Notes: A right circular cone is generated by revolving the region bounded by `y=3x/4, y=3, x=0` about the y-axis. Find the lateral surface area of the cone.

Sunday, 29 January 2017

A right circular cone is generated by revolving the region bounded by $\displaystyle{y}={3}\frac{{x}}{{4}},{y}={3},{x}={0}$ about the y-axis. Find the lateral surface area of the cone.

Surface area (S) obtained by rotating the curve x=g(y), c $\displaystyle\le$ y $\displaystyle\le$ d about y-axis is,

S= $\displaystyle\int{2}\pi{x}{d}{s}$

where, ds= $\displaystyle\sqrt{{{1}+{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{y}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$

We are given $\displaystyle{y}=\frac{{{3}{x}}}{{4}},{y}={3},{x}={0}$

$\displaystyle{y}=\frac{{{3}{x}}}{{4}}$

$\displaystyle\Rightarrow{x}=\frac{{{4}{y}}}{{3}}$

$\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{y}}}=\frac{{4}}{{3}}$

$\displaystyle{S}={\int_{{0}}^{{3}}}{\left({2}\pi\right)}{x}{d}{s}$

$\displaystyle{S}={2}\pi{\int_{{0}}^{{3}}}{\left(\frac{{{4}{y}}}{{3}}\right)}\sqrt{{{1}+{{\left(\frac{{4}}{{3}}\right)}}^{{2}}}}{\left.{d}{y}\right.}$

$\displaystyle{S}={2}\pi{\int_{{0}}^{{3}}}\frac{{{4}{y}}}{{3}}\sqrt{{{1}+\frac{{16}}{{9}}}}{\left.{d}{y}\right.}$

$\displaystyle{S}={2}\pi{\int_{{0}}^{{3}}}\frac{{{4}{y}}}{{3}}\sqrt{{\frac{{25}}{{9}}}}{\left.{d}{y}\right.}$

$\displaystyle{S}={2}\pi{\left(\frac{{4}}{{3}}\right)}{\left(\frac{{5}}{{3}}\right)}{\int_{{0}}^{{3}}}{y}{\left.{d}{y}\right.}$

$\displaystyle{S}=\frac{{{40}\pi}}{{9}}{{\left[\frac{{{y}}^{{2}}}{{2}}\right]}_{{0}}^{{3}}}$

$\displaystyle{S}=\frac{{{40}\pi}}{{9}}{\left[\frac{{{3}}^{{2}}}{{2}}-\frac{{{0}}^{{2}}}{{2}}\right]}$

$\displaystyle{S}=\frac{{{40}\pi}}{{9}}{\left(\frac{{9}}{{2}}\right)}$

$\displaystyle{S}={20}\pi$

So the Lateral surface area of the cone is $\displaystyle{20}\pi$