18.02 Line Integrals in the Plane (Solutions to Exercises)

4. Line Integrals in the Plane 4A. Plane Vector Fields 4A- 1 a) All vectors in the field are identical; continuously differentiable everywhere. b) The vector at P has its tail a t P and head at the origin; field is cont. diff. everywhere. c) All vectors have unit length and point radially outwards; cont. diff. except at (0,O). d) Vector at P has unit length, and the clockwise direction perpendicular to O P . 4B. Line Integrals in the Plane a) On Cl: y = 0, dy = 0; therefore (x2 - y) dx + 2%dy = x2 dx = l 3 2 =- -1 3' b) C: use the parametrization x = cos t , y = sin t; then dx = - sin t dt, dy = cost dt c) C = C l + C 2 + C 3 ; L C1:x=dx=O; C2: y = 1 - x ; L1o+1 1 ydx-xdy = (1-x)~x-x(-dx)+ d) C : x = 2 c o s t , y = s i n t ; e) C : x = t 2 , y = t 3 ; L 6y dx + x dy = 1 dx=-2sintdt dx=2tdt, dy=3t2dt 2 6t3(2t dt) + t2(3t2dt) = C3:y=dy=0 LO = Jdldx = 1. L y d x = J d Z T - 2 ~ i ~ ~= t d-t 2 ~ 1 2 2 (15t4) dt = 3t5] = 3 . 31. 1 4B-2 a) The field F points radially outward, the unit tangent t to the circle is always perpendicular to the radius; therefore F . t = 0 and Jc F . dr = Jc F . t ds = 0 b) The field F is always tangent to the circle of radius a, in the clockwise direction, and of magnitude a. Therefore F = -at, so that Jc F . dr = Jc F . t ds = - Jc a d s = -2aa2. E. 18.02 EXERCISES 2 i + j 4B-3 a) maximum if C is in the direction of the field: C = - JZ i + j b) minimum if C is in the opposite direction to the field: C = -- . i - j 4 . c) zero if C is perpendicular to the field: C = f- JZ d) max = 4, min = -4 by (a) and (b), fir the max or min F and C have respectively the same or opposite constant direction, so Jc F . dr = f IF1 . ICI = f a . 4C. Gradient Fields and Exact Differentials 4C-1 a) F = V f = 3x2yi + (x3 + 3y2)j b) (i) Using y as parameter, C1 is: x = Y 2 , y = y; thus dx = 2y dy, and b) (ii) Using y as parameter, C2 is: x = 1, y = y; thus dx = 0, and b) (iii) By the Fundamental Theorem of Calculus for line integrals, Vf . dl- = f (B) - f (A). P 4C-2 a) F = V f = (xyexY + exY)i + (x2exy)j. b) (i) Using x as parameter, C is: x = x, y = l / x , so dy = -dx/x2, and so F . dr = lo+ (e e) dx + (x2e)(-dx/x2) b) (ii) Using the F.T.C. for line integrals, Ic 0 = (2ex - ex)] = -e. F .dr = f (1,l) - f (0, co) = 0 - e = -e. 4C-3 a) F = V f = (cosxcosy)i - (sinxsiny)j. / F . dr is path-independent, for any C connecting A : (xo, yo) to B : (xi, yi), Jc we have by the F.T.C. for line integrals, b) Since b F . dr = sin XI cos yl - sin xo cos yo This difference on the right-hand side is maximized if sinxl cos yl is maximized, and sin xo cos yo is minimized. Since I sin x cos yl = I sin XI I cos yl 5 1, the difference on the right hand side has a maximum of 2, attained when sinxl cos yl = 1 and sinxo cos yo = -1. (For example, a C running from ( - ~ / 2 , 0 ) to (7r/2,0) gives this maximum value.) 4. LINE INTEGRALS IN T H E P L A N E 3 4C-5 a) F is a gradient field only if My = N,, that is, if 2y = ay, so a = 2. + By inspection, the potential function is f (x,y) = xY2 x2 +c; YOU can check that F = V f . b) The equation My = Nx becomes ex+Y(x+a) = xex+~+ex+Y,which = ex+Y(x+1). Therefore a = 1. To find the potential function f (x,y), using Method 2 we have fz = eYex(x + 1) + f (x, y) = eYxex + g(y). Differentiating, and comparing the result with N , we find fy = eYxex + gl(y) = xeX+y;therefore gl(y) = 0, so g(y) = c and f (a,y) = x ex+Y + c. 4C-6 a) ydx - xdy is not exact, since My = 1but N , = -1. b) (2x + y) dx + x(2y + x) dy is exact, since My = 2x + 2y = N,. Using Method 1 to find the potential function f (x,y), we calculate the line integral over the standard broken line path shown, C = 9 C2. + On C1 we have y = 0 and dy = 0, so L F . dr = 0. On C2, we have x = xl and dx = 0, so J c , ~ . d=~l y 1 x 1 ( 2 y+ x l ) d x = x l y : + x ? y l . Therefore, f (x,y) = x2y + xy2; to get all possible functions, add +c . 