Find the Area:

The portion of the sphere with equation$x^2+y^2+z^2=16$ that lies inside the cylinder with equation$x^2+y^2=9$

It is given that equation of sphere is $x^2+y^2+z^2=16$that lies inside cylinder with equation $x^2+y^2=9$.

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

The surface area of smooth surface is

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

As $x^2+y^2+z^2=16$

Differentiating partially wrt $x$

$2x+0+2x\frac{\partial z}{\partial x}=0$

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

Differentiating partially wrt $y$

role="math" localid="1653242172097" $0+2y+2z\frac{\partial z}{\partial y}=0$

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

Region of integration will be disk in $xy$sphere as given below:

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

Using values of partial derivative calculated above, we get

${\int }_{s}dS=\int {\int }_D\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^3+{\left(\frac{\partial z}{\partial y}\right)}^2+1}dA$

$=\int {\int }_D\sqrt{\frac{x^2}{z^2}+\frac{y^2}{z^2}+1}dA$

$={\iint }_D\sqrt{\frac{x^2+y^2+z^2}{z^2}}dA$

$={\iint }_D\sqrt{\frac{16}{z^2}}dA$

$={\iint }_D\frac{4}{z}dA$

As $z=\pm \sqrt{16-x^2-y^2}$

role="math" localid="1653242765680" ${\int }_{s}dS={\iint }_D\frac{4}{\sqrt{16-x^2-y^2}}dA$

$={\iint }_D\frac{4}{\sqrt{16-\left(x^2+y^2\right)}}dA$

Changing to polar coordinates, equation becomes

${\int }_{S}dS={\iint }_D\frac{4}{\sqrt{16-\left(x^2+y^2\right)}}dA$

$={\int }_0^{2\mathrm{\pi }}{\int }_0^3\frac{4}{\sqrt{16-r^2}}rdrd\theta$

$={\int }_0^{2\mathrm{\pi }}{\left[-4\sqrt{16-r^2}\right]}_0^{3}d\theta$
$={\int }_0^{2\mathrm{\pi }}\left[\left(-4\sqrt{16-3^2}\right)-\left(-4\sqrt{16-0^2}\right)\right]d\theta$

$=(16-4\sqrt{7}){\left[\theta \right]}_0^{2\mathrm{\pi }}$

$=2\pi (16-4\sqrt{7})$

$=(32-8\sqrt{7})\mathrm{\pi }$

Hence, surface area is$(32-8\sqrt{7})\mathrm{\pi }$sq units