JEE MAIN - Mathematics (2022 - 26th June Evening Shift - No. 6)

The area of the region bounded by y² = 8x and y² = 16(3 $$-$$ x) is equal to:

$${{32} \over 3}$$

$${{40} \over 3}$$

$${c_1}:{y^2} = 8x$$

$${c_2}:{y^2} = 16(3 - x)$$

JEE Main 2022 (Online) 26th June Evening Shift Mathematics - Area Under The Curves Question 73 English Explanation

Solving c₁ and c₂

$$48 - 16x = 8x$$

$$x = 2$$

$$\therefore$$ $$y = \pm \,4$$

$$\therefore$$ Area of shaded region

$$ = 2\int\limits_0^4 {\left\{ {\left( {{{48 - {y^2}} \over {16}}} \right) - \left( {{{{y^2}} \over 8}} \right)} \right\}dy} $$

$$ = {1 \over 8}\left[ {48y - {y^3}} \right]_0^4 = 16$$