Surface infinitesimals and its intuitive manipulation?

Question

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.