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  • Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 37 Maths.

Explain solution RD Sharma Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 37 Maths.

Answers (1)

Answer:

\frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units} 

Hints:

Given: Equation of circle x^2+y^2=32  exterior to parabola y^2=6x

Solution: First finding intersection point by solving the equation of two curves

\begin{aligned} &x^{2}+y^{2}=16 \ldots \text { (i) } \\ &\text { and } y^{2}=6 x \ldots \text { (ii) } \\ &\Rightarrow x^{2}+6 x=16 \\ &\Rightarrow x^{2}+6 x-16=0 \end{aligned}

\begin{aligned} &\Rightarrow x^{2}+8 x-2 x-16=0 \\ &\Rightarrow x(x+8)-2(x+8)=0 \\ &\Rightarrow(x+8)(x-2)=0 \\ & x=-8 \end{aligned}

(not possible y^2 cannot be -ve )

Or  x=2    (only allowed value)

y=\pm 2\sqrt{3}

\begin{aligned} &A_{1}=\int_{0}^{2} \sqrt{6 x} d x=\left[\frac{\sqrt{6} x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\ &=\frac{2 \sqrt{6}}{3} \times 2^{\frac{3}{2}}=\frac{2 \sqrt{3} \times \sqrt{2}}{3} \times 2 \sqrt{2} \\ &=\frac{8}{\sqrt{3}} \text { sq.units } \end{aligned}

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

\begin{aligned} &\text { Area }=2\left(A_{1}+A_{2}\right) \\ &=2\left(\frac{8}{\sqrt{3}}+\frac{8 \pi}{3}-\sqrt{12}\right) \\ &\text { Required area }=\pi \times 16-\frac{16 x}{3}+4 \sqrt{3}-\frac{16}{\sqrt{3}} \\ &=\frac{32 \pi}{3}-\frac{4}{\sqrt{3}} \end{aligned}\frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units}

Therefore, the area of the circle is \frac{32\pi}{3}-\frac{4}{\sqrt{3}} \; \text{sq.units} 

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