2
$\begingroup$

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?

$\endgroup$
  • 6
    $\begingroup$ $\{(x,y) : x \le 0, y = 0\}$ is a set of zero measure and won't change the integral $\endgroup$ – Cocopuffs Nov 24 '12 at 12:36
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.