6.4 Polar coordinates & examples

Polar coordinates

To compute the area of a circle. we use the transformation

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Actually, this map transforms a portion of the upper plane into the plane \(\mathbb{R}^2\) (not bijectively if we consider the line \(r = 0\))

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The transformation is given by the formula above \(u:\mathbb{R}_{\geq0}\times[0,2\pi]\rightarrow\mathbb{R}^{2},u(r,t)=(x(r,t),y(r,t ))\) and \(x(r,t) = r \cos(t)\), \(y(r,t) = r \sin(t)\)
The hypothesis of the theorem of change of variables are not valid in this case because the transformation is not bijective, as \(u\) sends all points with \(r=0\) to the point \((0,0)\).

Nevertheless, the formula in the theorem of change of variables remains valid in this case, because the sets of points where \(u\) is not injective lie on the boundary of \(D\), which is the graph of a curve (a continuous function on a closed interval) and has measure zero.
So, in terms of integration, one may not consider this set.

We can "forget" the sets of measure zero because
Given a set \(S\) with area and \(Z \subseteq S\) a subset of measure zero, then we write \(S = S' \cup Z\) (disjoint union), \(\int_{S} f = \int_{S'}f + \underbrace{\int_{Z} f}_{=0} = \int_{S'}f\)

As we already computed in last class, we have that \(J_{u}(r,t)=r,\forall r>0\)
So we get a general formula for the change of variables to polar coordinates \(\displaystyle\iint_{S} f(x,y) \, dx \, dy = \iint_{D} f(r \cos(t), r \sin(t)) \, r \, dr \, d t\)
Let us compute some examples to see how it works.

Examples

Compute \(\iint_S \ln(x^2+y^2) dx dy\) where \(S = \{(x,y) \in \mathbb{R}^2 \mid a^2 \le x^2+y^2 \le b^2\}\) with \(0 < a < b\) real numbers
\(a^2 = x^2+y^2\), \(x^2+y^2 = b^2\)

Reasons for taking polar coordinates:

1. It simplifies the function \(x^2+y^2 = (r\cos\theta)^2 + (r\sin\theta)^2 = r^2(\cos^2\theta+\sin^2\theta) = r^2\)

2. The shape of the integrating set ​

\(\displaystyle \iint_{S} \ln(x^{2}+y^{2}) dx dy = \int_{0}^{\pi/2}\int_{a}^{b} \ln((r\cos t)^{2} + (r\sin t)^{2}) r dr dt\) since Jacobian theorem
\(\displaystyle= \int_{0}^{\pi/2}\int_{a}^{b} \ln(r^{2}) \cdot r dr dt = \int_{a}^{b} \left( \int _{0}^{\pi/2}\ln(r^{2}) \cdot r dt \right) dr\)
\(\text{By Fubini}\) \(\displaystyle= \int_{a}^{b} \left( \ln(r^{2}) \cdot r \cdot t \bigg|_{0}^{\pi/2}\right) dr = \int_{a}^{b} \ln(r^{2}) \cdot r \cdot \frac{\pi}{2}dr = \frac{\pi}{2}\int_{a}^{b} \ln(r^{2}) \cdot r dr\)

If we substitute \(x\), \(u = r^2\), \(du = 2r \, dr\)​\(\implies \frac{1}{2} du = r \, du\), then \(= \frac{\pi}{2} \int_{a^2}^{b^2} \ln(u) \cdot \frac{1}{2} du \quad(?)\)

So we integrate by parts \(\displaystyle \frac{\pi}{2}\int_{a}^{b} \underbrace{\ln(r^2)}_{u}\cdot \underbrace{r \, dr}_{dv} = \frac{\pi}{2}\left( \left[ \ln(r^{2}) \frac{r^{2}}{2}\right]_{a}^{b} - \int_{a} ^{b} \frac{1}{r^{2}}\cdot 2r \frac{r^{2}}{2}dr \right)\) \(= \frac{\pi}{2} \left( \ln(b^2) \frac{b^2}{2} - \ln(a^2) \frac{a^2}{2} - \int_a^b r \, dr \right)\)
​\(= \frac{\pi}{2} \left( \ln(b^2) \frac{b^2}{2} - \ln(a^2) \frac{a^2}{2} - \frac{b^2}{2} + \frac{a^2}{2} \right).\)

Gaussian Integral

\(I\) is an improper integral. We have to check the convergence \(\displaystyle \int_{-\infty}^{0}e^{-x^2}dx \quad \text{and}\quad \int_{0}^{+\infty}e^{-x^2}dx\)
Since we have an even function, we need to check only one integral.

