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Find a parametrization of the surface $x^2-y^2=1$, where $x>0$, $-1 \leq y \leq1$ and $0 \leq z \leq 1$. Use your answer to express the area of the surface as an integral.

I'm confused because in the past I've only seen parametrizing as a change of variables... (or in the case of a graph $z=f(x,y)$, letting $x=u, y=v, z=f(u,v))$.

If I make $z=f(x,y)=x^2-y^2=1$, then I just have a hyperbola, which isn't a surface in $\mathbb R^3$.

I let $f(x,y,z)=x^2-y^2=1$ and constrained the graph to a section $x>0$, $-1\leq y \leq 1$, getting a portion of a sideways parabola on the $xy$ plane. This section of the curve is stacked from $0 \leq z \leq 1$.

I have an image for what the surface looks like but I don't see any obvious parametrization.

Bobby Lee
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    It is a surface because $z$ is arbitrary. Parametrize your curve $x^2-y^2=1$ in the $xy$-plane your favorite way, and then introduce $z$ as a second variable. – Ted Shifrin Apr 15 '14 at 22:38
  • Not sure if I understand... so could I do $u = x^2-y^2=1$ and $v=z$? Then integrate over $0 \leq v \leq 1$? But how would I form my interval for $u$? – Bobby Lee Apr 15 '14 at 22:46
  • Remember, the Hyperbolic Functions must be good for something! – Alan Apr 15 '14 at 22:56
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    No, either you need $x=\cosh u$, $y=\sinh u$ — if you know what those are — or you need $y=u$, $x=\sqrt{1+u^2}$ for part of it. – Ted Shifrin Apr 15 '14 at 23:00

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