Question

Let c be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. find the exact length of c from the origin to the point (2, 2, 4/3).

Answer 1

Parameterize the intersection by setting $x(t)=t$, so that

$x^2=2y\iff y=\dfrac{x^2}2\implies y(t)=\dfrac{t^2}2$
$3z=xy\iff z=\dfrac{xy}3\implies z(t)=\dfrac{t^3}6$

The length of the path $C$ is then given by the line integral along $C$,

$\displaystyle\int_C\mathrm dS$

where $\mathrm dS=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt$. We have

$\dfrac{\mathrm dx}{\mathrm dt}=1$
$\dfrac{\mathrm dy}{\mathrm dt}=t$
$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{t^2}2$

and so the line integral is

$\displaystyle\int_{t=0}^{t=2}\sqrt{1^2+t^2+\dfrac{t^4}4}\,\mathrm dt$

This result is fortuitous, since we can write

$1+t^2+\dfrac{t^4}4=\dfrac14(t^4+4t^2+4)=\dfrac{(t^2+2)^2}4=\left(\dfrac{t^2+2}2\right)^2$

and so the integral reduces to

$\displaystyle\int_{t=0}^{t=2}\frac{t^2+2}2\,\mathrm dt=\dfrac{10}3$