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  • Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 45 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 20 Areas of Bounded Regions Exercise 20.3 Question 45 Maths textbook solution.

Answers (1)

Answer:

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

Hints:

Use concept.

Given:

The curve y=\sqrt{4-x^2} ,x^2+y^2-4x=0  and the x-axis.

Solution:

Give, equation of a curves are

y=\sqrt{4-x^2} ……(1)

x^2+y^2-4x=0  ……(2)

Consider the curve

y=\sqrt{4-x^2}=y^2-4x=0

y^2+x^2=0 which represents a circle with centre (0, 0) and radius 2 units.

Now, consider the curve x^2+y^2-4x=0   which also represents a circle with centre (2,0) and radius 2 units.

Now Let us sketch the graph of given curves and find their points of intersection.

On substituting the value of y from Eq. (1) in Eq. (2), we get

x^2(4-x)^2-4x=0  or 4-4x=0  or x=1

On substituting  x =1 in Eq. (1), we get  y=\sqrt{3}

Thus, the point of intersection is \left ( 1.\sqrt{3} \right ) .

Clearly, required area = Area of shaded region OABO

\begin{aligned} &=\int_{0}^{1} y_{(\text {sacoricicic } k)} d x+\int_{1}^{2} y_{(f \text { trs cic } c i e)} d x \\ &=\int_{0}^{1} \sqrt{4 x-x^{2}} d x+\int_{1}^{2} \sqrt{4-x^{2}} d x \\ &=\int_{0}^{1} \sqrt{-\left(x^{2}-4 x\right)} d x+\int_{0}^{2} \sqrt{2^{2}-x^{2}} d x \end{aligned}

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

\begin{aligned} &=\left[\frac{(x-2)}{2} \sqrt{4 x-x^{2}}+2 \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{1}+\left[2 \cdot \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2 \cdot \frac{\pi}{6}\right] \\ &=\left[\left\{\frac{-\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{-1}{2}\right)\right\}-\left\{2 \sin ^{-1}(-1)\right\}\right]+\left(\pi-\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) \\ &=-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}(1)+\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\left[\therefore \sin ^{-1}(-x)=-\sin ^{-1} x\right] \end{aligned}

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

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