Is this parameterization correct?

  • Thread starter Thread starter andrassy
  • Start date Start date
andrassy
Messages
45
Reaction score
0
I have to parameterize the elliptic paraboloid z = 16 - x^2 - y^2 from z = 12 to z = 16. Is the correct parameterization just X(s,t) = (s*cos(t), s*sin(t), 16 - s^2 - t^2) where s ranges from 0 to 2 and t ranges from 0 to 2pi?
 
Physics news on Phys.org
I don't think so, since 16-(s \cos t)^2 - (s \sin t)^2 \neq 16 - s^2 - t^2.

Since you already have z=f(x,y), you can just let x,y be your parametric variables s,t.
 
Defennder's suggestion is the simplest.

If you really are determined to use polar coordinates (which is what you appear to be doing), then x= s cos t, y= s sin t and then

z= 16- x2- y2= what?
 
I had thought about doing it that way as well with (s, t, 16 - s^2 - t^2), but I don't know what s and t range from then. Is it just from 0 to 2 for both?

whoops also, before I meant to make my parameterization (s cost, s sint, 16 - s^2). Would this be right doing it that way? Because 16 - (s cost)^2 - (s sint)^2 = 16 - s^2(cos^2 + sin^2) = 16 - s^2. Then the s would range from 0 to 2. Is this right?
 
Last edited:
You said "z = 12 to z = 16" which would project to the xy-plane as the circle from r= 0 to r= 2. Yes, taking x= s cos(t), y= s sin(t), z= 16- s2, essentially polar coordinates, s ranges from 0 to 2 and t from 0 to 2\pi.
 
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...
Back
Top