# Differentiation under integral sign - one parameter family

1. Mar 3, 2013

### WannabeNewton

Hi guys! Let $\left \{ B_{t} \right \}_{t\in \mathbb{R}}$ be a one - parameter family of compact subsets of $\mathbb{R}^{3}$ with smooth (manifold) boundary (e.g. one - parameter family of closed balls). In my context, each $B_{t}$ belongs to a different constant time slice of Minkowski space - time. How does one compute $\frac{\partial }{\partial t}\int_{B_{t}}f(t,\mathbf{x})dV$? Again in my context, $f(t,\mathbf{x})$ happens to be $T_{00}(t,\mathbf{x})$, the time - time component of the stress energy tensor and as such this component is assumed to be a smooth scalar field.

Last edited: Mar 3, 2013
2. Mar 4, 2013

### Vargo

Isn't your derivative a Calc I style total derivative as opposed to a partial? Replace the first t by a u and the second t by a v: (the integral over Bu of the function f(v,x).)

Now to calculate the derivative, you set u=t, v=t and use the multivariable chain rule:
d/dt F(u,v) = d/du F (u=t) + d/dv F (v=t)
Sorry about the notation, my keyboard is broken and half the keys don't work :(

This is a simple but handy technical device to separate the two instances of t in the formula.

To account for the change in the domain, you could try to express the change in the domain Bt as B0 + s(t,x)n, where n is the unit outward normal vector on the boundary of B0, and s(t,x) depends on time t and the point x on the boundary of B0. Then when you differentiate with respect to the change in boundary, you will basically be integrating the function s(0,x)*f(0,x) over the boundary surface.

3. Mar 5, 2013

### WannabeNewton

Thanks Vargo. That is exactly what wiki does as well apparently. So if instead my integration regions were constant time slices of R^4 itself (so the integration regions would just be R^3 for each time slice) then could I just ignore the boundary integral and simply pull the time derivative inside? Again, thanks for the response I really appreciate it and I might have to ask more once this one is answered if that's ok =D (although they might be a bit physicsy? idk =p)

4. Mar 5, 2013

### Vargo

Yes, if your region of integration is constant, then you can just move the differentiation inside. Well, there is a condition that must be met. If, when you move the derivative inside the integral, the resulting integral is absolutely convergent, then the operation is valid. I think that's the rule. I'd have to look it up to be sure.