Compute \(\displaystyle\lim_{a \to \infty}\int_{0}^{a} e^{-x^2}dx\)

But \(\displaystyle\int_{0}^{a} e^{x^2}dx = \frac{1}{2}\cdot 2 \int_{0}^{a} e^{-x^2}dx = \frac{1}{2} \int_{-a}^{a} e^{-x^2}dx\). So, we compute \(\displaystyle \lim_{a \to \infty}\int_{-a}^{a} e^{-x^2}dx\)

"The trick"
Let \(\displaystyle J_{a} = \int_{-a}^{a} e^{-x^2}dx\), then \(\displaystyle J_{a}^{2} = \left( \int_{-a}^{a} e^{-x^2}dx \right) \left( \int_{-a}^{a} e^{x^2} dx \right) = \left( \int_{-a}^{a} e^{-x^2}dx \right) \left( \int_{-a}^{a} e^{-y^2} dy \right)\)
​\(\displaystyle = \int_{-a}^{a}\left( \int_{-a}^{a}e^{-x^2-y^2}dx \right) dy = \int _{-a}^{a}\int_{-a}^{a}e^{-(x^2+y^2)}dx dy\)

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Then we know \(f(x, y) = e^{-(x^2 + y^2)}\)

, then \(\lim_{a \to \infty} \int_{R_a} f = \lim_{a \to \infty} \int_{B_a} f\)

So, we compute \(\displaystyle \lim_{a \to \infty}\int_{B_a}e^{-(x^2 + y^2)}dx dy\)
​\({\displaystyle\int_{B_{a}}e^{-(x^2+y^2)}dxdy}\) (polar coordinates)\(\displaystyle=\int_{0}^{2\pi}\int_{0}^{a}e^{-r^2}\cdot rdrdt\) (Fubini) \(\displaystyle = \int_{0}^{a}\int_{0}^{2\pi}(e^{-r^2}\cdot r) dt dr = 2\pi \int_{0} ^{a}e^{-r^2}r dr=\pi - \pi e^{-a^2}\)

\(\lim_{a \to \infty} \pi - \pi e^{-a^2} = \pi = J^2\) \(\Rightarrow I = \sqrt{\pi}\) \(\Rightarrow \displaystyle \int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt{\pi}\)

Center of mass

1-dimension: mass and stick = "interval" ​

\(\displaystyle \text{mass stick}= \sum \text{masses}= \int \text{masse}\)

the mass of every particle is modelled by a function \(\displaystyle m = \int_{I} m(x)dx = \int_{a}^{b} m(x)dx\)

\(x_0 = \text{center of mass}\), then \(\displaystyle x_{0}m = \int_{a}^{b} xm(x)dx\) since the mean value has meaning

Example: take \(m(x)=1 \forall x \in [a,b]\)

\(\displaystyle m = \int_{a}^{b} 1dx = b-a\), then \(\displaystyle x_{0}(b-a) = x_{0}m = \int_{a}^{b} x \cdot 1dx = \frac{b^{2}-a^{2}}{2}= \frac{(b-a)(b+a)}{2}\)

\(\Rightarrow x_0 = \frac{b+a}{2}\) the center of mass is in the middle

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2-dimension: Mass of a something flat (a disk) D

\(\displaystyle m = \iint_{D} m(x,y) dx dy\), \(m(x,y) = \text{mass function}\)

Disk composed of one material with homogenous distribution \(\implies m(x,y) = 1\)

Center of mass \((x_0, y_0)\)

\(\displaystyle x_{0}m= \iint_{D}x m(x,y) dx dy = \iint_{D}x dx dy\), \(\displaystyle y_{0}m= \iint_{D}y m(x,y) dx dy = \iint_{D}y dx dy\) ​

Polar Coordinates \(m = \iint_{D} 1 = \int_{0}^{R}\int_{0}^{2\pi} r dr dt = 2\pi \frac{R^{2}}{2}= R^{2} \pi\)

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Exercise: Find \(x_0, y_0\)

\(\pi R^2 x_0 = \iint_D x \,dx\,dy\)

\(\pi R^2 y_0 = \iint_D y \,dx\,dy\)

(\(x_0=0\), \(y_0=0\))

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