Evaluating 3D Integral in Rectangular Coordinates

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SUMMARY

The discussion focuses on evaluating the triple integral \(\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} dz\,dy\,dx\) in rectangular coordinates. Participants suggest starting with the innermost integral, which simplifies to \(2\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{1-x^2-y^2}\,dy\,dx\). A substitution \(u = 1-x^2\) is proposed to facilitate further evaluation, although its application remains unclear to some contributors.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with rectangular coordinates
  • Knowledge of substitution methods in integration
  • Experience with evaluating square roots in integrals
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  • Study the method of evaluating triple integrals in rectangular coordinates
  • Learn about substitution techniques in multiple integrals
  • Explore the geometric interpretation of triple integrals
  • Investigate the use of polar coordinates as an alternative for evaluating integrals
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Students and professionals in mathematics, particularly those studying calculus and integral evaluation techniques, will benefit from this discussion.

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Problem: Evaluate (leave in rectangular coordinates):
[tex] <br /> \int_{-1}^{{1}}}\int_{-{\sqrt{1-x^2}}}^{{\sqrt{1-x^2}}}\int_{-{\sqrt{1-x^2-y^2}}}^{{\sqrt{1-x^2-y^2}}}\ \,dz\,dy\,dx[/tex]
 
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Well, start slow. Can you evaluate this integral?
[tex] \int_{-{\sqrt{1-x^2-y^2}}}^{{\sqrt{1-x^2-y^2}}}\ \,dz[/tex]
 
[tex] <br /> 2\int_{-1}^{{1}}}\int_{-{\sqrt{1-x^2}}}^{{\sqrt{1-x^2}}}\2{\sqrt{1-x^2-y^2}}}\,dy\,dx[/tex]

Supposedly it's easier to make a subtitution u = 1-x^2 at this point but I don't see how...
 
Last edited:

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