E&M proof for a integralnot sure whats it called

[Brown Arrow]

Homework Statement

Δ<--- the gradient X<---cross product
(o∫) closed integral

∫((ΔT)Xda = (o∫)-Tdl

btw what is the realtion ship above called?

Homework Equations

da= a_xdydz +a_ydxdz +a_zdxdy
dl=a_xdx + a_ydy + a_zdy

The Attempt at a Solution

not sure where to go on from here :/ I am kinda lost

 

[tiny-tim]

Hi Brown Arrow!

(have a del: ∇ and an integral: ∫ )

btw what is the realtion ship above called?

Stokes' theorem! (or Kelvin-Stokes theorem, see http://en.wikipedia.org/wiki/Stokes'_theorem#Kelvin.E2.80.93Stokes_theorem" )

da= a_xdydz +a_ydxdz +a_zdxdy
dl=a_xdx + a_ydy + a_zdy

The Attempt at a Solution

not sure where to go on from here :/ I am kinda lost

I don't understand what the question is.

 

[sgd37]

From what I can make out of the question you are supposed to prove that

$$\int \nabla T \times d\bold a = \oint T d\bold I$$

where $$d\bold a = (dydz,dxdz,dxdy) , d\bold I=(dx,dy,dz)$$

Am I right

 

[Brown Arrow]

yes, I am not sure how to go about it.
I am suppose to show that both of them are equal to each other
$$\int \nabla T \times d\bold a = - \oint T d\bold I$$

 

[sgd37]

its a bit tedious if you don't use the Levi-Civita tensor, but letting dxdy=-dydx (I know it's a bit weird but it's the only way this thing works) the LHS =

$$\int \int \frac{1}{2} \epsilon_{ijk} \epsilon_{klm} \partial_{j} T dx^l dx^m$$

$$= \int \int \frac{1}{2} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \partial_{j} T dx^l dx^m$$

$$= \int \int \frac{1}{2} (\partial_{m} T dx^i dx^m - \partial_{l} T dx^l dx^i)$$

$$= \int \int \frac{1}{2} (-\partial_{m} T dx^m dx^i - \partial_{l} T dx^l dx^i) = - \oint T dx^i$$

where we used the contracted epsilon identity and changed the last integral into a closed one because going from integrating with respect to an area to a line integral the boundaries change so that the integral becomes closed

Anyway hope this helped the way to do it using standard vector operations is to rewrite $$d\bold a = (dydz,dxdz,dxdy)$$ as $$d\bold a = \frac{1}{2} (dydz -dzdy,dxdz-dzdx,dxdy-dydx)$$ which makes sense because the area between two vectors is $$\left|\bold a \times \bold b \right|$$

 

[Brown Arrow]

thanks for the help, but i have not learned it the way you have shown above so its really hard for me to understand it. :/

is there a simpler way?

 

[sgd37]

well do you know what the cross product between two vectors is