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.

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 $\cos^2\theta$.
