JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Area Under Curves

The area of the region in the first quadrant inside the circle

x^{2} + y^{2} = 8

and outside the parabola

y^{2} = 2 x

is equal to :

Answer: (B)

Solution

Circle

x^{2} + y^{2} = 8

meets parabola

y^{2} = 2 x

where

x^{2} + 2 x - 8 = 0 ⟹ (x + 4) (x - 2) = 0 ⟹ x = 2, y = 2

(first quadrant).
For

0 \leq x \leq 2

the parabola lies below the circle (

2 x \leq 8 - x^{2}

), so the strip "inside circle, outside parabola" there is

8 - x^{2} - 2 x

.
For

2 \leq x \leq 22

the parabola lies *above* the circle, so every point under the circle in that range already satisfies

y^{2} < 2 x

(inside the parabola) — it contributes nothing to the outside-parabola area.
So the required area is just

\int_{0}^{2} (8 - x^{2} - 2 x) d x

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

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

Area = (2 + π) - \frac{8}{3} = π - \frac{2}{3}

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option