Compute the area bounded by an ellipse \(\frac{x^{2}}{4} +\frac{y^{2}}{9}=1\) in two different ways outlined. (a) By finding an appropriate change of co-ordinates and then using the Jacobian to compute the integral. (b) Using Green’s theorem.
15
Aug
Solution:- (a) (frac{x^{2}}{4} +frac{y^{2}}{9}=1) (a=2) and (b=3) from the above relation by changing the variable taking, (u= rcosTheta) and (v=rsinTheta) Jacobian ((J))(=) (=)(frac{du}{dr}) (frac{du}{dtheta}) (frac{dv}{dr}) (frac{dv}{dtheta}) (=)(cos) (-rsintheta) (sintheta) (rcostheta) (=) (rcos^{2}theta+rsin^{2}theta) (=)(r)((sin^{2}theta+cos^{2}theta)) (J)(=)(r).(1) (J)(=)(r) Limits are, (r=0) to(1) and 0 to (2pi) since radius is (1) Area (=) (ab) (int_{0}^{2pi}int_{0}^{1})(rdrdtheta) (=)(2.3int_{0}^{2pi}[r^{2}]_{0}^{1}) (=)(6int_{0}^{2pi}(1-0)dtheta) (=)(6.[theta]_{0}^{2pi}) (=)(frac{12Pi}{2}) […]
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