Lemma: $dA =r \, dr \,d\theta$. **Proof**: Follows from [[Change of variables formula]]. **Proof for closed curves**: Suppose $R$ is the region bounded by a closed curve $C_{R}$ traversed clockwise by $(x,y)$. Let $\underline{F}(r,\theta)=(r\cos\theta , r\sin \theta).$ Let $S$ be the region bounded by $C_{S}$ parametrised by $(r,\theta)$ such that $S=F^{-1}(R).$ Then by [[Green's theorem|Green's theorem in the plane]], $\begin{align} \text{Area of }R &= \int \int_{R} \, dx \, dy \\ &= \oint_{C_{R}} x\,dy = \oint_{C_{R}} x\, \frac{dy}{dt}\,dt \\ &= \int_{t_{1}}^{t_{2}} r\cos \theta \left[ \frac{ \partial y }{ \partial r } \frac{dr}{dt} +\frac{ \partial y }{ \partial \theta } \frac{d \theta}{dt} \right] \, dt &\text{by chain rule} \\ &=\int_{t_{1}}^{t_{2}} r\cos \theta \left[ \sin \theta \frac{dr}{dt} + r \cos \theta \frac{d\theta}{dt} \right] \, dt \\ &= \oint_{C_{S}} [r\cos \theta \sin \theta \, dr + r^{2} \cos^{2} \theta \, d \theta] \\ &= \int \int_{S} [2r\cos^{2}\theta - r\cos 2 \theta] \, dr \, d\theta & \text{using Green's theorem again} \\ \end{align}$Therefore $\int \int_{R} \, dx \, dy = \int \int_{S} r \, dr \, d\theta .$