Motivation: Fundamental Theorems of Vector Calculus

Transcription

1 Motivation: Fundamental Theorems of Vector Calculus Our goal as we close out the semester is to give several Fundamental Theorem of Calculus -type theorems which relate volume integrals of derivatives on a given domain to line and surface integrals about the boundary of the domain. The general form of these theorems, which we collectively call the fundamental theorems of vector calculus, is the following: The integral of a derivative-type object on a given domain D may be computed using only the function values along the boundary of D.

2 Motivation: Fundamental Theorems of Vector Calculus As an overview, we will roughly and informally summarize the content of the fundamental theorems of vector calculus. First let s start with the ones we have already seen: 1. FTC: An area integral of the form b a f (x)dx may be computed by evaluating f at the boundary points a and b of the 1-dimensional domain interval [a, b].

3 Motivation: Fundamental Theorems of Vector Calculus As an overview, we will roughly and informally summarize the content of the fundamental theorems of vector calculus. First let s start with the ones we have already seen: 1. FTC: An area integral of the form b a f (x)dx may be computed by evaluating f at the boundary points a and b of the 1-dimensional domain interval [a, b]. 2. Fundamental Theorem for Conservative Vector Fields: A line integral of the form C Vds may be computed by evaluating V at the boundary points of the 1-dimensional parametrized curve domain C.

4 Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green s Theorem: A volume integral of the form ( ) δf2 D δx δf1 d(x, y) δy may be computed by line integrating F = (F 1, F 2) along the boundary of the 2-dimensional domain D.

5 Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green s Theorem: A volume integral of the form ( ) δf2 D δx δf1 d(x, y) δy may be computed by line integrating F = (F 1, F 2) along the boundary of the 2-dimensional domain D. 2. Stokes Theorem: A surface integral of the form S curl( F )ds may be computed by line integrating F along the boundary of the 2-dimensional domain S. Note: The curl of F is a derivative-type object we will define later.

6 Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green s Theorem: A volume integral of the form ( ) δf2 D δx δf1 d(x, y) δy may be computed by line integrating F = (F 1, F 2) along the boundary of the 2-dimensional domain D. 2. Stokes Theorem: A surface integral of the form S curl( F )ds may be computed by line integrating F along the boundary of the 2-dimensional domain S. Note: The curl of F is a derivative-type object we will define later. 3. Divergence Theorem: A volume integral of the form W div( F )d(x, y, z) may be computed by surface integrating F along the boundary of the 3-dimensional domain W. Note: The divergence of F is another soon-to-be-introduced derivative-type object.

7 Definitions and Terminology Definition Let D be a region in R 2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δd. We call δd the boundary of D.

8 Definitions and Terminology Definition Let D be a region in R 2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δd. We call δd the boundary of D. Suppose C may be parametrized by a continuous one-to-one R 2 -valued function c with domain [a, b], where c(a) = c(b). Then we call C a simple closed curve.

9 Definitions and Terminology Definition Let D be a region in R 2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δd. We call δd the boundary of D. Suppose C may be parametrized by a continuous one-to-one R 2 -valued function c with domain [a, b], where c(a) = c(b). Then we call C a simple closed curve. If δd is a simple closed curve, then we choose to orient δd in the counterclockwise direction. This is called the boundary orientation.

10 Green s Theorem Theorem (Green s Theorem) Let D be a domain whose boundary δd is a simple closed curve. Let F = (F 1, F 2 ) be a vector field over R 2. Then δd F ds = D ( δf2 δx δf ) 1 d(x, y). δy

11 Example Verify Green s theorem for the line integral C (xy 2, x) ds about the unit circle C.

12 Example Verify Green s theorem for the line integral C (xy 2, x) ds about the unit circle C. If (F 1, F 2 ) = (xy 2, x), then δf 2 δx δf 1 δy = 1 2xy 2, and hence we are being asked to show that C(xy 2, x) ds = R (1 2xy)d(x, y), where R is the interior of the unit circle.

13 Solution: The Line Integral Goal: (xy 2, x) ds = (1 2xy)d(x, y) C R

14 Solution: The Line Integral Goal: (xy 2, x) ds = (1 2xy)d(x, y) C R We start with the line integral on the left. Parametrize C by We have c (t) = ( sin t, cos t). c(t) = (cos t, sin t) for 0 t 2π.

