Hey.

What's the answer for this:

$\displaystyle \int_{0}^{r}dr \int_{0}^{2\pi}rd\phi \int_{0}^{l}dl \rho (1-r/R)$

Printable View

- Feb 14th 2010, 01:36 PMjacektriple integral
Hey.

What's the answer for this:

$\displaystyle \int_{0}^{r}dr \int_{0}^{2\pi}rd\phi \int_{0}^{l}dl \rho (1-r/R)$ - Feb 14th 2010, 07:15 PMichoosetonotchoosetochoos
$\displaystyle \rho l2\pi r^{2}[\frac{1}{2}-\frac{r}{3R}]$

- Feb 15th 2010, 02:25 AMHallsofIvy
Actually, that is a very badly posed integral. It is never a good idea to use the same letter to represent the variable of integration

**and**a limit of integration and that is done twice here. Other than that, it's pretty close to trivial. We can "separate" those integrals and do them separately:

$\displaystyle \left(\int_0^r (1- r/R)r dr\right)\left(\int_0^{2\pi} d\phi\right)\left(\int_0^l dl\right)$.

And, of course, $\displaystyle \int_0^{2\pi} d\phi$ and $\displaystyle \int_0^l dl$**are**trivial!

$\displaystyle \int_0^r (1- r/R)r dr= \int_0^r rdr- \frac{1}{R}\int_0^r r dr$ should also be easy though not "trivial".