
#1
Jan1611, 08:21 PM


#2
Jan1611, 08:51 PM

PF Gold
P: 864

The "r" part of the integral is doable. I get
[tex]P=\int_{0}^{\pi}2\rho Gmdr^{2}a\sin{a}\dot ( \ln{\sqrt{r^{2}+d^{2}\cos{a}} +r} \frac{r}{\sqrt{r^{2}+d^{2}\cos{a}}})da\left^{r=6.275*10^6}_{r=6.175*10^6}[/tex] Integrating the "a" part seems like it should be a nightmare, though. You could do a substitution u=cos(a) if it wasn't for that a sitting outside. It's probably not expressible in elementary functions. Could you just evaluate it numerically? 



#3
Jan1611, 09:35 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,344

Starting with the a integral,
[tex]\int \frac{a \sin a\, da}{(b^2  \cos a)^{3/2}}[/tex] is an elliptic function of the 3rd kind according to http://integrals.wolfram.com. My knowledge of elliptic functions doesn't extend to knowing if your definite integral equals something nice, but at least that's a start. 


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