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 September 12th, 2017, 09:50 PM #1 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 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? Last edited by MATHEMATICIAN; September 12th, 2017 at 09:54 PM. September 12th, 2017, 10:04 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,505 Thanks: 1374 should this be $F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ? September 12th, 2017, 10:06 PM   #3
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 Originally Posted by romsek should this be $F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ?
Yes !! September 12th, 2017, 10:10 PM   #4
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 Originally Posted by romsek also is the volume bounded below by $z=0$ ?
Yes !!
The volume is in 1st octant September 12th, 2017, 10:20 PM   #5
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 Originally Posted by MATHEMATICIAN 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! September 12th, 2017, 10:22 PM   #6
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 Originally Posted by romsek $F=(2x^2 - 3z, -2xy, -4x)$ $\nabla \cdot F = 4x -2x+ 0 = 2x$ $\displaystyle \int_0^2 \int_0^{\frac{4-2x}{3}} \int_0^{4-2x-3y}~2x~dz~dy~dx = \dfrac{16}{9}$ You win!
Thank you so much romsek  September 12th, 2017, 10:31 PM #7 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Romsek What is the physical meaning of volume integration of divergence of a vector function? September 12th, 2017, 10:58 PM   #8
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 Originally Posted by MATHEMATICIAN 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. Tags integration, volume Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MATHEMATICIAN Pre-Calculus 13 September 2nd, 2017 09:21 PM MATHEMATICIAN Calculus 13 September 1st, 2017 11:12 AM xl5899 Calculus 2 December 10th, 2015 09:09 AM jiasyuen Calculus 9 March 29th, 2015 08:37 PM izseekzu Calculus 1 January 26th, 2010 06:34 PM

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