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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Area Under Curves

NDA 2015 · Mathematics · Area Under Curves

Consider the line x= 3 y and the circle x^2+y^2=4 . What is the...

Consider the line x=3yx=\sqrt{3} y and the circle x2+y2=4x^2+y^2=4. What is the area of the region in the first quadrant enclosed by the xx-axis, the line x=3x=\sqrt{3} and the circle?

Options

  1. Aπ3−32\frac{\pi}{3}-\frac{\sqrt{3}}{2}
  2. Bπ2−32\frac{\pi}{2}-\frac{\sqrt{3}}{2}
  3. Cπ3−12\frac{\pi}{3}-\frac{1}{2}
  4. DNone of these

Correct answer

A. π3−32\frac{\pi}{3}-\frac{\sqrt{3}}{2}

Step-by-step solution

x2+y2=4 and x=3yP(3,1)= First quadrant \begin{aligned} & x^2+y^2=4 \text{ and } x=\sqrt{3} y \\ & P(\sqrt{3}, 1)=\text{ First quadrant }\end{aligned} The point of intersection of the line and the circle in the first quadrant is (3,1)(\sqrt{3}, 1). Area of △OPA=\triangle \mathrm{OPA}= Area of △OPB+\triangle \mathrm{OPB}+ Area of PAB =12×OB×PB=12×3×1=32=\frac{1}{2} \times \mathrm{OB} \times \mathrm{PB}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2} Area of PAB ∫32ydx=∫324−x2dx=[x24−x2+42sin⁡−1x2]32[2×π2−32[4−3]−2sin⁡−132]=π−32−2π3=π3−32\begin{aligned} & \text{ Area of PAB } \int_{\sqrt{3}}^2 \mathrm{ydx} \\ & =\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x=\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^2 \\ & {\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}[\sqrt{4-3}]-2 \sin ^{-1} \frac{\sqrt{3}}{2}\right]} \\ & =\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{2} \end{aligned}$$
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