4. Line Integrals in the Plane

Transcription

1 4. Line Integrals in the Plane 4A. Plane Vector Fields 4A- a) All vectors in the field are identical; continuously differentiable everywhere. b) The vector at P has its tail at 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 (,). d) Vector at P has unit length, and the clockwise direction perpendicular to OP. 4A- a) ai +bj b) xi +yj r 4A-3 a) i +j b) r(xi +yj) c) 4A-4 k 4B- yi +xj r a) On : y =, dy = ; therefore On : y = x, dy = xdx; c) f xi +yj (r) r yi xj r 3 d) f(x,y)(i + j) 4B. Line Integrals in the Plane = (x y)dx+xdy = (x y)dx+xdy = [ ( x )dx = 3 x3 +x ] ] x dx = x3 = 3 3. (x )dx 4x dx = 4 3 = 3. b) : use the parametrization x = cost, y = sint; then dx = sintdt, dy = costdt ] xydx x dy = sin tcostdt cos tcostdt = costdt = sint =. π/ c) = ; : x = dx = ; : y = x; 3 : y = dy = ydx xdy = + ( x)dx x( dx)+ = dx =. 3 π d) : x = cost, y = sint; dx = sintdt ydx = sin tdt = π. e) : x = t, y = t 3 ; dx = tdt, dy = 3t dt 6ydx+xdy = 6t 3 (tdt)+t (3t dt) = f) (x+y)dx+xydy = + (x+)dx = x +x π/ (5t 4 )dt = 3t 5 ] = 3 3. ] = 5. 4B- a) The field F points radially outward, the unit tangent t to the circle is always perpendicular to the radius; therefore F t = and F dr = F tds = 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 F dr = F tds = ads = πa. π/

2 E. 8. EXEISES 4B-3 a) maximum if is in the direction of the field: = i + j b) minimum if is in the opposite direction to the field: = i + j c) zero if is perpendicular to the field: = ± i j d) max =, min = : by (a) and (b), for the max or min F and have respectively the same or opposite constant direction, so F dr = ± F = ±. 4. Gradient Fields and Exact Differentials 4- a) F = f = 3x yi +(x 3 +3y )j b) (i) Using y as parameter, is: x = y, y = y; thus dx = ydy, and F dr = 3(y ) y ydy+[(y ) 3 +3y ]dy = (7y 6 +3y )dy = (y 7 +y 3 ) ] = 4. b) (ii) Using y as parameter, is: x =, y = y; thus dx =, and F dr = (+3y )dy = (y +y 3 ) ] = 4. b) (iii) By the Fundamental Theorem of alculus for line integrals, f dr = f(b) f(a). Here A = (, ) and B = (,), so that f dr = (+) ( ) = a) F = f = (xye xy +e xy )i +(x e xy )j. b) (i) Using x as parameter, is: x = x, y = /x, so dy = dx/x, and so F dr = (e+e)dx+(x e)( dx/x ) = (ex ex) ] = e. b) (ii) Using the F.T.. for line integrals, F dr = f(, ) f(,) = e = e. 4-3 a) F = f = (cosxcosy)i (sinxsiny)j. b) Since F dr is path-independent, for any connecting A : (x,y ) to B : (x,y ), we have by the F.T.. for line integrals, F dr = sinx cosy sinx cosy This difference on the right-hand side is maximized if sinx cosy is maximized, and sinx cosy is minimized. Since sinxcosy = sinx cosy, the difference on the right hand side has a maximum of, attained when sinx cosy = and sinx cosy =. (For example, a running from ( π/,) to (π/,) gives this maximum value.)

