First of all thank you for reading and pardon my confusion. I am in Calc 2, aspiring math major here, and I've hit upon a tricky question

$$\begin{cases} x(t)=a\cos(\theta)\\ y(t)=b\sin(\theta)\end{cases} \quad\text{ with }\quad \theta \in [0, 2\pi ] $$

Find the Surface Area of Revolution from sweeping the above around the x-axis:

$$S = 2\pi \int_0^\pi b\sin(\theta) \sqrt{a^2\sin^2(\theta)+b^2\cos^2(\theta)} d\theta $$

[or equivalently $4 \pi$*...(same integral above)... from $0$ to $\pi/2$, etc...]

which comes from $dS = 2\pi y ds$, $ds = \sqrt{(x')^2+(y')^2}$.

The issue I am having is that our teacher says this problem is easy and I'm not finding the trick here.

The student solutions manual uses eccentricity $e = \frac{c}{a} = \frac{\sqrt{a^2-b^2}}{a}$ which he never taught us . But when they use their eccentricity formula , they simplify what's under the radical to $\sqrt{1-\frac{a^2-b^2}{a^2}\cos^2(\theta)}$ and go on-wards substituting "$e$" into the equation as $\frac{a^2-b^2}{a^2}$.

I cant get that fraction their substituting in e for. I keep getting $\frac{a^2+b^2}{a^2}$. Any way the solution the student manual comes up with for all of this

$$=2\pi b^2 + 2\pi(1b/e)\arcsin(e)$$

Which I'm just not able to reproduce algebraically using any trig identities. I originally thought this was a non-elementary problem but there are solutions evidently, so...advice? Any clarification I can offer?

P.S I just tried using u-substitution(implemented twice) and I'm getting an answer of $\frac{4 \pi a^2b}{3{(a^2+b^2)}}$.

But I've also gotten two similarly simplified answers in other approaches that looked good enough that I'd probably submit them if I was more certain because the math checks out. Even the math lab tutor hasn't been able to provide much insight. The only relevant question we've done in class was the arclength of this similar setup and the prof wrote it off as non-elementary and nothing more was said.

This is from Larson Single Variable Calculus 10e, page 711 question 70a. A hint, some insight, anything is appreciated.


  • $\begingroup$ You should learn how to format the math ( see this for a tutorial). I fixed it this time. there is a piece I don't know what you want to say (the part about condensing the solution down to), please fix it. $\endgroup$ – achille hui Nov 25 '16 at 23:01
  • $\begingroup$ @achillehui thanks, I saw what you meant and tried to clarify, also I appreciate the link $\endgroup$ – Dayten Sheffar Nov 26 '16 at 2:39

A hint: Rewrite $\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$ as $\sqrt{a^2-(a^2-b^2)\cos^2\theta}$.

  • $\begingroup$ you answered the following question and I provided an answer too, they both are similar in results, could you let me know if I am way off. I could not understand your answer. I would be happy if you could let me know how I my approach is wrong math.stackexchange.com/questions/2023923/… $\endgroup$ – Satish Ramanathan Nov 26 '16 at 10:36
  • $\begingroup$ Your answer to said question should be discussed over there. See Cato's comment. $\endgroup$ – Christian Blatter Nov 26 '16 at 11:14
  • $\begingroup$ I assume you meant $sin^2\theta$, regardless...thank you! I see my error now in reworking the identites $\endgroup$ – Dayten Sheffar Nov 26 '16 at 20:00

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