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*}
Best Answer
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.