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November 28th, 2017, 12:34 PM   #1
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Post Connection Between Line and Double Integrals in the Plane

Hi Everybody,

I really need to understand it, it is a very important concept in physics, mainly in electromagnetism. I searched many Calculus's books and only one of them propose to explain a bit more, but I didn't understand the explanation, and this doubt is the first part of the explanation. The complete explanation is in the chapter 5 of the book : Introduction to Calculus and Analysis, Vol. II (Richard Courant , John Fritz)

My doubt starts in the explanation about a connection between line and double integrals. There, we ned to consider two functions f(x,y) and g(x,y) as components of a vector field $\vec{A}$ in the plane. Considering any closed curve in this field, it considers the following equation:

\iint_R [f_x(x,y) + g_y(x,y)]dxdy = \int_{C^+}[f(x,y)dy, - g(x,y)dx]

Where C is the boundary of a set R in the plane, and $C^+$ is the orientation in wich R remains on the left. To this orientation we name positive orientation.

Taking the integrand of the right side of the equation and doing:

L &= f(x,y)dy - g(x,y)dx \\
dL &= (f_xdx + f_ydy)dy - (g_xdx + g_ydy)dx \\
&= f_xdxdy - g_ydydx = (f_x + g_y)dxdy \\

We have:

\iint_R dL = \int_{C^+} L
And it is beautiful

But, I do not understood this explanation because I was wondering that correct form to do this would be:

dL &= \frac{\partial L}{\partial x}dx + \frac{\partial L}{\partial y}dy \\ &= f_xdydx - g_xdxdx + f_ydydy - g_ydxdy \\ &= ( f_xdydx - g_ydxdy ) + (f_ydydy - g_xdxdx)

So, why the terms with $dxdx$ and $dydy$ vanished ? And why $-g_ydxdy$ became $g_ydxdy$ ?

With my best regards,
Danbakana is offline  
November 29th, 2017, 04:31 AM   #2
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Multiplication of "differential forms" is anti-commutative. That is, dxdy= -dydx. To understand exactly why that is true, you would really need to go deeply into the definitions of "differential forms" on surfaces. Essentially it has to do with the "Jacobian", $\displaystyle J(f(x,y), g(x, y))= \left|\begin{array} {cc}
\frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y}\end{array}\right|= \left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial g}{\partial y}\right)- \left(\frac{\partial g}{\partial x}\right)\left(\frac{\partial f}{\partial y}\right)$.

Swapping "f" and "g" changes the sign.
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