21. Multiple Integrals in Curvilinear Coordinates

a. Integrating in Polar Coordinates

5. Converting & Mixing, Rectangular & Polar Coordinates

Sometimes it is useful to use polar coordinates instead of rectangular coordinates when computing an integral. Other times it is best to use rectangular coordinates for one part of an integral and polar coordinates for another part in the same problem.

Converting Rectangular to Polar Coordinates

Sometimes the antiderivative of the integrand is not known, making the integral difficult if not impossible. By converting to polar coordinates, we can simplify the integral greatly.

Compute the integral \(\displaystyle I =\int_0^{3/\sqrt{2}}\int_y^{\sqrt{9-y^2}} \cos(x^2+y^2)\,dx\,dy\)

The upper limit for \(y\) is \(\dfrac{3}{\sqrt{2}}\) which is precisely the intersection point and the lower limit is \(y=0\) as shown in the plot.
We now convert to polar coordinates. Substituting \(x^2+y^2=r^2\), the integrand becomes \(\cos(r^2)\). From the plot, we see the bounds on \(r\) and \(\theta\) are: \[ 0 \le r \le 3 \qquad 0 \le \theta \le \dfrac{\pi}{4} \] Remember that the differentials become \(dA=dx\,dy=r\,dr\,d\theta\) so that the polar integral is computed as \[\begin{aligned} I&=\int_0^{\pi/4}\int_0^3\cos(r^2) r\,dr\,d\theta =\int_0^{\pi/4}\,d\theta\int_0^3 r\,\cos(r^2)\,dr \\ &=\dfrac{\pi}{4} \left[\dfrac{1}{2}\sin(r^2)\right]_0^3 =\dfrac{\pi}{8} \sin(9) \end{aligned}\]

In the following example, algebraic manipulation as well as polar coordinates are used to solve the problem.

Compute the improper integral \(\displaystyle I=\int_{-\infty}^\infty e^{-x^2}\,dx =\lim_{R\to\infty}\int_{-R}^R e^{-x^2}\,dx\) by using polar coordinates

We start by manipulating the integral algebraically. We first square \(I\), letting the dummy variable in the second \(I\) become \(y\), so that we don't confuse the two differentials: \[\begin{aligned} I^2&=\lim_{R\to\infty}\int_{-R}^R e^{-x^2}\,dx\int_{-R}^R e^{-y^2}\,dy \\ &=\lim_{R\to\infty}\int_{-R}^R\int_{-R}^R e^{-x^2-y^2}\,dx\,dy \end{aligned}\] From here, we switch to polar coordinates by substituting \(x^2+y^2=r^2\) and \(dx\,dy=r\,dr\,d\theta\) in the integral. The integration is performed over the entire \(xy\) plane, so the bounds on \(r\) and \(\theta\) become \[ 0 \le r \le \infty \qquad 0 \le \theta \le 2\pi \] So \(I^2\) in polar coordinates is \[\begin{aligned} I^2&=\lim_{R\to\infty}\int_0^{2\pi}\int_0^R e^{-r^2}r\,dr\,d\theta \\ &=\int_0^{2\pi} \,d\theta\lim_{R\to\infty}\int_0^R e^{-r^2}r\,dr =2\pi\lim_{R\to\infty}\left[-\,\dfrac{1}{2}e^{-r^2}\right]_0^R \\ &=-\pi\lim_{R\to\infty}\left[e^{-r^2}\right]_0^R =-\pi\left[\lim_{R\to\infty}e^{-R^2}-1\right] =\pi \end{aligned}\] Thus, taking the square root, \[ I=\sqrt{\pi} \]

This result is essential in statistics, where the function \(p(x)=\dfrac{1}{\sqrt{\pi}}e^{-x^2}\) is called the Gaussian or normal probability density function, which gives the probability density that a continuous random variable takes on the value \(x\). In particular, if you have an interval \([a,b]\), then the integral \(\displaystyle \int_a^b p(x)\,dx\), gives the probability that the random variable lies within the interval. The function \(e^{-x^2}\) is divided by \(I=\sqrt{\pi}\) so that the total probability equals \(1\), i.e. \(\displaystyle \int_{-\infty}^\infty p(x)\,dx=1\).

Mixing Rectangular and Polar Coordinates

Sometimes it is useful to use polar coordinates on part of the integral and rectangular coordinates on another part, as in the example below.

This integral can be computed using either rectangular or polar coordinates. However, it is easier to compute the integral over a quarter circle in polar coordinates and subtract the integral over a triangle. Mathematically, we express this as: \[ \iint_R xy\,dA=\iint_\text{quarter circle}xy\,dA-\iint_\text{triangle}xy\,dA \] First, we use polar coordinates to integrate of \(f=xy\) inside the quarter circle \(y=\sqrt{9-x^2}\) in the first quadrant: \[\begin{aligned} &\iint_\text{quarter circle} xy\,dA =\iint (r\cos\theta)(r\sin\theta)\,r\,dr\,d\theta \\ &=\int_0^{\pi/2}\int_0^3 r^3\sin\theta\cos\theta\,dr\,d\theta =\int_0^3 r^3\,dr\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta \\ &=\left[\dfrac{1}{4}r^4\right]_{r=0}^3 \left[\dfrac{\sin^2\theta}{2}\right]_0^{\pi/2} =\dfrac{3^4}{4}\left(\dfrac{1}{2}-0\right) =\dfrac{81}{8} \end{aligned}\] Next, we use rectangular coordinates to integrate the function \(f=xy\) under the line \(y=3-x\) from \(x=0\) to \(x=3\) \[\begin{aligned} &\iint_\text{triangle}xy\,dA =\int_0^3\int_0^{3-x} xy\,dy\,dx =\int_0^3 \left[x\dfrac{y^2}{2}\right]_{y=0}^{3-x}\,dx \\ &=\dfrac{1}{2} \int_0^3 x(3-x)^2\,dx =\dfrac{1}{2}\int_0^3 (9x-6x^2+x^3)\,dx \\ &=\dfrac{1}{2}\left[\dfrac{9}{2}x^2-2x^3+\dfrac{1}{4}x^4\right]_0^3 =\dfrac{1}{2}\left(\dfrac{9}{2}(3)^2-2(3)^3+\dfrac{1}{4}(3)^4\right) =\dfrac{27}{8} \end{aligned}\] Thus, the net integral is \[\begin{aligned} \iint xy\,dA &=\iint_\text{quarter circle}xy\,dA-\iint_\text{triangle}xy\,dA \\ &=\dfrac{81}{8}-\,\dfrac{27}{8} =\dfrac{27}{4} \end{aligned}\]