4D. Green's Theorem 4D-1a) Evaluating the line integral first, we have C : x = cost, y = sin t , so si;2t)]r h2ydx+xdy= (-2sinZt+cos2t)dt= ( 1 - 3 s i n2t ) d t = t - 3 ( 12" - -s. For the double integral over the circular region R inside the C , we have JJ,(N. - M.) d~ = (1- 2) d~ = - area of R = -s. b) Evaluating the line integral, over the indicated path C = C1 d x 2 d x + x 2 d v = / 2 ~ z d x +/ ' 4 d y + + C2 + C3 + C4, C? R [Ox2dx+ /OOdy = 4, since the first and third integrals cancel, and the fourth is 0. For the double integral over the rectangle R , I I C, 1 2, E. 18.02 EXERCISES 4 c) Evaluating the line integral over C = C1 + C2, we have Therefore, i xy dx 7 -2 = -1. + y dy = 12 3 12 2 Evaluating the double integral over the interior R of C , we have -x dA = evaluating: Inner: -xy I = -x2+x3; li: -x dydx; Outer: :] Y=x y=xZ 4D-2 By Green's theorem, h 4x3y dx + x4 dy = 1 1 1 = - - + - = -- --+- JJ 3 4 12. (4x3 - 4x3) dA = 0. This is true for every closed curve C in the plane, since M and N have continuous derivatives for all x, y. 4D-3 We use the symmetric form for the integrand since the parametrization of the curve does not favor x or y; this leads to the easiest calculation. Using sin2 t cos2 t = (sin 2t)2 = 4D-4 By Green's theorem, . $ (1 - cos 4t), the above i -y3dx +x3dy = is always positive outside the origin. JS, 3 t sin4t =s(Z-a)o + (3x2 3y2)dA 2x 3Ir : 8 . > 0, since the integrand 4D-5 Let C be a square, and R its interior. Using Green's theorem, i xy2dx + (x2y + 22) dy = 4E. JS, (2xy + 2 - 2xy) dA = JS, 2 dA = 2(area of R) Two-dimensional Flux 4E-1 The vector F is the velocity vector for a rotating disc; it is at each point tangent to the circle centered at the origin and passing through that point. a) Since F is tangent to the circle, F . n = 0 a t every point on the circle, so the flux is 0. b) F = x j at the point (x,O) on the line. So if xo > 0, the flux a t xo has the same magnitude as the flux at -xo but the opposite sign, so the net flux over the line is 0. 4. LINE INTEGRALS IN T H E PLANE 4E-2 All the vectors of F have length segment C of length 1 will be f i and point 5 northeast. So the flux across a line a) maximal, if C points northwest; b) minimal, if C point southeast; c) zero, if C points northeast or southwest; d) -1, if C has the direction and magnitude of i or - j ; the corresponding normal vectors are then respectively - j and - i , by convention, so that F . n = ( i j ) . - j = -1. or ( i j ) . - i = -1. e) respectively f i and -fi, since the angle 9 between F and n is respectively 0 and 7r, so that respectively F . n = IF1 cos 9 = ffi. + + 4E-4 Taking the curve C = Cl + C2 + C3 + C4 as shown, ~ ~ d y - ~ d ~ = ~ , O + ~ ' - d ~ + ~ ~ d y + ~ O = - 2 . 4E-5 Since F and n both point radially outwards, F . n = IF1 = am, at every point of the circle C of radius a centered at the origin. a) The flux across C is am .27ra = 27ram+l. b) The flux will be independent of a if m = -1. 4F. Green's Theorem in Normal Form 4F-1 a) both are 0 b) div F = 22 4F-2 a) div F = (-wy), + 2y; + (wx), = 0; curl F = 0 c) div F = x + y; curl F = y - x curl F = (wx), - (-wy), = 2w. b) Since F is the velocity field of a fluid rotating with constant angular velocity (like a rigid disc), there are no sources or sinks: fluid is not being added to or subtracted from the flow at any point. c) A paddlewheel placed a t the origin will clearly spin with the same angular velocity w as the rotating fluid, so by (15), the curl should be 2w at the origin. It is less obvious that the curl is 2w at all other