Green's Theorem with Parameterized CurvesDate: 10/18/2003 at 23:26:14 From: Nicole Subject: Green's Theorem My calculus professor worked on a problem in class and I don't understand it very well. Can you explain how to do this type of problem? Find the area enclosed by the ellipse parameterized by the curve x = 3cos(x), y = sin(x), for o <= x <= 2pi by applying Green's theorem to the vector field F = -(1/2)yi + (1/2)xj I understand what Green's theorem states and how to use it, but the parameterizations are confusing me. Date: 10/19/2003 at 08:12:30 From: Doctor Anthony Subject: Re: Green's Theorem Hi Nicole, Green's theorem allows you to use double integrals over a region to replace a line integral round a closed curve or vice-versa. Using Green's theorem, we can write INT[P dx + Q dy] = INT INT[(part(dQ/dx) - part(dP/dy))dx dy] C A F = Pi + Qj = (-y/2)i + (x/2)j where P = -y/2 Q = x/2 part[dP/dy] = -1/2 part[dQ/dx] = 1/2 So we get INT[P dx + Q dy] = INT INT[(1/2 + 1/2)dx dy] C A = area of the ellipse Therefore our problem reduces to integrating round the curve as given by the line integral on the left. We parameterize x = 3cos@ y = sin@ dx = -3sin@ d@ dy = cos@ d@ P = -sin@/2 Q = 3cos@/2 and so INT[P dx + Q dy] 2pi = INT[(-sin@/2)(-3sin@ d@) + (3cos@/2)cos@ d@] 0 2pi = (3/2)INT[[sin^2(@) + cos^2(@)] d@] 0 2pi = (3/2)INT[d@] 0 = (3/2)[@] from 0 to 2pi = (3/2)[2pi] = 3pi - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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