How to calculate volume of a rotated polar function

  • #1

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Hello,

I've just found a book which mentions the formula for calculating the volume of a rotated polar function:

[tex]\int_{\theta_1}^{\theta_2} \frac{2}{3} \pi r^3 sin(\theta) d\theta [/tex]



How does one calculate this? In an https://www.physicsforums.com/showthread.php?t=457896", I calculated that the volume would be

[tex]\int_{\theta_1}^{\theta_2} \pi r^2 sin(\theta) d\theta [/tex]


if one just added up cones with the side of the cone being the function [tex]f(\theta)[/tex]. This method would be similar to shells, but apparently I'm [tex]\frac{2}{3} r [/tex] off.

If anyone could help me understand how to calculate volume I would be eternally grateful.

Thanks for taking the time to read this!
 
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  • #2
mathman
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Your formula (the second one) is dimensionally an area, not a volume.
 
  • #3
I'm actually working on this as we speak, and realized, of course, that my formula was not correct. However, I decided to go back to spherical sectors.

If you look at the 1st image http://mathworld.wolfram.com/SphericalSector.html" [Broken], imagine that the initial line was the line going directly through the sphere. Then you can assume that the sphere is actually a rotation of the polar function r=1 over the initial line. You can calculate the infinitesimally small Volume (dV) by finding the spherical sectors.

That would essentially mean that the cones are infinitely close together (essentially if you had a [tex] d\theta [/tex] separating them), then you could calculate [tex] dV[/tex]. However, it would seem that the h-value for dV would be something like

[tex] r*cos(\theta) - r*cos(\theta + d\theta) [/tex]. That doesn't really make sense in the context of an integral...

Could you please help me set up such an integral?

Thank you so much for your time!!
 
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  • #4
SammyS
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Hello,

I've just found a book which mentions the formula for calculating the volume of a rotated polar function:

[tex]\int_{\theta_1}^{\theta_2} \frac{2}{3} \pi r^3 \sin(\theta) d\theta [/tex]
...

Thanks for taking the time to read this!

The formula you found is the correct formula! I plan to write up the derivation on the other site on which you posted this question.

If [tex]\display r=f(\theta)[/tex], then:

[tex]\display V=\int_{\theta_1}^{\theta_2} \frac{2}{3} \pi \left(f(\theta)\right)^3 \sin(\theta) d\theta \, ,[/tex]

where [tex]\display 0\le\ \theta_1,\,\theta_2\le\pi.[/tex]

 
  • #5
Thank you very much! I look forward to your derivation!
 
  • #6
SammyS
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Thank you very much! I look forward to your derivation!
I've drawn a figure of a narrow sector of a circle. The circle is centered at the origin and has a radius of [tex]\displaystyle r[/tex]. The sector is oriented at an angle of [tex]\displaystyle \theta[/tex] with respect to the positive x-axis and subtends an angle of [tex]\displaystyle \text{d}\theta[/tex]. The figure is an attachment to this post.

As I understand it, your proposal is to obtain a volume element which can be used to calculate the volume of a solid of revolution, where the cross-section of the solid is given in polar coordinates as, [tex]\displaystyle r=f(\theta).[/tex] To accomplish this, we will find the volume of the solid generated by revolving this circular sector about the x-axis.

In the figure, we have an element of area, [tex]\displaystyle \text{d}A={{\text{d}x}\over{\cos \theta}}[/tex] [tex]\cdot {{x\ \text{d}\theta}\over{\cos \theta}}[/tex]. The volume swept out by the element of area, [tex] \text{d}A,[/tex] is:
[tex]\displaystyle 2\pi y\ \text{d}A =[/tex] [tex]\displaystyle 2\pi x\tan\theta\ \text{d}A .[/tex]

This will be integrated over x, from 0 to [tex]\displaystyle r\cos\theta ,[/tex] leaving an expression containing [tex]\displaystyle \text{d}\theta[/tex].

[tex]\displaystyle \text{d}V=[/tex][tex]\displaystyle \int_{0}^{r\cos\theta} 2\pi x\tan\theta\ {{x\,\text{d}x \,\text{d}\theta}\over{\cos^2 \theta}}[/tex]

[tex]\displaystyle =\int_{0}^{r\cos\theta} {{2\pi x^2\sin\theta}\over{\cos^3 \theta}}\ \text{d}x \,{d}\theta[/tex]

[tex]\displaystyle ={{2\pi \sin\theta \,{\text{d}\theta}\over{\cos^3 \theta}}\int_{0}^{r\cos\theta} x^2\text{d}x[/tex]

[tex]\displaystyle ={{2\pi r^3 \sin\theta \,{\text{d}\theta}\over{3 }}[/tex]

Check this for a sphere of radius, R. Integrate over [tex]\displaystyle \theta[/tex] from 0 to [tex]\displaystyle \pi[/tex].

 
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  • #7
Your derivation is very interesting! I hope someday I can get to this level of derivation- I've only just completed the BC track and will be entering multivariable calc next semester.

Thank you for you help, on both forums!
 
  • #8
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Thank you progressive for posing this question and finding an answer, and thanks SammyS for such a thorough explanation. (I'm new to physicsforums.com, so haven't found your diagram attached to your derivation of the formula.
This is a relief, because I've been mulling out this very problem for several days.

SammyS, how would this formula change if we stated theta as a function of r? Sometimes it's easier/possible to separate variables this way.
 
  • #9
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By using this formula, I was able to find the volume of the solid obtained by revolving the cardioid r = a(1+cos(theta)) about the initial line. V = (8/3)*pi*a^3, twice the volume of the sphere r = a.
 

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