To solve this problem, we need to find the area of the region bounded by the curve \( y^2 = x^3 \), the y-axis, and the horizontal lines \( y = 1 \) and \( y = 8 \).
- First, interpret the given curve equation \( y^2 = x^3 \). Solving for \( x \) in terms of \( y \), we have \( x = y^{2/3} \).
- Since the region is bounded by the y-axis, we consider only the part of the graph where \( x = y^{2/3} \) from \( y = 1 \) to \( y = 8 \).
- We need to integrate \( x = y^{2/3} \) with respect to \( y \) from \( y = 1 \) to \( y = 8 \) to find the area. This is because the area under the curve from the y-axis involves integrating along \( y \), not \( x \).
- Set up the integral to find the area:
\(A = \int_{1}^{8} y^{2/3} \, dy\)
- Compute the integral:
\(\int y^{2/3} \, dy = \frac{y^{5/3}}{5/3} = \frac{3}{5} y^{5/3}\)
- Evaluate this from \( y = 1 \) to \( y = 8 \):
\(A = \left[\frac{3}{5} y^{5/3}\right]_{1}^{8} = \frac{3}{5} \left( 8^{5/3} - 1^{5/3} \right)\)
- Simplify the expression:
Calculate \( 8^{5/3} \):
- \( 8^{1/3} = 2 \), because the cube root of 8 is 2.
- \( 2^5 = 32 \). Therefore, \( 8^{5/3} = 32 \).
Now calculate:
\(A = \frac{3}{5} \left( 32 - 1 \right) = \frac{3}{5} \times 31 = \frac{93}{5}\)
- This value matches the correct answer in the options.
Therefore, the area of the region bounded by the given curve, the y-axis, and the lines \( y = 1 \) and \( y = 8 \) is \(\frac{93}{5}\) square units.