1
$\begingroup$

I'm trying to integrate $$ F_0a^2(-\frac{a}{2})\frac{1}{(\rho^2+\frac{a^2}{4})^{3/2}} $$ This is in cylindrical coordinates, so $\rho$ represents a radius and I want to integrate the expression over a disk with radius a. The correct answer is $$ -2\pi F_0a^2(1-\frac{1}{\sqrt{5}}) $$ How can I find this answer? Any pointers would be much appreciated. I don't think sharing my attempts would do any good, I don't have much faith in them. But basically I've tried using a table of indefinite integral because I don't know how to start from scratch on this.

$\endgroup$
2
$\begingroup$

Make a substitution $u=\rho^2+\tfrac{a^2}{4}$. Don't forget the polar Jacobian $dA=\rho\,d\rho d\varphi$.

$\endgroup$
  • $\begingroup$ With that substition, and the Jacobian, I'll have to integrate $\frac{ \sqrt{u-\frac{a^2}{4}}} {u^{3/2}}$ which is not easy either, I think. $\endgroup$ – user3680 Jan 6 '13 at 1:06
  • $\begingroup$ the integrand becomes $du/(2u^{3/2})$ $\endgroup$ – Jonathan Jan 6 '13 at 1:07
  • $\begingroup$ Yes you are right of course. Solved it now, thanks a lot. $\endgroup$ – user3680 Jan 6 '13 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy