Question

Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 2

Answer 1

For the given curve, the line element is

`ds = \sqrt{\left((dx)/(dt)\right)^2 + \left((dy)/(dt)\right)^2} \, dt = √(9t^4 + 16t^6) \, dt`

since

`x = t^3 \implies (dx)/(dt) = 3t^2`

`y = t^4 \implies (dy)/(dt) = 4t^3`

Then the line integral is

`\displaystyle \int_C \frac xy \, ds = \int_1^2 (t^3)/(t^4) \, ds = \int_1^2 \frac{√(9t^4+16t^6)}t \, dt`

Simplify the integrand to

`\displaystyle \int_1^2 t √(9 + 16t^2) \, dt`

and substitute

`u = 9 + 16t^2 \implies du = 32t \, dt`

Then the integral is

`\displaystyle \int_1^2 t √(9+16t^2) \, dt = \frac1{32} \int_(25)^(73) √(u) \, du = (73^(\frac32) - 25^(\frac32))/(48) = \boxed{(73√(73) - 125)/(48)}`