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  • Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 40 Maths.

Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 40 Maths.

Answers (1)

Answer:

\frac{\pi}{8} sq. Units

Hints:

Use concept of definite integrals.

Given:

The curve y=\sqrt{1-x^2} , line y =x  and the positive x-axis

Solution:

\begin{aligned} &y=\sqrt{1-x^{2}} \\ &\Rightarrow y^{2}=1-x^{2} \\ &\Rightarrow x^{2}+y^{2}=1 \end{aligned}

Hence,

y=\sqrt{1-x^2} represents the upper half of the circle x^2+y^2=1  a circle with centre O(0,0) and radius 1 unit.y = x  represents equation of a straight line passing through O(0,0)

Point of intersection is obtained by solving two equations

\begin{aligned} &y=x \\ &y=\sqrt{1-x^{2}} \\ &\Rightarrow x=\sqrt{1-x^{2}} \\ &\Rightarrow x^{2}=1-x^{2} \\ &\Rightarrow 2 x^{2}=1 \\ &\Rightarrow x=\pm \frac{1}{\sqrt{2}} \\ &\Rightarrow y=\pm \frac{1}{\sqrt{2}} \end{aligned}

D\left (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) And D\left (\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right )  are two points of intersection between the circles and the straight line.

And D\left (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) is the intersection point of y=\sqrt{1-x^2} and y =x .

Required area = Shaded area (ODAEO) = Area (ODEO) + Area (EDAE)…..(1)

Now, area (ODEO) =  \int_{0}^{\frac{1}{\sqrt{2}}} x d x

\begin{aligned} &=\left[\frac{x^{2}}{2}\right]_{0}^{\frac{1}{\sqrt{2}}} \\ &=\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right) \\ &=\frac{1}{4} \text { sq.units } \ldots . .(2) \end{aligned}

Area (EDAE) = \int_{\frac{1}{\sqrt{2}}}^{1} \sqrt{1-x^{2}} d x

\begin{aligned} &=\left[\frac{1}{2} x \sqrt{1-x^{2}}+\times \frac{1}{2} \times \sin ^{-1}\left(\frac{x}{1}\right)\right]_{\frac{1}{\sqrt{2}}}^{1}\\ &=0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ &=\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{4}-\frac{1}{2} \times \frac{\pi}{4} \ldots \ldots \ldots .\left\{\text { using, } \sin ^{-1}(1)=\frac{\pi}{2} \text { and } \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\right\}\\ &=\frac{\pi}{4}-\frac{\pi}{8}-\frac{1}{4}\\ &=\frac{\pi}{8}-\frac{1}{4} \text { sq.units } \ldots &\text { (3) } \end{aligned}

Area (ODAEO) = \frac{1}{4}+\frac{\pi}{4}-\frac{1}{4}=\frac{\pi}{8}  sq.units

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