Question

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

Answer 1

Parameterizing $\mathcal C$ by

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

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

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt=\sqrt{(3t^2)^2+1^2}\,\mathrm dt=\sqrt{9t^4+1}\,\mathrm dt$

So

$\displaystyle\int_{\mathcal C}y^3\,\mathrm ds=\int_{t=0}^{t=2}t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_0^236t^3\sqrt{9t^4+1}\,\mathrm dt$
$=\displaystyle\frac1{36}\int_{u=1}^{u=145}\sqrt u\,\mathrm du$
$=\dfrac{145^{3/2}-1}{54}$