Find the Area:

The portion of the plane $2x+3y+4z=12$that lies inside the cylinder with equation $x^2+y^2=9$.

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

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

The surface is given by

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

$\Rightarrow z=3-\frac{1}{2}x-\frac{3}{4}y$

Solving $\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$

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

$\frac{\partial z}{\partial x}=-\frac{1}{2}$

and

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

$\frac{\partial z}{\partial y}=-\frac{3}{4}$

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

In polar coordinates, region of integration D is

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

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

role="math" localid="1653240997410" ${\int }_{s}ds=\iint \sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial x}\right)}^2+l}dA$

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

$=\iint \frac{\sqrt{29}}{4}dA$

$=\frac{\sqrt{29}}{4}\iint dA$

Surface Area is

${\int }_{s}dS=\frac{\sqrt{29}}{4}{\iint }_{\rho }dA$

$=\frac{\sqrt{29}}{4}{\int }_0^{2\mathrm{\pi }}{\int }_0^{3}rdrd\theta$

$=\frac{\sqrt{29}}{4}{\int }_0^{2\mathrm{\pi }}\left[{\int }_0^{3}rdr\right]d\theta$

$=\frac{\sqrt{29}}{4}{\int }_0^{2\mathrm{\pi }}\left[\frac{3^2}{2}-\frac{0^2}{2}\right]d\theta$

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

$=\frac{9\pi }{4}\sqrt{29}$

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