How to Parameterize a Plane Between Two Given Planes

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Homework Statement

Find parametric equations for the portion of the plane x+y = 1 that extends between the planes z = -1 and z = 1

The attempt at a solution

z = u -1\lequ\leq1

x = ?
y = ?
 
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Can you write x + y = 1 in parametric form? I.e., x = an expression in t (the parameter), y = another expression in t.

For z, all you need is -1 \leq z \leq 1. No parameter needed.
 
What do you mean?

I can have x and y equal anything as long as they both add up to 1
 
Like if I made x = 3t + t^2, then y = 1 - 3t - t^2
 
Why not let x = t? Why would you pick x = 3t + t^2?
 
To parameterize a surface you need two parameters. They can't be x and y in this problem because one of them determines the other. So try perhaps x and z as your parameters.

R(x,z) = < ?, ?, ?>

where you express the x, y, and z components in terms of x and z. It's really easy...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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