points as well. 4F-3 The line integral for flux is x dy - y dx; its value is 0 on any segment of the x-axis since y = dy = 0; on the upper half of the unit semicircle (oriented counterclockwise), F . n = 1, so the flux is the length of the semicircle: 7r. Letting R be the region inside C , 4F-4 For the flux integral S L d i v F d A = S L 2 d A = 2(7r/2) = 7r. x2dy - xy dx over C = C1 we get for the four sides respectively + C2 + C3 + C4, fh 1 - 1 - \, 1 E. 18.02 EXERCISES 6 For the double integral, 4~-5 = (x2 + y2)1/2 IS, +. div F d A = l L 3 ~ d A = J,lJ,l T, 3 l 3 3xdydx = -x2]o = 5. 2 x Y -. T T = $(x2 + Y2)-1/2 . 2 x = -; by symmetry, ry = To calculate div F, we have M = r n x and N = rny; therefore by the chain rule, and the above values for T, and T,, we have x M, = T n + n ~ n - l x. - = Tn + nTn-2 x ; similarly (or by symmetry), yr Ny = rn + n r n - l y . - = rn + nTn-2 y2 , T div F = Mx sothat + Ny = 2rn + nrnP2(x2 + Y2)= ~ ~+ n), ( 2which = 0 if n = -2. To calculate curl F , we have by the chain rule x N, = nTn-' . - . y; T Y My = nTn-l . - . x, T SO that curl F = N, - My = 0, for all n . 4G. Simply-connected Regions 4G-1 Hypotheses: the region R is simply connected, F = M i derivatives in R, and curl F = 0 in R. Conclusion: F is a gradient field in R (or, M dx +N j has continuous + N dy is an exact differential). a) curl F = 2y - 2y = 0, and R is the whole xy-plane. Therefore F = Vf in the plane. b) curl F = -y sin x - x sin y # 0, so the differential is not exact. c) curl F = 0, but R is the exterior of the unit circle, which is not simply-connected; criterion fails. d) curl F = 0, and R is the interior of the unit circle, which is simply-connected, so the differential is exact. e) curl F = 0 and R is the first quadrant, which is simply-connected, so F is a gradient field. c) Using Method 1, we take the origin as the starting point and use the straight line to (XI,yl) as the path C. In polar coordinates, xl = TI cosO1, yl = TI sin el; we use T as the parameter, so the path is C : x = T cos el, y = T sin el, 0 5 T 5 1-1. Then Therefore, xdx + y d y = d ( - d g ) . dF-7 Another approach: x dx+y dy = $d(r2); therefore xdx + ydy - -- d(r2) = d ( - d s ) . dC-7 2 d C - 7 (Think of r2 as a new variable u , and integrate.) 4. LINE INTEGRALS IN T H E PLANE 4G-3 By Example 3 in Notes V5, we know that hl,l) -I: (3>4) Therefore, = 4G-4 By Green's theorem 1 fi = A- F= xi +yj 7.3 = V(-:). 1 %. ~xydx+x2dy= For any plane region of density 1, we have & x d A = %.(area of R), where Z is the x-component of its center of mass. Since our region is symmetric with respect to the y-axis, its center of mass is on the y-axis, hence % = 0 and so J& x dA = 0. a) Yes b) no (a circle surrounding the line segment lies in R, but its interior does not) c) yes (no finite curve could surround the entire positive x-axis) d) no (the region does not consist of one connected piece) e) yes if 00 < 27r; no if 00 2 27r, since then R is the plane with (0,O) removed f) no (a circle between the two boundary circles lies in R, but its interior does not) g) Yes a) continuously differentiable for x, y > 0; thus R is the first quadrant without the two axes, which is simply-connected. b) continuous differentiable if r < 1; thus R is the interior of the unit circle, and is simply-connected. c) continuously differentiable if r > 1; thus R is the exterior of the unit circle, and is not simply-connected. d) continuously differentiable if r # 0; thus R is the plane with the origin removed, and is not simply-connected. e) continuously differentiable if T # 0; same as (d). 4H. Multiply-connected Regions 4H-2 In each case, the winding number about each of the points is given, then the value of the line integral of F around the curve.