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 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
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Quote:
 Originally Posted by MATHEMATICIAN 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?
So

$\displaystyle \int_V ~4 - 4y+2z~dV$

you can do this

$\displaystyle \int_0^3 \int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}~4 - 4y+2z~dy~dx~dz$

but I think I'd do

$\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$ August 31st, 2017, 01:35 PM   #5
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Quote:
 Originally Posted by MATHEMATICIAN 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?
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 xy-plane, is a circle with center at (0, 0) and radius 2. In three dimensions, it is a cylinder with axis along the z-axis 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
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Quote:
 Originally Posted by romsek but I think I'd do $\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$
What is the value of the integration?
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
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Quote:
 Originally Posted by MATHEMATICIAN $\displaystyle F = 4xi -2y^{2}j +z^{2}k$ I need to integrate divergence of this function, I prefer using rectangular coordinate system
I get $\displaystyle 84\pi$ using cylindrical polar coordinates.

$\displaystyle \nabla \cdot F = 4 - 4y +2z = 4 - 4r\sin\theta + 2z$

$\displaystyle \int \int \int \nabla \cdot F dV = \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$

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
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 Originally Posted by Benit13 I get $\displaystyle 84\pi$ using cylindrical polar coordinates. $\displaystyle \nabla \cdot F = 4 - 4y +2z = 4 - 4r\sin\theta + 2z$ $\displaystyle \int \int \int \nabla \cdot F dV = \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$ 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... $\displaystyle = \left[ 21\pi r^2 \right]^2_0$ $\displaystyle = 84 \pi$
thanks. i get the same 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
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Quote:
 Originally Posted by romsek $\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$
I will attempt to integrate this. It's been a long time since I've done any integration so please don't be too harsh 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. 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 xl5899 Calculus 2 December 10th, 2015 09:09 AM jiasyuen Calculus 9 March 29th, 2015 08:37 PM PhysicsWizard Calculus 2 March 25th, 2014 06:47 AM pjlloyd100 Calculus 3 February 18th, 2013 11:51 AM izseekzu Calculus 1 January 26th, 2010 06:34 PM

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