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.
Best Answer
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
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2} = 2 \pi \, a \, (a^2-b^2)^{-3/2}$$