Triple Integration of a Sphere in Cylindrical Coordinates

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Homework Help Overview

The problem involves finding the volume enclosed by a sphere of radius "a" centered at the origin using a triple integral in cylindrical coordinates.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up a triple integral with specific bounds but expresses uncertainty about their correctness. Some participants question the inclusion of the volume element in the integral setup.

Discussion Status

Contextual Notes

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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