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April 17th, 2017, 05:13 PM   #1
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3D polar coordinates problem

Find the volume below z = r, above the x-y plane, and inside .

Answer: 4/9

Is there anyone online tool I could use to plot the 3d polar coordinates? I couldn't approach the problem without any visual clues...
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April 17th, 2017, 06:00 PM   #2
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You can use desmos graphics calculator at
https://www.desmos.com/calculator

If you click on the wrench icon ("Graph Settings") then "Grid" and click on the polar coordinates icon.
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April 17th, 2017, 06:01 PM   #3
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I use the Mathematica function ParametricPlot3D to plot all these things.

There is limited access to it at wolframalpha.com

You'll have to read up on how to use it.

this one is

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ~\int_0^{\cos(\theta)} \int_0^r~r~dz~dr~d\theta$
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April 18th, 2017, 05:04 PM   #4
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Quote:
Originally Posted by romsek View Post
I use the Mathematica function ParametricPlot3D to plot all these things.

There is limited access to it at wolframalpha.com

You'll have to read up on how to use it.

this one is

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ~\int_0^{\cos(\theta)} \int_0^r~r~dz~dr~d\theta$

How did you come up the boundaries (why -pi/2 and pi/2)? It would be great if you could explain a little...

Last edited by zollen; April 18th, 2017 at 05:07 PM.
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April 18th, 2017, 05:07 PM   #5
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Quote:
Originally Posted by zollen View Post
How did you come up the boundaries? It would be great if you could explain a little...
do you understand what $r = \cos(\theta)$ is?

and do you understand what

$z = r$ is?

see if you can sketch the first one in 2d and the 2nd one in 3D
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April 18th, 2017, 05:10 PM   #6
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I don't know how to plot z=r polar coordinate in 3D...
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April 18th, 2017, 05:12 PM   #7
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Quote:
Originally Posted by zollen View Post
I don't know how to plot z=r polar coordinate in 3D...
bah.

forget about plotting it, can you sketch it? can you at least visualize it?

it's a pretty familiar shape
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April 18th, 2017, 05:29 PM   #8
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File Type: jpg Clipboard02.jpg (46.4 KB, 6 views)
Thanks from greg1313, topsquark and zollen
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April 18th, 2017, 05:38 PM   #9
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Thanks. I am going to study your example closely..
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April 18th, 2017, 05:54 PM   #10
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this will probably also help

$r = \cos(\theta)$

$r^2 = r \cos(\theta)$

wave the magic wand

$x^2 + y^2 = x$

$x^2 - x + y^2 = 0$

$\left(x-\dfrac 1 2\right)^2 +y^2 = \dfrac 1 4$

i.e. circle at $\left( \dfrac 1 2 , 0\right)$ with radius $\dfrac 1 2$
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