Is there another way to find the flux of the gradient of a scalar field through a shape than just taking the line integral for a vector field?

- anonymous

- chestercat

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- amistre64

thats was going to be my next chapter that i read to the kids for a bedtime story :)

- anonymous

The function i have gives me an unsolvable integral when i use a line integral to find the flux of the gradient..

- amistre64

what does flux of a gradient mean; i know what a gradient is; but flux?

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- amistre64

sounds like soldering :)

- anonymous

im given a scaler field im told to find the gradient and integrate it around a shape

- anonymous

the gradient that is.

- amistre64

whats the function?

- amistre64

im gonna read vector fields tonight; i hope its steamy ;)

- anonymous

xe^(yz)
my shape is r(t)=<-2rcos(t),rcos(t),2sqrt(2)sin(t)>
t is between 0 and pi and my r is 2

- amistre64

Ok; the gradient should be the derivatives of your function in vector format..right?

- amistre64

gF(x,y,z) = right?

- anonymous

yeah

- anonymous

i cant take the strait line integral and stokes doesnt work because the curl of the grad is zero

- amistre64

gF = the gradient vector right?

- anonymous

yup

- amistre64

thats the extent of my abilities with that :)

- anonymous

...

- amistre64

what is a flux?

- anonymous

you cant help me

- amistre64

prolly not :)....give me a week ;)

- anonymous

maybe a little longer

- anonymous

No he's taking the same class you are (calc 3) so he'll probably be covering this later in the term

- anonymous

found it
http://en.wikipedia.org/wiki/Line_integral#Path_independence

- anonymous

im taking vectors

- anonymous

didnt cover this in calc 3

- anonymous

really? It was covered in my 3rd semester calc class along with surface integrals, etc. Stoke's, Green's, etc

- anonymous

flux but not this indepth of line integral theory

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