Parameterizing a Cone between Z=2 and Z=3 | r(u,v) = (ucos(v), u(sin(v), u) • Physics Forums

Lancelot59 · 2011-03-22T20:20:29+00:00

I'm given a problem where I need to parameterize a cone, but only the segment between two planes, being z=2 and z=3. This is what I ended up with: r(u,v)=(ucos(v),u(sin(v),u) u:[2,3] v:[0,2\pi] Is this right?

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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Parameterizing a Cone between Z=2 and Z=3 | r(u,v) = (ucos(v), u(sin(v), u)

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

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