(a) We split the square up into four pieces, parametrizing and integrating one a time. Right side: C 1 is parametrized by r 1 (t) = (1, t), 0 t 1.

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1 Thursda, November 5 Green s Theorem Green s Theorem is a 2-dimensional version of the Fundamental Theorem of alculus: it relates the (integral of) a vector field F on the boundar of a region to the integral of a suitable derivative of F over the whole of.. Let be the unit square with vertices (,), (,), (,), and (,) and consider the vector field F(, ) = P(, ), Q(, ) =, +. See below right for a plot. (a) For the curve = oriented counterclockwise, directl evaluate F dr. Hint: to speed things up, have each group member focus on one side of. (b) Now compute da. (c) heck that Green s Theorem works in this eample. F (a) We split the square up into four pieces, parametrizing and integrating one a time. Right side: is parametrized b r (t) = (, t), t. (t, + t) (,)dt = Top side: 2 is parametrized b r 2 (t) = ( t,), t. 2 ( t, t) (,)dt = Left side: 3 is parametrized b r 3 (t) = (, t), t. 3 (, t) (, )dt = Bottom side: 4 is parametrized b r 4 (t) = (t,), t. 4 (, t) (,)dt = [ + tdt = t + ] 2 t 2 = 3 2 [ ] t dt = 2 t 2 t = 2 [ ] t dt = 2 t 2 t = 2 dt = So, the line integral around the entire boundar going counterclockwise is = = 2 (b) da = dd = [ 2 2 ] = = d = 2 d = 2

2 (c) The result for either computation was, demonstrating Green s theorem for this eample ompute the line integral of F(, ) = 3, 4 along the path shown at right against a grid of unit-sized squares. To save work, use Green s Theorem to relate this to a line integral over the vertical path joining B to A. Hint: Look at the region bounded b these two paths. heck our answer with the instructor. A B Let L be the line segment going from B to A. Then, we can now appl Green s theorem to combination of and L. Let be the region bounded b these two paths. Then, b Green s theorem, since we are oriented correctl, da = (4) (3 ) da = 4 da = 4 Area(A) = 6 because the area of the region is made of eactl 4 unit squares. The boundar of is and L: F dr + 6 L The line integral along L is easier: parametrizing L b r(t) = (, t) for t, we get L (, 4) (,)dt = 4dt = 4 Putting it together, 6 6 ( 4) = 2 L

3 3. onsider the quarter circle shown below and the vector field F(, ) = 2e, + 2 e. The goal of this problem is to compute the line integral I = F dr. B = (,4) A = (4,) (a) Parameterize and start directl epanding out I into an ordinar integral in t until ou are convinced that finding I this wa will be a highl unpleasant eperience. (b) heck that F is not conservative, so we can t use that trick directl to compute I. (c) Find a function f (, ) such that F = G + f, where G is the vector field,. (d) Argue geometricall that G integrates to along an line segment contained in either the -ais or the -ais. (e) Use part (d) with Green s Theorem to show that G dr = 4π. (f) ombine parts (c e) with the Fundamental Theorem of Line Integrals to evaluate I. heck our answer with the instructor. (a) (b) Look at the partials: P = 2e + 2e =. For conservative vector fields, these two partials have to be the same (since mied partial commute). (c) F G = 2e, 2 e, which now is conservative, with potential function f (, ) = 2 e (d) Along the -ais, =, so G =,. Along the -ais, G =,, is alwas eactl vertical, hence perpendicular to an portion of the -ais. So, the line integral will be. (e) Take to be region bounded b the arc and the two aes. B part (d), G dr = G dr Now, Green s Theorem tells us that G dr = da = 4 (π 42 ) = 4π ombining these tells us that G dr = 4π

4 (f) F = G + f, so taking the line integrals, G dr + f dr In (e), we computed the first integral, and we can immediatel evaluate the other using the Fundamental Theorem of alculus for Line Integrals: 4π + (f (,4) f (4,)) = 4π 6 4. onsider the shaded region V shown, bounded b a circle of radius 5 and two smaller circles 2 and 3 of radius. Suppose F(, ) = P, Q is a vector field where = 2 on V. Assuming in addition that F dr = 3π and F dr = 4π, 2 3 compute F dr. heck our answer with the instructor. 2 V 3 Note that the outer curve is oriented as we would want it to be to use for Green s theorem, but the other two are oriented backwards (walking along 2 or 3, our left arm points outside V ). So, V = 2 3, and Green s theorem gives F dr + 2 F dr + 3 V The area is just that of the larger disk minus the other two: Area(V ) = π 5 2 π π = 23π da = 3π + 4π + 2 Area(V ) and so 53π 5. Suppose is a region in the plane bounded b a closed curve. Use Green s Theorem to show that both d and d are equal to Area(). Using the alternate notation for line integrals, Green s theorem sas P d +Q d = da So, appling this two the given vector fields: d = da = d = da = da = Area() da = Area()

5 6. The curve satisfing = 3 is called the Folium of escartes and is shown at right. (a) Let be the bulb part of this folium, more precisel, the part in the positive quadrant. Show that an line = t for t > meets in eactl two points, one of which is the origin. Use this fact to parameterize b taking the slope t as the parameter. (b) Use part (a) and Problem 5 to compute the area bounded b. heck our answer with the instructor. (a) Our curve is defined b =. If we consider a line of slope t > through the origin, it meets in one other point. Let = t ; then, 3 + t = 2 ( + t 3 ) = We can divide b as we are looking for the solution with > : This gives a parametrization for t <. = 2 = = + t 3 + t 3 (b) We use the first integral in the previous problem to compute the area: Area() = d = + t 3 ( 6t( + t 3 ) 2 ( 2 ) ) 2 (6 3 ) ( + t 3 ) 2 dt = ( + t 3 ) 3 dt We can compute this integral with a u-substitution, with u = + t 3, du = 2 : 9 3u u 3 du = 9u 3 3u 2 du = [ 92 ] u= u 2 + 3u = ( 9 u= 2 + 3) = 3 2