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