Surface element of inverted cone in cylindrical coordinates. I have to integrate over the curved surface of a inverted cone, since the surface is of a cone I think all three $s, \phi$ and $z$ are varying under the integral sign.
Now this causes the problem, I don't know surface element when all three are varying nor I can find a solution on internet,  I thought I will take $\displaystyle|d\vec a| = s\ ds\ d\phi\ dz$ but that is volume element. 
What should I do ? Drop $s$ from $\displaystyle |d\vec a| = s\ ds\ d\phi \ dz$ ?
 A: A surface integral needs two parameters although the surface element vector needs three components. We can write those components in cartesian coordinates and the surface parameters can be adopted from the description in cylindrical coordinates.
For an inverted cone: $z^2=K(x^2+y^2)$ (with $K$ some constant), that in cylindrical is, $z=\pm K s$, dropping the minus sign as it's stated that it's an inverted cone, $z=K s$. We can parametrize the surface with the help of the cylindrical coordinates $s$ and $\phi$ as $z$ is determined by the equation for the cone: In cartesian coordinates, the parametric equations for this cone are:
$$C:r=(x,y,z)=(s\cos\phi,s\sin\phi,Ks)$$
Now, to find the pointing outwards, normal vector, we need two vectors tangent to the surface, then the former is the cross product of the laters:
$$\begin{cases}
 \dfrac{\partial r}{\partial\phi}=(-s\sin\phi,s\cos\phi,0)\\
 \dfrac{\partial r}{\partial s}=(\cos\phi,\sin\phi,K)
\end{cases}$$
$$\dfrac{\partial r}{\partial\phi}\times\dfrac{\partial r}{\partial s}=(Ks\cos\phi,Ks\sin\phi,-s\sin^2\phi-s\cos^2\phi)=$$
$$=s(K\cos\phi,K\sin\phi,-1)$$
And
$$\left|\dfrac{\partial r}{\partial\phi}\times\dfrac{\partial r}{\partial s}\right|=s\sqrt{K^2+1}$$
The integral is in cylindrical coordinates,
$$\iint_S f(s,\phi)s\sqrt{K^2+1}\,dsd\phi$$
A: From [1], consider a right solid circular cone with height $h$ and aperture $2\, \phi $, whose axis is the $z$ coordinate axis and whose apex is the origin.  By $\omega$, I denote the differential area element of this cone in cylindrical coordinates. I find that

$$
\omega = z\,\tan {(\phi)} \,\sqrt{\tan^2{(\phi)}  + 1}\,dz\,d\theta
$$

Following from the sphere example in [2],  this cone can be parametrized using cylindrical coordinates with the map
$$\Phi{(\theta, z)} = \left( z\,\tan{(\phi)}\,\cos{(\theta)}, z\,\tan{(\phi)}\,\sin{(\theta)}, z \right).$$
Then,
\begin{align}
g_{11} 
&= 
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{\theta}}
\frac{\partial{\Phi_k}}{\partial{\theta}} 
\\
&= 
\frac{\partial{\Phi_1}}{\partial{\theta}}
\frac{\partial{\Phi_1}}{\partial{\theta}} 
+ 
\frac{\partial{\Phi_2}}{\partial{\theta}}
\frac{\partial{\Phi_2}}{\partial{\theta}} 
+ 
\frac{\partial{\Phi_3}}{\partial{\theta}}
\frac{\partial{\Phi_3}}{\partial{\theta}} 
\\
&=
\left[-z\,\tan{(\phi)}\,\sin{(\theta)}\right]^2
+
\left[+z\,\tan{(\phi)}\,\cos{(\theta)}\right]^2
+ 0^2
\\
&=
\left[z\,\tan{(\phi)} \right]^2
\\
g_{12} 
&= 
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{\theta}}
\frac{\partial{\Phi_k}}{\partial{z}} 
\\
&= 
\frac{\partial{\Phi_1}}{\partial{\theta}}
\frac{\partial{\Phi_1}}{\partial{z}} 
+ 
\frac{\partial{\Phi_2}}{\partial{\theta}}
\frac{\partial{\Phi_2}}{\partial{z}} 
+ 
\frac{\partial{\Phi_3}}{\partial{\theta}}
\frac{\partial{\Phi_3}}{\partial{z}} 
\\
&=
\left[-z\,\tan{(\phi)}\,\sin{(\theta)}\right]\,\left[\tan{(\phi)}\,\cos{(\theta)}\right]
+
\left[+z\,\tan{(\phi)}\,\cos{(\theta)}\right]\,\left[\tan{(\phi)}\,\sin{(\theta)}\right]
+ (0)\,(1) 
\\
&=
z\,\left[ - \,\tan^2{(\phi)}\,\sin{(\theta)} \,  \cos{(\theta)} +
 \tan^2{(\phi)}\,\cos{(\theta)} \, \sin{(\theta)}\right]
\\
&=
0
\\
g_{21} 
&= 
0
\\
g_{22} 
&= 
\sum\limits_{k=1}^{3}
\frac{\partial{\Phi_k}}{\partial{z}}
\frac{\partial{\Phi_k}}{\partial{z}} 
\\
&= 
\frac{\partial{\Phi_1}}{\partial{z}}
\frac{\partial{\Phi_1}}{\partial{z}} 
+
\frac{\partial{\Phi_2}}{\partial{z}}
\frac{\partial{\Phi_2}}{\partial{z}} 
+
\frac{\partial{\Phi_3}}{\partial{z}}
\frac{\partial{\Phi_3}}{\partial{z}} 
\\
&= 
\left[\tan{(\phi)}\,\cos(\theta)\right]^2
+
\left[\tan{(\phi)}\,\sin(\theta)\right]^2
+
1^2
\\
&= 
\tan^2{(\phi)}  + 1
\end{align}
Now the Jacobian, $g$, is
\begin{align}
g
&=
\begin{vmatrix}
z^2\,\tan^2{(\phi)}
&
0
\\
0
&
\tan^2{(\phi)}  + 1
\end{vmatrix}
\\
&=
\left|z^2\,\tan^2{(\phi)} \,\left(\tan^2{(\phi)}  + 1\right)
\right|
\\
&=
z^2\, \tan^2{(\phi)} \,\left(\tan^2{(\phi)}  + 1\right)
\end{align}
Finally, I write the differential area element as
\begin{align}
\omega = \sqrt{g} \,dz\,d\theta
\end{align}
Billiography
[1] https://en.wikipedia.org/wiki/Cone#Equation_form
[2] https://en.wikipedia.org/wiki/Volume_element
