The problem presented is asking to evaluate a line integral of the form:

$\int_C y^3 \, ds$

where the curve $C$ is given parametrically as $x = t^3$ and $y = t$, with $0 \leq t \leq 5$.

Steps to solve the line integral:

1. Calculate the differential of arc length $ds$: The formula for $ds$ for a curve parametrized by $t$ is:

   $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

   Given that $x = t^3$ and $y = t$, let's calculate the derivatives:

   $\frac{dx}{dt} = 3t^2, \quad \frac{dy}{dt} = 1$

   Therefore,

   $ds = \sqrt{(3t^2)^2 + (1)^2} \, dt = \sqrt{9t^4 + 1} \, dt$

2. Substitute the values into the integral:

   The integrand is $y^3$, and since $y = t$, we have $y^3 = t^3$. Thus, the line integral becomes:

   $\int_0^5 t^3 \sqrt{9t^4 + 1} \, dt$

3. Evaluate the integral:

   This is a non-trivial integral that likely requires numerical integration. I will now compute this integral.

Let me calculate that for you.The correct value of the line integral is approximately:

$7814.56$

This result differs slightly from the provided value $7808.46$, possibly due to rounding or a calculation error in the original attempt.

Would you like any additional details on the steps, or do you have further questions?

Here are five related questions to consider:

1. How does the formula for $ds$ change for different types of parameterized curves?
2. How do you calculate the differential arc length for a curve in three dimensions?
3. Can you explain the physical interpretation of a line integral in vector fields?
4. What other numerical methods could be used for evaluating more complex integrals?
5. How do line integrals relate to work done by a force field?

Tip: When working with parametric curves, always check the bounds of the parameter carefully to ensure you're integrating over the correct portion of the curve.