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We are tasked with finding the area bounded by the curve \( y = x^3 \), the line \( y = 8 \), and the vertical line \( x = 0 \).

### Step-by-step Solution:

1. **Intersection Points:**
   - The curve given is \( y = x^3 \). For the line \( y = 8 \), we find the corresponding \( x \) values by solving:
     \[
     y = x^3 = 8 \implies x = 2.
     \]
   - So, the area is bounded between \( x = 0 \) and \( x = 2 \).

2. **Area Calculation:**
   The area between the curve \( y = x^3 \) and the line \( y = 8 \) can be calculated using the integral:
   \[
   \text{Area} = \int_0^2 \left( 8 - x^3 \right) dx.
   \]
   
3. **Compute the Integral:**
   \[
   \int_0^2 \left( 8 - x^3 \right) dx = \int_0^2 8 \, dx - \int_0^2 x^3 \, dx.
   \]
   We now calculate each term:
   - \( \int_0^2 8 \, dx = 8x \Big|_0^2 = 8(2) - 8(0) = 16. \)
   - \( \int_0^2 x^3 \, dx = \frac{x^4}{4} \Big|_0^2 = \frac{(2)^4}{4} - \frac{(0)^4}{4} = \frac{16}{4} = 4. \)
   
4. **Final Area:**
   \[
   \text{Area} = 16 - 4 = 12 \, \text{square units}.
   \]

Therefore, the correct answer is **(C) 12 square units**.

Would you like more details on this or have any other questions?

### Related Questions:
1. How is the area under a curve calculated using integrals?
2. What is the geometric significance of setting \( x = 0 \) in area problems?
3. How can you visualize the region bounded by \( y = x^3 \) and \( y = 8 \)?
4. What is the process to calculate definite integrals?
5. How does the power rule for integration apply in this example?

**Tip:** Always double-check the limits of integration, as they define the boundaries of the region under consideration.

Area bounded by y = x^3, y = 8, and x = 0 is:

Find the area bounded by the curve y = x^3, the line y = 8, and x = 0 using definite integrals. This detailed solution includes step-by-step calculations suitable for students in Grades 11-12.

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