Area (in sq. units) of the region bounded by curves $y^2 = x$ and $x = 4$ is

$\frac{32}{3}$

The correct answer is Option (2) → $\frac{32}{3}$

Given curves: $y^2 = x$ and $x = 4$

To find the area enclosed between the parabola and the vertical line $x = 4$, express $x$ as a function of $y$:

$x = y^2$

We need the area between $x = y^2$ and $x = 4$ from $y = -2$ to $y = 2$, since $x = y^2 = 4 \Rightarrow y = \pm 2$.

So, area = $\int_{-2}^{2} \left( 4 - y^2 \right) dy$

$= \left[ 4y - \frac{y^3}{3} \right]_{-2}^{2}$

$= \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right)$

$= \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right)$

$= \left( \frac{24 - 8}{3} \right) - \left( \frac{-24 + 8}{3} \right)$

$= \frac{16}{3} - \left( \frac{-16}{3} \right) = \frac{32}{3}$