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Luminosity integral help

  • #1
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If the total luminosity is given by

[tex]L_T = \int_0^{\infty} e^{-(r/a)^{1/4}} r^2 dr[/tex]

estimate the radius [tex]r(a)[/itex] corresponding to half of the total luminosity.

This would be to integrate from zero to r and get a function [itex]r(a,L)[/itex] but this is impossible so anyone got an idea on how to estimate this?
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
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Well the primitive of the integrand exists, you knew that right?

The difficult part is solving for [itex]r_{1/2}[/itex] in [tex]\int_0^{r_{1/2}} e^{-(r/a)^{1/4}} r^2 dr=L_T/2[/tex]. But computers are good with this.
 
Last edited:

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