Does Green's Theorem hold for polar coordinates?

I was working on a proof of the formula for the area of a region $\mathcal R$ of the plane enclosed by a closed, simple, regular curve $\mathcal C$, where $\mathcal C$ is traced out by the function (in polar coordinates) $r = f(\theta)$. My concern was that the last application of Green's Theorem (towards the end of the proof) was invalid since I'm not using it over Cartesian coordinates. But I get the same answer as would come from the change of coordinates formula, so maybe it's right after all?

By Green's Theorem, we can evaluate the area inside of the curve as \begin{align*} A & = \int_{\mathcal C} x\, dy = \int_{\mathcal C} {f(\theta)\cos\theta (f(\theta)\cos\theta + f'(\theta)\sin\theta )\,d\theta} \\ & = \int_{\mathcal C} (f(\theta)^2\cos^2\theta + f(\theta) f'(\theta)\sin\theta\cos\theta )\,d\theta \\ & = \int_{\mathcal C} r^2\cos^2\theta\,d\theta + r\sin\theta\cos\theta \, dr \\ & = \int_{\mathcal C} Q\, d\theta + P \, dr, \end{align*} where $Q(r,\theta) = r^2\cos^2\theta$ and $P(r,\theta) = r\sin\theta\cos\theta$. Then \begin{align*} \frac{\partial Q}{\partial r} = 2r\cos^2\theta \hspace{0.5cm} \text{and} \hspace{0.5cm} \frac{\partial R}{\partial \theta} = -r\sin^2\theta + r\cos^2\theta. \end{align*} Finally, Green's Theorem tells us that \begin{multline*} A = \int_{\mathcal C} P\,dr + Q\,d\theta = \iint_{\mathcal R} \left(\frac{\partial Q}{\partial r} - \frac{\partial P}{\partial \theta} \right) \,dr\,d\theta \\ = \iint_{\mathcal R} (r\cos^2\theta + r\sin^2\theta) \, dr \,d\theta = \iint_{\mathcal R} r \,dr\,d\theta. \end{multline*}

• I don't know what your question has to do with polar coordinates. For me it looks like a simple renaming $(x,y)\mapsto (r,\theta)$ and as variables which are integrated over can be renamed, the statement is correct. – Fabian Oct 2 '15 at 20:49
• @Fabian I guess I was wondering if there is anything going on with the difference in the coordinate system that would cause it to no longer work. – justin Oct 2 '15 at 20:50
• As Fabian stated, you’ve just relabeled the axes. If you convert the Green’s Thm equation from Cartesian to polar coordinates, on the other hand, it’s going to look a bit different from what you’ve got. – amd Oct 2 '15 at 20:52
• @amd I just edited the question to show how I'm trying to use Green's theorem. Is it okay to use it that way at the end of the proof? – justin Oct 2 '15 at 21:08
• Yes. You’re just applying it in the $r\theta$-plane instead of the $xy$ plane. Strictly speaking, $\mathcal C$ and $\mathcal R$ should be replaced by their preimages under the polar to Cartesian transformation. You could instead apply Green’s Thm immediately, then convert the resulting double integral to polar coordinates. If you know the change of variables formula for double integrals, it’s fairly easy. – amd Oct 2 '15 at 21:45

Your application of Green’s Theorem is justified. You can think of $r$ and $\theta$ as the labels of axes in a different Cartesian plane. You have to be a little careful about $\mathcal C$ and $\mathcal R$ with this point of view, though—they need to be replaced by their preimages under the polar-to-Cartesian map.
You are justified. If you apply Stokes theorem with cylindrical polar coordinates to $$F(\mathbf{x})=P\mathbf{e}_r + \frac{Q}{r}\mathbf{e}_\theta$$ over a flat surface orientated in the positive $z$ direction you can prove this.