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Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 29 maths textbook solution

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Answer:

6\pi \; s q \cdot { unit }

Hint:

Use x=3 \cos t, y=2 \sin t

Given:

Find area enclosed by curve x=3 \cos t, y=2 \sin t

Solution:

The given curve x = 3cot\; t, y = 2sin\; t represents the parametric equation of ellipse.

Eliminating the parameter t, we get,

\frac{x^{2}}{9}+\frac{y^{2}}{4}=\cos ^{2} t+\sin ^{2} t=1

This represent the Cartesian equation of the ellipse with centre \left ( 0,0 \right ). The co ordinates of the vertices are \left ( \pm 3,0 \right ) and \left ( 0,\pm 2 \right ).

\therefore Required area = Area of the shaded region

                         = Area of regionOABO

                         \begin{aligned} &=4 \times \int_{0}^{3} y_{\text {ellipse }} d x \\\\ &=4 \times \int_{0}^{3} \sqrt{4\left(1-\frac{x^{2}}{9}\right)} d x \\\\ &=4 \times \frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \end{aligned}

                         \begin{aligned} &=\frac{8}{3}\left(\frac{x}{2} \sqrt{9-x}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right)_{0}^{3} \\\\ &=\frac{8}{3}\left[\left(0+\frac{9}{2} \sin ^{-1} 1\right)-(0+0)\right] \\\\ &=\frac{8}{3} \times \frac{9}{2} \times \frac{\pi}{2} \end{aligned}

                         =6\pi \; s q \cdot { unit }

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