August 31st, 2017, 11:57 AM  #1 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित  Volume integration
If I am supposed to do volume Integration over a region bounded by $\displaystyle x^{2}+y^{2}=4$ $\displaystyle z=0$ $\displaystyle z=3$ What limits should I take for x, y and z? Last edited by skipjack; August 31st, 2017 at 12:17 PM. 
August 31st, 2017, 12:19 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,978 Thanks: 2229 
In what order will you be integrating with respect to $x$, $y$ and $z$?

August 31st, 2017, 12:24 PM  #3 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित 
$\displaystyle F = 4xi 2y^{2}j +z^{2}k$ I need to integrate divergence of this function, I prefer using rectangular coordinate system 
August 31st, 2017, 01:16 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  Quote:
$\displaystyle \int_V ~4  4y+2z~dV$ you can do this $\displaystyle \int_0^3 \int_{2}^2 \int_{\sqrt{4x^2}}^{\sqrt{4x^2}}~4  4y+2z~dy~dx~dz$ but I think I'd do $\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(44 r \sin(\theta)+2z)r~dr~dz~d\theta$  
August 31st, 2017, 01:35 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723  If you convert (x,y) to polar coordinates, it is quite simple. R goes from 0 to 2, angle is complete circle and z goes from 0 to 3.

August 31st, 2017, 03:34 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
$\displaystyle x^2+ y^2= 4$, in the xyplane, is a circle with center at (0, 0) and radius 2. In three dimensions, it is a cylinder with axis along the zaxis with radius 2. Given that z goes from 0 to 3, it is a cylinder with radius 2 and height 3. Its volume is $\displaystyle \pi (2)^2(3)= 12\pi$.

September 1st, 2017, 02:24 AM  #7  
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित  Quote:
I am trying to prove divergence theorem but I am not able to get correct answer Last edited by MATHEMATICIAN; September 1st, 2017 at 02:58 AM.  
September 1st, 2017, 06:50 AM  #8  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
$\displaystyle \nabla \cdot F = 4  4y +2z = 4  4r\sin\theta + 2z$ $\displaystyle \int \int \int \nabla \cdot F dV = \int_0^{2\pi} \int_0^3 \int_0^2~(44 r \sin(\theta)+2z)r~dr~dz~d\theta$ Integrate firstly over angle... $\displaystyle = \int_0^3 \int_0^2~\left[(4\theta + 4 r\cos(\theta) + 2z\theta)r\right]^{2\pi}_0~dr~dz$ $\displaystyle = \int_0^3 \int_0^2~(8\pi + 4 \pi z)r~dr~dz$ Integrate secondly over z... $\displaystyle = \int_0^2~\left[(8\pi z + 2 \pi z^2)r\right]^3_0~dr$ $\displaystyle = \int_0^2~(24\pi + 18 \pi )r~dr$ $\displaystyle = \int_0^2~42\pi r~dr$ Integrate lastly over r... $\displaystyle = \left[ 21\pi r^2 \right]^2_0$ $\displaystyle = 84 \pi$  
September 1st, 2017, 06:57 AM  #9  
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित  Quote:
what about the surface integration over the curved surfaces, taking integration function $\displaystyle F\cdot \hat{n}$ where, $\displaystyle \hat{n}$ is unit vector perpendicular to the curved surface i.e. $\displaystyle \frac{x\vec{i} + y\vec{j}}{\sqrt{2}}$ Last edited by MATHEMATICIAN; September 1st, 2017 at 07:02 AM. Reason: spelling error  
September 1st, 2017, 08:10 AM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
From the inside out simplify the parenthesis , multiply by r $ 4r 4r^2 \sin(\theta) + 2zr $ Integrating with respect to $ \ r \ $ treating $ \ z \ $ and $ \ \sin(\theta) \ $ as constants $ \frac{4r^2}{2}  \frac{4r^3}{3} \sin(\theta) + \frac{2zr^2}{2} =$ $ 2r^2  \frac{4r^3}{3} \sin(\theta) + zr^2 $ Plug in the limits of integration for $ \ r \ $ , note lower limit evaluates to $ \ 0 \ $ $ 2(2)^2  \frac{4(2)^3}{3} \sin(\theta) + z(2)^2 = $ $ 8  \frac{32}{3} \sin(\theta) + 4z $ Intgrating with respect to $ \ z \ $ keeping $ \ \sin(\theta) \ $ constant $ 8z  \frac{32z}{3} \sin(\theta) + \frac{4z^2}{2} $ Plug in limits of integration for $ \ z \ $ , note lower limit evaluates to $ \ 0 \ $ $ 24  32 \sin(\theta) + 18 = $ $ 42  32 \sin(\theta) $ Intgrating with respect to $ \ \theta \ $ $ 42 \theta + 32 \cos(\theta) $ Plug in the limits of integration $ 42(2 \pi) + 32 \cos (2 \pi) \{ 42(0) + 32 \cos (0) \} = $ $84 \pi + 32  ( 0 + 32) $ Final answer $ \ 84 \pi $ If I made errors I apologise. [EDIT] Benit13 beat me to it. Took me 2 hours to type this in LaTeX and proof read it [ENDEDIT] Last edited by agentredlum; September 1st, 2017 at 08:13 AM.  

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