Is the Gradient Always Normal to the Flux?

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Is it always true that the gradient of a function is normal to the flux coming out of the surface represented by the function?
 
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The gradient of a function is normal to the surfaces on which the function is constant. To see why this is, note that if you move in the plane tangent to such a surface, the function does not change value (to first order), so you must be moving perpendicular to the gradient.
 
The flux of what? Depending on the formula for the flux, the flux may be at any angle to the surface and, if your surface is a level surface for some function, then the gradient of that function is normal to the surface. That has nothing to do with the flux which may depend on an entirely different function.
 

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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