Question

Evaluate the line integral, where c is the given curve. c y3 ds, c.x = t3, y = t, 0 ≤ t ≤ 4

Answer 1

The path $C$ is parameterized by

$\mathbf r(t)=\langle x(t),y(t)\rangle=\langle t^3,t\rangle$

with $0\le t\le4$.

We have

$\dfrac{\mathrm d\mathbf r}{\mathrm dt}=\langle3t^2,1\rangle$
$\left\|\dfrac{\mathrm d\mathbf r}{\mathrm dt}\right\|=\sqrt{9t^4+1}$

So the line integral is

$\displaystyle\int_Cy^3\,\mathrm dS=\int_{t=0}^{t=4}t^3\sqrt{9t^4+1}\,\mathrm dt$

Substitute $u=9t^4+1$, so that $\mathrm du=36t^3\,\mathrm dt$.

$=\displaystyle\frac1{36}\int_{u=1}^{u=2305}\sqrt u\,\mathrm du$
$=\dfrac{2305^{3/2}-1}{54}$
$\approx2049.31$