Science:Math Exam Resources/Courses/MATH200/April 2012/Question 10

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MATH200 April 2012

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• April 2012 • December 2011 •

Question 10
Evaluate $\iiint \limits _{\mathbb {R} ^{3}}\left[1+(x^{2}+y^{2}+z^{2})^{3}\right]^{-1}{\textrm {d}}V.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The combination $x^{2}+y^{2}+z^{2}$ suggests a change to spherical coordinates.

Hint 2
We integrate over the entire space, so what are the integration limits for ρ, θ and φ?

Hint 3
Don't forget to account for the differential volume element $\displaystyle {}dV$ .

Hint 4
Luckily, the integrals with respect to φ and θ are easy to solve directly. For the final integration with respect to ρ use the substitution u = ρ³.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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Solution
The combination $x^{2}+y^{2}+z^{2}$ suggests a change to spherical coordinates. The domain is compatible with this: in spherical coordinates, the whole space is covered when: θ runs from 0 to 2π, φ runs from 0 to π, and ρ runs from 0 to +∞. In spherical coordinates, the differential volume element is: $dV=\rho ^{2}\sin(\phi )\,d\rho \,d\theta \,d\phi$ . And the function to integrate becomes: $(1+(x^{2}+y^{2}+z^{2})^{3})^{-1}=(1+(\rho ^{2})^{3})^{-1}={\frac {1}{1+\rho ^{6}}}$ . Thus the integral of interest becomes: $I=\int _{\rho =0}^{\infty }\int _{\phi =0}^{\pi }\int _{\theta =0}^{2\pi }{\frac {\rho ^{2}\sin(\phi )}{1+\rho ^{6}}}d\theta \,d\phi \,d\rho =\int _{\rho =0}^{\infty }{\frac {\rho ^{2}}{1+\rho ^{6}}}\,d\rho \underbrace {\int _{\phi =0}^{\pi }\sin(\phi )\,d\phi } _{=2}\underbrace {\int _{\theta =0}^{2\pi }d\theta } _{=2\pi }$ Hence $I=4\pi \int _{\rho =0}^{\infty }{\frac {\rho ^{2}}{1+\rho ^{6}}}\,d\rho$ Substitute: $\displaystyle u=\rho ^{3}$ $\displaystyle du=3\rho ^{2}\,d\rho$ And you obtain: $I={\frac {4\pi }{3}}\int _{u=0}^{\infty }{\frac {du}{1+u^{2}}}={\frac {4\pi }{3}}{\Bigl .}\tan ^{-1}(u){\Bigr \vert }_{u=0}^{\infty }={\frac {2\pi ^{2}}{3}}.$