MATH 55 Green s Theorem Fall 16 Here is a statement of Green s Theorem. It involves regions and their boundaries. In order have any hope of doing calculations, you must see the region as the set of points described bnequalities involving at least level curves and often by graphs. A smooth region is a region given by a finite collection of inequalities of differentiable functions. With a little algebra one can always assume that these inequalities are of the form f i x, y. The boundary of is the set of points where at least one of the inequalities is an equality. The precise generalitn which Green s Theorem holds is still an area of theoretical research but you have no hope of doing any calculations except for smooth bounded regions. Let be a smooth bounded region in the plane. Let M, N be a field defined on such that N x and M GT are continuous on. Then M dx + N dy = N x M da The left hand side of GT depends on an orientation of the curve; the right hand side does not. Hence for the result to be true, must be oriented correctly. The correct orientation is the usual one: if you stand at a point on the boundary with the region on your left, you are facing in the preferred direction. Hereafter, denotes the boundary curve with this orientation. Remark. It is important that the needed partials exist on all of. y For the field v = x + y, x, N x + y x M = on the punctured plane but not on all of the plane, even though the right hand side is a perfectly good function on all of R. If you forget this issue you may think π = v r = da = which is just wrong. Here is one way to remember the function in the double integral. Look at the determinant below. It makes no sense without further interpretation. Write det a b c d = ad bc, being careful with the order of the terms. Then N N is defined to be x x and M M is defined to be. Then we have so Green s Theorem says det x M M dx + N dy = N = N x M det x M da N This is not particularly useful for calculations but if you are suddenly not sure what goes under the double integral sign, may help you remember especially after we get to Stokes Theorem. 1
Example. Replace a line integral by a double integral: xy, x over the region above the parabola y = x and below y =. Note the yellow region can be described as y x and y so we have a smooth region which is clearly bounded and the boundars the part of y x = in union the part of y = in. Then xy dx + x dy = x x da = x da =. You should be able to explain why I know that the double integral is without doing any work. lick here for why. To check Green s Theorem, let us do two line integrals 1 xy dx + x dy and xy dx + x dy, where 1 is the line segment along the top and is the parabola. Then 1 is parametrized by r 1 t = t,, t and is parametrized by r t = t, t, t xy dx + x dy = t, t 1 1, dt = t dt = t = 9 9 = xy dx + x dy = t, t 1, t dt = t dt = t4 4 = 81 4 81 4 = Strictly speaking, xy dx + x dy = xy dx + x dy + xy dx + x dy = xy dx + x dy + xy dx + x dy 1 1 but since both line integrals are you can t really tell in this example. Another way to apply Green s Theorem is to do a line integral along a non-closed path, say. The trick is to find another path 1 such that 1 = for some region and then Q P, Q dr = x P da P, Q dr and sometimes the two integrals on the right are easier than the original.
Example. Integrate x 5 + y, x y along the quarter circle of radius centered at the origin and lying in the first quadrant, beginning at,. 1 Here is our quarter circle, 1 goes from the origin to, and goes from the origin to,. Let be the quarter disk so = 1. You can set up x 5 + y, x y π/ dr = cos 5 t + sint, 4 cost 8 sin t sint, cost dt which looks messy. Now x y x5 + y x = 1 = 1, so x y x 5 + y, x y dr = t 5 +, t 1, dt = 1 x 5 + y, x y dr = + t, t, 1 dt = Hence x 5 + y, x y dr = π 44 + 4 = π x x5 + y da = π. t 5 dt = 64 6 = t dt = 16 4 = 4
4 1. Area The other obvious direction is to replace a double integral with a line integral. Given F x, y da if you can find a field M, N such that N x M = F x, y for every point in then you can do M, N dr instead. Example. Find sinx da where is the region below y = sin x and above the x axis between x = and x = π. Take v = y sinx,. sinx da = y sinx, 1, cos x dx + y sinx, 1, dx where 1 is the 1 curve x, sinx, x π and is the line segment x,, x π. π π sinx da = sin t dt + dt = t cost π = π 1 1 = π The granddaddy of this sort of application is to find area and less often centroids. Recall that the centroid of a region is the center of mass of that region assuming uniform density. If v = 1 y, x or y, or, x, 1 y dx + x dy = y dx = x dy is the area of. Note 1, x dr is the moment about the y axis and 1 y, dr is the moment about the x axis. Other choices for the field will occur to you.
