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

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

Answers (1)

Answer:

\left (4 \pi-\frac{32}{3} \right ) \text{sq. units}

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 .

Given:

Given curves are x^2+y^2=8x  and  y^2=4x .

Solution:

We have given equations

x^2+y^2=8x            ...............(1)

y^2=4x            ........(2)

Equation (1) can be written as

(x-y)^2+y^2=4^2

So equation (i) represents a circle with centre (4, 0) and radius 4.

 Again, clearly equation (ii) represents parabola with vertex (0, 0) and axis as x -axis. The curve (i) and (ii) are shown infigure and the required region is shaded. On solving equation (i) and (ii) we have points of intersection 0(0, 0) and A(4, 4), C(4, - 4)

 Now, we have to find the area of region bounded.

 

By(i) and (ii) & above  x - axis.

So requires region is OBAO.

\begin{aligned} A &=\int_{0}^{4}\left(\sqrt{8 x-x^{2}}-\sqrt{4 x}\right) d x \\ &=\int_{0}^{4}\left(\sqrt{\left(4^{2}\right)-(x-4)^{2}}-2 \sqrt{x}\right) d x \\ &=\left[\frac{(x-4)}{2} \sqrt{(4)^{2}-(x-4)^{2}}+\frac{16}{2} \sin -1 \frac{(x-4)}{4}-2\left(\frac{2 x}{3}\right)^{3 / 2}\right]_{0}^{4} \\ &=\left[8 \sin -{ }^{1} 0-\frac{4}{3} \times 8\right]-\left[8 \times\left(\frac{-\pi}{2}\right)\right] \\ &=\frac{-32}{3}+4 \pi \\ &=\left(4 \pi-\frac{32}{3}\right) \text { sq.units } \end{aligned}

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