JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q76

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MathematicsSingle correct+4 / -1Question 76 of 90

The area (in sq. units) of the smaller portion enclosed between the curves, $x^2+y^2=4$ and $y^2=3x$ , is :

\dfrac1{2\sqrt3}+\dfrac\pi3

\dfrac1{\sqrt3}+\dfrac{2\pi}3

\dfrac1{2\sqrt3}+\dfrac{2\pi}3

\dfrac1{\sqrt3}+\dfrac{4\pi}3

Answer and worked solution

Official answer: D

The curves meet at $x=1$ , $y=\pm\sqrt3$ . The region inside both is the smaller one:
$2\left[\int_0^1\sqrt{3x}\,dx+\int_1^2\sqrt{4-x^2}\,dx\right]=2\left[\frac2{\sqrt3}+\frac{2\pi}3-\frac{\sqrt3}2\right]=\frac1{\sqrt3}+\frac{4\pi}3.$

Topics:Area between curves

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 76