January 9th, 2017, 01:21 PM  #1 
Newbie Joined: Jan 2017 From: Costa Rica Posts: 3 Thanks: 0  What exactly is a line integral?
As in many math courses, I can do the exercises, but I dont really understand the concepts of the theory. What exactly is a line integral, conceptually speaking? Also, what is the difference between a double and triple integral? I remember that you can calculate volumes and areas with double integral, so what is a triple integral? Volumes plus more abstract concepts such as mass maybe? Thank you. 
January 9th, 2017, 01:24 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,481 Thanks: 744 
the simplest concept is that of the length of a curve in space. if you add a force field in then the line integral becomes the total amount of force required to move some particle (that is affected by this force) along the path of the space curve. the difference between a double and triple integral is simply the number of dimensions you integrate over. You can't calculate volumes with a double integral unless they are simple rotations of a planar area through space. You can calculate areas though. And similarly triple integrals calculate volumes. Last edited by romsek; January 9th, 2017 at 01:28 PM. 
January 9th, 2017, 04:15 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,338 Thanks: 532  https://en.wikipedia.org/wiki/Line_integral. Double and triple integrals (as well as integrals over higher dimensions) are integrals over functions of 2, 3, or more variables. 
January 9th, 2017, 06:57 PM  #4  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
A line integral occurs when the pieces are summed over a line. A contour integral occurs when the pieces are summed over a contour. A double integral occurs when the pieces are summed over two dimensions. A triple integral occurs when the pieces are summed over three dimensions.  
January 11th, 2017, 08:52 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 
Definition of line integrals: $\displaystyle \int_{C}H(x,y,z)ds=\sum H(x_{k},y_{k},z_{k})\Delta s_{k}$ $\displaystyle \int_{C}H(x,y,z)dx=\sum H(x_{k},y_{k},z_{k})\Delta x_{k}$ $\displaystyle \int_{C}H(x,y,z)dy=\sum H(x_{k},y_{k},z_{k})\Delta y_{k}$ $\displaystyle \int_{C}H(x,y,z)dz=\sum H(x_{k},y_{k},z_{k})\Delta z_{k}$ 
January 11th, 2017, 10:56 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 
In rectangular coordinates: $\displaystyle \int_{A}\int fdxdy=\lim\sum f(xk,yk)\Delta Ak$ $\displaystyle \int_{V} \int \int fdxdydz=\lim\sum f(xk,yk,zk)\Delta Vk$ More generally (curvilinear coordinates): $\displaystyle \int_{A}\int fdA=\lim\sum f(Pk)\Delta Ak$ $\displaystyle \int_{V} \int \int fdV=\lim\sum f(Pk)\Delta Vk$ 
January 11th, 2017, 07:27 PM  #7  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,590 Thanks: 936 Math Focus: Elementary mathematics and beyond  Quote:
You sure have a lot to say about higher mathematics. Off the top I'll say that I understand very little of it (higher mathematics) but your posts seem to be generally inaccurate and therefore misleading. I urge you to explain yourself instead of posting arcane equations and the like.  
January 11th, 2017, 08:05 PM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 
Draw a rectangle in the x,y plane and suppose f(x,y)=x+y Divide the rectangle into four parts with a horizantal and vertical line. Evaluate f(x,y) at the center of each rectangle, Let the area of each rectangle be A1, A2, A3, A4 and the center of each rectangle be P1, P2, P3, P4, respectively. Then the double integral of f over A is very approximately f(P1)A1 + f(P2)A2 + f(P3)A3 + f(P4)A4 Do exactly the same thing for a cube and f(x,y,z) to get the volume integral. As the divisions become infinitesimal, the summation procedure leads to the rules for evaluating multiple integrals: Add all the little boxes times f at a point in the box in a vertical (or horizantal) strip, and then add the strips. See your textbook for that. When I get confused, I always go back to the basic definitions though. 

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