Find the area:

The portion of the paraboloid with equation $z=9-x^2-y^2$that lies above $xy$ plane.

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

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

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

Solving, we get

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

$\frac{\partial z}{\partial x}=-2x$

Similarly

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

$\frac{\partial z}{\hat{\partial }y}=-2y$

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

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

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

$\Rightarrow x^2+y^2=9$

Hence, the region of integration D is

$D=\left\{\right(r,\theta )\mid 0\le r\le 3,0\le \theta \le 2\pi \}$

Using values of partial derivative, we get

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

$={\iint }_D\sqrt{4x^2+4y^2+1}dA$

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

Changing to polar coordinates, the required area is:

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

$={\int }_0^{2\mathrm{\pi }}\left[{\int }_0^3\sqrt{4r^2+1}dr\right]d\theta$

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

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

$={\int }_0^{2\mathrm{\pi }}\frac{1}{12}(37\sqrt{37}-1)d\theta$

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

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

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