The area of the region \(\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2} \leq \mathrm{y} \leq 8-\mathrm{x}^{2}, \mathrm{y} \leq 7\right\}\) is

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Area Under Curves

The area of the region

{(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}

Answer: (D)

The region lies above

y = x^{2}

and below the smaller of

y = 8 - x^{2}

and

y = 7

. The parabolas meet where

x^{2} = 8 - x^{2}

, i.e.

x = \pm 2

(with

y = 4

), and

y = 8 - x^{2}

meets

y = 7

where

x = \pm 1

.

For

∣ x ∣ \leq 1

the upper boundary is

y = 7

; for

1 \leq ∣ x ∣ \leq 2

it is

y = 8 - x^{2}

. By symmetry about the

y

-axis,

Area = \int_{- 1}^{1} (7 - x^{2}) d x + 2 \int_{1}^{2} [(8 - x^{2}) - x^{2}] d x .

Now

\int_{- 1}^{1} (7 - x^{2}) d x = 2 (7 - \frac{1}{3}) = \frac{40}{3}

and

\int_{1}^{2} (8 - 2 x^{2}) d x = [8 x - \frac{2 x ^{3}}{3}]_{1}^{2} = (16 - \frac{16}{3}) - (8 - \frac{2}{3}) = \frac{10}{3} .

Therefore the area is

\frac{40}{3} + 2 \cdot \frac{10}{3} = \frac{60}{3} = 20.

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option