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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 19.

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 19.

Answers (1)

Answer:

\frac{4a^2}{3}9(\sqrt{3}+4\pi)

Hint:

\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -\frac{1}{a}+c \text {. } .

Given:

Given curves x^2+y^2=16a^2 and y^2=6ax .

Solution:

The point of intersection of circle

x^2+y^2=16a^2

And parabola,  y^2=6ax is

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

The required common area ,A=2[APOA]

\begin{aligned} &=2 \int_{0}^{2 a} y d x+2 \int_{2 a}^{4 a} y d y \\ &=2 \int_{0}^{2 a} \sqrt{6 a} \sqrt{x} d x+2 \int_{2 a}^{4 a} \sqrt{(4 a)^{2}-x^{2}} d x \\ &=2 \cdot \sqrt{6 a}\left[\frac{2}{3} x 3 / 2\right]_{0}^{2 a}+2\left[\frac{x}{2} \sqrt{(4 a)^{2}-x^{2}}+\frac{1}{2}(4 a)^{2} \sin -1 \frac{x}{4 a}\right]_{2 a}^{4 a} \\ &=2 \cdot \sqrt{6 a} \frac{2}{3}(2 a)^{3 / 2}+2\left[(0-2 a \sqrt{3 a})+8 a^{2}\left(\sin -^{1}-\sin -1 \frac{1}{2}\right)\right] \\ &=2 \cdot 2 \sqrt{3} \cdot \frac{4}{3} a^{2}+2\left[-2 \sqrt{3} a^{2}+8 a^{2}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\right] \\ &=\frac{16}{3} \sqrt{3} a^{2}-4 \sqrt{3} a^{2}+16 a^{2} \frac{\pi}{3} \\ &=\frac{4 \sqrt{3}}{3} a^{2}+\frac{16 \pi a^{2}}{3} \\ &=\frac{4 a^{2}}{3}(\sqrt{3}+4 \pi) \text { sq.units. } \end{aligned}

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