How do you find the area of the surface generated by rotating the curve about the y-axis x=t^2, y=1/3t^3, 0<=t<=3?

Jul 30, 2018

For infinitesimal arc length, $\mathrm{ds}$:

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}}$

• $\left\{\begin{matrix}x = {t}^{2} & \mathrm{dx} = 2 t \setminus \mathrm{dt} \\ y = \frac{1}{3} {t}^{3} & \mathrm{dy} = {t}^{2} \setminus \mathrm{dt}\end{matrix}\right.$

$\therefore \mathrm{ds} = \sqrt{4 {t}^{2} + {t}^{4}} \setminus \mathrm{dt} = t \sqrt{4 + {t}^{2}} \setminus \mathrm{dt}$

Spinning round the y-axis, creating infinitesimal surface area $\mathrm{dS}$, where:

• $\mathrm{dS} = 2 \pi \setminus x \setminus \mathrm{ds}$

$= 2 \pi \setminus {t}^{3} \setminus \sqrt{4 + {t}^{2}} \setminus \mathrm{dt}$

$\therefore S = 2 \pi {\int}_{0}^{3} \setminus \mathrm{dt} q \quad {t}^{3} \sqrt{4 + {t}^{2}}$

$= \frac{2 \pi}{15} \left(64 + 247 \sqrt{13}\right)$

I have skipped the mechanical integration steps. you can start with a sub:

• $u = 4 + {t}^{2} q \quad \mathrm{du} = 2 t \setminus \mathrm{dt}$

$\therefore S = 2 \pi {\int}_{4}^{13} \setminus \frac{\mathrm{du}}{2 t} q \quad {t}^{3} \setminus \sqrt{u}$

$= \pi {\int}_{4}^{13} \setminus \mathrm{du} q \quad \left(u - 4\right) \setminus \sqrt{u}$

$= \pi {\int}_{4}^{13} \setminus \mathrm{du} q \quad {u}^{\frac{3}{2}} - 4 {u}^{\frac{1}{2}}$

Which is trivial but protracted