The area (in sq. units) of the region $\left\{(x, y): y^{2} \geq 2 x\right.$ and $\left.x^{2}+y^{2} \leq 4 x, x \geq 0, y \geq 0\right\}$ is:

(B) $\pi-\frac{4 \sqrt{2}}{3}$

(C) $\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}$

JEE Main · 2016 · 03 Apr 2016 · Offline · Area Under Curves

The area (in sq. units) of the region

{(x, y) : y^{2} \geq 2 x

and

x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}

is:

(A) $π - \frac{8}{3}$ ✓ Correct
(B) $π - \frac{4 2}{3}$
(C) $\frac{π}{2} - \frac{2 2}{3}$
(D) $π - \frac{4}{3}$

Answer: (A)

Solution

The circle

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

has centre

(2, 0)

and radius 2, and it meets the parabola

y^{2} = 2 x

at the origin and at

(2, 2)

in the first quadrant. The region wanted lies inside the circle but on the far side of the parabola, so its area is the first-quadrant quarter-circle minus the strip under the parabola:

\frac{1}{4} π (2)^{2} - \int_{0}^{2} 2 x d x = π - 2 \cdot \frac{2}{3} [x^{3/2}]_{0}^{2} = π - \frac{2 2}{3} \cdot 22 = π - \frac{8}{3}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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JEE Main · 2016 · 03 Apr 2016 · Offline · Area Under Curves

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