Question

Evaluate the line integral, where c is the given curve. c y3 ds,

c.x = t3, y = t, 0 ≤ t ≤ 2

Answer 1

Parameterizing
`\mathcal C` by


`\mathbf r(t)=(t^3,t)`

with
`0\le t\le2`, we have


`\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=√((3t^2)^2+1^2)\,\mathrm dt=√(9t^4+1)\,\mathrm dt`

So


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

`=\displaystyle\frac1{36}\int_0^236t^3√(9t^4+1)\,\mathrm dt`

`=\displaystyle\frac1{36}\int_(u=1)^(u=145)\sqrt u\,\mathrm du`

`=(145^(3/2)-1)/(54)`