0 (0, b) Video Example ) (a,0) EXAMPLE 2 or a + and so SOLUTION Solving...

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0 (0, b) Video Example ) (a,0) EXAMPLE 2 or a² + and so SOLUTION Solving the equation of the ellipse for y, we get 22²2= Find the area enclosed by the ellipse = 1 y=: = 1. = + 2/( -3²2- Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (see the figure). The part of the ellipse in the first quadrant is given by this function. a² -(%( = 0≤x≤a dx. de. To change the sin = 1, so To evaluate this integral we substitute x = a sin 8. Then dx = limits of integration we note that when x = 0, sin 0 = 0, so 0 = 0; when x = a, a²-x² since 0 ≤ 0 ≤ 1/2. Therefore [ A = 4 = 11 11 = = = a Jo 420 2ab Also 2ab 0 + r/2 a² - a² sin² 0 = a cos 8. *л/2 10 r/2 a²(cos² e) = a|cos 8| = a cos 8 dx cos² e de (1 + cos 20) de 7/2 0 de +0-0 We have shown that the area of an ellipse with semiaxes a and b is лab. In particular, taking a = b = r, we have proved the famous formula that the area of a circle with radius ris ². 0 (0, b) Video Example ) (a,0) EXAMPLE 2 or a² + and so SOLUTION Solving the equation of the ellipse for y, we get 22²2= Find the area enclosed by the ellipse = 1 y=: = 1. = + 2/( -3²2- Because the ellipse is symmetric with respect to both axes, the total area A is four times the area in the first quadrant (see the figure). The part of the ellipse in the first quadrant is given by this function. a² -(%( = 0≤x≤a dx. de. To change the sin = 1, so To evaluate this integral we substitute x = a sin 8. Then dx = limits of integration we note that when x = 0, sin 0 = 0, so 0 = 0; when x = a, a²-x² since 0 ≤ 0 ≤ 1/2. Therefore [ A = 4 = 11 11 = = = a Jo 420 2ab Also 2ab 0 + r/2 a² - a² sin² 0 = a cos 8. *л/2 10 r/2 a²(cos² e) = a|cos 8| = a cos 8 dx cos² e de (1 + cos 20) de 7/2 0 de +0-0 We have shown that the area of an ellipse with semiaxes a and b is лab. In particular, taking a = b = r, we have proved the famous formula that the area of a circle with radius ris ².