I need to find the differential element dS in terms of dx and dy for a shape drawn in the x-y plane. The shape consists of a triangle with one of its edges rounded. This arc at the edge is a part of a circle whose center is at the origin. The triangle also has one of the vertices at the origin (shown in the diagram).
The figure is symmetric about the x-axis. I have the equations that define the boundaries of the figure, but I need to find the differential surface element of this shape. I am trying to use this in an electromagnetics problem for finding the vector potential along this surface. The equation has one of the parts as a surface integral - $\iint\limits_S{J*G.dS}$ along the surface S given by the boundaries in the diagram (J and G may or may not be constants).
If I use the property $\iint\limits_S{F.dS} = \iint\limits_S{(F.\hat{n})dS}$ where $\hat{n}$ is the unit vector along the normal of the plane, I encounter a problem:
$dS = \frac{dxdy}{|\hat{n}.\hat{k}|}$ and $\hat{n} = \frac{\nabla{f}}{|\nabla{f}|}$ where f defines the boundary of the surface. However, since $\nabla{f}$ does not contain a z component, there isn't any $\hat{k}$ component in $\hat{n}$. This means the denominator in the RHS of dS is 0.
How do I find dS?