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  • Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 13.

Provide solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Region Exercise 20.3 question 13.

Answers (1)

Answer:

\begin{aligned} &{\left[\frac{4}{\sqrt{3}} a^{3 / 2}+\left[\frac{8 \pi}{3}-a \sqrt{\frac{16}{3}-a^{2}}-\frac{16}{3} \sin ^{1}-\left(\frac{\sqrt{3 a}}{4}\right)\right]\right] \text { square unit. }} \\ &\text { where } a=\frac{-9+\sqrt{273}}{9} \end{aligned}

Hint: Using the identity formula \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin -{ }^{1} \frac{x}{a}+c . .

Given: Here we know that R = {(x, y): y^2 \leq 3x, 3x^2 + 3y^2\leq 16}

Solution:

Here we know that R = {(x, y): y^2 \leq 3x, 3x^2 + 3y^2\leq 16}

We can write it as

R = {(x, y): y^2 \leq 3x} \cap {(x, y): 3x^2 + 3y^2 \leq 16} = R_1 \cap R_2

 Here R_1 = {(x, y): y^2 \leq 3x}  which represents the region inside the parabola, y2 = 3x with vertex (0, 0) and x-axis as it axis

R = {(x, y): 3x^2 + 3y^2 \leq 16}  represents the interior of 3x^2 + 3y^2 = 16 circle having (0, 0) as centre and \frac{4}{\sqrt{3}} as radius

So the region R which is intersection of points R1 and R2 is shaded in the figure

 3x^2 + 3y^2 = 16 …… (1)

y^2 = 3x…… (2)

Solving both the equations

 3x2 + 9x – 16 = 0

 So we get

x = \frac{(-9 \pm \sqrt{2}73)}{}6

 Here x = \frac{(-9 \pm \sqrt{2}73)}{}6  is the rejecting negative value

 Substituting y = 0 in equation (1)

 x=\frac{4}{\sqrt{3}}

 We know that the circle (1) cuts x-axis at P \left (\frac{4}{\sqrt{3}},0 \right ) and P’ (\left (-\frac{4}{\sqrt{3}},0 \right )

 So the required area can be written as

 Required area = 2 [area of ODPAO] = 2 [area of ODAO + area of ADPA] .

\text { Re quired area }=2\left[\int_{0}^{a} \sqrt{3 x} d x+\int_{a}^{4, \sqrt{\beta}} \sqrt{\sqrt{3}-x^{2}} d x\right]

Intergrating w.r.t

=2\left(\sqrt{3}\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{a}+\left[\frac{x}{2} \sqrt{\frac{16}{3}-x^{2}}+\frac{16}{3} \sin ^{-1}\left(\frac{x}{4 / \sqrt{3}}\right)\right]_{a}^{4 / \sqrt{s}}\right)

Substituting the value of x

\begin{aligned} &{\left[\frac{4}{\sqrt{3}} a^{3 / 2}+\left[\frac{8 \pi}{3}-a \sqrt{\frac{16}{3}-a^{2}}-\frac{16}{3} \sin ^{1}-\left(\frac{\sqrt{3 a}}{4}\right)\right]\right] \text { square unit. }} \\ &\text { where } a=\frac{-9+\sqrt{273}}{9} \end{aligned}

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