derivative of domain of integration

Question

Suppose $f(x,y)$ is a function $\mathbb{R}^2\times \mathbb{R}^2 \to\mathbb{R}$ and $\Omega(x)$ is a family of compact regions of the plane whose boundary curve $\gamma(s,x)$ varies smoothly in $x$. I need to take the derivative of $$\int_{\Omega(x)} f(x,y)\,dy$$ with respect to $x$.

Intuitively I expect to get something like

$$\int_{\Omega(x)} \frac{\partial f}{\partial x}(x,y)\,dy + \int_{\gamma} f(x,\gamma(s,x))\frac{\partial \gamma}{\partial x}(s,x)\cdot n(s)\,ds$$ where $n(s)$ is the normal to $\gamma$... but how can I see this formally?

score 2 · Accepted Answer · answered Dec 20 '14 at 20:03

2

I've found the pertinent result -- my integral above is exactly the statement of the Reynolds transport theorem and a short proof can be found on page 22 of Advanced Transport Phenomena: Fluid Mechanics and Convective Transport Processes by Leal.

answered Dec 20 '14 at 20:03

user7530

50,625

score -1 · Answer 2 · edited Apr 13 '17 at 12:21

-1

Combine three standard results:

If $x$ appears in an expression multiple times, we can differentiate the expression with respect to $x$ by differentiating with respect to each $x$ individually, holding the others constant, and adding up the results. (This is just the ordinary multivariate chain rule, applied to a function $G(x_1, x_2, \ldots x_n)$ by taking $x_i(x) = x$.)
Differentiation under the integral (https://math.stackexchange.com/a/12911/3296)
Stokes's theorem

edited Apr 13 '17 at 12:21

Community

1

answered Dec 20 '14 at 07:56

Daniel McLaury

25,340

Sure, I'm aware of all three, but don't see how Stokes applies here. – user7530 Dec 20 '14 at 07:59
Apply Stokes's theorem to rewrite the integral as an integral over the boundary, then differentiate under the integral sign. – Daniel McLaury Dec 20 '14 at 08:20
Sorry, I'm not seeing it. I can integrate up a $g$ with $\nabla_y \cdot g(x,y) = f(x,y)$ but I don't see how Stokes helps. – user7530 Dec 20 '14 at 08:30

derivative of domain of integration

2 Answers2