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

Open in App

Solution

The required surface area $S . A = \int_{0}^{a} 2 π y d x$
$= 2 π \int_{0}^{a} y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$
Here $y^{2} = 4 a x \Rightarrow$ $2 y y^{'} = 4 a$ $\Rightarrow y^{'} = \frac{2 a}{y}$
So, $1 + {(\frac{d y}{d x})}^{2} = 1 + {(\frac{2 a}{y})}^{2} = 1 + \frac{4 a^{2}}{y^{2}} = 1 + \frac{4 a^{2}}{4 a x}$
$= 1 + \frac{a}{x} = \frac{x + a}{x}$ $y^{2} = 4 a x \Rightarrow y = \sqrt{4 a x} = 2 \sqrt{a} \sqrt{x}$
$∴ S . A = \int_{0}^{a} 2 π (2 \sqrt{a} \sqrt{x}) \sqrt{\frac{x + a}{x}} d x$
$= 4 π \sqrt{a} \int_{0}^{a} {(x + a)}^{\frac{1}{2}} d x = 4 π \sqrt{a} {⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \frac{{[x + a]}^{\frac{3}{2}}}{\frac{3}{2}} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭}_{0}^{a}$
$= \frac{8 π \sqrt{a}}{3} {(2 a)}^{3 / 2} - {(a)}^{3 / 2}$
$= \frac{8 π a^{2}}{3} (2 \sqrt{2} - 1)$ sq.units

Suggest Corrections

Similar questions

y^{2} = 4 a x,

y^{2} = 4 a x

y^{2} = 4 a x

x = 6

x = a (t + sin t), y = a (1 + cos t)

x

y = x^{2}

x = y^{2}

Join BYJU'S Learning Program