Green's theorem
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From: aegis on 1 May 2010 00:11 I have the line integral y - ln(x^2 + y^2)dx + 2arctan(y/x)dy where C is the ellipse: [(x-1)^2]/9 + [(y+2)^2]/4 = 1 Applying green's theorem, it boils down to the iterated integral int( int( -1 ) ) dA, however, how should this be evaluated in terms of C? -- aegis
From: David Bernier on 1 May 2010 00:48 aegis wrote: > I have the line integral y - ln(x^2 + y^2)dx + 2arctan(y/x)dy where C > is the ellipse: [(x-1)^2]/9 + [(y+2)^2]/4 = 1 > > Applying green's theorem, it boils down to the iterated integral > int( int( -1 ) ) dA, however, how should this be evaluated in > terms of C? > > -- > aegis If you integrate 1 over the interior of an a by b rectangle in R^2, the value is ab. Also, int_A( f(x,y) + g(x, y))dx dy = int_A f(x,y) dx dy + int_A g(x,y)dx dy even if f or g is somewhat discontinuous, say along a finite set of segments of finite length. And int_A( c f(x,y))dxdy = c int_A( f(x,y))dxdy , c being a constant. Or, maybe think of integration over a planar domain as calculating a volume. David Bernier
From: aegis on 1 May 2010 09:03 On Apr 30, 11:48 pm, David Bernier <david...(a)videotron.ca> wrote: > aegis wrote: > > I have the line integral y - ln(x^2 + y^2)dx + 2arctan(y/x)dy where C > > is the ellipse: [(x-1)^2]/9 + [(y+2)^2]/4 = 1 > > > Applying green's theorem, it boils down to the iterated integral > > int( int( -1 ) ) dA, however, how should this be evaluated in > > terms of C? > > > -- > > aegis > > If you integrate 1 over the interior of an a by b rectangle in R^2, > the value is ab. > > Also, > int_A( f(x,y) + g(x, y))dx dy = int_A f(x,y) dx dy + int_A g(x,y)dx dy > even if f or g is somewhat discontinuous, say along a finite set of segments of > finite length. > > And int_A( c f(x,y))dxdy = c int_A( f(x,y))dxdy , c being a constant. > > Or, maybe think of integration over a planar domain as calculating a volume. > > David Bernier So Green's theorem gives me the area of the region in R^2 and in R^3 i can find the volume of regions, yeah? In the case above, I end up with: int( int(-1) ) dA and if I orient that in the clockwise direction, I would get int( int( 1 ) ) dA over the given ellipse, which would then be the area of the ellipse, yes? -- aegis
From: David Bernier on 1 May 2010 11:01 aegis wrote: > On Apr 30, 11:48 pm, David Bernier <david...(a)videotron.ca> wrote: >> aegis wrote: >>> I have the line integral y - ln(x^2 + y^2)dx + 2arctan(y/x)dy where C >>> is the ellipse: [(x-1)^2]/9 + [(y+2)^2]/4 = 1 >>> Applying green's theorem, it boils down to the iterated integral >>> int( int( -1 ) ) dA, however, how should this be evaluated in >>> terms of C? >>> -- >>> aegis >> If you integrate 1 over the interior of an a by b rectangle in R^2, >> the value is ab. >> >> Also, >> int_A( f(x,y) + g(x, y))dx dy = int_A f(x,y) dx dy + int_A g(x,y)dx dy >> even if f or g is somewhat discontinuous, say along a finite set of segments of >> finite length. >> >> And int_A( c f(x,y))dxdy = c int_A( f(x,y))dxdy , c being a constant. >> >> Or, maybe think of integration over a planar domain as calculating a volume. >> >> David Bernier > > So Green's theorem gives me the area of the region in R^2 and in > R^3 i can find the volume of regions, yeah? In the case above, > I end up with: If you look at the introduction in < http://en.wikipedia.org/wiki/Green's_theorem > , a line integral (where we turn counterclockwise) is stated to be equal to an integral over a planar region. I didn't check the computation you did of the two partial derivatives. The orientation (anticlockwise) carries over to the + and - in the partials ... If we integrated clockwise, we'd change the signs in the partials ... It relates here to an area only because we integrate the constant -1 in: int_A( -1 )dxdy = (-1) int_A( 1 )dxdy. > int( int(-1) ) dA and if I orient that in the clockwise direction, > I would get int( int( 1 ) ) dA over the given ellipse, which would > then be the area of the ellipse, yes? Not quite. In Green's Theorem as appears in the Introduction of < http://en.wikipedia.org/wiki/Green's_theorem > , it's assumed that the contour C is followed in a counterclockwise way. int( int(-1) ) dA = (-1) int( int(1) ) dA isn't an oriented integral. David Bernier
From: Stephen Montgomery-Smith on 1 May 2010 14:27 aegis wrote: > I have the line integral y - ln(x^2 + y^2)dx + 2arctan(y/x)dy where C > is the ellipse: [(x-1)^2]/9 + [(y+2)^2]/4 = 1 > > Applying green's theorem, it boils down to the iterated integral > int( int( -1 ) ) dA, however, how should this be evaluated in > terms of C? > > -- > aegis I think this question is harder than perhaps even the creator of the question intended. You cannot use Green's Theorem when the functions are discontinuous in the domain. Fortunately, (0,0) is not contained on the ellipse, so no problems with ln(x^2+y^2). However the ellipse does contain two points where x=0. Using the usual conventional meaning of arctangent (remember tan x = y has many solutions x for a given y), arctan(y/x) will be discontinuous at those two points. This makes the problem quite a bit harder.
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