In my old Calculus 3 notes, I see that I have written the following:
$$ \int_0^{2 \pi \it}\int_0^1 e^{-r^2} r dr d \theta = 2 \pi \int_0^1 e^{-r^2} r dr d \theta$$
I am puzzled because I believe there should be only one differential per integral. In particular, the second integral should not have a $d \theta$ at the end of it (since this $d \theta$ belongs to the first integral from 0 to 2).
Can anyone confirm which way is correct?
I am especially puzzled since I seem to repeat this same error a second time in my notes.
It would seem that the person, in the time far far ago, needed a corrector of records and all things important.
It would seem that the correction team took good care of the $ r \, dr \, d\theta$ and would not let $d\theta$ be independent of its friend $r \, dr$ once used.
Otherwise \begin{align} \int_{0}^{2 \pi} \, \int_{0}^{a} e^{- r^2} \, r \, dr \, d\theta &= \int_{0}^{2 \pi} d\theta \, \int_{0}^{a} r \, e^{-r^2} \, dr \\ &= 2 \pi \, \left(\frac{-1}{2}\right) \, \int_{0}^{a} \frac{d}{dr} \, \left( e^{- r^2} \right) \, dr \\ &= - \pi \left[ e^{-r^2} \right]_{0}^{a} = \pi \, \left(1 - e^{-a^2}\right). \end{align}