Triple Integration of a Sphere in Cylindrical Coordinates

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SUMMARY

The discussion centers on calculating the volume of a sphere with radius "a" using a triple integral in cylindrical coordinates. The correct integral setup involves the volume element r dr dθ dz, which was initially omitted by the user. The accurate volume of the sphere is determined to be 4πa³/3, contrasting with the user's incorrect result of 4/3a²π. The bounds for the integral were confirmed to be correct, focusing on the region defined by the sphere.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with cylindrical coordinates
  • Knowledge of volume elements in integration
  • Basic geometry of spheres
NEXT STEPS
  • Study the derivation of volume elements in cylindrical coordinates
  • Learn how to set up and evaluate triple integrals
  • Explore the relationship between Cartesian and cylindrical coordinates
  • Practice solving integrals involving spherical regions
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Students in calculus, particularly those studying multivariable calculus, as well as educators teaching integration techniques and geometric applications of integrals.

Zarlucicil
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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



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

\iiint\limits_E dV : Triple integral for finding volume of a region E.

The Attempt at a Solution



I solved the triple integral (but I don't think it's right) and got this: \frac{4}{3}a^{2}\pi ---> 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:

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} \ \}

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

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
 
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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:)
Zarlucicil said:
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.

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


erm :redface: … it isn't dz dr dθ, it's … ? :smile:
 
Well, your bounds are in fact correct. What's missing is a volume element, namely, you need to have r drd\theta dz. 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 4\pi a^3/3).

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

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

Gives the same answer ofcourse.
 
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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