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Unformatted text preview: SECTION 16.3: Green’s Theorem in the Plane 865 18. 3 Suppose div F = 0 in a domain D any point P of which can by joined to the origin by a straight line segment in D . Let r = tx i + ty j + tz k , (0 ≤ t ≤ 1), be a parametrization of the line segment from the origin to ( x , y , z ) in D . If G ( x , y , z ) = Z 1 t F ( r ( t )) × d r dt dt , show that curlG = F throughout D . Hint: It is enough to check the first components of curlG and F . Proceed in a manner similar to the proof of Theorem 4. H I 19. Use the Maple VectorCalculus package to verify the identities (a)–(f) of Theorem 3. Hint: For expressions of the form ( F • ∇ ) G you will have to use > F*diff(G,x)+F*diff(G,y) > +F*diff(G,z) because Del cannot be applied to a vector field except via a dot or cross product. 16.3 Green’s Theorem in the Plane The Fundamental Theorem of Calculus, Z b a d dx f ( x ) dx = f ( b ) − f ( a ), expressesthe integral, taken overthe interval [ a , b ], of the derivativeof a single-variable function, f , as a sum of values of that function at the oriented boundary of the interval [ a , b ], that is, at the two endpoints a and b , the former providinga negative contribution and the latter a positive one. The line integral of a conservative vector field over a curve C from A to B , Z C ∇ φ • d r = φ( B ) − φ( A ), has a similar interpretation; ∇ φ is a derivative, and the curve C , although lying in a two- or three-dimensional space, is intrinsically a one-dimensional object, and the points A and B constitute its boundary. Green’s Theorem is a two-dimensional version of the Fundamental Theorem of Calculus that expresses the double integral of a certain kind of derivative of a two- dimensional vector field F ( x , y ) , namely, the k-component of curlF , over a region R in the xy-plane as a line integral (i.e., a “sum”) of the tangential component of F around the curve C which is the oriented boundary of R : Z R curlF • k d A = I C F • d r , or, more explicitly, Z R ⎩ ∂ F 2 ∂ x − ∂ F 1 ∂ y dx dy = I C F 1 ( x , y ) dx + F 2 ( x , y ) dy . For this formula to hold, C must be the oriented boundary of R considered as a surface with orientation provided by ˆ N = k . Thus, C is oriented with R on the left as we move around C in the direction of its orientation. We will call such a curve positively oriented with respect to R . In particular, if C is a simple closed curve bounding R , then C is oriented counterclockwise. Of course, R may have holes, and the boundaries of the holes will be oriented clockwise. In any case, the unit tangent ˆ T and unit exterior (pointing out of R ) normal...
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- Spring '08
- Vector Calculus, Vector field, Stokes' theorem