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I have two questions about integration with polar coordinates.

1.

If I have a indefinite integral like $$\int \int e^{x^2 + y^2} \mathrm{d}x \mathrm{d}y$$

Do I still have indefinite integrals after polar transformation? $$\int \int re^{r^2} \mathrm{d}r \mathrm{d}\varphi=\frac{e^{x^2}}{2}+C$$

2.

If I have an definite integral like $$\int\limits_{a}^b \int\limits_{m}^n e^{x^2 + y^2} \mathrm{d}x \mathrm{d}y$$

What exactly happens with the integration limits when I use polar transformation? $$\int\limits_?^? \int\limits_?^? re^{r^2} \mathrm{d}r \mathrm{d}\varphi$$ I know that when $[-\infty,\infty]\times [-\infty,\infty] \implies [-0,2\pi]\times [0,\infty]$

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    I can't understand what "indefinite integral in more than one variable" can possibly be, or what its goal could be. There exists "potential function" for vector fields, though... – 2017-01-15
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    @DonAntonio So are indefinite integrals only used when we talk about integration with one variable? – 2017-01-15
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    As far as I know, yes. And this is due to the fact that "indefinite integral" is a lousy name for what we really do there: antidifferentiation. Something similar can be done for some vector fields (conservative ones, to be sure) in two or more dimensions, but "antiderivative" of a function with more than one variable is not a very clear concept: would we mean differential (nabla = gradient), partial derivatives, only real functions or also vectorial ones, etc.? – 2017-01-15
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    @DonAntonio Ok thanks, any thoughts on question 2? – 2017-01-15
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    It all depends on the first limits. If you have a circle, for example, you can easily take $\;0\le r\le a\;,\;\;a=$ the circle's radius, and $\;0\le\phi\le2\pi\;$ , say. Otherwise it could be hard or almost impossible (think of a weird rectangle: how would you handle both polar coordinates in this case?) – 2017-01-15
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    @DonAntonio So if I have $x^2 + y^2 \leq a$ then integral limits become $\int\limits_0^{2\pi} \int\limits_0^a$? But If I have $[b,c]\times[d,c]$ there is no clear answer? – 2017-01-15
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    In general, yes. That's the main reaosn we have several sets of coordinates to our disposal: some are better suit for some purposes than other ones. – 2017-01-15
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    @DonAntonio Ok thank you very much, this was extremely helpful! – 2017-01-15

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