15 Solution: The Line Integral Goal: (xy 2, x) ds = (1 2xy)d(x, y) C R We start with the line integral on the left. Parametrize C by We have c (t) = ( sin t, cos t). Now compute: C (xy 2, x) ds = c(t) = (cos t, sin t) for 0 t 2π. = 2π 0 2π 0 (cos t sin 2 t, sin t) ( sin t, cos t)dt ( cos t sin 3 t + cos 2 t)dt = [ 14 sin4 t + 12 t + 14 ] 2π sin(2t) = (0 + π + 0) ( ) = π. 0

16 Solution: The Volume Integral Goal: C (xy 2, x) ds = R (1 2xy)d(x, y) Known: C (xy 2, x) ds = π

17 Solution: The Volume Integral Goal: (xy 2, x) ds = (1 2xy)d(x, y) C R Known: (xy 2, x) ds = π C Now we compute the integral on the right and hope we get π:

18 Solution: The Volume Integral Goal: (xy 2, x) ds = (1 2xy)d(x, y) C R Known: (xy 2, x) ds = π C Now we compute the integral on the right and hope we get π: R (1 2xy)d(x, y) = So Green s theorem is true in this case! = = x 2 (1 2xy)dydx 1 x 2 [y xy 2 1 x ] 2 y= 2 1 x 2 dx = [2 arcsin x] 1 1 = π. 1 x 2

19 Example Compute the circulation of F = (sin x, x 2 y 3 ) about the path C = δd, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2).

20 Example Compute the circulation of F = (sin x, x 2 y 3 ) about the path C = δd, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green s theorem instead.

21 Example Compute the circulation of F = (sin x, x 2 y 3 ) about the path C = δd, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green s theorem instead. The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that δf 2 δx = 2xy 3 and δf 1 δy = 0.

22 Example Compute the circulation of F = (sin x, x 2 y 3 ) about the path C = δd, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green s theorem instead. The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that δf 2 δx = 2xy 3 and δf 1 δy = 0. So by Green s theorem we get: δd (sin x, x 2 y 3 ) ds = (2xy 3 0)d(x, y) D 2 x = 2xy 3 dydx = [ xy 4 ] x y=0 dx = 1 2 x 5 dx 2 0 = = 16 3.

23 Corollary (Area Formula) Let D be a region in R 2 and assume C = δd is a simple closed curve. Then the area of D is equal to 1 2 ( y, x) ds. C

24 Corollary (Area Formula) Let D be a region in R 2 and assume C = δd is a simple closed curve. Then the area of D is equal to 1 2 ( y, x) ds. C Proof. Set F = (F 1, F 2 ) = 1 δf2 2 ( y, x) and check that δx δf1 δy = 1 2 ( 1 2 ) = 1. Now apply Green s theorem: Area of D = 1d(x, y) D = ( δf 2 D δx δf 1 )d(x, y) δy = F ds C = 1 ( y, x) ds. 2 C

25 Some Quick Terminology We wish to give some indications toward the proof of Green s Theorem. For this we need a little more notation and a lemma.

26 Some Quick Terminology We wish to give some indications toward the proof of Green s Theorem. For this we need a little more notation and a lemma. Definition Let F = (F 1, F 2 ) be a vector field over R 2 and let c(t) = (x(t), y(t)) be a smooth parametrization of a curve in R 2 with domain [a, b]. Define the x- and y-components of the vector field line integral of F over C to be, respectively, C F 1dx = b a F 1( c(t))x (t)dt C F 2dy = b a F 2( c(t))y (t)dt.

27 Some Quick Terminology We wish to give some indications toward the proof of Green s Theorem. For this we need a little more notation and a lemma. Definition Let F = (F 1, F 2 ) be a vector field over R 2 and let c(t) = (x(t), y(t)) be a smooth parametrization of a curve in R 2 with domain [a, b]. Define the x- and y-components of the vector field line integral of F over C to be, respectively, C F 1dx = b a F 1( c(t))x (t)dt C F 2dy = b a F 2( c(t))y (t)dt. It follows from the definitions that b F ds = (F 1 ( c(t)), F 2 ( c(t))) (x (t), y (t))dt C a = F 1 dx + F 2 dy. C C

28 Some Quick Terminology We wish to give some indications toward the proof of Green s Theorem. For this we need a little more notation and a lemma. Definition Let F = (F 1, F 2 ) be a vector field over R 2 and let c(t) = (x(t), y(t)) be a smooth parametrization of a curve in R 2 with domain [a, b]. Define the x- and y-components of the vector field line integral of F over C to be, respectively, C F 1dx = b a F 1( c(t))x (t)dt C F 2dy = b a F 2( c(t))y (t)dt. It follows from the definitions that b F ds = (F 1 ( c(t)), F 2 ( c(t))) (x (t), y (t))dt C a = F 1 dx + F 2 dy. C C Lemma The values of the x- and y-components of a vector field line integral are independent of the choice of parametrization c.

