I'd like to calculate, according to the question of an exam, the "length of the perimeter" for the surface S : $z = x^2-2y^2$ with both $x$ and $y$ between $-1$ and $1$.
I searched on the forum and internet in general and I don't see what is being asked. For me a perimeter is on a 2D surface ...
On
You have to find length of four parabolic arcs formed by intersection of planes
$$ x=\pm 1, y= \pm 1$$
with a hyperbolic paraboloid
$$ z= x^2 - 2 y^2 $$
( There is an error in the above given example, but you can take coefficient correctly $2$ for $y^2$ ).
The parabola orthogonal projections are of form:
$$ z = a + b x^2,\, z = c + d y^2;\, $$
Perimeter length is of four arcs
$$ 2 (AB+BC) $$
Hope you can take it further.
The question is asking you to find the length of the boundary of the surface described by the given equation with the bounds $x=-1,x=1,y=-1,y=1$. Here is a plot of this bounded surface:
There are four curves that make up the boundary, whose equations can be derived by substituting the bound definitions into the surface equation. Two of the curves (the ones "pointing" upwards above) have equation $z=1-2y^2$ and the other two have equation $z=x^2-2$. We now need to find the arc length of these curves over $y\in[-1,1]$ or $x\in[-1,1]$ respectively, so the final answer is (using the arc length definition for functions) $$2\left(\int_{-1}^1\sqrt{1+((1-2y^2)')^2}\,dy+\int_{-1}^1\sqrt{1+((x^2-2)')^2}\,dx\right)$$ $$=2\left(\int_{-1}^1\sqrt{1+(4y)^2}\,dy+\int_{-1}^1\sqrt{1+(2x)^2}\,dx\right)$$ $$=2\left(\left[\frac{y\sqrt{16y^2+1}}2+\frac{\sinh^{-1}4y}8\right]_{-1}^1+\left[\frac{x\sqrt{4x^2+1}}2+\frac{\sinh^{-1}2x}4\right]_{-1}^1\right)$$ $$=2\left(\frac{\sinh^{-1}4}4+\sqrt{17}+\frac{\sinh^{-1}2}2+\sqrt5\right)$$ $$=\frac{\sinh^{-1}4}2+2\sqrt{17}+\sinh^{-1}2+2\sqrt5$$