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  • Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 13 maths textbook solution

Answers (1)

Answer:

\frac{\pi a^{2}}{4} s q \cdot \text { unit }

Hint:

y=\sqrt{a^{2}-x^{2}} \text { at line } x=0, x=a

Given: Determine the area under curve y=\sqrt{a^{2}-x^{2}} include between line x=0 and x=a

Solution:

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

Required area =  area of shaded region

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

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

=\frac{\pi a^{2}}{4} s q \cdot u n i t

      

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