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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 11

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 11

Answers (1)

Answer:

6\pi \; sq\cdot unit

Hint:

9 x^{2}+4 y^{2}=36

Given:

Sketch the region (x, y): 9 x^{2}+4 y^{2}=36 and find the area of region enclosed by using integration.

Solution:

We have,

9 x^{2}+4 y^{2}=36                                    .........(i)

\begin{aligned} &4 y^{2}=36-9 x^{2} \\\\ &y^{2}=\frac{9}{4}\left(4-x^{2}\right) \\\\ &y=\frac{3}{2} \sqrt{4-x^{2}} \end{aligned}                                    ...........(ii)

From (i), we get

\frac{x^{2}}{4}+\frac{y^{2}}{9}=1

Since \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 all power of x and y are equal                 

\begin{aligned} &A=\text { Area of enclosed curve }=4 \int_{0}^{2}|y| d x \\\\ &A=4 \int_{0}^{2} \frac{3}{2} \sqrt{4-x^{2}} d x \\\\ &A=4 \times \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} d x \end{aligned}

\begin{aligned} &A=6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} d x \\\\ &A=6\left[\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{1}{2} 2^{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2} \\\\ &A=6\left[0+\frac{1}{2} 4 \sin ^{-1} 1\right] \end{aligned}

\begin{aligned} &A=6\left[\frac{1}{2} \times 4\left(\frac{\pi}{2}\right)\right] \\\\ &A=6 \pi \mathrm{sq} \cdot \text { unit } \\\\ &{\left[\because \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\right] \quad \text { and }\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]} \end{aligned}             

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