Question

Evaluate the line integral c y3 ds,
c.x = t3, y = t, 0 ≤ t ≤ 3

Answer 1

`\displaystyle\int_(\mathcal C)y^3\,\mathrm dS=\int_(t=0)^(t=3)t^3√(1^2+(3t^2)^2)\,\mathrm dt=\int_0^3t^3√(1+9t^4)\,\mathrm dt`

Take
`u=1+9t^4` so that
`\mathrm du=36t^3\,\mathrm dt`. Then


`\displaystyle\int_(\mathcal C)y^3\,\mathrm dS=\frac1{36}\int_(u=1)^(u=730)\sqrt u\,\mathrm du=(730^(3/2)-1)/(54)`