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- Compute the integral of the "curl" of the vector field ##A_i(x_i)## over a 2-dimensional surface.

Hi,

So my goal is to compute the integral of the "curl" of the vector field ##A_i(x_i)## over a 2-dimensional surface. Following a physics book that I am reading, I introduce the antisymmetric 2-nd rank tensor ##\Omega_{ij}##, defined as:

$$\Omega_{ij} = \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i}$$

then, from what I gather reading the Wikipedia page (https://en.wikipedia.org/wiki/Surface_integral), if the surface can be parametrized by a set of 2 variables ##u## and ##v## I should be able to write the surface element as ##dudv## times the determinant of the jacobian:

$$ \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v}$$

Putting all together I get:

$$\int_S \left( \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i} \right) \left( \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v} \right) dudv$$

Using the chain rule, I can simplfy the expression to yield:

$$ \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} - \frac {\partial A_j}{\partial u} \frac {\partial x_j}{\partial v} + \frac {\partial x_j}{\partial u} \frac {\partial A_j}{\partial v} \right) dudv = 2 \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} \right) dudv

$$

But I am pretty sure there should not be a 2 in the last equation. Is my reasoning wrong?

Thanks

Ric

So my goal is to compute the integral of the "curl" of the vector field ##A_i(x_i)## over a 2-dimensional surface. Following a physics book that I am reading, I introduce the antisymmetric 2-nd rank tensor ##\Omega_{ij}##, defined as:

$$\Omega_{ij} = \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i}$$

then, from what I gather reading the Wikipedia page (https://en.wikipedia.org/wiki/Surface_integral), if the surface can be parametrized by a set of 2 variables ##u## and ##v## I should be able to write the surface element as ##dudv## times the determinant of the jacobian:

$$ \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v}$$

Putting all together I get:

$$\int_S \left( \frac {\partial A_i}{\partial x_j} - \frac {\partial A_j}{\partial x_i} \right) \left( \frac {\partial x_i}{\partial u} \frac {\partial x_j}{\partial v} - \frac {\partial x_j}{\partial u} \frac {\partial x_i}{\partial v} \right) dudv$$

Using the chain rule, I can simplfy the expression to yield:

$$ \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} - \frac {\partial A_j}{\partial u} \frac {\partial x_j}{\partial v} + \frac {\partial x_j}{\partial u} \frac {\partial A_j}{\partial v} \right) dudv = 2 \int_S \left( \frac {\partial A_i}{\partial v} \frac {\partial x_i}{\partial u} - \frac {\partial x_i}{\partial v} \frac {\partial A_i}{\partial u} \right) dudv

$$

But I am pretty sure there should not be a 2 in the last equation. Is my reasoning wrong?

Thanks

Ric

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