Find the area of the surface formed by revolving the given curve about $(i)x$-axis and $(i)y$-axis 
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.
 A: 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 $\sqrt{b^2-a^2}u=a\tan\phi$.
A: The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $a\implies\theta$ ranges from $\pi\rightarrow 2\pi$, not $0\rightarrow 2\pi$. For the surface area about the $y$-axis, $\theta$ ranges from $-\pi/2 \rightarrow +\pi/2$, or from $3\pi/2\rightarrow 2\pi$ and $0\rightarrow\pi/2$. 
For the surface area about $x$-axis, take $t=\cos\theta \implies dt=-\sin\theta\ d\theta$
$S_x=2\pi\int_\pi^{2\pi}|b\sin\theta| \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}\ d\theta\\ \ \ \ \ =2\pi b\int_\pi^{2\pi}|\sin\theta| \sqrt{a^2(1-\cos^2\theta)+b^2\cos^2\theta}\ d\theta\\\\ \ \ \ \ =2\pi b\int_\pi^{2\pi}(-\sin\theta) \sqrt{a^2+(b^2-a^2)\cos^2\theta}\ d\theta\\\\ \ \ \ \ =2\pi b\int_{-1}^{1}\sqrt{a^2+(b^2-a^2)t^2}\ dt\\\\ \ \ \ \ =4\pi b\int_0^{1}\sqrt{a^2+(b^2-a^2)t^2}\ dt\\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $\int \sqrt{a^2-x^2}\ dx$ or $\int \sqrt{a^2+x^2}\ dx$.
For the surface area about $y$-axis, because we have $\cos\theta\ d\theta$ outside the square root, take $t=\sin\theta$, and try to get the argument of the square root in terms of $\sin\theta$ alone, this time by substituting for $\cos^2\theta$.
