My Math Forum Volume integration

 Calculus Calculus Math Forum

 September 12th, 2017, 10:50 PM #1 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 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 10:54 PM.
 September 12th, 2017, 11:04 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,601 Thanks: 816 should this be $F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ?
September 12th, 2017, 11:06 PM   #3
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 872
Thanks: 60

Math Focus: सामान्य गणित
Quote:
 Originally Posted by romsek should this be $F=(2x^2 - 3z)\hat{i} - 2xy \hat{j} - 4x \hat{k}$ ?
Yes !!

September 12th, 2017, 11:10 PM   #4
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 872
Thanks: 60

Math Focus: सामान्य गणित
Quote:
 Originally Posted by romsek also is the volume bounded below by $z=0$ ?
Yes !!
The volume is in 1st octant

September 12th, 2017, 11:20 PM   #5
Senior Member

Joined: Sep 2015
From: Southern California, USA

Posts: 1,601
Thanks: 816

Quote:
 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, 11:22 PM   #6
Math Team

Joined: Jul 2013
From: काठमाडौं, नेपाल

Posts: 872
Thanks: 60

Math Focus: सामान्य गणित
Quote:
 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, 11:31 PM #7 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित Romsek What is the physical meaning of volume integration of divergence of a vector function?
September 12th, 2017, 11:58 PM   #8
Senior Member

Joined: Sep 2015
From: Southern California, USA

Posts: 1,601
Thanks: 816

Quote:
 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 Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MATHEMATICIAN Pre-Calculus 13 September 2nd, 2017 10:21 PM MATHEMATICIAN Calculus 13 September 1st, 2017 12:12 PM xl5899 Calculus 2 December 10th, 2015 10:09 AM jiasyuen Calculus 9 March 29th, 2015 09:37 PM izseekzu Calculus 1 January 26th, 2010 07:34 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top