Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=a\cos\theta ,y=b\sin\theta,0\le\theta\le2\pi$$
About $x-$axis is, $S=2\pi\int_0^{2\pi}b\sin\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$
About $y-$axis is, $S=2\pi\int_0^{2\pi}a\cos\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$
from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.
Thanks in advance.
Best Answer
Hint: For the first one let $\cos\theta=u$ $$ S=2\pi\int_0^{\pi}b\sin\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta =2\pi\int_{-1}^1 b\sqrt{a^2+(b^2-a^2)u^2}\ du $$ and then let substitution $\sqrt{b^2-a^2}u=a\tan\phi$.