Introduction to Line IntegralsDate: 02/08/2004 at 00:31:54 From: Mike Subject: Line integrals Dr. Math, I don't understand what line integrals are. It's never made clear in my textbook. I have a problem where its C int(x dy + y dx). The problem says that: C is the broken path line that goes from (1,1,2) to (2,0,-1) by following a line parallel to the x axis from (1,1,2) to (2,1,2) then parallel to the y axis from (2,1,2) to (2,0,2) then parallel to the z axis from (2,0,2) to (2,0,-1). What does all this mean? Date: 02/08/2004 at 08:39:55 From: Doctor Jerry Subject: Re: Line integrals Hello Mike, I can't undertake teaching you about line integrals, other than to say that int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ] is evaluated by finding a parameterization x = p(t) y = q(t), z = r(t), a <= t <= b, for C and then int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ] = int(t=a,t=b,[ f(p(t),q(t),r(t))*p'(t) + g(p(t),q(t),r(t))*q'(t)]*dt + h(p(t),q(t),r(t))*r'(t)]*dt) If the curve is in pieces then one just parameterizes each piece. On a piece parallel to the x-axis dy = dz = 0 and things simplify. On the first part of the curve you have been given, from (1,1,2) to (2,0,-1), use x = p(t) = 1 + (2-1)t y = q(t) = 1 + (0-1)t z = r(t) = 2 + (-1-2)t, t in [0,1]. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 02/08/2004 at 09:49:01 From: Mike Subject: Thank you (Line integrals) Thanks for your help. It's a little clearer now. However, I'm wondering if you could help me understand what exactly a line integral is. I know that integrals are areas under a given curve but what are line integrals? Thank you again. Date: 02/08/2004 at 10:44:11 From: Doctor Jerry Subject: Re: Thank you (Line integrals) Hello Mike, There are several interpretations. One of the most easily grasped is the following. If we have a curve C and a vector force F(x,y,z) at each point (x,y,z) of space, where F(x,y,z) = i*f(x,y,z) + j*g(x,y,z) + k*h(x,y,z), and an object moves from one end of C to the other, then the work dW done by F as the object moves from (x,y,z) on C a distance of ds is approximately dW = F(x,y,z).T(x,y,z) * ds, where . means dot product, T(x,y,z) is the tangent vector to the curve C at (x,y,z), and ds is the arc length. Because (now thinking of the curve as parameterized with a parameter t), T(x,y,z) = [1/sqrt(x'(t)^2 + y'(t)^2 + z'(t)^2)] (x'(t)*i + y'(t)*j + z'(t)*k ) and ds/dt = sqrt(x'(t)^2 + y'(t)^2 + z'(t)^2), we see that (now letting x'(t) = dx/dt, etc) dW = F(x,y,z).T(x,y,z) * ds = f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz. Adding these up on the entire curve we see that W = int_C [ f(x,y,z)*dx + g(x,y,z)*dy + h(x,y,z)*dz ] - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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