tutorial9_sol_1011-final

tutorial9_sol_1011-final - 3. Let n denote the outward unit...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3. Let n denote the outward unit normal vector to the closed curve C parametrized by r ( t ) = ( x ( t ) ,y ( t )), a ≤ t ≤ b , such that the area inside C is on the left when one traverses C . Then n is given by n = ( y ( t ) ,- x ( t )) k ( y ( t ) ,- x ( t )) k . Therefore the flux of F = r across C is given by I C F · n ds = Z b a ( x i + y j ) · y i- x j p ( y ) 2 + ( x ) 2 p ( x ) 2 + ( y ) 2 dt = Z b a x dy dt- y dx dt dt = I C- y dx + xdy = ZZ D ∂ ∂x x- ∂ ∂y (- y ) dA (by Green’s theorem, where D is region enclosed by C ) = ZZ D 2 dA = 2(area inside C ) . 4. (a) Since ∂N ∂x = 2 x cos y = ∂M ∂y , F is conservative. To find a potential f for F , we solve for ∂f ∂x = 2 x sin y ∂f ∂y = x 2 cos y . Integrating the first equation wrt x , we obtain: f = Z 2 x sin y dx + g ( y ) = x 2 sin y + g ( y ) for some function g ( y ). Then from the second equation we have x 2 cos y = ∂f ∂y = x 2 cos y + g ( y ) , which implies g ( y ) = 0, or g ( y ) = C . By taking C = 0 we find a potential f for F as f = x 2 sin y . (b) Since ∇ × F =- k 6 = , we conclude that F is not conservative....
View Full Document

This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

Page1 / 4

tutorial9_sol_1011-final - 3. Let n denote the outward unit...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online