Multivariable Calculus Review Session 4

Multivariable Calculus: Review Session 4 6 . Line Integrals ( from Stewart , Calculus, Chapter 16) Vector Fields A vector field is a function F that assigns a vector F to each point . F ( x , y ) = ( P( x , y ) , Q ( x , y ) ) F { x , y, z ) = ( P ( x , y , z ) , Q ( x , y , z ) , R( x , y , z ) ) Conservative / Gradient Vector Fields: F = V / for some / . The function / is called a potential for F . Ex. F ( x , y ) = ( 2 x y , x 2 - 3 y 2 ) is a gradient vector field . A potential function is f ( x , y ) = x 2 y y3. — A gradient vector field F = V / is perpendicular to the level sets f ( x , y ) = k of the potential. Line Integrals of Vector Fields Line integral of F along C : To compute: d r = r' ( t ) d t I/ J df = P d x 4- Q d y d x = x' ( t ) d t L F dr = f J a ’ d y = y ( t )d t F ( r ( t ) ) • r' ( t ) d t Rmk . Similarly for vector fields of three variables. Interpretation: The line integral of F along C is the work performed by the force field in moving along C . -. Ex. 1. Evaluate the line integral fc x 2 d x - x y d y where C is the arc x \\ x j r ^ -A C Tlx o tawcW aaS o T i l. j - cas 1 ' ~ ^ '' t “ Css UL tfk \ - 1 cos ^ t s;^ t dt - U (o ) 0 \ “ f l \K ~ ^ Z ' .m = cos ( t ) , y = sin ( f ) . 0 < t < TT / 2 . % 5 ~ ' 1 " ' The Fundamental Theorem for Line Integrals. Let C be a smooth ( or piecewise smooth ) curve given by the vector function r ( t ) r a < t < b. Let / be a differentiable function of two or three variables whose gradient vector V / is continuous on C . Then L V / . df = / ( r ( 6 )) - / (f (a ) ) f t Independence of Path: By the fundamental theorem , if F = V / is a gradient vector field , then the line integral depends only on the endpoints of C, not the path itself . The figure at right shows the level sets of a function f { x , y ) and its gradient field. C, Ci , and C2 are three paths from the point P to the point Q . f JC V f d f= [ = J[C V/ dr • JCi V/ • d f = f {Q ) - f ( P ) 2 - ^ -0 Ex. 2. Let F { x , y ) = (3 + 2 x y , x 2 - 3 y 2 ) . = <( £ % ) ) ( a ) Show that the line integral fc F • dr depends only on the endpoints of the path C and not on the path taken between those endpoints. ^ ^ ( ^ TV»i , VC / JC 6 fc F • d r * ^ : - where F( x , y , z ) - ACv W ) KoriMC sr - , ^ \ e = ( t/ z, xz , xy + 2z ) and “ tX ' < , ,f >e r (cT) - o ? CWW 0 - C is the line segment from (1, 0, - 2 ) to ^ 1 1u Aor f sa i-ft ei V4 - P >X ^4 4 (AOi = lA ) - Ex. 3. Evaluate ) Kb) ( 4, 6, 3' ^ I \9 \ -- -, - xN + 2 = $ 7Wf ~ 4WA F d r where C is the curve given by r ( t ) = ( e 1 sin £ , ef cost ) , 0 < t < 7r . §c cVA F ^4 #«*)) i * t h, s« tWt ( b ) Evaluate F s\\>i -\ Vm-V Ak FILL , 4 s 'Vj - 7 Z -^ 1 -( IA C ^ 71 Green’s Theorem. Let C be a positively oriented , piecewise-smooth , simple closed curve in the plane and let D be the region bounded by C. Let F { x , y ) = ( P ( x , y ) , Q ( x , y ) ) be a smooth vector field on D . Then L P dx + Q dy = [ ( 09. _ ££ D \9 X dA dy Let C be a curve traversed once by r ( t ) pob\ for a < t < b. C is closed if the initial and terminal points of C coincide, i .e. r( a ) = r( b ) . not simple, simple, HOC dosed = ( x( i ) , y( t ) ) not dosed C is simple if C has no self-intersections except at the endpoints, i.e. r (fi ) r ( / 2 ) for all a < t\ < t i < b. ^ V simple, closed C is positively oriented if it is traversed in the counterclockwise direction as t increases from a to b . The region D is always to the left of not simple , c. closed Rmk. The line integral of a vector field along a curve depends on the orientation of the curve as follows: denotes the curve C traversed in the opposite direction , then f c / (x, y ) d x = f c / ( x, y ) d x . _ ^ Ex. 4. Evaluate 3y -e sin C p SK I x ) d x 4- ( 7x + \J y 4 + 1 ) d y where C is the circle x 2 4- y 2 = 9. 3 d /V a* - J) ' P « - — '* H A *** OS) H f IT 3*0 3 r drei © 0 \ X J3 bK — 7 b A$ I T A -ros © 3 ^ ^ /3 =1 Ex. 6. Let C be the triangle with vertices ( 0, 0 ) , ( 1, 0 ) , and ( 0 , 1) , oriented counterclockwise and let F ( x , y ) ( x3 , x y ) . x 3 d x + x y d y is equal to a certain double ( a ) According to Green’s Theorem , the line integral fr F - d r = integral. Set up and evaluate this double integral. = Jc a* -*It *' 0> A U F • c 0 ^ " D \ * ' X TT^ “ $ 7 Cx 0, o ) iK'j - o'l V* vp O r v| ; A* 0 0 1 . 'J 4 c- x + i'ix r O r 1 o [ 1 - '/b . ( b ) Verify Green 's Theorem by evaluating the line integral directly. cr- 0 X’ t / -o> r r° - ^ < J Ie. o Hi\ v = - it - if cIf > * r Cx ife * ^4 * 4> o -o , vj * o it H i x 0, ' F rr t . < 0; 0> < O -OjU $ (l4 - (W -v + 0 1 - —^ Ti 4 ID 0 fr* * i K 0+ \ . + o') o 4