18.3 Ex 26 - 32

# 18.3 Ex 26 - 32 - 1260 C H A P T E R 18 F U N D A M E N TA...

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

1260 CHAPTER 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) 25. Prove that div ( f ×∇ g ) = 0. SOLUTION We compute the cross product: f g = ± f x , f y , f z ® × ± g x , g y , g z ® = ¯ ¯ ¯ ¯ ¯ ¯ ijk f x f y f z g x g y g z ¯ ¯ ¯ ¯ ¯ ¯ = ± f y g z f z g y , f z g x f x g z , f x g y f y g x ® We now compute the divergence of this vector. Using the Product Rule for scalar functions and the equality of the mixed partials, we obtain div ( f g ) = x ( f y g z f z g y ) + y ( f z g x f x g z ) + z ( f x g y f y g x ) = f yx g z + f y g zx f g y f z g + f zy g x + f z g xy f g z f x g + f xz g y + f x g yz f g x f y g = ( f f ) g z + ( g g ) f y + ( f f ) g y + ( g g ) f z + ( f f ) g x + ( g g ) f x = 0 In Exercises 26–28, let 1 denote the Laplace operator deFned by 1 ϕ = 2 x 2 + 2 y 2 + 2 z 2 26. Prove the identity curl ( curl ( F )) =∇ ( div ( F )) 1 F where 1 F denotes h 1 ± 1 ,1 ± 2 ± 3 i . We compute the left-hand side of the identity. We have curl ( F ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z ± 1 ± 2 ± 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¿ ± 3 y ± 2 z , ± 1 z ± 3 x , ± 2 x ± 1 y À Hence, curl ( curl ( F )) = ¿ y µ ± 2 x ± 1 y z µ ± 1 z ± 3 x , z µ ± 3 y ± 2 z x µ ± 2 x ± 1 y , x µ ± 1 z ± 3 x y µ ± 3 y ± 2 z ¶À = * 2 ± 2 y x 2 ± 1 y 2 2 ± 1 z 2 + 2 ± 3 z x , 2 ± 3 z y 2 ± 2 z 2 2 ± 2 x 2 + 2 ± 1 x y , 2 ± 1 x z 2 ± 3 x 2 2 ± 3 y 2 + 2 ± 2 y z + (1) We now compute the right-hand side of the given identity: ( div ( F )) µ ± 1 x + ± 2 y + ± 3 z = ¿ x µ ± 1 x + ± 2 y + ± 3 z , y µ ± 1 x + ± 2 y + ± 3 z , z µ ± 1 x + ± 2 y + ± 3 z ¶À = * 2 ± 1 x 2 + 2 ± 2 x y + 2 ± 3 x z , 2 ± 1 y x + 2 ± 2 y 2 + 2 ± 3 y z , 2 ± 1 z x + 2 ± 2 z y + 2 ± 3 z 2 + Therefore, ( div ( F )) 1 F ( div ( F )) * 2 ± 1 x 2 + 2 ± 1 y 2 + 2 ± 1 z 2 , 2 ± 2 x 2 + 2 ± 2 y 2 + 2 ± 2 z 2 , 2 ± 3 x 2 + 2 ± 3 y 2 + 2 ± 3 z 2 +

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

View Full Document
SECTION 18.3 Divergence Theorem (ET Section 17.3) 1261 = * 2 F 2 x y + 2 F 3 x z 2 F 1 y 2 2 F 1 z 2 , 2 F 1 y x + 2 F 3 y z 2 F 2 x 2 2 F 2 z 2 , 2 F 1 z x + 2 F 2 z y 2 F 3 x 2 2 F 3 y 2 + (2) Since the mixed partials are equal, the expressions obtained in (1) and (2) are the same. This proves the desired identity.
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 8

18.3 Ex 26 - 32 - 1260 C H A P T E R 18 F U N D A M E N TA...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online