Question
Find the area enclosed by the curve $\text{x}=3\cos\text{t}, \text{y}=2\sin\text{t}.$
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Answer
The given curve $\text{x}=3\cos\text{t}, \text{y}=2\sin\text{t}$in t respesent the parmetric of the elipse. Eliminating the t, we get $\frac{\text{x}^{2}}{9}+\frac{\text{y}^{2}}{4}=\cos^{2}\text{t}+\sin^{2}\text{t}=1$ This represents the cartesion eqution of the elioes with center (0, 0). The coordinates of the vertics are (3, 0) and (0, 2)
Required area = Area of the shaded region = 4 × Area of the region OABO $=4\times \int\limits_{0}^{3}\text{y}\text{dx} $ $=4\times \int\limits_{0}^{3}\sqrt{4(1-\frac{\text{x}^{2}}{9})}\text{dx}$ $=4\times \frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^{2}}{})\text{dx}$ $=\frac{8}{3}\Big[(0+\frac{9}{2}\sin^{-1}1)-(0+0)\Big] $ $=\frac{8}{3}\times\frac{9}{2}\times\frac{\pi}{2}$ $=6\pi\ \text{sq.}\ \text{units}$
Required area = Area of the shaded region = 4 × Area of the region OABO $=4\times \int\limits_{0}^{3}\text{y}\text{dx} $ $=4\times \int\limits_{0}^{3}\sqrt{4(1-\frac{\text{x}^{2}}{9})}\text{dx}$ $=4\times \frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^{2}}{})\text{dx}$ $=\frac{8}{3}\Big[(0+\frac{9}{2}\sin^{-1}1)-(0+0)\Big] $ $=\frac{8}{3}\times\frac{9}{2}\times\frac{\pi}{2}$ $=6\pi\ \text{sq.}\ \text{units}$
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