Suppose I have a cone, and I want to find the surface area. I could use integration to solve this. However, there are two ways of going about this.
I could either take a surface area element and integrate it over the surface, or I could take a circular strip and integrate it. Both of them give me the same numerical answer which is what one would expect. Suppose, I care about area vectors now, instead of the numeric value. Suppose I have some electric field $\vec{E}=E(r)\hat{r}$, and I care about the flux of this electric field through the curved surface of the cone.
This is given by $$\Phi=\int \vec{E}.d\vec{S}$$
My problem is regarding how $d\vec{S}$ is defined in this case.
According to Wikipedia, in spherical coordinates, the curved surface area element of the cone is simply $dS_\theta=r\sin\theta d\phi dr$. However, what is the unit vector associated with this ? My guess is $\hat{\theta}$, but I don't think that is correct, as that would give me a $0$ total flux since $\hat{r}.\hat{\theta}=0$, which I don't think is correct.
On the other hand, I can take a strip of width $dl$, with the area $\pi r^2dl$. In this case, $d\vec{l}=dl \hat{l}$ which has a unit vector along the surface of the cone, away from the vertex.
So, there are two possible surface area elements $\pi r^2dl\hat{l}$ and $r\sin\theta d\phi dr \hat{\theta}$, which if integrated for the total area, gives me the same exact answer. However, once I involve the unit vectors, and calculate vectors such as flux, I run into trouble.
Can anyone point out to me, which one is correct and why is it so?
Best Answer
You should take a look at the formulas/definitions given in Wikipedia's page on Surface Integrals again.
In general, for any surface, we should always parametrize it (or atleast a certain portion of it) and then calculate things accordingly. Since you want to do things in spherical coordinates, I'll assume for convenience we have an inverted right-cone $C$ with vertex at the origin, and its axis along the positive $z$-axis. Suppose also that the opening angle of the cone is $0<\alpha<\frac{\pi}{2}$. Suppose the slant length of the cone is $L$. Then, we can consider the following parametrization of the most of the cone: $f:(0,L)\times (0,2\pi)\to \Bbb{R}^3$ \begin{align} f(r,\phi)&:= (r\sin\alpha\cos\phi, r\sin\alpha\cos\phi,r\cos\alpha) \end{align} Note that because the cone's surface is $2$-dimensional, we only need two parameters to describe it. Here, because we assumed the cone is inverted, this just amounts to fixing the spherical coordinate angle $\theta$ to equal $\alpha$, the opening angle of the cone.
Now, the vector area element is equal, up to sign, to \begin{align} \left(\frac{\partial f}{\partial r}\times \frac{\partial f}{\partial \phi}\right)\,dr\,d\phi&= \begin{pmatrix} -r\sin\alpha\cos\alpha\cos\phi\\ -r\sin\alpha\cos\alpha\sin\phi\\ r\sin^2\alpha \end{pmatrix}\,dr\,d\phi= \begin{pmatrix} -\cos\alpha\cos\phi\\ -\cos\alpha\sin\phi\\ \sin\alpha \end{pmatrix}\,r\sin\alpha\,dr\,d\phi. \end{align} Why does this give the formula for the surface element? Briefly, it is because $f$ being a parametrization for the surface of the cone means that the vectors $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \phi}$ are tangent to the cone (in fact they form a basis for the tangent space of the cone at the point $f(r,\phi)$). The obvious way to get something orthogonal to the surface of the cone is to take the cross product of two vectors tangent to the cone. Also, recall that geometrically, magnitudes of cross products yield the area of parallelograms; that it why multiplying it by the differentials yields the area element of the cone.
Now, we come to the tricky part: orientation. The fact that the $z$-component is positive (since $\alpha$ is actue) means that the vector points "into the cone". Typically, we define the normal vector so that it points "outside", so that tells us we should consider the negative of what we found. Hence, \begin{align} d\mathbf{S}&=\begin{pmatrix} \cos\alpha\cos\phi\\ \cos\alpha\sin\phi\\ \sin\alpha \end{pmatrix}\,r\sin\alpha\,dr\,d\phi = (r\sin\alpha\,dr\,d\phi)\,\mathbf{e}_{\theta}(r,\alpha,\phi), \end{align} where the last equal sign is simply by observation that the vector equals to $\mathbf{e}_{\theta}$ at coordinates $r, \theta=\alpha,\phi$ (and again, we're fixing $\theta=\alpha$ because we're on the surface of the cone).