Mathematics
Mathematics, 08.11.2019 20:31, pnpampanoa

Suppose we have an ellipse parameterized by x = a cos(θ) and y = b sin(θ) with a and b positive. this is the ellipse described by the equation x 2/a2 + y 2/b2 = 1. it is known that the area of the ellipse is πab. the following argument shows that there is only one positive real number. what is wrong with the argument? suppose a and b are positive real numbers. we shall compute the area of the ellipse parameterized by x = a cos(θ) and y = b sin(θ) for 0 ≤ θ ≤ 2π by using polar coordinates. since r 2 = x 2 + y 2 , we have r 2 = a 2 cos2 (θ) + b 2 sin2 (θ) = b 2 + (a 2 − b 2 ) cos2 (θ). thus, πab = z 2π 0 z √ b 2+(a 2−b 2) cos2(θ) 0 r dr dθ = 1 2 z 2π 0 (b 2 + (a 2 − b 2 ) cos2 (θ)) dθ = 1 2 2π 0 (b 2 θ + (a 2 − b 2 )(θ/2 + sin(2θ)/4) = (a 2 + b 2 )π 2 so, 2ab = a 2 + b 2 , and hence (a − b) 2 = 0 which means a = b. since a and b were arbitrary positive real numbers, we conclude that there is only one positive real number

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Suppose we have an ellipse parameterized by x = a cos(θ) and y = b sin(θ) with a and b positive. thi...

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