# Evaluate the line integral, where C is the given curve. \int y3\ ds,\ C\div x=t3,\ y=t,\ 0?\ t?\ 3

Evaluate the line integral, where C is the given curve.
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Step 1

$ds=|{r}^{\prime }\left(t\right)|dt$
$ds=\sqrt{\left(3{t}^{2}{\right)}^{2}+{1}^{2}}dt$
$ds=\sqrt{\left(9{t}^{4}\right)+1}dt$
Integral ${y}^{3}ds$
Integral

Let $\left(9{t}^{4}\right)+1=u$
$⇒9×4{t}^{3}dt+0=du⇒{t}^{3}dt=\frac{dt}{36}$
$⇒{\int }^{3}\cdot \sqrt{\left(9{t}^{4}\right)+1}dt$
$⇒\int \sqrt{u}\frac{du}{36}$
$⇒\left(\frac{1}{36}\right)\frac{{u}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+c$
$⇒\left(\frac{1}{54}\right){u}^{\frac{3}{2}}+c$
$⇒\left(\frac{1}{54}\right)\left[\left(9{t}^{4}\right)+1{\right]}^{\frac{3}{2}}+c$
${\int }_{0}^{3}{t}^{3}\cdot \sqrt{\left(9{t}^{4}\right)+1\right]}dt={\int }_{0}^{3}\left(\frac{1}{54}\right)\left[\left(9{t}^{4}\right)+1{\right]}^{\frac{3}{2}}+c$
$⇒\left(\frac{1}{54}\right)\left[\left(9×{3}^{4}\right)+1{\right]}^{\frac{3}{2}}+c-\left(\frac{1}{54}\right)\left[\left(9×{0}^{4}\right)+1{\right]}^{\frac{3}{2}}-c$
$⇒\left(\frac{1}{54}\right)\left[730{\right]}^{\frac{3}{2}}-\left(\frac{1}{54}\right)$
$⇒\left(\frac{1}{54}\right)\left[{730}^{\frac{3}{2}}-1\right]$
$⇒365.23$