Multiple integrals: Double integrals
Double integrals in polar coordinates
For regions in the \(xy\)-plane that can be described well in polar coordinates \((r,\varphi)\) it is useful to also calculate the double integrals in these coordinates. You then cover the region of integration with small sub-regions as shown in the figure below:
What we still have to determine is the area of the shaded sub-region for infinitesimal differences \(\dd r\) and \(\dd\varphi\). This area is equal to \[ \begin{aligned}\frac{1}{2}\dd\varphi\cdot (r+\dd r)^2- \frac{1}{2}\dd\varphi\cdot r^2&= \dd\varphi\cdot r\,\dd\varphi + \frac{1}{2}\dd\varphi\cdot (\dd r)^2\\[0.25cm] &= r\,\dd r\,\dd\varphi\qquad\blue{\text{infinitesimal case}}\end{aligned}\] This gives us the following computational rule:
Double integral in polar coordinates In polar coordinates, a double integral of a function \(f(x,y)\) over a region \(R\) in the \(xy\)-plane can be written as an iterated integral in polar coordinates in the following way: \[\iint_R f(x,y)\,\dd(x,y)=\int_Rf(r\cos\varphi, r\sin\varphi)\,r\,\dd r\,\dd\varphi\]
Calculate the improper double integral \[\iint_R e^{-x^2-y^2}\,\dd(x,y)\] over the right half-plane \(R\).
Solution
The double integral is easy to calculate in polar coordinates \((r,\varphi)\) with \(r\ge 0\) and \(-\tfrac{1}{2}\pi\le \varphi\le -\tfrac{1}{2}\pi\) as iterated integral: \[\begin{aligned}\iint_R e^{-x^2-y^2}\,\,\dd(x,y)&= \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\int_{0}^{\infty}e^{-r^2}\,r\,\dd r\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\lim_{\rho\to\infty}\biggl[-\frac{1}{2}e^{-r^2}\biggr]_{0}^{\rho}\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\lim_{\rho\to\infty}\frac{1}{2}\Bigl(1-e^{-\rho^2}\Bigr)\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{1}{2}\,\dd\varphi\\[0.25cm] &= \biggl[\frac{1}{2}\varphi\biggr]_{\varphi=-\frac{1}{2}\pi}^{\varphi=\frac{1}{2}\pi}\\[0.25cm] &= \frac{1}{2}\pi\end{aligned}\]
Due to symmetry of the integrand, the double integral over the left half-plane is also equal to \(\tfrac{1}{2}\pi\) and so the improper double-integral of the integrand over the entire \(xy\)-plane is equal to \(\pi\). Since \[\begin{aligned}\pi &=\iint_{\mathbb{R}^2} e^{-x^2+y^2}\,\dd(x,y)\\[0.25cm] &= \int_{x=-\infty}^{x=\infty}\int_{y=-\infty}^{y=\infty}e^{-x^2-y^2}\,\dd x \, \dd y\\[0.25cm]&=\int_{y=-\infty}^{y=\infty}e^{-y^2}\left( \int_{x=-\infty}^{x=\infty} e^{-x^2}\,\dd x\right) \dd y \\[0.25cm] &=\left(\int_{y=-\infty}^{y=\infty}e^{-y^2}\,\dd y\right)\cdot\left(\int_{x=-\infty}^{x=\infty}e^{-x^2}\,\dd x\right)\\[0.25cm] &=\left(\int_{-\infty}^{\infty}e^{-x^2}\,\dd x\right)^2\end{aligned}\] we also determined the following improper integral in one variable: \[\int_{-\infty}^{\infty}e^{-x^2}\,\dd x=\sqrt{\pi}\]
