I have a curve C, which is the intersection between the plane
$$ 3z = xy $$ and the cylinder $$ x^2 = 2y $$ I come to the conclusion that i can parameterize it by $$ x = t $$ $$ y = t^2/2 $$ $$ z = t^3/6 $$ which, finally, gives me this integral to get the lenght of the curve between the origin and the point (6,18,36): $$ \int_{0}^{6} \sqrt{1+ t + t^2/2} \, \mathrm{d}t $$ But then I am lost... I am not sure if it is my parametrization which is not done right (or in the easiest way), or I am just not seeing an easy way to integrate this. An hint would be really appreciated.
Frequently, when simplifying integrals containing quadratic expressions, a useful substitution is to define a variable equal to the derivative of the quadratic expression.
$$ \frac{d}{dt}\left(1-t+\frac{t^2}{2}\right)=t-1$$
so let $s=t-1$
Then
$$ \int_0^6\sqrt{1-t+\frac{t^2}{2}}\,dt=\frac{\sqrt{2}}{2}\int_1^7\sqrt{s^2+1}\,ds$$
To complete the integral you will need the trigonometric substitution $s=\tan\theta$ and will need to use the integral of $\sec^3\theta$. By this point you should be familiar with both of these.