Example. Find the area of the polygon, x, y, x 1, y 1,..., x n, y n, where we list the points in order starting somewhere and traversing the polygon counterclockwise. Assume that the points are distinct and that the curve formed by connecting the points with line segments in the order given and then returning from the last listed point to the starting point is a polygon. Equivalently, none of the line segments intersect except possible in one point which is an end point for each segment. all the area A. Let i be the line segment from 1, 1 to,, 1 i n and let n+1 be the line segment from x n, y n to x, y needed to closeup the polygon. The notation works out better if we define x n+1, y n+1 = x, y and then n+1 has the same definition as all the others. By Green s Theorem A = 1 y, x dr i Next let us work out the line integral for an arbitrary line segment. 1 = 1 t a, b + t c, d = a at + ct, b bt + dt y, x dr = 1 1 bt dt b, a + ct at a + c, b + d dt = b dc at bc a + c ad bt + ad b dt = b dc a + c ad b t bc a + ad b dt = 5 1 b dc a + c ad b t + bc + ad dt = 1 a b bc + ad dt = bc + ad = det c d Hence, if x, y, x 1, y 1,..., x n, y n, x n+1, y n+1 = x, y is a polygon in the plane traversed counterclockwise, n+1 Area = det 1 1 Hence d b 1 1 1 1, x dr = 1 1 a + c at d b dt = d b a + ac at + c a t dt = d b ac + c a, a + c at c a, d b dt = 1 d b a + c at dt = a + ac a + = d ba + ac + c 6 c a = 6 M y = 1 x i + 1 +x i 1 = x i + 1 +x i 1 x i 1 1 1 x i 1 1 The monomial x i will be canceled by the same monomial with a minus sign in the next term so 6 M y = 1 x i + 1 + x i 1 = 1 + x i 1 x i 1 1 1
6 Now check n+1 6 M y = + 1 det 1 You may use any of a number of arguments to see 1 so x = 1 6 M x = + 1 det 1 + 1 det 1 det 1 1 1 ȳ = 1 1 + 1 det 1 det 1 These formulas can be found in old books on surveying and new books on computer graphics. There is an instrument called a planimeter which is used to compute the area enclosed by a plane curve. The book and the web site have additional material on planimeters showing that they are just mechanical manifestations of Green s method for computing area. There are also some apps calling themselves planimeters which measure area of a region on a google map after you put some pins around the region. I have not reverse engineered the software but I strongly suspect that these apps just compute the area of the polygon using the formula above. 1 1. Further applications Going from line integrals around a boundary to double integrals is straightforward, just compute N x M. Going from a double integral to a line integral is more interesting since you need to find a field such that N x M is your function and there are always lots of choices. There are some other sorts of applications which are more subtle. If M, N and M 1, N 1 have continuous partials on and if N 1 N M 1 M =, then M, N dr = x N M 1, N 1 dr since they are both equal to x M da. Example. Integrate y + e sinx dx + xy + e cosy dy counterclockwise around the unit circle. Note + esinx = xy + ecosy and = xy x x so y+e sinx dx+xy+e cosy dy = y dx+xy dy = π cos t sint sin tdt = cos t π t sint sint sint +cost sint cost dt = π = 1 π 1 = π If you try to do the line integral directly you will get nowhere. In this case you could also do the double integral y 1 da = Moment about x-axis Area.
. hange of curve Given a region with a conservative field P, Q defined on minus some points, line integrals along complicated curves can be replaced by line integrals along simpler curves. y Example. Below is a graph of sin xx 4 + y 4 = 16. Let F = P, Q = x + y, x and x + y compute F dr around this curve. We already checked that Q x P =. 7 We can not applied Green s Theorem directly because F is defined only on R {, } which is not the entire interior of the curve. However is we let be the light blue region, then F is defined on. In this case is the outer curve sin xx 4 +y 4 = 16 and the circle x +y = 1. The outer curve 8 is oriented counterclockwise and the inner circle is oriented clockwise. If 1 is sin xx 4 + y 4 = 16 oriented counterclockwise and if is the circle x + y = 1 oriented counterclockwise, then 8 F dr = F dr 1 F dr
8 We checked F dr = π so 1 F dr = π even though we have no hope of doing the integral directly. id you see the fast one in that last bit. We actually checked F dr = π where is the unit circle, not the one of radius 1. But another application of the above principal shows that 8 F dr = π for every circle centered at the origin. 4. The normal form of Green s Theorem Another way to write Green s Theorem is P, Q T ds = When discussing line integrals, we found that Q x P da P, Q N ds has important applications to fluid flow. Fluid flow includes such unlikeltems as Maxwell s Equations in addition to more conventional possibilities. Hence it is important to realize that there is a Green s Theorem for this integral too: P, Q N ds = P x + Q da The proof is easy. By formula in the line integral handout P, Q N ds = Qdx + P P dy = x Q P da = x + Q da. If P, Q is incompressible then P, Q N ds = for any closed curve. It follows that P, Q is incompressible if and onlf P x + Q =. This follows since if P Q a, b+ a, b > at some point a, b, then it is positive in some disk x P centered at the point and hence x + Q da > which contradicts P, Q N ds = plus Green s Theorem. It follows that P x Q P a, b, so x Q a, b + a, b =. Q P a, b+ a, b. Similarlt follows that a, b+ x Suppose P, Q is both conservative and incompressible. Let p be a potential function for it in some region. Then p is an harmonic function on, which just means that p satisfies Laplace s equation p x + p y = Harmonic functions are extremelnteresting, useful and well-studied. The book on page 9 shows e y sinx is harmonic. So are e y cosx, e x siny and e x cosy. Linear functions are harmonic.
Any linear combination of harmonic functions is harmonic. x y and xy are harmonic. So is x xy. 9
1 5. Why was our double integral zero? Whs x da = if is the yellow region below? Here is the answer and, more importantly, what you need to write if you are doing this for partial credit. You don t need to write your explanation in italics. The integral x da is the moment about the y axis of the region. In this case, is symmetric about the y axis so the moment about the y-axis is. In general, if L is any line cutting through a region such that is symmetric about L then the centroid of lies on that line. If you know the y coordinate of the centroid, ȳ, you know x da = Area ȳ.