3 4. LINE INTEGALS IN THE PLANE a) F is a gradient field only if M y = N x, that is, if y = ay, so a =. By inspection, the potential function is f(x,y) = xy +x +c; you can check that F = f. b)theequationm y = N x becomese x+y (x+a) = xe x+y +e x+y, which= e x+y (x+). Therefore a =. To find the potential function f(x,y), using Method we have f x = e y e x (x+) f(x,y) = e y xe x +g(y). Differentiating, and comparing the result with N, we find f y = e y xe x +g (y) = xe x+y ; therefore g (y) =, so g(y) = c and f(x,y) = xe x+y +c. 4-6 a) ydx xdy is not exact, since M y = but N x =. b) y(x+y)dx+x(y +x)dy is exact, since M y = x+y = N x. Using Method to find the potential function f(x,y), we calculate the line integral over the standard broken line path shown, = +. (x,y ) f(x,y ) = F dr = y(x+y)dx+x(y +x)dy. (,) On we have y = and dy =, so F dr =. On, we have x = x and dx =, so F dr = y (x,y ) x x (y +x )dx = x y +x y. Therefore, f(x,y) = x y +xy ; to get all possible functions, add +c. 4D. Green s Theorem 4D- a) Evaluating the line integral first, we have : x = cost, y = sint, so π π ( t ydx+xdy = ( sin t+cos t)dt = ( 3sin t)dt = t 3 sint 4 For the double integral over the circular region inside the, we have (N x M y )da = ( )da = area of = π. b) Evaluating the line integral, over the indicated path = , x dx+x dy = x dx+ 4dy + x dx+ dy = 4, since the first and third integrals cancel, and the fourth is. For the double integral over the rectangle, ] xda = xdydx = x = 4. )] π 4 = π. 3

4 4 E. 8. EXEISES c) Evaluating the line integral over = +, we have : x = x, y = x ; xydx+y dy = x x dx+x 4 xdx = x4 4 + x6 3 : x = x, y = x; xydx+y dy = (x dx+x dx) = ] 3 x3 = 3. Therefore, xydx+y dy = 7 3 =. Evaluating the double integral over the interior of, we have x xda = xdydx; x ] y=x ] evaluating: Inner: xy = x +x 3 ; Outer: x3 y=x 3 + x4 = =. 4D- By Green s theorem, 4x 3 ydx+x 4 dy = (4x 3 4x 3 )da =. ] = 7 This is true for every closed curve 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. Area= ydx+xdy = π 3sin 4 tcos tdt+3sin tcos 4 tdt = 3 π sin tcos tdt Using sin tcos t = 4 (sint) = 4 ( cos4t), the above = 3 ( t 8 sin4t ) π = 3π D-4 By Green s theorem, is always positive outside the origin. y 3 dx+x 3 dy = (3x +3y )da >, since the integrand 4D-5 Let be a square, and its interior. Using Green s theorem, xy dx+(x y +x)dy = (xy + xy)da = da = (area of ). 4E. Two-dimensional Flux 4E- 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 = at every point on the circle, so the flux is. b) F = xj at the point (x,) on the line. So if x >, the flux at x has the same magnitude as the flux at x but the opposite sign, so the net flux over the line is. c) n = j, so F n = xj j = x. Thus F nds = xdx =.

5 4. LINE INTEGALS IN THE PLANE 5 4E- All the vectors of F have length and point northeast. So the flux across a line segment of length will be a) maximal, if points northwest; b) minimal, if point southeast; c) zero, if points northeast or southwest; d), if 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 =. or (i + j) i =. e) respectively and, since the angle θ between F and n is respectively and π, so that respectively F n = F cosθ = ±. 4E-3 M dy N dx = x dy xydx = (t+) tdt (t+)t dt = (t 3 +3t +t)dt = t4 4 +t3 +t ] = E-4 Taking the curve = as shown, xdy ydx = + dx+ dy + =. 4 4E-5 Since F and n both point radially outwards, F n = F = a m, at every point of the circle of radius a centered at the origin. a) The flux across is a m πa = πa m+. b) The flux will be independent of a if m =. 4F. Green s Theorem in Normal Form 4F- a) both are b) div F = x+y; curl F = c) div F = x+y; curl F = y x 4F- a) div F = ( ωy) x +(ωx) y = ; curl F = (ωx) x ( ωy) y = ω. 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 at the origin will clearly spin with the same angular velocity ω as the rotating fluid, so by Notes V4,(), the curl should be ω at the origin. (It is much less clear that the curl is ω at all other points as well.) 4F-3 The line integral for flux is xdy ydx; its value is on any segment of the x-axis since y = dy = ; on the upper half of the unit semicircle (oriented counterclockwise), F n =, so the flux is the length of the semicircle: π. - Letting be the region inside, div FdA = da = (π/) = π. 4F-4 For the flux integral we get for the four sides respectively x dy xydx over = , + dy+ xdx+ = (,) 3 (,)

