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Integral - what does it represent?

by Benny
Tags: integral, represent
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Benny
#1
Nov30-05, 12:02 AM
P: 585
Hi, I am given the following integral.

[tex]
\int\limits_{}^{} {\int\limits_S^{} {\mathop F\limits^ \to } } \bullet d\mathop S\limits^ \to = \int\limits_{}^{} {\int\limits_D^{} {\mathop F\limits^ \to \bullet \mathop n\limits^ \to } } dS
[/tex]

The n vector is an outward unit normal. So does the RHS of the above represent how much 'stuff' is coming out of a surface? Or does it have something to do with what is happening on a surface? The book says that the integral is called the flux of F across S. In many examples, there are little arrows point out of the surface so I'm not sure what the integral is supposed to represent.

Any help would be good thanks.
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HallsofIvy
#2
Nov30-05, 06:30 AM
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PF Gold
P: 39,568
Yes, both of those can be interpreted to represent the total "flow" of something across the surface (so that flow in opposite directions at different parts of the surface cancel). I am, however, puzzled over what "S" and "D" are. For those to be equal, in general, S and D must be the same.
benorin
#3
Nov30-05, 09:12 AM
HW Helper
P: 1,025
D and S are right next to each other on one's keyboard.

Benny
#4
Nov30-05, 07:51 PM
P: 585
Integral - what does it represent?

Oops, I need to be more careful when copying out the definitions. Thanks for the help.

Edit: In many of the questions I've been doing, the procedure is generally to rewrite the integral so that the dS is repalced by ndA where n is a normal to the surface. After the replacement, instead of integrating over a surface, is the integration done over the x-y plane?
benorin
#5
Dec3-05, 05:04 AM
HW Helper
P: 1,025
The form you put above is still a surface integral on either side, further work is required to change such into a double integral over a domain D in the xy-plane; one such way is to parameterize the surface by, say [itex]\vec{r}(u,v)[/itex] where the parameters u and v range over values in D, the formula is then

[tex]\int\limits_{}^{} {\int\limits_S^{} {\mathop F\limits^ \to } } \bullet d\mathop S\limits^ \to = \int\limits_{}^{} {\int\limits_D^{} {\mathop F\limits^ \to \left( \vec{r}(u,v) \right) \bullet \left( \vec{r}_{u} \times \vec{r}_{v}\right) } } dA[/tex]
Benny
#6
Dec3-05, 07:02 AM
P: 585
Ok thanks for the clarification. I've been working through some questions using the procedure you've presented during the last few days so I'm fairly comfortable with them now.
HallsofIvy
#7
Dec3-05, 08:03 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,568
[tex]\int\limits_{}^{} {\int\limits_S^{} {\mathop F\limits^ \to } } \bullet d\mathop S\limits^ \to = \int\limits_{}^{} {\int\limits_S^{} {\mathop F\limits^ \to \bullet \mathop n\limits^ \to } } dS[/tex]
Just different ways of writing the same thing:
[tex]d\mathop S\limits^ \to [/tex] is defined as [tex]\mathop n\limits^ \to } } dS[/tex]
Yes, it can be interpreted as "flux of F across S"- that is, the rate at which some property (whatever F represents) is flowing through surface S. One way to interpret the Divergence theorem:
[tex]\int\int\int\limits_B^{ } (\nabla \bullet \mathop v\limits^ \to )dV= \int\int_S \mathops v\limit^ \to \bullet d\mathop \sigma\limit^ \to [/tex]
is that the amount of something inside region B depends upon how much is flowing in or out through the surface S.


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