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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20 3 Question 10 Maths textbook solution

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 Question 10 Maths textbook solution.

Answers (1)

Answer:

\left (\frac{4\sqrt{3}}{3}+\frac{16}{3}\pi \right ) Square unit

Hint:

Given:

x^{2}+y^{2}=16

y^2=6x

Solution:

https://www.sarthaks.com/?qa=blob&qa_blobid=3302140514589017426

x^{2}+y^{2}=16     …(1)

Equation of circle with centre (0,0) and radius 4

Equation of parabola:

y^2=6x     ….(2)

Intersecting piont of (1) and (2)

\begin{aligned} &x^{2}+6 x-16=0 \\ &x^{2}+8 x-2 x-16=0 \\ &x(x+8)-2(x+8)=0 \\ &(x+8)(x-2)=0 \\ &x=-8,2 \\ &\text { but }, x \neq-8, x=2 \\ &y^{2}=12 \\ &y=\pm \sqrt{12} \\ &y=\pm 2 \sqrt{3} \end{aligned}

So points are (2,2\sqrt{3}) and (2,-2\sqrt{3})

Area of shaded region

\begin{aligned} &=2\left[\int_{0}^{3} \sqrt{6} \sqrt{x} d x+\int_{2}^{4} \sqrt{16-x^{2}} d x\right] \\ &=2 \sqrt{6}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{2}+2 \int_{2}^{4} \sqrt{16-x^{2}} d x \\ &=\frac{4}{3} \sqrt{6}(2)^{3 / 2}+2\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \sin ^{-1} \frac{x}{4}\right]_{2}^{4} \\ &=\frac{4}{3} \sqrt{6} \times 2 \sqrt{2}+16 \sin ^{-1} 1-2 \sqrt{12}-16 \sin ^{-1} \frac{1}{2} \\ &=\frac{8}{3} \times 2 \sqrt{3}+16 \frac{\pi}{2}-4 \sqrt{3}-16 \frac{\pi}{6} \end{aligned}

\frac{4\sqrt{3}}{3}+8 \pi -\frac{8 \pi}{3}

=\left (\frac{4\sqrt{3}}{3}+\frac{16}{3}\pi \right ) square unit

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