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August 31st, 2017, 11:57 AM   #1
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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.
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August 31st, 2017, 12:19 PM   #2
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In what order will you be integrating with respect to $x$, $y$ and $z$?
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August 31st, 2017, 12:24 PM   #3
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$\displaystyle F = 4xi -2y^{2}j +z^{2}k$

I need to integrate divergence of this function, I prefer using rectangular coordinate system
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August 31st, 2017, 01:16 PM   #4
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Quote:
Originally Posted by MATHEMATICIAN View Post
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

$\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$
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August 31st, 2017, 01:35 PM   #5
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Quote:
Originally Posted by MATHEMATICIAN View Post
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.
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August 31st, 2017, 03:34 PM   #6
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$\displaystyle x^2+ y^2= 4$, 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 $\displaystyle \pi (2)^2(3)= 12\pi$.
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September 1st, 2017, 02:24 AM   #7
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Originally Posted by romsek View Post

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.
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September 1st, 2017, 06:50 AM   #8
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Quote:
Originally Posted by MATHEMATICIAN View Post
$\displaystyle F = 4xi -2y^{2}j +z^{2}k$

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...

$\displaystyle = \left[ 21\pi r^2 \right]^2_0$

$\displaystyle = 84 \pi$
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September 1st, 2017, 06:57 AM   #9
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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
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September 1st, 2017, 08:10 AM   #10
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Quote:
Originally Posted by romsek View Post
$\displaystyle \int_0^{2\pi} \int_0^3 \int_0^2~(4-4 r \sin(\theta)+2z)r~dr~dz~d\theta$
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]
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Last edited by agentredlum; September 1st, 2017 at 08:13 AM.
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