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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 15

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 15

Answers (1)

Answer:

\pi a^{2}\; s q \cdot \text { unit }

Hint:

Use definite integral

Given:

Using definite integral, find area of circle x^{2}+y^{2}=a^{2}

Solution:

x^{2}+y^{2}=a^{2}

Centre=\left ( 0,0 \right )

Radius=a

Hence,  OA=OB=Radius=a

               A=(a,0),B=(0,a)

Area of circle=4\times area of region OBAO

                   =4 \int_{0}^{a} y d x

We know that

                    \begin{aligned} &x^{2}+y^{2}=a^{2} \\\\ &y^{2}=a^{2}-x^{2} \\\\ &y=\pm \sqrt{a^{2}-x^{2}} \end{aligned}

Since  AOBA lies in first quadrant, value of y is positive

                    y=\sqrt{a^{2}-x^{2}}

Now,

Area of circle   =4 \int_{0}^{a} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]

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

                      \begin{aligned} &=4\left[0+\frac{a^{2}}{2} \sin ^{-1} 1\right]_{0}^{a} \\\\ &=4 \times \frac{a^{2}}{2} \times \frac{\pi}{2} \\\\ &=\pi a^{2} \text { sq.units } \end{aligned}

 

 

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