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📅 May 4, 2026 👤 scraface

May 04, 2026

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Let \(r = \sqrt{x^2 + y^2} = c \sin^2\phi\)
But from \(z = c \sin\phi \cos\phi = c \sin\phi \sqrt{1 - \sin^2\phi}\...
Set \(u = \sin\phi\), then:
r = c u^2, \quad z = c u \sqrt{1 - u^2}
u = rac{r^{1/2}}{\sqrt{c}}, \quad z = c \cdot rac{r^{1/2}}{\sqrt{c}}...
z^2 = c \cdot r \left(1 - rac{r^2}{c^2}
ight) = c r - rac{r^3}{c}
But \(r = rac{\sqrt{x^2 + y^2}}{1}\), so:
z^2 = c \sqrt{x^2 + y^2} - rac{(x^2 + y^2)^{3/2}}{c}
This is messy. Instead, return to the standard geometric interpretatio...

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