- #1
bjnartowt
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Homework Statement
The expression [itex]{\textstyle{{3{a^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}}[/itex] has a volume integral equal to [tex]4\pi [/tex] for arbitrary "a".
Homework Equations
The Attempt at a Solution
[tex]\int_0^R {\int_0^{\pi /2} {\int_0^\pi {{\textstyle{{3{a^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}} \cdot dr \cdot r \cdot d\theta \cdot r\sin \theta \cdot d\phi } } } = 12\pi {a^2}\int_0^R {{\textstyle{{{r^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}} \cdot dr} [/tex]
Maple claims this integral, before evaluating at endpoints, is,
[tex]\left( {\frac{{{r^3}}}{{3{a^2}{{({r^2} + {a^2})}^{3/2}}}}} \right)_{r = 0}^{r = R} = \left( {\frac{{{R^3}}}{{3{a^2}{{({R^2} + {a^2})}^{3/2}}}} - \frac{{{r^3}}}{{3{a^2}{{({r^2} + {a^2})}^{3/2}}}}} \right)[/tex]
I can't evaluate the second term, because the possibility of a --> 0 comes up later in the problem, and as a --> 0 and r --> 0, the second term clearly diverges, as would a Dirac delta function. How am I to argue that the integral does boil down to 4*pi? It just seems a farfetched claim...