find the lenght of the curve at u= t^3/6, v= t where t is from -1 to 1
and E = 1, F=0 and G= 1+v^2.
do we find it from t =0 to 1 and then multiply it by 2? because if i were to integrate it from -1 to 1, then it would cancel each other out.
find the lenght of the curve at u= t^3/6, v= t where t is from -1 to 1
and E = 1, F=0 and G= 1+v^2.
do we find it from t =0 to 1 and then multiply it by 2? because if i were to integrate it from -1 to 1, then it would cancel each other out.
The length of a vector, <u, v>, on a surface with that first fundamental form would be
$\displaystyle \sqrt{\begin{bmatrix}u & v \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1+v^2\end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix}}$
$\displaystyle = \sqrt{u^2+ v^4+ v^2}= \sqrt{t^3/6+ t^4+ t^2}$
So you want to integrate
$\displaystyle \int_{-1}^1\sqrt{t^4+ t^3/6+ t^2}dt$
You can factor out a $\displaystyle t^2$ to get
$\displaystyle \int_{-1}^1 \sqrt{t^2+ t/6+ 1}tdt$
Because that, like any "metric" integral involves only even powers, the integral from -1 to 1 cannot be 0. The integrals from -1 to 0 and from 0 to 1 will add, not cancel.