How to formulate the "derivative" of an area integral with respect to the boundary curve?

Question

I'm working on an analysis problem and I've reached a roadblock. I have a smooth function $f$ on $\mathbb{R}^2$ and a simple, closed, plane curve $\gamma$. I'm interested in a quantity which is the integral of $f$ over the interior of $\gamma$: $I = \int_{\operatorname{int}(\gamma)} f \, dA$. Where I'm getting stuck is how to define the functional describing how $I$ changes with respect to changes in the boundary curve $\gamma$. Can anyone help me out?

Answer 1

If the boundary curve is smooth and is moving smoothly, one can zoom in on an infinitely small segment of it (which looks like a straight line since it's an infinitely small arc of a smooth curve) and ask how fast is it moving in a direction orthogonal to the curve. That rate of motion, times the value of $f$ at that point, times the infinitely small length of that arc, equals the infinitely small contribution of that small part of the picture, to the change in the value of the integral.

In a sense, that is why the derivative of the area of a circle with respect to the radius is the circumference: the rate at which the boundary moves is the rate at which the radius changes. The size of the boundary times the rate of motion of the boundary equals the rate of change of size of the bounded region.