Evaluate the line integral, where C is the given curve.
C y^3 ds, C: x = t^3, y = t, 0 ≤ t ≤ 2
To evaluate the line integral ∫C y^3 ds, we need to parameterize the given curve C: x = t^3, y = t, where 0 ≤ t ≤ 2.
First, let's express ds in terms of t. The arc length formula tells us that ds = sqrt((dx/dt)^2 + (dy/dt)^2) dt.
Taking the derivatives, we have dx/dt = 3t^2 and dy/dt = 1. Substituting these values into ds, we get ds = sqrt((3t^2)^2 + 1^2) dt = sqrt(9t^4 + 1) dt.
Now, we can rewrite the line integral as ∫C y^3 ds = ∫(0 to 2) (t^3)(sqrt(9t^4 + 1)) dt.
To evaluate this integral, you can use integration techniques like substitution or integration by parts. In this case, substitution is a good choice.
Let u = 9t^4 + 1. Then, du = 36t^3 dt.
Now, we can rewrite the integral as ∫(0 to 2) ((t^3)(sqrt(9t^4 + 1)) dt = ∫(0 to 2) ((sqrt(u))/36) du.
After substituting the limits of integration, the integral becomes ∫(1 to 145) ((sqrt(u))/36) du.
Now, evaluate this integral using antiderivatives and the fundamental theorem of calculus to get the final answer.