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
$$\int_0^{2 \pi} \frac{d\theta}{(a-b \cos{\theta})^2} = 2 \pi \, a \, (a^2-b^2)^{-3/2}$$
Ill answer for spherical coordinates.
Lets say the maximum radius of the cone(in spherical coordinates!) is $R$. If you dont have it then:
$$R=\sqrt{h^2+b^2}$$
Where $h$ is the cone height, and $b$ is the base radius.
To clarify the coordinates: $\varphi$ is the angle on the xy plane(azimuth, I believe?), $\theta$ the height angle, and $r$ the radius in spherical coordinates.
The integral form will be:
$$S_1 = \int_{\varphi = 0}^{2\pi}\int_{r=0}^{R} r\ \sin\theta\ d\varphi\ dr$$
Solving the integral we get:
$$S_1 = 2\pi [{{r^2}\over {2}}]|^{R}_{0}\sin\theta = \pi R^2\sin\theta $$
Now we'll notice that geometrically: $\sin\theta = \frac bR$
So finally we get:
$$S_1 = \pi bR$$
All that remains is adding the surface area of the base, which is a circle:
$$S_2 = \pi b^2$$
And finally our result is:
$$ S = S_1 + S_2 = \pi b^2 + \pi b R = \pi b(b + R)$$
Best Answer
Consider your cone lying in a xyz-space. Put the vertex of the cone at the origin, and imagine your base of cone lying in the plane $z=h$. Let $\alpha$ be apex angle so that $\tan\alpha=\cfrac{R}{h}$.
At height $z$ above the origin(vertex of cone), the radius of the base is $z\tan\alpha$. So lateral surface of the cone is represented by the region in $\mathbb{R}^3$, $R=\{(x,y,z):x^2+y^2=(z\tan\alpha)^2, 0\le z\le h\}$. To do the surface integral, we need to parameterize this, as follows:
$\gamma(r,\theta)=(r\cos\theta,r\sin\theta,\cfrac{r}{\tan\alpha})$, $r\in [0,h\tan\alpha]$, $\theta\in [0,2\pi)$
(the parameterization question asked for)
Now we find:
$\cfrac{\partial{\gamma}}{\partial{r}}=(\cos\theta,\sin\theta,\cfrac{1}{\tan\alpha})$
$\cfrac{\partial{\gamma}}{\partial{\theta}}=(-r\sin\theta,r\cos\theta,0)$
Then $\left|\cfrac{\partial{\gamma}}{\partial{r}}\wedge\cfrac{\partial{\gamma}}{\partial{\theta}}\right|=\left|\big(\cfrac{-r\cos\theta}{\tan\alpha},\cfrac{-r\sin\theta}{\tan\alpha},r\big)\right|=\cfrac{r\sec\alpha}{\tan\alpha}$.
Now do the surface integral:
$\iint\limits_R \,dx\,dy=\int\limits_{0}^{2\pi}\int\limits_{0}^{h\tan\alpha}\cfrac{r\sec\alpha}{\tan\alpha} \,dr \,d\theta=2\pi \cfrac{\sec\alpha}{\tan\alpha} \left[\cfrac{r^2}{2}\right]_0^{h\tan\alpha}=\pi h^2 \sec\alpha \tan\alpha=\pi h^2 \cfrac{R}{h} \cfrac{\sqrt{h^2+R^2}}{h}=\pi R\sqrt{h^2+R^2}$