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Dear student,

x = 3 \cos t and y = 2 \sin t 2 x = 6 \cos t . . . . (1) and 3 y = 6 \sin t . . . . (2) Squaring and adding (1) and (2), we get 4 x^{2} + 9 y^{2} = 36 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 This is the equation of the ellipse . Now we find the area enclosed by the ellipse Area = 4 \int_{0}^{3} \frac{2}{3} \sqrt{9 - x^{2}} d x = \frac{8}{3} \int_{0}^{3} \sqrt{9 - x^{2}} d x = \frac{8}{3} {[\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{x}{3})]}_{0}^{3} = \frac{8}{3} [\{\frac{3}{2} \sqrt{9 - 9} + \frac{9}{2} \sin^{- 1} (\frac{3}{3})\} - \{\frac{0}{2} \sqrt{9} + \frac{9}{2} \sin^{- 1} (\frac{0}{3})\}] = \frac{8}{3} \times \frac{9 π}{4} = 6 π sq units

Regards!

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