Q. 34 F(x,y,z)=ln(xy+1)+5x3y2z,4xz2, a... [FREE SOLUTION]

Chapter 14: Q. 34 (page 1141)

$F (x, y, z) = <\ln (x y + 1) + 5^{x} 3^{y} 2^{z}, 4 x z^{2}>, and C is the$ $b o u n d a r y o f t h e s q u a r e i n t h e p l a n e z = 6 a n d w i t h v e r t i c e s a t (2, 0, 6), (- 2, 0, 6), (2, 4, 6), a n d (- 2, 4, 6) .$

Short Answer

Expert verified

$L e t S b e a n o r i e n t e d, s m o o t h o r p i e c e w i s e - s m o o t h s u r f a c e b o u n d e d b y a c u r v e C . S u p p o s e t h a t i s a n o r i e n t e d u n i t n o r m a l v e c t o r o f S a n d C h a s a p a r a m e t r i z a t i o n t h a t t r a v e r s e s C i n t h e o u n t e r c l o c k w i s e d i r e c t i o n w i t h r e s p e c t t o n .$

Step by step solution

stokes theorem

$S t o k e s' T h e o r e m s t a t e s t h a t, " L e t S b e a n o r i e n t e d, s m o o t h o r p i e c e w i s e - s m o o t h s u r f a c e b o u n d e d b y a c u r v e C . S u p p o s e t h a t $ \ m a t h b f {n} $ i s a n o r i e n t e d u n i t n o r m a l v e c t o r o f S a n d C h a s a p a r a m e t r i z a t i o n t h a t t r a v e r s e s C i n t h e c o u n t e r c l o c k w i s e d i r e c t i o n w i t h r e s p e c t t o n . I f a v e c t o r f i e l d F i s d e f i n e d o n S, t h e n,$

hypothesis

${鈭�}_{C} F (x, y, z) 路 d r = {鈭�}_{S} curl F (x, y, z) 路 n d S =$ $H e n c e, t h e h y p o t h e s i s o f S t o k e s' T h e o r e m i s a s f o l l o w s, (1) T h e v e c t o r f i e l d F i s d e f i n e d a n d c o n t i n u o u s l y d i f f e r e n t i a b l e o n a n o p e n s e t c o n t a i n i n g t h e s u r f a c e S b o u n d e d b y C . (2) T h e b o u n d a r y c u r v e C i s s i m p l e, s m o o t h a n d c l o s e d . (3) T h e s u r f a c e S i s o r i e n t e d a n d s m o o t h .$

consider

$C o n s i d e r t h e f o l l o w i n g s u r f a c e : T h e s u r f a c e S i s t h e p y r a m i d w i t h v e r t i c e s a t (0, 0, 6), (2, 0, 0), (- 2, 0, 0), (0, 3, 0) a n d (0, - 3, 0) . T h e o b j e c t i v e i s t o s h o w t h a t S t o k e s' T h e o r e m d o e s n o t a p p l y i n t h i s c a s e . H e r e, n o t i c e t h a t, t h i s s u r f a c e S i s n o t s m o o t h o r p i e c e - w i s e s m o o t h s u r f a c e . T h i s i m p l i e s t h a t, S t o k e s' T h e o r e m d o e s n o t a p p l y b y (3) .$

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91影视

$F (x, y, z) = <\ln (x y + 1) + 5^{x} 3^{y} 2^{z}, 4 x z^{2}>, and C is the$ $b o u n d a r y o f t h e s q u a r e i n t h e p l a n e z = 6 a n d w i t h v e r t i c e s a t (2, 0, 6), (- 2, 0, 6), (2, 4, 6), a n d (- 2, 4, 6) .$

Short Answer

Step by step solution

stokes theorem

hypothesis

consider

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91影视

F(x,y,z)=ln(xy+1)+5x3y2z,4xz2, andCis theboundaryofthesquareintheplanez=6andwithverticesat(2,0,6),(-2,0,6),(2,4,6),and(-2,4,6).

Short Answer

Step by step solution

stokes theorem

hypothesis

consider

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Logic and Functions

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Pure Maths

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$F (x, y, z) = <\ln (x y + 1) + 5^{x} 3^{y} 2^{z}, 4 x z^{2}>, and C is the$ $b o u n d a r y o f t h e s q u a r e i n t h e p l a n e z = 6 a n d w i t h v e r t i c e s a t (2, 0, 6), (- 2, 0, 6), (2, 4, 6), a n d (- 2, 4, 6) .$