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In my textbook, the author states that $$ \iint_R f(r\cos\theta,r\sin \theta)r\,dr\,d\theta =\iint_{G(R)} f(x,y)\,dy\,dx, $$ where $G(r,\theta)=(r\cos\theta, r\sin \theta)$ and he says that this is the formula for the change of variables between rectangular coordinates and polar ones.

But for me, it doesn't look right, if we want to change coordinates then the formula should be like this $$ \iint_R f(x,y)\,dy\,dx=\iint_{G(R)} f(r\cos\theta,r\sin \theta)r\,dr\,d\theta. $$

Am I right? or are both formulas the same thing?

This is a picture from the textbook, it says that the map $G$ takes rectangular to polar enter image description here

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    $\begingroup$ What is crucial here is that you said $G(r,\theta) = (r\cos\theta,r\sin\theta) = (x,y),$ i.e. $G$ takes polar coordinates as its input and the output is rectangular coordinates. $\qquad$ $\endgroup$ Commented Jun 17, 2021 at 23:31
  • $\begingroup$ No, i didn't say that, $G$ is a map that takes rectangular coordinates to polar coordinates @MichaelHardy $\endgroup$
    – PNT
    Commented Jun 17, 2021 at 23:55
  • $\begingroup$ I never said you said that. I merely pointed out that that is the crucial central fact to be noted here. $\endgroup$ Commented Jun 17, 2021 at 23:56
  • $\begingroup$ You didn't understand me, I mean that $G$ is actually a map that takes rectangular coordinates to polar coordinates, this is what i think, is it right? @MichaelHardy $\endgroup$
    – PNT
    Commented Jun 18, 2021 at 0:06
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    $\begingroup$ Since $G(r,\theta) = (x,y),$ $G$ takes polar coordinates to rectangular coordinates. Are you asking whether it goes in both directions? $\endgroup$ Commented Jun 18, 2021 at 0:14

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No, of course both cannot be right. The formula in the textbook is right. The way I "remember" the formula is that if $(r,\theta)\in R$ then this is my "polar coordinate space" so $(x,y)=(r\cos\theta,r\sin\theta)=G(r,\theta)$ are the corresponding cartesian coordinates so they have to lie in $G(R)$.


In general (under appropriate hypotheses) the change of variables formula is $$ \int_{G(R)} f(\xi)\,d\xi = \int_{R}f(G(\zeta))\cdot \left|\det DG(\zeta)\right|\,d\zeta $$ Here, we interpret it as $\zeta\in R$ and the "substitution" is $\xi=G(\zeta)$ so we must have $\xi\in G(R)$.

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The first formula is right, but you need to undestand what the symbols mean.

To simplify, the change of variable to polar coordinates works when you are integrating over a polar rectangle; that is, $$ G(R)=\{(r\cos\theta,r\sin\theta):\ (r,\theta)\in R\}. $$ Then you have $$ \iint_{G(R)} f(x,y)\,dy\,dx=\iint_R f(r\cos\theta,r\sin \theta)r\,dr\,d\theta $$ and your integral became a double integral over a rectangle.

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