Vector fields, line integrals, and Green's Theorem

Transcription

1 Vector fields, line integrals, and Green's Theorem Line integral strategy suggested problems solutions Bunch of assorted line integral of vector filed problems here. In the solutions, I ll show the way I d have solved it, and why. You might have chosen a different way. You may have even chosen a faster way! Strategies include:. Parameterize the curves and work straight from F d r. Work directly in the M d + N dy form. Find a potential function and apply the Fundamental Theorem of Integrals. Replace the path with a simpler path if appropriate 5. Using Green s Theorem Make sure your approach is appropriate for the problem (check for conservative, simply connected region, etc.). P: Find the value of y d + ydy on each of the paths shown below:

2 First move is to check for conservative, especially since multiple paths this may simplify things considerably. y y = = = y = y = onservative. That simplifies things considerably. The value of the line integral on the paths shown in (c) and in (d) is immediately known to be zero, since they are closed paths. Looks fairly easy to find a potential function, so that s what I d do to evaluate (a) and (b): f ( y, ) f =< y,y > = y and so f = y d = y + g y ( ) f = y dy = y + h ( ) (a) This path is from (,) to (,). (b) This path is from (,) to (, ) y d + ydy = f(,) f(,) = () () = 6 y d + ydy = f(,) f(,) = () ( )() =

3 P: Find the value of ( + ) + ( + + ) y d y dyon each of the paths shown below: = ( y + ) = = ( + y + ) = Not conservative. That limits options somewhat. In order of least annoying to most: (d) Is a good candidate for Green s Theorem border of simply connected region, and easy in polar. If it were a standalone problem, that s probably how I d do it for this eercise in particular, I ll note that we ll have to do (b) as a parameterized path anyway, and it d be quicker to just build on that result, However, let s take it as if it were a standalone = ( ) = r Region is π π θ π / = / y d y dy rdrdθ

4 Inner: rdr r π Outer: / d [ ] (c) Is easy to work in y = e dy = e d = = / π / / θ = θ = π y + d + + y + dy = π M d + N dy form, since the path is a single y = f( ) y+ d+ + y+ dy = e + d + + e + e d ( ) = ( e + + e + e + e ) d = ( + e + + e + e ) d e e e e e e = = = + + e + e e = e + e u = dv= e d In the above, there was an IBP on the e d: du = d v = e And a u sub on the e e d= e e e d: = = u du d (b) I d parameterize as a sine/cosine, since circular. = sin t d= costdt y = cost dy = sin tdt π π t

5 π / ( + ) + ( + + ) = ( sin cos + )( cos ) + ( sin + cos + )( sin ) y d y dy t t t dt t t t dt / π / / π / / ( sin cos cos cos sin sin cos sin ) ( cos t sin t cost sin t) dt = t t t+ t t t t t dt = + t sin( t) t sin( t) = + + sint + cost t sin( t) t sin( t) = + + sin t+ cost sin( t) = t + sint+ cost π / / π sinπ π π = + sin + cos π / / π / / sin( ) π π + sin + cos = + + π + [ ] [ ] = + *Integrals of sin t and cos t use half angle formula, or you can just pull them off an integral table. (a) Is annoying no matter what you can Green s Theorem it, but the region is neither horizontally nor vertically simple, so that means two integrals. Or you can just write equations of the lines and go piece by piece, ending up with three integrals. Either way, need to start by getting equations of lines. : y = : 5 : y = *Note to get the direction right, you want to go from to.

6 On : y = dy = d y+ d+ + y+ dy = d d = 6 d On : y = 5 dy = d = d + = + = ( ( ) ) ( ( ) ) y+ d+ + y+ dy = 5 + d ( ) d On : y = dy = d Total: ( 5 ) = d ( ) 6 8 d 8 = = = y+ d+ + y+ dy = d d = d = d + = + = 8 6 y + d + + y + dy = =

7 Now, try it using Green s Theorem. The integral over the region has to be split at = Region one: y Region two: y 5 Integral on region one: Inner: Outer: dy d dy = = 5 5 5d= = = ( ) = Integral on region two: Inner: 5 5 dy d dy = 5 = d= = Outer: ( ) Total is, but then note curve is clockwise oriented; reverse sign, and by Green s Theorem, y + d + + y + dy = = I think Green s Theorem was less annoying!

8 P: Evaluate F d rwhere F(, y, z ) = yz i+ z j+ y k and is given by r() t = t i+ tj+ t k, t. This is a place to recognize that sometimes you just want to go straight by the definition it s a single curve, already parameterized, and it s in space (so we don t have that many options anyway). I wouldn t even bother to check for conservative, since it s fast enough to just do it t d tdt = = y = t dy = dt = = z t dz t dt ()( ) ( t )( t) ( t ( ), yt ( ), zt ( )) = t t + t t + = t + t + t 5 F i j k i j k F 5 dr =< t, t, t > < t,,t > ( + + ) = t t t dt = 6t dt dt t dt t F dr = = =

9 P: Evaluate F d rwhere F(, yz, ) = yz i+ yz j+ yzk and is given by any smooth curve from (,,) to (,, ). This is a hope it s conservative since we don t have a path problem. heck curl: i j k P P curlf= = i j+ k z z z M N P curlf= ( yz) ( yz) i ( yz) ( y z) j+ ( yz) ( y z) k z z ( yz yz) ( 8y z 8y z ) ( 6y z 6y z ) = i j+ k = i+ j+ k Find a potential function: f = y d = + g y z z y z (, ) f = dy = + yz yz hz (, ) f = d= yz + k y yz (, ) f =< y z, y z, y z > f(, y, z) y = z and ( )( ) F d r = f = = (,,) f(,,) 6

10 P5: Evaluate F d rwhere F (, yz, ) = yi + j + zk and is the line segment from (,,) to (,, ). heck curl: i j k P P curlf= = i j+ k z z z M N P curlf= ( z) ( ) i ( z) ( y) j+ ( ) ( y) k z z = i j+ k ( ) ( z ) ( ) = i zj k Not conservative. Now what? In this case, we DO have a path that s not just any curve from (,,) to (,, ), but specifically a line segment. So we can write a parameterization and work from the definition. Writing the equation of the line segment: Point: (,,), parallel vector <,, >=<,, > = t d= dt y = dy = dt z = t dz = dt t r () t = < yz,, >=<,,> + t<,,> F( ( t), y( t), z( t)) = ( t)() i+ j+ ( t)( t) k = i+ j+ 8t k F r =< > < > = d,,8t,, dt 6t dt 6 6 F dr = 6t dt = t =