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

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

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best_answer

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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