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=\frac{x^2}2\implies y(t)=\frac{t^2}2`

`3z=xy\iff z=\frac{xy}3\implies z(t)=\frac{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((\mathrm dx)/(\mathrm dt)\right)^2+\left((\mathrm dy)/(\mathrm dt)\right)^2+\left((\mathrm dz)/(\mathrm dt)\right)^2}\,\mathrm dt`. We have


`(\mathrm dx)/(\mathrm dt)=1`

`(\mathrm dy)/(\mathrm dt)=t`

`(\mathrm dz)/(\mathrm dt)=\frac{t^2}2`

and so the line integral is


`\displaystyle\int_(t=0)^(t=2)\sqrt{1^2+t^2+\frac{t^4}4}\,\mathrm dt`

This result is fortuitous, since we can write


`1+t^2+\frac{t^4}4=\frac14(t^4+4t^2+4)=\frac{(t^2+2)^2}4=\left(\frac{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=\frac{10}3`