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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: $$ \begin{equation} \iint_R [f_x(x,y) + g_y(x,y)]dxdy = \int_{C^+}[f(x,y)dy,  g(x,y)dx] \end{equation} $$ 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: $$ \begin{equation} \iint_R dL = \int_{C^+} L \end{equation} $$ 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 anticommutative. 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  
Newbie Joined: Nov 2017 From: Brasil Posts: 3 Thanks: 0  Quote:
Last edited by skipjack; April 27th, 2018 at 11:53 PM.  

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