Triple Integration of a Sphere in Cylindrical Coordinates

  • Thread starter Zarlucicil
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  • #1
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Homework Statement



The problem was to find the volume enclosed by a sphere of radius "a" centered on the origin by crafting a triple integral and solving for it using cylindrical coordinates.

Homework Equations



[tex]x^{2}+y^{2}+z^{2}=a^{2}[/tex] : Equation for a sphere of radius "a" centered on the origin.

[tex]\iiint\limits_E dV[/tex] : Triple integral for finding volume of a region [tex]E[/tex].

The Attempt at a Solution



I solved the triple integral (but I don't think it's right) and got this: [tex]\frac{4}{3}a^{2}\pi[/tex] ---> Actually, I think I solved the integral right, but I think my bounds are incorrect.

I used the following as my bounds and subsequent iterated integral:

[tex]E=\{ \ (r,\theta,z) \ | \ 0\leq r\leq a, \ 0\leq \theta\leq 2\pi, \ -\sqrt{a^2-r^2}\leq z\leq \sqrt{a^2-r^2} \ \}[/tex]

[tex]\int^{2\pi}_{0}\int^{a}_{0}\int^{\sqrt{a^2-r^2}}_{\sqrt{a^2-r^2}} dz dr d\theta[/tex]

If my proposed answer isn't right could the problem lie within my bounds? I'm not really great at determining the bounds for iterated integrals yet >.<'

Thanks :D
 

Answers and Replies

  • #2
tiny-tim
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Hi Zarlucicil! :smile:

(have a pi: π and a theta: θ and try using the X2 tag just above the Reply box :wink:)
The problem was to find the volume enclosed by a sphere of radius "a" centered on the origin by crafting a triple integral and solving for it using cylindrical coordinates.

[tex]\int^{2\pi}_{0}\int^{a}_{0}\int^{\sqrt{a^2-r^2}}_{\sqrt{a^2-r^2}} dz dr d\theta[/tex]


erm :redface: … it isn't dz dr dθ, it's … ? :smile:
 
  • #3
525
7
Well, your bounds are in fact correct. What's missing is a volume element, namely, you need to have [tex]r drd\theta dz[/tex]. But maybe you just missed it, since without it you probably won't get the final answer you mentioned (which isn't correct by the way - you need [tex]4\pi a^3/3[/tex]).

You can also write [tex]r[/tex] in terms of [tex]z[/tex], which is in my opinion a bit more intuitive. In that case you get:

[tex]\int^{2\pi}_{0}\int^{-a}_{a}\int^{0}_{\sqrt{a^2-z^2}} r dr dz d\theta[/tex]

Gives the same answer ofcourse.
 
  • #4
13
2
Ahhh I see. Thanks for the replies, I understand what I did wrong now :D. I can't believe I missed the volume element, ughhh. O well.
 

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