<s>[INST] Finding surface area of a cone I will describe the problem then show what I tried to solve it.
I need to find the area of the cone defined as follows:
$$z^2=a^2(x^2+y^2)$$
$$0\leq z\leq bx+c$$
where $a,b,c>0$ and $b<a$.
For this I considered the parametrization $x=r\rm{cos}\theta$, $y=r\rm{sin}\theta$, $z=ar$, with $0\leq\theta\leq2\pi$. To find the interval which $r$ varies, I used the fact that $0\leq z\leq bx+c$, then $0\leq ar\Rightarrow 0\leq r$, and $ar\leq br\rm{cos}\theta+c\Rightarrow  r\leq c/(a-b\rm{cos}\theta)$, therefore $0\leq r\leq c/(a-b\rm{cos}\theta)$.
The area of this surface is $$\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}\bigg\lvert(\rm{cos}\theta, \rm{sin}\theta, a)\times(-r\rm{sin}\theta,r\rm{cos}\theta, 0)\bigg\lvert dr \ d\theta=$$
$$=\int_0^{2\pi}\int_0^{\frac{c}{a-b\rm{cos}\theta}}r\sqrt{a^2+1} dr \ d\theta=\sqrt{a^2+1}\int_0^{2\pi}\bigg[\frac{r^2}{2}\bigg]_0^{\frac{c}{a-b\rm{cos}\theta}} \ d\theta=$$
$$=\sqrt{a^2+1}\int_0^{2\pi}\frac{c^2}{2(a-b\rm{cos}\theta)^2} \ d\theta=\frac{c^2\sqrt{a^2+1}}{2}\int_0^{2\pi}\frac{1}{(a-b\rm{cos}\theta)^2} \ d\theta .$$
Well, this last integral is quite difficult, in fact i can't integrate this, and wolframaplha showed me a terrible solution. Now I think I've made some mistake but don't know where. Any help is very welcome. Thanks. 
PS: I tried to find something like a^2(x^2+y^2) or a^2\cdot(x^2+y^2) in the search but that was useless. Later I did some tests and verified that this search is not that good for TeX search, is there some better way to better searching when using TeX language? Again, thanks.
 [/INST] I have not looked at the complete problem in detail yet, but you did mention difficulty with the integral
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2}$$
Evaluation of this integral is not too bad using the Residue theorem.  Let $z=e^{i \theta}$, $d\theta = dz/(i z)$, $\cos{\theta} = (z + z^{-1})/2$, and the integral becomes
$$\oint_{|z|=1} \frac{dz}{i z} \frac{1}{(a-b(z+z^{-1})/2)^2} = -i \frac{4}{b^2} \oint_{|z|=1} dz \: \frac{z}{(z^2 - 2 (a/b) z + 1)^2}$$
By the residue theorem, this integral is $i 2 \pi$ times the sum of the residues of the poles inside the unit circle.  The poles of the integrand are at
$$z_{\pm} = \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b} \right )^2 - 1}$$
Note that only $z_-$ is inside the unit circle, so we need only evaluate the residue at that pole to get the value of the integral.  Note also that this is a double pole, so the residue calculation looks like
$$\begin{align}\text{Res}_{z=z_-} \frac{z}{(z^2 - 2 (a/b) z + 1)^2} &= \lim_{z \rightarrow z_-} \frac{d}{dz} \left [(z-z_-)^2 \frac{z}{(z^2 - 2 (a/b) z + 1)^2} \right ] \\ &= \left[\frac{d}{dz} \frac{z}{(z-z_+)^2}\right]_{z=z_-} \\ &= \frac{1}{(z_- -z_+)^2} - \frac{2 z_-}{(z_--z_+)^3}\\ &= \frac{z_++z_-}{(z_+-z_-)^3}\\ &=\frac{2 (a/b)}{(2 \sqrt{(a/b)^2-1})^3}\\ &= \frac{1}{4} \frac{(a/b)}{[(a/b)^2-1]^{3/2}}\end{align}$$
We then multiply this residue by $(i 2 \pi)(-i 4/b^2)$ and the result is
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2} = 2 \pi \, a \, (a^2-b^2)^{-3/2}$$
</s>