- #1
Castilla
- 241
- 0
Please I need help about polar coordinates. I have this expression:
[tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu.[/tex]. Now they say:
"Let's convert to polar coordinates. Define [tex] t = r cos \theta, u = r sin \theta[/tex]. Then [tex] t^2 + u^2 = r^2[/tex] (this is OK) and [tex] dtdu = rd \theta dr[/tex] (first why?) and the limits of integration become [tex] 0 < r < \infty, 0 < \theta < \frac{\pi}{2}[/tex] (second why??). We now have:
[tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu = \int_0^{\infty}\int_0^{\infty}re^\frac{-r^2}{2}d \theta dr [/tex] (third why?)".
I know this may be a boring question for you. Please don't just tell me "sustitute these differentials in the integrand by those other and you are done". I would like to understand and not only to apply a mecanism.
As Borges said (in better words, obviously): "Oh happiness of understanding, greater than that of imagining or sensing".
Thanks.
[tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu.[/tex]. Now they say:
"Let's convert to polar coordinates. Define [tex] t = r cos \theta, u = r sin \theta[/tex]. Then [tex] t^2 + u^2 = r^2[/tex] (this is OK) and [tex] dtdu = rd \theta dr[/tex] (first why?) and the limits of integration become [tex] 0 < r < \infty, 0 < \theta < \frac{\pi}{2}[/tex] (second why??). We now have:
[tex] \int_0^{\infty}\int_0^{\infty}e^\frac{-(t^2 + u^2)}{2}dtdu = \int_0^{\infty}\int_0^{\infty}re^\frac{-r^2}{2}d \theta dr [/tex] (third why?)".
I know this may be a boring question for you. Please don't just tell me "sustitute these differentials in the integrand by those other and you are done". I would like to understand and not only to apply a mecanism.
As Borges said (in better words, obviously): "Oh happiness of understanding, greater than that of imagining or sensing".
Thanks.