solution to assignment 6

# solution to assignment 6 - THE UNIVERSITY OF HONG KONG...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1211 Multivariable Calculus 2010-11 First Semester: Solution to Assignment 6 1. Let ( x ( t ) ,y ( t )), where a ≤ t ≤ b , be a parametrization of the curve C . Then T = x i + y j √ x 2 + y 2 , and I C F · T ds = Z b a ( M ( x,y ) i + N ( x,y ) j ) · x i + y j p x 2 + y 2 ! q x 2 + y 2 dt = Z b a ( M ( x,y ) i + N ( x,y ) j ) · ( x i + y j ) dt = I C Mdx + Ndy. Hence I C F · T ds = I C Mdx + Ndy = ZZ D ∂N ∂x- ∂M ∂y dA by Green’s theorem, where D is the region bounded by C . 2. (a) For the double integral RR D ( N x- M y ) dA , we have ZZ D ∂ ∂x x x 2 + y 2- ∂ ∂y- y x 2 + y 2 dA = ZZ D- x 2 + y 2 ( x 2 + y 2 ) 2-- x 2 + y 2 ( x 2 + y 2 ) 2 dA = ZZ D dA = 0 . ∂D consists of two closed curves: the outer curve x 1 ( t ) = (cos t, sin t ) , ≤ t ≤ 2 π , and the inner curve x 2 ( t ) = ( a cos t,- a sin t ) , ≤ t ≤ 2 π ( note the orientation of the two curves. ) Thus I ∂D M dx + N dy = I x 1- y dx + xdy + I x 2- y dx + xdy = Z 2 π (sin 2 t + cos 2 t ) dt + Z 2 π- a 2 sin 2 t a 2- a 2 cos 2 t a 2 dt = Z 2 π dt- Z 2 π dt = 0 . Thus the two answers agree. (b) If D is the unit disk (centered at the origin): D = { ( x,y ) : x 2 + y 2 ≤ 1 } , then F is not defined at the origin, and hence the double integral RR D ( N x- M y ) dA is actually not well-defined. Thus the Green’s theorem cannot apply. ( However we will still be able to compute the vector line integral H ∂D M dx + N dy , where ∂D is only the unit circle; the vector line integral can be computed ( work it out yourself ) as 2 π . ) 1 (c) Let E be the region lying between C and C a . Choose orientation of C as counterclockwise and C a as clockwise. Then ∂E = C ∪ C a and the orientations of...
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## This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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solution to assignment 6 - THE UNIVERSITY OF HONG KONG...

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