When is $\int \int_D f(r, \theta) dD = \int h(\theta) d \theta \int g(r) dr$?
where $g(r) h(\theta) = f$, i.e. the integral has been split because $f(r, \theta)$ contains a multiplication in $r$s and $\theta$s.
When is $\int \int_D f(r, \theta) dA = \int f_{\theta} d \theta \int f_r dr$?
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multiple-integral
iterated-integrals
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0@cjohnson No, but as I described $f=g(r) h(\theta)$, which means that $f$ is a function of multiplied $r$s, $\theta$s and some constants. E.g. $f(r, \theta) = 2 r \cos(\theta) r \sin(\theta) = 2 r^2 \cos(\theta) \sin(\theta) = 2 g(r) h(\theta) $. – 2017-02-09
1 Answers
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If $f(r, \theta)$ can be written as $f(r, \theta) = g(r) \cdot h(\theta)$ and if $D$ is a polar rectangle $$ D = \left\{ (r, \theta) \, \big| a \leq r \leq b, \, \alpha \leq \theta \leq \beta \right\}, $$ then you have $$ \iint\limits_D f(r, \theta) \, dA = \int_\alpha^\beta \int_a^b g(r) \cdot h(\theta) r \, dr \, d\theta = \int_\alpha^\beta h(\theta) \, d\theta \, \cdot \, \int_a^b g(r) r \, dr. $$ Where you run into trouble in trying to write the integral like this is if the function can't be factored in this way (e.g., $f(r, \theta) = r + \theta$), or if the region you want to integrate over is more complicated than a polar rectangle and so your limits of integrations for $r$ are functions of $\theta$, or vice versa.
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0Why exactly does the last equality hold? – 2017-02-09
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0In the iterated integral in the middle, the inner-most integral is with respect to $r$, meaning that for the purposes of that integral $h(\theta)$ can be thought of as a constant, so you could pull it outside of the innermost integral. Now the integral of $g(r) \cdot r$ with respect to $r$ is some number -- whatever it happens to be. So, that's just some constant and you can pull it out of the integral with respect to $\theta$. – 2017-02-09