$=\frac{\mathrm{Ï€}}{6}(37\sqrt{37}-1)$

Hence, the area is$\frac{\mathrm{Ï€}}{6}(37\sqrt{37}-1)$sq units.