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**Unformatted text preview: **18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS 18.1 Green’s Theorem (ET Section 17.1) Preliminary Questions 1. Which vector field F is being integrated in the line integral I x 2 dy − e y dx ? SOLUTION The line integral can be rewritten as H − e y dx + x 2 dy . This is the line integral of F = D − e y , x 2 E along the curve. 2. Draw a domain in the shape of an ellipse and indicate with an arrow the boundary orientation of the boundary curve. Do the same for the annulus (the region between two concentric circles). SOLUTION The orientation on C is counterclockwise, meaning that the region enclosed by C lies to the left in traversing C . C For the annulus, the inner boundary is oriented clockwise and the outer boundary is oriented counterclockwise. The region between the circles lies to the left while traversing each circle. 3. The circulation of a gradient vector field around a closed curve is zero. Is this fact consistent with Green’s Theorem? Explain. SOLUTION Green’s Theorem asserts that Z C F · d s = Z C P dx + Q dy = ZZ D µ ∂ Q ∂ x − ∂ P ∂ y ¶ d A (1) If F is a gradient vector field, the cross partials are equal, that is, ∂ P ∂ y = ∂ Q ∂ x ⇒ ∂ Q ∂ x − ∂ P ∂ y = ( 2 ) Combining (1) and (2) we obtain R C F · d s = 0. That is, Green’s Theorem implies that the integral of a gradient vector field around a simple closed curve is zero. 4. Which of the following vector fields possess the following property: For every simple closed curve C , Z C F · d s is equal to the area enclosed by C ? (a) F = h− y , i (b) F = h x , y i (c) F = sin ( x 2 ), x + e y 2 ® SOLUTION By Green’s Theorem, Z C F · d s = ZZ D µ ∂ Q ∂ x − ∂ P ∂ y ¶ dx dy (1) S E C T I O N 18.1 Green’s Theorem (ET Section 17.1) 637 D C We compute the curl of each one of the given fields. (a) Here, P = − y and Q = 0, hence ∂ Q ∂ x − ∂ P ∂ y = − ( − 1 ) = 1. Therefore, by (1), Z C F · d s = ZZ D 1 dx dy = Area ( D ) (b) We have P = x and Q = y , therefore ∂ Q ∂ x − ∂ P ∂ y = − = 0. By (1) we get Z C F · d s = ZZ D dx dy = 6= Area ( D ) (c) In this vector field we have P = sin ( x 2 ) and Q = x + e y 2 . Therefore, ∂ Q ∂ x − ∂ P ∂ y = 1 − = 1 . By (1) we obtain Z C F · d s = ZZ D 1 dx dy = Area ( D ). Exercises 1. Verify Green’s Theorem for the line integral I C xy dx + y dy , where C is the unit circle, oriented counterclockwise. SOLUTION Step 1. Evaluate the line integral. We use the parametrization γ ( θ ) = h cos θ , sin θ i , 0 ≤ θ ≤ 2 π of the unit circle. Then dx = − sin θ d θ , dy = cos θ d θ and xy dx + y dy = cos θ sin θ ( − sin θ d θ ) + sin θ cos θ d θ = ³ − cos θ sin 2 θ + sin θ cos θ d θ The line integral is thus Z C xy dx + y dy = Z 2 π ³ − cos θ sin 2 θ + sin θ cos θ d θ = Z 2 π − cos θ sin 2 θ d θ + Z 2 π sin θ cos θ d θ = − sin 3 θ 3 ¯ ¯ ¯ ¯ 2 π − cos 2 θ 4 ¯ ¯ ¯ ¯ 2 π = ( 1 ) x y C D Step 2. Evaluate the double integral. SinceEvaluate the double integral....

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