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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 42 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 42 Maths textbook solution.

Answers (1)

Answer:

\left(6 \pi-\frac{9 \sqrt{3}}{2}\right)  sq. units

Hints:

Use concept.

Given:

The two curves x^2+y^2=9  and (x-3)^2+y^2=9

Solution:

To find area enclosed by

x^2+y^2=9 ……. (1)

(x-3)^2+y^2=9  …….. (2)

Equation (1) represents a circle with centre (0,0) and meets axis at \left ( \pm 3,0 \right )  , \left ( 0,\pm 3 \right )

Equation (2) is a circle with center (3,0) and meets axis at (0,0) , (6,0)

they intersect each other at \left ( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right ) and \left ( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right ) . A rough sketch of the curves is as under:

Shaded region is the required area.

Required area = Region OABCO

A = 2(Region OBCO)

   = 2(Region ODCO + Region DBCD)

=2\left[\int_{0}^{\frac{3}{2}} \sqrt{9-(x-3)^{2}} d x+\int_{\frac{3}{2}}^{3} \sqrt{9-x^{2}} d x\right]

\begin{aligned} &\left.=2\left[\left\{\frac{(x-3)}{2} \sqrt{9-(x-3)^{2}}+\frac{9}{2} \sin ^{-1} \frac{(x-3)}{3[]}\right\}_{0}^{\frac{3}{2}}+\left\{\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right\}_{\frac{3}{2}}\right]_{2}^{3}\right] \\ &=2\left[\left\{\left(-\frac{3}{4} \sqrt{9-\frac{9}{4}}+\frac{9}{2} \sin ^{-1}\left(-\frac{3}{6}\right)\right)-\left(0+\frac{9}{2} \sin ^{-1}(-1)\right)\right\}+\left\{\left(0+\frac{9}{2} \sin ^{-1}(1)\right)-\left(\frac{3}{4} \sqrt{9-\frac{9}{4}}+\frac{9}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right) .\right.\right. \end{aligned}

\begin{aligned} &=2\left[\left\{-\frac{9 \sqrt{3}}{8}-\frac{9}{2}-\frac{\pi}{6}+\frac{9}{2}-\frac{\pi}{2}\right\}+\left\{\frac{9}{2}-\frac{\pi}{2}-\frac{9 \sqrt{3}}{8}-\frac{9}{2}-\frac{\pi}{6}\right\}\right] \\ &=2\left[-\frac{9 \sqrt{3}}{8}-\frac{3 \pi}{4}+\frac{9 \pi}{4}+\frac{9 \pi}{4}-\frac{9 \sqrt{3}}{8}-\frac{3 \pi}{4}\right] \\ &=2\left[\frac{12 \pi}{4}-\frac{18 \sqrt{3}}{8}\right] \\ &=\left(6 \pi-\frac{9 \sqrt{3}}{2}\right) \text { sq. units } \end{aligned}

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