1
$\begingroup$

Let $C=(0,1)^{N}$ be the $N$-dimensional unit cube. Let $f:\mathbb{R}^{N}\to\mathbb{R}$ be a sufficiently regular function for what follows. What does the following notation mean:

$$\int_{C}\nabla f(x)\text{d}x$$

How can I interpret the integral of a vector field over a volume? As it may help, here is the context: let $\mathbf{a}\in\mathbb{R}^{N}$ be a constant vector. We are interested in the following integral (where $\cdot$ denotes the natural scalar product on $\mathbb{R}^{N}$):

$$\int_{C}\mathbf{a}\cdot\nabla f(x)\text{d}x$$

And it is written that

$$\int_{C}\mathbf{a}\cdot\nabla f(x)\text{d}x=\mathbf{a}\cdot\int_{C}\nabla f(x)\text{d}x=-\mathbf{a}\cdot\int_{\partial C}f(x)\mathbf{n}\,\text{d}\sigma(x)$$

(where $\mathbf{n}$ is the normal) which I understand intuitively as a variant of

$$\int_{C}\left(g\text{ div}\mathbf{F}+\mathbf{F}\cdot\nabla g\right)\text{d}V=\int_{\partial C}g\mathbf{F}\cdot\mathbf{n}\,\text{d}S$$

where $g:\mathbb{R}^{N}\to\mathbb{R}$ and $\mathbf{F}:\mathbb{R}^{N}\to\mathbb{R}^{N}$ is a vector field.

$\endgroup$
4
  • 1
    $\begingroup$ It is just the integral of each component. And your intuition is correct. This is a very nice little book on these things: goodreads.com/book/show/703104.DIV_Grad_Curl_and_All_That $\endgroup$ Feb 28, 2016 at 12:18
  • $\begingroup$ @GiuseppeNegro Thank you, I supposed it was, but wasn't sure. Do you know any reference work on the subject? $\endgroup$ Feb 28, 2016 at 12:19
  • $\begingroup$ Edited my previous comment $\endgroup$ Feb 28, 2016 at 12:20
  • $\begingroup$ @GiuseppeNegro Thanks a lot. $\endgroup$ Feb 28, 2016 at 12:21

0

You must log in to answer this question.

Browse other questions tagged .