124 lines
2.2 KiB
TeX
124 lines
2.2 KiB
TeX
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\subsection{Green's theorem}
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\index{Green's theorem}
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Green's theorem tells us that
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$$\oint P\,dx+Q\,dy=\int\!\!\!\int
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\left({\partial Q\over\partial x}-{\partial P\over\partial y}\right)
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dx\,dy$$
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\medskip
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\noindent
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Example 1.
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Evaluate $\oint (2x^3-y^3)\,dx+(x^3+y^3)\,dy$ around the circle
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$x^2+y^2=1$ using Green's theorem.\footnote{
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Wilfred Kaplan, {\it Advanced Calculus, 5th Edition,} 287.}
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\medskip
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\noindent
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It turns out that Eigenmath cannot solve the double integral over
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$x$ and $y$ directly.
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Polar coordinates are used instead.
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\medskip
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\verb$P=2x^3-y^3$
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\verb$Q=x^3+y^3$
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\verb$f=d(Q,x)-d(P,y)$
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\verb$x=r*cos(theta)$
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\verb$y=r*sin(theta)$
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\verb$defint(f*r,r,0,1,theta,0,2pi)$
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$${3\over2}\pi$$
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\medskip
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\noindent
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The $defint$ integrand is $f{*}r$ because $r\,dr\,d\theta=dx\,dy$.
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\medskip
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\noindent
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Now let us try computing the line integral side of Green's theorem
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and see if we get the same result.
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We need to use the trick of converting sine and cosine to exponentials
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so that Eigenmath can find a solution.
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\medskip
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\verb$x=cos(t)$
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\verb$y=sin(t)$
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\verb$P=2x^3-y^3$
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\verb$Q=x^3+y^3$
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\verb$f=P*d(x,t)+Q*d(y,t)$
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\verb$f=circexp(f)$
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\verb$defint(f,t,0,2pi)$
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$${3\over2}\pi$$
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\newpage
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\noindent
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Example 2.
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Compute both sides of Green's theorem for
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$F=(1-y,x)$ over the disk $x^2+y^2\le4$.
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\medskip
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\noindent
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First compute the line integral along the boundary of the disk.
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Note that the radius of the disk is 2.
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\medskip
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\verb$--Line integral$
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\verb$P=1-y$
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\verb$Q=x$
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\verb$x=2*cos(t)$
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\verb$y=2*sin(t)$
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\verb$defint(P*d(x,t)+Q*d(y,t),t,0,2pi)$
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$$8\pi$$
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\verb$--Surface integral$
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\verb$x=quote(x) --remove parametrization of x$
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\verb$y=quote(y) --remove parametrization of y$
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\verb$h=sqrt(4-x^2)$
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\verb$defint(d(Q,x)-d(P,y),y,-h,h,x,-2,2)$
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$$8\pi$$
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\verb$--Bonus point: Compute the surface integral using polar coordinates.$
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\verb$f=d(Q,x)-d(P,y) --do before change of coordinates$
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\verb$x=r*cos(theta)$
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\verb$y=r*sin(theta)$
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\verb$defint(f*r,r,0,2,theta,0,2pi)$
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$$8\pi$$
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\verb$defint(f*r,theta,0,2pi,r,0,2) --try integrating over theta first$
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$$8\pi$$
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\medskip
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\noindent
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In this case, Eigenmath solved both forms of the polar integral.
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However, in cases where Eigenmath fails to solve a double integral, try
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changing the order of integration.
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