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.

2 Answers

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 $$\sqrt{b^2-a^2}u=a\tan\phi$$.

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 $$\cos^2\theta$$.

• Thanks @Shubham Johri Sir for your nice explanation :) but I have a question:How could I know the sign of $(b^2-a^2)$? Because the question isn't provide anything.Can I show both standard form for the seek of general answer??! Thanks again Nov 23, 2018 at 13:34
• You may use both the forms to get a general answer that takes care of both the cases $a>b$ and $b>a$. Nov 23, 2018 at 13:50