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. Fundamental Theorem for Conservative Vector Fields: A line integral of the form R C rVds may be computed by evaluating V at the boundary points of the 1-dimensional parametrized curve domain C. 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 R b 0 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]. 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 R b 0 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 R C rVds may be computed by evaluating V at the boundary points of the 1-dimensional parametrized curve domain C. 2. Stokes' Theorem: A surface integral of the form RR ~ 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 RRR ~ 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. Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green's Theorem: A volume integral of the form ZZ δF2 δF1 − d(x; y) D δx δy may be computed by line integrating F~ = (F1; F2) along the boundary of the 2-dimensional domain D. 3. Divergence Theorem: A volume integral of the form RRR ~ 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. Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green's Theorem: A volume integral of the form ZZ δF2 δF1 − d(x; y) D δx δy may be computed by line integrating F~ = (F1; F2) along the boundary of the 2-dimensional domain D. 2. Stokes' Theorem: A surface integral of the form RR ~ 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. Fundamental Theorems of Vector Calculus, contd. Now here are the new ones. 1. Green's Theorem: A volume integral of the form ZZ δF2 δF1 − d(x; y) D δx δy may be computed by line integrating F~ = (F1; F2) along the boundary of the 2-dimensional domain D. 2. Stokes' Theorem: A surface integral of the form RR ~ 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 RRR ~ 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. Suppose C may be parametrized by a continuous one-to-one R2-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. Definitions and Terminology Definition Let D be a region in R2. 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. If δD is a simple closed curve, then we choose to orient δD in the counterclockwise direction. This is called the boundary orientation. Definitions and Terminology Definition Let D be a region in R2. 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 R2-valued function ~c with domain [a; b], where ~c(a) = ~c(b). Then we call C a simple closed curve. Definitions and Terminology Definition Let D be a region in R2. 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 R2-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. Green's Theorem Theorem (Green's Theorem) Let D be a domain whose boundary δD is a simple closed curve. 2 Let F = (F1; F2) be a vector field over R . Then I ZZ δF δF F~ · ds = 2 − 1 d(x; y). δD D δx δy 2 If (F1; F2) = (xy ; x), then δF2 δF1 2 δx − δy = 1 − 2xy , and hence we are being asked to show that H 2 RR C(xy ; x) · ds = R (1 − 2xy)d(x; y), where R is the interior of the unit circle. Example H 2 Verify Green's theorem for the line integral C(xy ; x) · ds about the unit circle C. Example H 2 Verify Green's theorem for the line integral C(xy ; x) · ds about the unit circle C. 2 If (F1; F2) = (xy ; x), then δF2 δF1 2 δx − δy = 1 − 2xy , and hence we are being asked to show that H 2 RR C(xy ; x) · ds = R (1 − 2xy)d(x; y), where R is the interior of the unit circle. We start with the line integral on the left. Parametrize C by ~c(t) = (cos t; sin t) for 0 ≤ t ≤ 2π. We have ~c 0(t) = (− sin t; cos t). Now compute: I Z 2π (xy 2; x) · ds = (cos t sin2 t; sin t) · (− sin t; cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0 1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π: Solution: The Line Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) Now compute: I Z 2π (xy 2; x) · ds = (cos t sin2 t; sin t) · (− sin t; cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0 1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π: Solution: The Line Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) We start with the line integral on the left. Parametrize C by ~c(t) = (cos t; sin t) for 0 ≤ t ≤ 2π. We have ~c 0(t) = (− sin t; cos t). Solution: The Line Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) We start with the line integral on the left. Parametrize C by ~c(t) = (cos t; sin t) for 0 ≤ t ≤ 2π. We have ~c 0(t) = (− sin t; cos t). Now compute: I Z 2π (xy 2; x) · ds = (cos t sin2 t; sin t) · (− sin t; cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0 1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π: Now we compute the integral on the right and hope we get π: p ZZ Z 1 Z 1−x2 (1 − 2xy)d(x; y) = p (1 − 2xy)dydx R −1 − 1−x2 1 p Z 2 = [y − xy 2] 1−px 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π: So Green's theorem is true in this case! Solution: The Volume Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) H 2 Known: C(xy ; x) · ds = π p ZZ Z 1 Z 1−x2 (1 − 2xy)d(x; y) = p (1 − 2xy)dydx R −1 − 1−x2 1 p Z 2 = [y − xy 2] 1−px 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π: So Green's theorem is true in this case! Solution: The Volume Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) H 2 Known: C(xy ; x) · ds = π Now we compute the integral on the right and hope we get π: Solution: The Volume Integral H 2 RR Goal: C(xy ; x) · ds = R (1 − 2xy)d(x; y) H 2 Known: C(xy ; x) · ds = π Now we compute the integral on the right and hope we get π: p ZZ Z 1 Z 1−x2 (1 − 2xy)d(x; y) = p (1 − 2xy)dydx R −1 − 1−x2 1 p Z 2 = [y − xy 2] 1−px 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π: So Green's theorem is true in this case! Solution.