29 Proof of a Special Case of Green s Theorem Goal: Show δd F ds = D ( δf2 δx δf ) 1 d(x, y). δy We are unable to give a proof of Green s theorem in its full generality, but we can prove it if we make the following very special simplifying assumptions: the boundary δd may be described as the union of two graphs of the form y = T (x) with B(x) T (x) for a x b; and δd may also be described as the union of two graphs of the form x = L(y) and x = R(y) with L(y) R(y) for c y d. (Picture on whiteboard!)

30 Proof of a Special Case, contd. Goal: Show δd F ds = D ( δf2 δx δf ) 1 d(x, y). δy In order to observe Green s theorem, we will break it up into two parts. Since F ds = δd δd F 1dx + δd F 2dy, it suffices for us to show the following two equalities: δf 1 d(x, y) δd F 1dx = D δd F 2dy = D δf 2 δx δy d(x, y).

31 Proof of a Special Case, contd. Goal: Show δd F ds = D ( δf2 δx δf ) 1 d(x, y). δy In order to observe Green s theorem, we will break it up into two parts. Since F ds = δd δd F 1dx + δd F 2dy, it suffices for us to show the following two equalities: δf 1 d(x, y) δd F 1dx = D δd F 2dy = D δf 2 δx δy d(x, y). We begin by parametrizing the bottom C 1 and top C 2 of δd. C 1 : c 1 (t) = (x 1 (t), y 1 (t)) = (t, B(t)) C 2 : c 2 (t) = (x 2 (t), y 2 (t)) = (t, T (t)) Note c 1 traverses C 1 along the boundary orientation but c 2 goes clockwise, i.e. c 2 parametrizes C 2.

32 Proof of a Special Case, contd. Mid-Goal: Show that δd F1dx = D δf 1 d(x, y) δy

33 Proof of a Special Case, contd. Mid-Goal: Show that δd F1dx = D Now we compute D D δf 1 d(x, y) = δy δf 1 d(x, y). δy = b T (x) a b a b B(x) δf 1 d(x, y) δy δf 1 δy dydx [F 1(x, T (x)) F 1(x, B(x))]dx = F 1(c 2(t))x 2(t)dt a = F 1dx F 1dx C 2 C 1 = F 1dx. This completes the proof of the mid-goal. δd b a F 1(c 1(t))x 1(t)dt

34 Proof of a Special Case, contd. Mid-Goal: Show that δd F2dy = D δf 2 d(x, y) δx

35 Proof of a Special Case, contd. Mid-Goal: Show that δd F2dy = D δf 2 d(x, y) δx Now let C 3 and C 4 denote the left and right boundaries, and parametrize again: C 3: c 3(t) = (x 3(t), y 3(t)) = (L(t), t) C 4: c 4(t) = (x 4(t), y 4(t)) = (R(t), t). This time c 3 parametrizes C 3 but c 4 parametrizes C 4.

36 Proof of a Special Case, contd. Mid-Goal: Show that δd F2dy = D δf 2 d(x, y) δx Now let C 3 and C 4 denote the left and right boundaries, and parametrize again: C 3: c 3(t) = (x 3(t), y 3(t)) = (L(t), t) C 4: c 4(t) = (x 4(t), y 4(t)) = (R(t), t). This time c 3 parametrizes C 3 but c 4 parametrizes C 4. D δf d 2 d(x, y) = δx = c d c d R(y) L(y) δf 2 δx dxdy F 2(R(y), y)dy d = F 2( c 4(t))y 4(t)dt c = F 2dy F 2dy C 4 C 3 = F 2dy. δd c d c F 2(L(y), y)dy F 2( c 3(t))y 3(t)dt

37 Proof of a Special Case, contd. We conclude by observing that, as promised: δd F ds = F 1 dx + F 2 dy δd δd δf 1 δf 2 = d(x, y) + d(x, y) D δy D δx ( δf2 = δx δf ) 1 d(x, y). δy D