Question

Evaluate the line integral, where c is the given curve. C y3 ds, c: x = t3, y = t, 0 ≤ t ≤ 5.

Answer 1

We have

x(t) = t³   ⇒   dx/dt = 3t²

y(t) = t   ⇒   dy/dt = 1

so that the line element is

ds = √((dx/dt)² + (dy/dt)²) dt = √(9t⁴ + 1) dt

and the line integral is

$\displaystyle \int_C y^3 \, ds = \int_0^5 t^3 \sqrt{9t^4 + 1} \, dt$

Substitute u = 9t⁴ + 1 and du = 36t³ dt :

$\displaystyle \int_C y^3 \, ds = \frac1{36} \int_1^{5626} \sqrt u \, du = \frac1{36} \cdot \frac23 \left((5626^{\frac32} - 1^{\frac32}\right) = \boxed{\frac{5626^{\frac32} - 1}{54}}$