{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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...

This preview shows pages 1–3. Sign up to view the full content.

1064 C H A P T E R 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 ) = 0 in D , in particular ϕ x ( x , b ) = 0 for a x c . Therefore, for a x c we have ϕ ( x , b ) = x a ϕ u ( u , b ) du + ϕ ( a , b ) = x a 0 du + ϕ ( a , b ) = k + ϕ ( a , b ) Substituting x = a determines k = 0. Hence, ϕ ( x , b ) = ϕ ( a , b ) for a x c In particular, ϕ ( c , b ) = ϕ ( a , b ) ϕ ( R ) = ϕ ( P ) (1) Similarly, since ϕ y ( x , y ) = 0 in D , we have ϕ y ( c , y ) = 0 for b y d . Therefore for b y d we have ϕ ( c , y ) = y b ϕ ∂v ( c , v) d v + ϕ ( c , b ) = y b 0 d v + ϕ ( c , b ) = k + ϕ ( c , b ) Substituting y = b gives ϕ ( c , b ) = k + ϕ ( c , b ) or k = 0. Therefore, ϕ ( c , y ) = ϕ ( c , b ) for b y d In particular, ϕ ( c , d ) = ϕ ( c , b ) ϕ ( Q ) = ϕ ( R ) (2) Combining (1) and (2) we obtain the desired equality ϕ ( P ) = ϕ ( Q ) . Since P and Q are any two points in D , we conclude that ϕ 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 C 1 ds = 5, we have C 10 ds = 10 C 1 ds = 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 = x , 0 (d) F = y , 0 (e) F = 0 , x (f) F = 0 , y SOLUTION The vertical segment from ( 0 , 0 ) to ( 0 , 1 ) has the parametrization c ( t ) = ( 0 , t ), 0 t 1 Therefore, c ( t ) = 0 , 1 and c ( t ) = 1. The line integrals are thus computed by C f ( x , y ) ds = 1 0 f ( c ( t )) c ( t ) dt (1) C F · ds = 1 0 F ( c ( t )) · c ( t ) dt (2) (a) We have f ( c ( t )) = x = 0. Therefore by (1) the line integral is zero.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
S E C T I O N 17.2 Line Integrals (ET Section 16.2) 1065 (b) By (1), the line integral is C f ( x , y ) ds = 1 0 t · 1 dt = 1 2 t 2 1 0 = 1 2 = 0 (c) This vector line integral is computed using (2). Since F ( c ( t )) = x , 0 = 0 , 0 , the vector line integral is zero. (d) By (2) we have C F · d s = 1 0 t , 0 · 0 , 1 dt = 1 0 0 dt = 0 (e) The vector integral is computed using (2). Since F ( c ( t )) = 0 , x = 0 , 0 , the line integral is zero. (f) For this vector field we have C F · d s = 1 0 F ( c ( t )) · c ( t ) dt = 1 0 0 , t · 0 , 1 dt = 1 0 t dt = t 2 2 1 0 = 1 2 = 0 So, we conclude that (a), (c), (d), and (e) have an integral of zero. 3. State whether true or false. If false, give the correct statement. (a) The scalar line integral does not depend on how you parametrize the curve. (b) If you reverse the orientation of the curve, neither the vector nor the scalar line integral changes sign. SOLUTION (a) True: It can be shown that any two parametrizations of the curve yield the same value for the scalar line integral, hence the statement is true. (b) False: For the definition of the scalar line integral, there is no need to specify a direction along the path, hence reversing the orientation of the curve does not change the sign of the integral. However, reversing the orientation of the curve changes the sign of the vector line integral. 4. Let C be a curve of length 5. What is the value of C F · d s if (a) F ( P ) is normal to C at all points P on C ?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern