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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 17.

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 17.

Answers (1)

Answer: \frac{\pi}{3}

Hint:

use basic concepts.

Given: The equations of the curves are

x^2+y^2=4

x=\sqrt{3}y

Solution: Obviously x^2+y^2=4  is a circle having centre at (0, 0) and radius 2 units. Forgraph of line

x=\sqrt{3}y.

x 0 1
y 0 0.58

 

For intersecting point of given circle and line

Putting  x=\sqrt{3}  in x^2+y^2=4  we get

\begin{aligned} &(\sqrt{3} y)^{2}+y^{2}=4 \\ &3 y^{2}+y^{2}=4 \\ &4 y^{2}=4 \\ &y=\pm 1 \\ &x=\pm \sqrt{3} \end{aligned}

Intersecting points are (\sqrt{3},1 ), (-\sqrt{3},-1)

Shaded region is required region.

 

Now required area =\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{0}^{\sqrt{3}} \sqrt{4-x^{2}}

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

\begin{aligned} &=\frac{\sqrt{3}}{2}+\left[2 \frac{\pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right] \\ &=\frac{\sqrt{3}}{2}+\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3} \\ &=\pi-\frac{2 \pi}{3} \\ &=\frac{\pi}{3}(\text { proved }) \end{aligned}

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