Question

Find the area of the surface generated by revolving the curve
x= 2at, y = at^2,(0<=t<=a) about y-axis.​

Answer 1

Our curve satisfies the condition,

$\longrightarrow x=2at$

$\longrightarrow t=\dfrac{x}{2a}$

and,

$\longrightarrow y=at^2$

$\longrightarrow y=a\left(\dfrac{x}{2a}\right)^2$

$\longrightarrow y=\dfrac{x^2}{4a}$

$\longrightarrow x^2=4ay\quad\quad\dots(1)$

This is the Cartesian equation of our curve.

The curve is revolved around y axis, then we get a surface from which an elemental ring, whose radius is $x$ and width is $dy,$ can be considered.

The area of this elemental ring is,

$\longrightarrow dA=2\pi x\ dy\quad\quad\dots(2)$

But when we differentiate (1) wrt $x,$

$\longrightarrow x^2=4ay$

$\longrightarrow 2x\ dx=4a\ dy$

$\longrightarrow dy=\dfrac{x\ dx}{2a}$

Then (2) becomes,

$\longrightarrow dA=2\pi x\cdot\dfrac{x\ dx}{2a}$

$\longrightarrow dA=\dfrac{\pi}{a}\, x^2\ dx\quad\quad\dots(3)$

But,

$\longrightarrow x=2at$

$\longrightarrow dx=2a\ dt$

Then (3) becomes,

$\longrightarrow dA=\dfrac{\pi}{a}(2at)^2\ 2a\ dt$

$\longrightarrow dA=8\pi a^2t^2\ dt$

Hence the whole area of the surface generated is, since $0\leq t\leq a,$

$\displaystyle\longrightarrow A=8\pi a^2\int\limits_0^at^2\ dt$

$\displaystyle\longrightarrow A=\dfrac{8}{3}\pi a^2\left[t^3\right]_0^a$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{8}{3}\pi a^5}}$