Polar Coordinates Formula
Theorem (Polar Coordinates Formula)
Suppose \(f(x,y)\) is a Riemann integrable function on the disk \(D_R := \{(x,y) \ : \ x^2 + y^2 \leq R^2\}\). Then \(r f (r \cos \theta,r \sin \theta)\) is Riemann integrable on the box \(B_R := \{ (r,\theta) \ : \ r \in [0,R], \theta \in [0,2 \pi]\}\) and
\[ \int_{D_R} f = \int_{B_R} f(r \cos \theta,r\sin \theta) r dr d\theta. \]
Note
The formula isn't restricted to integrating over disks. Recall that we can integrate over any Jordan region \(E\) by integrating \(f\) times the characteristic function of \(E\). The appearance of \(D_R\) here is simply a technical issue and not a significant limitation.
Step 1 (Prepare the region for change of variables)
For each \(\epsilon > 0\), let
\[{}R_{\epsilon,+}{}\]
\[{}:= \{(x,y) \in D_R \ : y \geq 0, \ x^2 + y^2 \geq \epsilon \},{}\]
\[{}R_{\epsilon,-}{}\]
\[{}:= \{(x,y) \in D_R \ : y \leq 0, \ x^2 + y^2 \geq \epsilon \}, {}\]
\[{}R_{\epsilon,0}{}\]
\[{}:= \{(x,y) \in D_R \ : x^2 + y^2 \leq \epsilon \}. {}\]
These regions are nonoverlapping and have union \(D_R\) so
\[ \int_{D_R} f = \int_{R_{\epsilon,+}} f + \int_{R_{\epsilon,-}} f + \int_{R_{\epsilon,0}} f. \]
Also note that
\[ \lim_{\epsilon \rightarrow 0^+} \int_{R_{\epsilon,0}} f = 0 \]
because \(f\) is necessarily bounded and the area of the ball of radius \(\epsilon\) tends to zero.
Step 2 (Change of variables)
Let \(\Phi(r,\theta) := (r \cos \theta, r \sin \theta)\). Observe that \(\Phi\) is 1-1 with nonvanishing Jacobian determinant on \(U_+ := (0,\infty) \times (-\frac{\pi}{2},\frac{3 \pi}{2})\) and also on \(U_- := (0,\infty) \times (\frac{\pi}{2},\frac{5 \pi}{2})\). Note that the upper half plane minus the origin is contained in the image \(\Phi(U_+)\) and the lower half plane minus the origin is contained in \(\Phi(U_{-})\). The change of variables formula then implies that
\[ \int_{R_{\epsilon,+}} f = \int_{[\epsilon,R] \times [0,\pi]} \! \! \! r f(r \cos \theta,r \sin \theta) dr d \theta\]
\[ \int_{R_{\epsilon,-}} f = \int_{[\epsilon,R] \times [\pi,2\pi]} \! \! \! r f(r \cos \theta,r \sin \theta) dr d \theta\]
where the factor of \(r\) arises as the absolute value of the Jacobian determinant of \(\Phi\). Because the regions on the right-hand sides of the two formulas above are nonoverlapping, we may express their sum as the integral over the union, i.e.,
\[{}\int_{R_{\epsilon,+}} f + \int_{R_{\epsilon,-}} f{}\]
\[{}= \int_{[\epsilon,R] \times [0,2\pi]} r f(r \cos \theta,r \sin \theta) dr d \theta.{}\]
In particular, the change of variables formula guarantees Riemann integrability of \(r f(r \cos \theta, r \sin \theta)\) on \([\epsilon,R] \times [0,2 \pi]\).
Step 3 (Approximation and Riemann Integrability)
Next we show that \(r f(r \cos \theta, r \sin \theta)\), which for convenience we denote by \(\tilde f\), is Riemann integrable on \(B_R\). Let \(M\) be a positive upper bound for \(|f|\). We build a partition \({\mathcal P}\) of \(B_R\) as follows. Fix a small \(\epsilon\) and let \(R' := [0,\epsilon/(4MR)] \times [0,2 \pi]\) belong to \({\mathcal P}\), and let all remaining elements of \({\mathcal P}\) be taken from a partition \({\mathcal P}'\) of \([\epsilon/(2MR),R] \times [0,2 \pi]\) chosen so that
\[{}\mathcal{U}(\tilde f, {\mathcal P}') - \mathcal{L}(\tilde f, {\mathcal P}') < \frac{\epsilon}{2},{}\]
which we know must exist thanks to Step 2. We then have that \(\mathcal{U}(\tilde f,{\mathcal P}) - \mathcal{L}(\tilde f,{\mathcal P}')\) equals
\[{}\mathcal{U}(\tilde f,{\mathcal P}) - \mathcal{L}(\tilde f,{\mathcal P}'){}\]
\[{}+ \frac{\epsilon}{4M} \left[ \sup_{D_{\epsilon/(4MR)}} f - \inf_{D_{\epsilon/(4MR)}} f \right]{}\]
because the supremum and infimum of \(\tilde f\) on the box \([0,\epsilon/(4MR)] \times [0,2 \pi]\) must simply equal the supremum and infimum, respectively, of \(f\) on the disk \(D_{\epsilon/(4MR)}\). Using the bound \(|f| \leq M\) gives that \(\mathcal{U}(\tilde f,{\mathcal P}) - \mathcal{L}(\tilde f,{\mathcal P}') < \epsilon\). So \(\tilde f\) must be Riemann integrable on \(B_R\). By essentially the same argument as in Step 1, it also follows that
\[ \int_{B_R} \tilde f = \lim_{\epsilon \rightarrow 0^+} \int_{[\epsilon,R] \times [0,2 \pi]} \tilde f. \]
Step 4 (Conclusion)
By Step 1,
\[ \int_{D_R} f = \lim_{\epsilon \rightarrow 0^+} \left( \int_{R_{\epsilon,+}} f + \int_{R_{\epsilon,-}} f \right). \]
By Step 2,
\[{}\lim_{\epsilon \rightarrow 0^+} \left(\int_{R_{\epsilon,+}} f + \int_{R_{\epsilon,-}} f \right){}\]
\[{}= \lim_{\epsilon \rightarrow 0^+} \int_{[\epsilon,R] \times [0,2 \pi]} \tilde f{}\]
\[ \lim_{\epsilon \rightarrow 0^+} \int_{[\epsilon,R] \times [0,2 \pi]} \tilde f = \int_{B_R} \tilde f. \]
This completes the proof.