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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 11.

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 11.

Answers (1)

Answer:

\left (\frac{8 \pi }{3}-2\sqrt{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{x}{a}+c .

Given: The equation of the given curves is

x^2+y^2=4        ........(I)

(x-2)^2+y^2=4        .........(II)

Solution:  The equation of the given curves is

x^2+y^2=4        ........(I)

(x-2)^2+y^2=4        .........(II)

Clearly x^2+y^2=4  represents a circle with center (0, 0) and radius 2. Also, (x-2)^2+y^2=4 represents a circle with centre (2, 0) and radius 2. To find the point of intersection of the given curves, we solve (i) and (ii). Simultaneously, we find the two curves intersect at A(1, \sqrt{3}) and D(1, -\sqrt{3}).

Since both the curves are symmetrical about x-axis, So, the required area = 2(Area OABCO) Now, we slice the area OABCO into vertical strips. We observe that the vertical strips change their character at A(1,√3). So, Area OABCO = Area OACO + Area CABC.

When area OACO is sliced in the vertical strips, we find that each strip has its upper end on the circle (x - 2)2 + (y - 0)2 = 4 and the lower end on x-axis. So, the approximating rectangle shown in figure has length = y1 width = \Delta x and area = y1\Delta x. As it can move from x = 0 to x = 1

\begin{aligned} &\therefore \text { Area } O A C O=\int_{0}^{1} y_{1} d x \\ &\therefore \text { AreaO } A C O=\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x \end{aligned}

Similarly, approximating rectangle in the region CABC has length = y2 , width = \Delta x and area = y2\Delta x

As it can move from x =1 to x =2

\text { Area } C A B C=\int_{1}^{2} y_{2} d x=\int_{1}^{2} \sqrt{4-x^{2}} d x

Hence,required area  A  is given by

\begin{aligned} &A=2\left[\int_{0}^{1} \sqrt{4-(x-2)^{2}} d x+\int_{1}^{2} \sqrt{4-x^{2}} d x\right] \\ &\Rightarrow A=2\left[\left[\frac{(x-2)}{2} \sqrt{4-(x-2)^{2}}+\frac{4}{2} \sin ^{-1} \frac{(x-2)}{2}\right]_{0}^{1}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{1}^{2}\right] \\ &\Rightarrow A=2\left\{-\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(-\frac{1}{2}\right)-2 \sin ^{-1}(-1)+2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin -1 \frac{1}{2}\right\} \end{aligned}

\begin{aligned} &=2\left[-\sqrt{3}-2\left(\frac{\pi}{6}\right)+2\left(\frac{\pi}{2}\right)+2\left(\frac{\pi}{2}\right)-2\left(\frac{\pi}{6}\right)\right] \\ &=2\left(-\sqrt{3}-\frac{2 \pi}{3}+2 \pi\right) \\ &=2\left(\frac{4 \pi}{3}-\sqrt{3}\right) \\ &=\left(\frac{8 \pi}{3}-2 \sqrt{3}\right) \text { sq.units. } \end{aligned}

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