Vector CalculusDate: 02/26/98 at 23:57:40 From: Keith Britton Subject: Vector Calculus We're working on Green's Theorem right now in my Calculus class. The vector field, F = xj, is given. I'm asked to show that the line integral of F around any closed curve in the xy-plane, oriented as in Green's Theorem, is equal to the area of the region enclosed by the curve. First, I don't understand the wording of the problem. Second, if I pick the unit circle as the arbitrary curve, how do I translate it into the proper form for the integral in Green's Theorem: / | (pdF2/pdx - pdF1/pdy) dxdy / R (I'm using "pdF.." to indicate the partial derivative). The problem, much to my enjoyment, goes on and requests I use the results to calculate the area of the ellipse x^2/a^2 + y^2/b^2 = 1 parametrized by x = acost, y = bsint; 0 <= t <= 2pi I think the "dxdy" indicates the area of the curve, but I thought the only way to compute an area in 3-space was to use a double or triple integral! I am really confused on this problem and on the topic of line integrals in general. - Keith Date: 02/27/98 at 12:30:26 From: Doctor Anthony Subject: Re: Vector Calculus Green's Theorem states: INT.INT[partdP/dy - partdQ/dx]dx.dy = -INT(path)(P.dx+Q.dy) That is, a double integral over the plane region can be transformed into a line integral over the boundary c of the region. If F = P.i + Q.j then we are given P = 0 and Q = x and if we take the path round the circle x^2+y^2 = 1 = -INT[path)[x.dy] where x = +-sqrt(1-y^2) = -INT(path)[sqrt(1-y^2)dy] put y = sin(theta) dy = cos(theta).d(theta) = -INT(path)[cos(theta).cos(theta)d(theta)] = -INT(path)[cos^2(theta).d(theta)] = -(1/2)INT(path)[1 + cos(2theta).d(theta)] = -(1/2)[theta + (1/2)sin(2theta)] (from pi to 0) = (-1/2)[0 - pi] = pi/2 If we carry out the same operation again, but take x =-sqrt(1-y^2) and integrate theta from 0 to pi we get another pi/2. So the total integral = pi = the area of the given region. If we use the double integral version, we have: I = -INT.INT[partdP/dy - partdQ/dx)dx.dy] P = 0 and Q = x so partdQ/dx = 1 I = -INT.INT[0-1)dx.dy = INT.INT(region)[dx.dy] = area enclosed by c. = pi You should be able to apply the same technique now to the ellipse you quote below, using the line integral round the boundary to show that area enclosed is pi.ab -Doctor Anthony, The Math Forum Check out our Web site http://mathforum.org/dr.math/ |
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