The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is

The equation of the curve is given by $y^2 = x^3$, which can be written as $x = y^{2/3}$.

The required area is bounded by the curve $x = y^{2/3}$, the $y$-axis ($x = 0$), and the horizontal lines $y = 1$ and $y = 8$.

The area $A$ is given by the definite integral with respect to $y$:

$A = \int_{1}^{8} x \, dy$

Substituting $x = y^{2/3}$:

$A = \int_{1}^{8} y^{2/3} \, dy$

Integrating the function:

$A = \left[ \dfrac{y^{5/3}}{5/3} \right]_{1}^{8}$

$A = \dfrac{3}{5} \left[ y^{5/3} \right]_{1}^{8}$

Evaluating the limits:

$A = \dfrac{3}{5} \left( 8^{5/3} - 1^{5/3} \right)$

Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:

$A = \dfrac{3}{5} (32 - 1)$

$A = \dfrac{3}{5} \times 31 = \dfrac{93}{5}$ sq. units.

Answer: $\dfrac{93}{5}$ sq. units