My Math Forum Connection Between Line and Double Integrals in the Plane

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 November 28th, 2017, 12:34 PM #1 Newbie   Joined: Nov 2017 From: Brasil Posts: 3 Thanks: 0 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: \begin{align} 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 \\ \end{align} 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: \begin{align} 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) \end{align} So, why the terms with $dxdx$ and $dydy$ vanished ? And why $-g_ydxdy$ became $g_ydxdy$ ? With my best regards, Daniel.
 November 29th, 2017, 04:31 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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.
November 29th, 2017, 05:56 PM   #3
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 Originally Posted by Country Boy 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.
I thought about it for a long time and never understood. It was very good advice Country Boy. So, for the same reason the multiplication by equals is zero, is it right? One more question, do you have a good tip about a good bibliography about differential forms on surface? I really thank you!

Last edited by skipjack; April 27th, 2018 at 11:53 PM.

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