Surface infinitesimals and its intuitive manipulation? The excess pressure in the concave side of any liquid bubble or drop with surface tension of the liquid being $T$ is $\frac {4T}r$ and $\frac {2T}r$ respectively. I wanted to derive it using a parametrized sphere and then considering the equilibrium of an infinitesimal area element, but this heavily depended on the intuition behind surface integrals and was unconventional. I was hoping for some support here:-
Alternative Method
Consider a spherical drop (the bubble case is almost similar) centred at $(0,0,0)$ and characterised by $$\vec r=r \cos \theta \hat z+r \sin \theta \cos \phi \hat x+ r \sin \theta \sin \phi \hat y$$ 
$$0<\theta<\pi\text{ };\text{ }0<\phi<2\pi$$
 Where $\theta$ is the angle of the position vector with $z$ axis and $\phi$ the angle of the projection of the position vector in x-y plane with x axis.
The infinitesimal area element will be $$|\frac {d\vec r}{d\theta}\times \frac{d\vec r}{d\phi}|d\theta d\phi\hat r=r^2\sin\theta d\theta d\phi \hat r=\vec {da}$$ 
This is, intuitively( or so i believe), actually an area bound by the parallelogram having its two sides as $\frac {d\vec r}{d\theta}d\theta(=r_\theta d\theta)$ and $\frac{d\vec r}{d\phi}d\phi(=r_\phi d\phi)$[Fig(a)]. Thus, to calculate the downward force by the surface tension, we consider the net force in the downward ($-\hat r$) direction and equate it to $\Delta P\vec {da}$ to get the value of $\Delta P$. To evaluate the force of surface tension, consider the side $r_\theta$ of the infinitesimal area parallelogram. Considering $T$ to be per unit length, the force on this (sidewards) is $2T \sin\frac{d\theta}{2}$ which is $Td\theta$ per unit length[Fig(b)]. This is valid for the length of $|r_\phi|d\phi$ [Fig(c)]which evaluates to net downward force as $T|r_\phi|d\theta d\phi$. Considering the side of $r_\phi$, the same arguments apply and the net downward force on either side of this will be $T|r_\theta|d\theta d\phi$. hence:-$$\Delta P\vec {da}=Td\theta d\phi (|r_\theta|+|r_\phi|)(\hat {-r})$$ 
But this does not yield the answer. It comes close but not quite the answer. In fact it predicts non-uniform pressure for different spherical co-ordinates, which is obviously not correct. Is there something wrong with my intuition or my calculations? Sorry, if this has a trivial mistake. 

 A: You compute the downward force as the sum of two terms. The error is in the second term, which you compute as $Tr_{\theta}\;d\theta \;d\phi$. That can't possibly be right, since $r_{\theta}\equiv r$ for all $\theta,\phi$, but as $\theta\to 0$ or $\theta\to \pi$, the area $r_{\theta} r_{\phi} d\theta d\phi$ goes to $0$, so the downward force better goes to $0$ as well.
So where did you go wrong in that computation? The downward force comes from $T$ acting along $2$ sides  of the parallelogram of length $r_{\theta}d\theta$ each. You got that part right. But you multiply by a factor $\sin(d\phi/2)$ to account for the angle between the forces on these two sides. That factor can't be right: as $\theta\to 0$ or $\pi$, these sides get closer and closer, so the angle between the forces on these sides must go to $0$, even with $d\phi$ fixed.
So what should the correction factor be? The angle between the forces on these two sides is the same as the angle with vertex at the center of the sphere of an arc from one side of the parallelogram to the other. The distance from one of side to the other is $r \sin\theta \; d\phi \equiv r_{\phi} d\phi$. The angle is then gotten by dividing by $r$, i.e., it is $\sin\theta \; d\phi$. The downward force for this term is then $2 T r_{\theta} d\theta \; \sin(\sin\theta\; d\phi/2) = Tr_{\theta} \sin\theta\; d\theta\;d\phi=T r_{\phi} d\phi d\theta$. This is the same as the downward force from the other two sides. Adding them up and dividing by $da=r_{\phi}r_{\theta}d\phi d\theta$ gives $2T/r$.