6 6 E. 8. EXEISES For the double integral, div FdA = 3xdA = 3xdydx = 3 x ] = 3. 4F-5 r = (x +y ) / r x = (x +y ) / x = x r ; by symmetry, r y = y r. To calculate div F, we have M = r n x and N = r n y; therefore by the chain rule, and the above values for r x and r y, we have M x = r n +nr n x x r = rn +nr n x ; similarly (or by symmetry), N y = r n +nr n y y r = rn +nr n y, so that div F = M x +N y = r n +nr n (x +y ) = r n (+n), which = if n =. To calculate curl F, we have by the chain rule N x = nr n x r y; M y = nr n y r x, so that curl F = N x M y =, for all n. 4G. Simply-connected egions 4G- Hypotheses: the region is simply connected, F = M i + N j has continuous derivatives in, and curl F = in. onclusion: F is a gradient field in (or, M dx+n dy is an exact differential). a) curl F = y y =, and is the whole xy-plane. Therefore F = f in the plane. b) curl F = ysinx xsiny, so the differential is not exact. c) curl F =, but is the exterior of the unit circle, which is not simply-connected; criterion fails. d) curl F =, and is the interior of the unit circle, which is simply-connected, so the differential is exact. e) curl F = and is the first quadrant, which is simply-connected, so F is a gradient field. 4G- a) f(x,y) = xy +x b) f(x,y) = 3 x3/ + 3 y3/ c) Using Method, we take the origin as the starting point and use the straight line to (x,y ) as the path. In polar coordinates, x = r cosθ, y = r sinθ ; we use r as the parameter, so the path is : x = rcosθ, y = rsinθ, r r. Then f(x,y ) = Therefore, xdx+ydy r xdx+ydy r = = r r rcos θ +rsin θ r dr r r dr = r ] r = d( r ). = r +. Anotherapproach: xdx+ydy = d(r ); therefore xdx+ydy r (Think of r as a new variable u, and integrate.) = d(r ) = d( r ). r

7 4. LINE INTEGALS IN THE PLANE 7 4G-3 By Example 3 in Notes V5, we know that F = Therefore, (3,4) (,) = r 4G-4 By Green s theorem ] 5 = 5. xydx+x dy = xda. xi +yj r 3 ( = ). r For any plane region of density, we have xda = x (area of ), where x 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 x = and so xda =. 4G-5 a) yes b) no (a circle surrounding the line segment lies in, 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 θ < π; no if θ π, since then is the plane with (,) removed f) no (a circle between the two boundary circles lies in, but its interior does not) g) yes 4G-6 a) continuously differentiable for x,y > ; thus is the first quadrant without the two axes, which is simply-connected. b) continuous differentiable if r < ; thus is the interior of the unit circle, and is simply-connected. c) continuously differentiable if r > ; thus is the exterior of the unit circle, and is not simply-connected. d) continuously differentiable if r ; thus is the plane with the origin removed, and is not simply-connected. e) continuously differentiable if r ; same as (d). 4H. Multiply-connected egions 4H- a) ; b) ; 4π c) ; π d) ; 4π 4H- 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. a) (,,); + 3 b) (,,); + 3 c) (,,); d) (,,); + 3

8 8. Notes and Exercises by A. Mattuck with the assistance of T.Shifrin and S. LeDuc, and including a section on non-independent variables by Bjorn Poonen. c M.I.T. -4