Question

Find the surface area of the solid generated by revolving the arc of the parabola $y^2=4ax$ bounded by its latus rectum about $x$-axis.

Answer 1

The required surface area S.A. $={\int }_0^{a}2\pi yds$
$=2\pi {\int }_0^{a}y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$
Here $y^2=4ax\Rightarrow 2y{y}^{\prime }=4a\Rightarrow y^{\prime }=\frac{2a}{y}$
So $1+{\left(\frac{dy}{dx}\right)}^2=1+{\left(\frac{2a}{y}\right)}^2=1+\frac{4a^2}{y^2}$
$=1+\frac{4a^2}{4ax}=1+\frac{a}{x}=\frac{x+a}{x}$
$y^2=4ax\Rightarrow y=\sqrt{4ax}=2\sqrt{a}\sqrt{x}$
$\therefore$ S.A. $={\int }_0^{a}2\pi \left(2\sqrt{a}\sqrt{x}\right)\sqrt{\frac{x+a}{x}}ds$
$=4\pi \sqrt{a}{\int }_0^a{(x+a)}^{\frac{1}{2}}dx=4\pi \sqrt{a}{\left\{\frac{[x+a]\frac{3}{2}}{\frac{3}{2}}\right\}}_0^a$
$=\frac{8\pi \sqrt{a}}{3}{\left(2a\right)}^{\frac{3}{2}}=\frac{8\pi a^2}{3}(2\sqrt{2}-1)$sq.units.