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JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Area Under Curves

The area of the region bounded by

y^{2} = 8 x

and

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

is equal to :-

(A) $\frac{32}{3}$
(B) $\frac{40}{3}$
(C)16✓ Correct
(D)19

Answer: (C)

Solution

Find intersection points:

8 x = 16 (3 - x)

gives

8 x = 48 - 16 x

, so

24 x = 48

and

x = 2

. At

x = 2

,

y = \pm 4

.

The area is:

A = 2 \int_{0}^{4} (x_{R} - x_{L}) d y = 2 \int_{0}^{4} ((3 - \frac{y ^{2}}{16}) - \frac{y ^{2}}{8}) d y

= 2 \int_{0}^{4} (3 - \frac{y ^{2}}{16} - \frac{y ^{2}}{8}) d y = 2 \int_{0}^{4} (3 - \frac{3 y ^{2}}{16}) d y

= 2 [3 y - \frac{y ^{3}}{16}]_{0}^{4} = 2 (12 - \frac{64}{16}) = 2 (12 - 4) = 16

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option

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