Surface area of parabolic sheet

cscott
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Am I correct in saying the surface parameterized by r = (sin v, u, cos v), v = [-pi/2, pi/2], u = [-1, 1] has an area of 2pi ?

I get something different by computing the arc length of the parabola within the bounds and multiplying by 2.

Which method is wrong?
 
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Sure, but that is not a parabolic sheet, it is the half-shell of a cylinder with radius 1..
 
Hmm.. I see. Thanks.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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