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September 12th, 2017, 09:50 PM   #1
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Volume integration

$\displaystyle F=(2x^{2}-3z)i-2xy-4xk$
Evaluate volume integration of $\displaystyle \bigtriangledown \cdot F$ over a vloume bounded by the planes $\displaystyle x=0$, $\displaystyle y=0$, $\displaystyle z=0$ and $\displaystyle 2x+3y+z =4$

I got answer as $\displaystyle 16/9$, is it correct?

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September 12th, 2017, 10:04 PM   #2
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should this be

$F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ?
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September 12th, 2017, 10:06 PM   #3
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should this be

$F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ?
Yes !!
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September 12th, 2017, 10:10 PM   #4
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also is the volume bounded below by $z=0$ ?
Yes !!
The volume is in 1st octant
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September 12th, 2017, 10:20 PM   #5
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Yes !!
The volume is in 1st octant
$F=(2x^2 - 3z, -2xy, -4x)$

$\nabla \cdot F = 4x -2x+ 0 = 2x$

$\displaystyle \int_0^2 \int_0^{\frac{4-2x}{3}} \int_0^{4-2x-3y}~2x~dz~dy~dx = \dfrac{16}{9}$

You win!
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September 12th, 2017, 10:22 PM   #6
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Originally Posted by romsek View Post
$F=(2x^2 - 3z, -2xy, -4x)$

$\nabla \cdot F = 4x -2x+ 0 = 2x$

$\displaystyle \int_0^2 \int_0^{\frac{4-2x}{3}} \int_0^{4-2x-3y}~2x~dz~dy~dx = \dfrac{16}{9}$

You win!
Thank you so much romsek
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September 12th, 2017, 10:31 PM   #7
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What is the physical meaning of volume integration of divergence of a vector function?
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September 12th, 2017, 10:58 PM   #8
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Romsek

What is the physical meaning of volume integration of divergence of a vector function?
This comes up when applying Gauss's theorem to compute flux integrals.

Roughly the divergence of a field produces a "charge" density, and the volume integral of this "charge" density is equal to the surface integral of the field through the boundary surface of the volume.

I put "charge" in quotes because it can be any sort of field, not just an electric field.
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