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BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Area Under Curves

BITSAT 2015 · Mathematics · Area Under Curves

The area of the region R=\ (x, y):|x| ≤|y|. and x^ 2 .+y^ 2 ≤ 1\ is

The area of the region

\mathrm{R}=\left\{(x, y):|x| \leq|y|\right.\text{ and } x^{2}\left.+y^{2} \leq 1\right\}

is $The area of the region R=\ (x, y):|x| ≤|y|. and x^ 2 .+y^ 2 ≤ 1\ is$

Options

A $\frac{3 \pi}{8}$ sq units
B $\frac{5 \pi}{8} \mathrm{sq}$ units
C $\frac{\pi}{2}$ sq units
D $\frac{\pi}{8}$ sq unit

Correct answer

C. $\frac{\pi}{2}$ sq units

Step-by-step solution

$The area of the region R=\ (x, y):|x| ≤|y|. and x^ 2 .+y^ 2 ≤ 1\ is$
Required area

=4

(Area of the shaded region in first quadrant) =4 \int_{0}^{1

\sqrt{2}}\left(

y_{1}-y_{2}\right) d x=4 \int_{0}^{1

\sqrt{2}}\left(\sqrt{1-x^{2}}-

x\right) d x

=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}

sq units

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