Evaluating Integral F dot dA using Divergence Theorem

joemama69
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Homework Statement



use the divergence theorem to evaluate the integral F dot dA

F = (2x-z)i + x2yj + xz2k

s is the surface enclosing the unit cube and oriented outward

Homework Equations





The Attempt at a Solution



is the the region from -1 to 1 for x y and z

div F = x2 + 2xz + 2

x2 + 2xz + 2 dx = x3/3 + x2z + 2x from -1 to 1

14/3 dy = 14y/3 from -1 to 1 = 28/3

28/3 dz = 28z/3 from -1 to 1 = 56/3

is this correct
 
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No, the "unit cube" is defined by 0\le x\le 1, 0\le y\le 1, 0\le z\le 1.
 

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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