Subject: Calculus

Greens Theorem

Calculus.GreensTheorem History

July 03, 2011 by matthew_suan -
July 03, 2011 by matthew_suan -
'''NEXT TOPIC''': [[Divergence theorem]]
March 20, 2011 by matthew_suan -
Changed lines 23-27 from:
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'''Solution'''. We can actually solve this by having three different integrals of the functions from (0,0) to (1,0) , from (1,0) to (0,1), and from (0,1) to (0,0). Here, we would like to show a different approach using the Green’s Theorem.
We let {$P(x,y)= x^4$} and {$Q(x,y)= xy$}. By Green’s theorem, we have
March 20, 2011 by matthew_suan -
March 20, 2011 by matthew_suan -
Changed lines 13-23 from:
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to:
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!!!Example#1

Evaluate

{$$\int_{C} x^4 dx+xydy$$}

where {$C$} is the triangular curve consisting of the line segments from (0,0) to (1,0) , from (1,0) to (0,1), and from (0,1) to (0,0) .

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.png
March 20, 2011 by matthew_suan -
Changed lines 7-13 from:
Take for example the figure shown below. {$C$} here corresponds to an oriented close path or a path where the starting point is the same as the endpoint. {$D$} here is the interior region of closed path {$C$}. Green’s Theorem transforms a line integral at a closed curve {$C$} and a double integral over the plane region {$D$} bounded by {$C$}.
to:
Take for example the figure shown below. {$C$} here corresponds to an oriented close path or a path where the starting point is the same as the endpoint. {$D$} here is the interior region of closed path {$C$}. Green’s Theorem transforms a line integral at a closed curve {$C$} and a double integral over the plane region {$D$} bounded by {$C$}.

%center%Attach:green1.png

In stating Green’s Theorem, we use the convention that the positive orientation of a simple closed {$C$} refers to a single counterclockwise traversal of {$C$}. This means that if {$C$} is given by the vector function of {$r(t)$}, {$a \leq t \leq b$}, then the region {$D$} is always on the left as the point {$r(t)$} go across {$C$} as shown in the figure below.

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March 20, 2011 by matthew_suan -
Changed lines 4-7 from:
%cframe%{$\int_{C}Pdx+Qdy=\int_{D}\int \left (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right )dA$}
to:
%cframe%{$\int_{C}Pdx+Qdy=\int_{D}\int \left (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right )dA$}

!!Discussion
Take for example the figure shown below. {$C$} here corresponds to an oriented close path or a path where the starting point is the same as the endpoint. {$D$} here is the interior region of closed path {$C$}. Green’s Theorem transforms a line integral at a closed curve {$C$} and a double integral over the plane region {$D$} bounded by {$C$}.
March 20, 2011 by matthew_suan -
%cframe%{$\int_{C}Pdx+Qdy=\int_{D}\int \left (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right )dA$}