17.2 Prel 1-4 Ex 1-18

# 17.2 Prel 1-4 Ex 1-18 - 1064 C H A P T E R 17 L I N E A N D...

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1064 CHAPTER 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) x y D P = ( a , b ) R ( c b ) Q ( c d ) Since ϕ x ( x , y ) = 0in D , in particular x ( x , b ) = 0for a x c . Therefore, for a x c we have ( x , b ) = Z x a u ( u , b ) du + ( a , b ) = Z x a 0 + ( a , b ) = k + ( a , b ) Substituting x = a determines k = 0. Hence, ( x , b ) = ( a , b ) for a x c In particular, ( c , b ) = ( a , b ) ( R ) = ( P ) (1) Similarly, since y ( x , y ) = D ,wehave y ( c , y ) = b y d . Therefore for b y d we have ( c , y ) = Z y b ∂v ( c ,v) d v + ( c , b ) = Z y b 0 d v + ( c , b ) = k + ( c , b ) Substituting y = b gives ( c , b ) = k + ( c , b ) or k = 0. Therefore, ( c , y ) = ( c , b ) for b y d In particular, ( c , d ) = ( c , b ) ( Q ) = ( R ) (2) Combining (1) and (2) we obtain the desired equality ( P ) = ( Q ) .S ince P and Q are any two points in D ,we conclude that is constant on D . 17.2 Line Integrals (ET Section 16.2) Preliminary Questions 1. What is the line integral of the constant function f ( x , y , z ) = 10 over a curve C of length 5? SOLUTION Since the length of C is the line integral R C 1 ds = 5, we have Z C 10 = 10 Z C 1 = 10 · 5 = 50 2. Which of the following have a zero line integral over the vertical segment from ( 0 , 0 ) to ( 0 , 1 ) ? (a) f ( x , y ) = x (b) f ( x , y ) = y (c) F = h x , 0 i (d) F = h y , 0 i (e) F = h 0 , x i (f) F = h 0 , y i The vertical segment from ( 0 , 0 ) to ( 0 , 1 ) has the parametrization c ( t ) = ( 0 , t ), 0 t 1 Therefore, c 0 ( t ) = h 0 , 1 i and k c 0 ( t ) k= 1. The line integrals are thus computed by Z C f ( x , y ) = Z 1 0 f ( c ( t )) k c 0 ( t ) k dt (1) Z C F · = Z 1 0 F ( c ( t )) · c 0 ( t ) (2) (a) We have f ( c ( t )) = x = 0. Therefore by (1) the line integral is zero.

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SECTION 17.2 Line Integrals (ET Section 16.2) 1065 (b) By (1), the line integral is Z C f ( x , y ) ds = Z 1 0 t · 1 dt = 1 2 t 2 ¯ ¯ ¯ ¯ 1 0 = 1 2 6= 0 (c) This vector line integral is computed using (2). Since F ( c ( t )) = h x , 0 i = h 0 , 0 i , the vector line integral is zero. (d) By (2) we have Z C F · d s = Z 1 0 h t , 0 i · h 0 , 1 i = Z 1 0 0 = 0 (e) The vector integral is computed using (2). Since F ( c ( t )) = h 0 , x i = h 0 , 0 i , the line integral is zero. (f) For this vector feld we have Z C F · d s = Z 1 0 F ( c ( t )) · c 0 ( t ) = Z 1 0 h 0 , t i · h 0 , 1 i = Z 1 0 tdt = t 2 2 ¯ ¯ ¯ ¯ 1 0 = 1 2 0 So, we conclude that (a), (c), (d), and (e) have an integral o± zero. 3. State whether true or ±alse. I± ±alse, give the correct statement. (a) The scalar line integral does not depend on how you parametrize the curve. (b) I± you reverse the orientation o± the curve, neither the vector nor the scalar line integral changes sign. SOLUTION (a) True: It can be shown that any two parametrizations o± the curve yield the same value ±or the scalar line integral, hence the statement is true. (b) False: For the defnition o± the scalar line integral, there is no need to speci±y a direction along the path, hence reversing the orientation o± the curve does not change the sign o± the integral. However, reversing the orientation o± the curve changes the sign o± the vector line integral. 4. Let C be a curve o± length 5. What is the value o± Z C F · d s (a) F ( P ) is normal to C at all points P on C ?
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## This homework help was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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17.2 Prel 1-4 Ex 1-18 - 1064 C H A P T E R 17 L I N E A N D...

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