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