# Polar coordinates in simple integral

Please help me clarify the following: I want to verify the area of the unit disk $D$ in $\mathbb{R^2}$ by means of the integral $\int_D 1\,dxdy$.

The polar coordinates provide a $C^1$-diffeomorphism from $D-\{x\leq0\}$ to $\{(r,\phi) \mid 0< r< 1,\, 0< \phi < 2\pi\}$, i.e. not the complete domain $D$.

Does this mean that I can't use polar coordinates here since they can't be used on the whole domain $D$? Or is it possible to argue that the integral is unchanged when $\{x\leq0\}$ is removed?

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$\{(x,y) : x \le 0, y = 0\}$ is a set of zero measure and won't change the integral – Cocopuffs Nov 24 '12 at 12:36

It's just a matter of simply checking that we get the desired result:

$$\int\int_D dx\,dy\stackrel{?}=\int_0^1\int_0^{2\pi}r\,d\theta\,dr=2\pi\int_0^1r\,dr=\pi$$

The reason why we get the correct result was already stated by Cocopuffs in his comment above.

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