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etotheipi

I learned that, because ##du \, dv = \frac{\partial(u,v)}{\partial(x,y)} dx \, dy##, if you set ##u=y## and ##v=x## then you get that ##dx \, dy = - dy \, dx##. And that the product of two differentials is a wedge product, which is antisymmetric. If coordinates are orthogonal, then ##dx \, dy = dx \wedge dy = - dy \wedge dx = - dy \, dx##.

Now take a double integral, over the unit square:$$\int_0^1 \left( \int_0^1f(x,y) dy \right) dx = \int_C f(x,y) (dx \wedge dy) = \int_0^1 \left( \int_0^1f(x,y) dx \right) dy$$The two integrals on either side have the same orientation; what does this mean? I'm guessing it has to do with whether you go around the integration path clockwise or anti-clockwise in the ##x##-##y## space, but how do you judge this from the limits on the integral?

Now take a double integral, over the unit square:$$\int_0^1 \left( \int_0^1f(x,y) dy \right) dx = \int_C f(x,y) (dx \wedge dy) = \int_0^1 \left( \int_0^1f(x,y) dx \right) dy$$The two integrals on either side have the same orientation; what does this mean? I'm guessing it has to do with whether you go around the integration path clockwise or anti-clockwise in the ##x##-##y## space, but how do you judge this from the limits on the integral?

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