91Ó°ÊÓ

The equation of plane is $2x+3y+4z=12$and the equation of cylinder is localid="1653379627078" role="math" $x^2+y^2=9$.

Formula for finding area of surface is

$z=f(x,y)$

Surface area of smooth surface is

role="math" localid="1653240309778" ${∫}_{3}dS={∫}_s\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}dA$

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

The surface is given by

$2x+3y+4z=12$

$⇒z=3-\frac{1}{2}x-\frac{3}{4}y$

Solving $\frac{∂z}{∂x},\frac{∂z}{∂y}$

$\frac{∂z}{∂x}=\frac{∂}{∂x}\left(3-\frac{1}{2}x-\frac{3}{4}y\right)$

$\frac{∂z}{∂x}=-\frac{1}{2}$

and

$\frac{∂z}{∂y}=\frac{∂}{∂y}\left(3-\frac{1}{2}x-\frac{3}{4}y\right)$

$\frac{∂z}{∂y}=-\frac{3}{4}$

Region of Integration is the disk in cartesian plane determined by $x^2+y^2=9$

In polar coordinates, region of integration D is

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

Using value of $\frac{∂z}{∂x},\frac{∂z}{∂y}$, we get

role="math" localid="1653240997410" ${∫}_{s}ds=âˆ\neg \sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂x}\right)}^2+l}dA$

role="math" localid="1653240989403" $=âˆ\neg \sqrt{{\left(-\frac{1}{2}\right)}^2+{\left(-\frac{3}{4}\right)}^2+1}dA$

$=âˆ\neg \frac{\sqrt{29}}{4}dA$

$=\frac{\sqrt{29}}{4}âˆ\neg dA$

Surface Area is

${∫}_{s}dS=\frac{\sqrt{29}}{4}{âˆ\neg }_{\mathrm{ÒÏ}}dA$

$=\frac{\sqrt{29}}{4}{∫}_0^{2\mathrm{π}}{∫}_0^{3}rdrd\mathrm{θ}$

$=\frac{\sqrt{29}}{4}{∫}_0^{2\mathrm{π}}\left[{∫}_0^{3}rdr\right]d\mathrm{θ}$

$=\frac{\sqrt{29}}{4}{∫}_0^{2\mathrm{π}}\left[\frac{3^2}{2}-\frac{0^2}{2}\right]d\mathrm{θ}$

$=\frac{\sqrt{29}}{4}·\frac{9}{2}[\mathrm{θ}{]}_0^{2\mathrm{π}}$

$=\frac{9\mathrm{Ï€}}{4}\sqrt{29}$

Hence, area is $\frac{9\mathrm{Ï€}}{4}\sqrt{29}$sq units.