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Consider the integral $$\iint\limits_R dxdy$$ in Cartesian coordinates. I know that we can use Jacobian to switch this integral to polar coordinates, which will give us

$$\iint\limits_S rdrd\theta$$

But how can we do this transformation without using Jacobians? Is it true that if $r^2 = x^2 + y^2$ then $2rdr = 2xdx + 2ydy$? Then $xdx = rdr - ydy$ or $dx = rdr/x - ydy/x=\frac{dr}{\cos\theta}-\frac{\sin\theta dy}{\cos\theta}=\frac{dr}{\cos\theta}-\tan\theta dy$.

Also, $dy=rdr/y-xdx/y=\frac{dr}{\sin\theta}-\cot\theta dx$.

So how come $dxdy = rdrd\theta$? That is, what is it that I'm not doing right here?

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    $\begingroup$ I’ve done some research, and I believe this night answer your question! Let me know what you think. $\endgroup$ Mar 13, 2018 at 1:40
  • $\begingroup$ @ChaseRyanTaylor Thanks, this helped! $\endgroup$
    – sequence
    Mar 13, 2018 at 6:02

2 Answers 2

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You can use the exterior derivative operator d to figure out this problem where for a function f $df = f_xdx+f_ydy+f_zdz$ We can use the other definitions of polar coordinates to solve for dx and dy in terms of r and $\theta$. Note that in this derivation we are using exterior products (Think cross product) of differentials not the usual point wise product and note: $dx\wedge dx = 0 $ and $dx\wedge dy = -dy \wedge dx$ $$ x = r\cos(\theta) \\ y = r\sin(\theta) \\ dx = \frac{\partial}{\partial r} r\cos(\theta)dr + \frac{\partial}{\partial \theta} r\cos(\theta)d\theta \\ dx = \cos(\theta)dr - r\sin(\theta)d\theta \\ dy = \frac{\partial}{\partial r} r\sin(\theta)dr + \frac{\partial}{\partial \theta} r\sin(\theta)d\theta \\ dy = \sin(\theta)dr + r\cos(\theta)d\theta \\ \text{Multiplication yields} \\ dxdy = (\cos(\theta)dr - r\sin(\theta)d\theta) * (\sin(\theta)dr + r\cos(\theta)d\theta) \\dxdy = \cos(\theta)dr*\sin(\theta)dr - r^2\sin(\theta)d\theta*\cos(\theta)d\theta-r\sin^2(\theta)d\theta dr+r\cos^2(\theta)drd\theta \\dxdy = 0 -0+r(\cos^2(\theta)drd\theta-\sin^2(\theta)d\theta dr) \\dxdy = r(\cos^2(\theta)drd\theta+\sin^2(\theta)dr d\theta) \\dxdy = rdrd\theta( \cos^2(\theta)+\sin^2(\theta))=rdrd\theta \\\text{More accurately} \\dx\wedge dy = rdr\wedge d\theta \\\text{You can then integrate the differential 2-form} \\ \int dx\wedge dy = \int rdr\wedge d\theta = \iint rdrd\theta $$

Also when you differentiate both sides of an equation you have to differentiate with respect to a variable you cant differentiate each variable with a different variable. The exterior derivative is different however because it is a derivative with respect to every variable. $$ r^2 = x^2 + y^2 \equiv\frac{d}{dx}r^2 = \frac{d}{dx}(x^2 + y^2) \\r^2 = x^2 + y^2 \not \equiv 2rdr = 2xdx + 2ydx $$

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It is important to understand that multiplication of differentials is "anti-commutative". That is, dxdy= -dydx and, of course, dxdx= dydy= 0. Given that $dx= cos(\theta)dr- rsin(\theta)d\theta$ and that $dy= sin(\theta)dr+ rcos(\theta)d\theta$, $dxdy= (cos(\theta)dr- rsin(\theta)d\theta)(sin(\theta)dr+ rcos(\theta)d\theta)= cos(\theta)sin(\theta)drdr- rsin^2(\theta)d\theta dr+ rcos^2(\theta)drd\theta- r^2 sin(\theta)cos(\theta)drdr$.

The first and last terms are 0 because drdr and $d\theta d\theta$ are both 0. And $- rsin^2(\theta)d\theta dr= rsin^2(\theta)drd\theta$ so the second and third terms sum to $r(sin^2(\theta)+ cos^2(\theta)) drd\theta= r drd\theta$.

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  • $\begingroup$ Multiplication of differentials is not anti commutative, the wedge product of differential forms is anti-commutative, while this seems like a trivial distinction its very important since in general. $\iint dxdy = \iint dydx$ The differential 1-forms only become differentials when they are integrated, $\int dx\wedge dy = \iint dxdy$ $\endgroup$
    – Sam
    Jul 9, 2019 at 15:39
  • $\begingroup$ How do you deduce $dx=cos(\theta)dr-r\sin(\theta)d\theta$? $\endgroup$
    – sequence
    Jul 19, 2019 at 15:37
  • $\begingroup$ That's the product rule from Calculus I. Let $x= r cos(\theta)$ and let r and $\theta$ be functions of some other variable, t. Then $\frac{dx}{dt}= \frac{dr}{dt}cos(\theta)- r sin(\theta)\frac{d\theta}{dt}$. Going back to "differential form" removes the "t": $dx= dr cos(\theta)- r sin(\theta)d\theta$. $\endgroup$
    – user247327
    Jul 19, 2019 at 17:44

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