91Ó°ÊÓ

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

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

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

Differentiating partially wrt $x$

$2x+0+2x\frac{∂z}{∂x}=0$

$\frac{∂z}{∂x}=-\frac{x}{y}$

Differentiating partially wrt $y$

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

$\frac{∂z}{∂y}=-\frac{y}{z}$

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

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

Using values of partial derivative calculated above, we get

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

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

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

$={âˆ\neg }_D\sqrt{\frac{16}{z^2}}dA$

$={âˆ\neg }_D\frac{4}{z}dA$

As $z=Â\pm \sqrt{16-x^2-y^2}$

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

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

Changing to polar coordinates, equation becomes

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

$={∫}_0^{2\mathrm{π}}{∫}_0^3\frac{4}{\sqrt{16-r^2}}rdrd\mathrm{θ}$

$={∫}_0^{2\mathrm{π}}{\left[-4\sqrt{16-r^2}\right]}_0^{3}d\mathrm{θ}$
$={∫}_0^{2\mathrm{π}}\left[\left(-4\sqrt{16-3^2}\right)-\left(-4\sqrt{16-0^2}\right)\right]d\mathrm{θ}$

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

$=2\mathrm{Ï€}(16-4\sqrt{7})$

$=(32-8\sqrt{7})\mathrm{Ï€}$

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