[b][size=150]Polar coordinates and polar curves[/size][/b][br][br]Besides the usual Cartesian coordinates [math](x,y)[/math], we can also specify a point [math]P[/math] in 2D space using [b]polar coordinates[/b] [math](r,\theta)[/math], where [math]r[/math] is the [b]signed distance[/b] from the origin [math](0,0)[/math], usually called the [b]pole[/b], to the point and [math]\theta[/math] is the angle measured (anti-clockwise if [math]\theta>0[/math] and clockwise if [math]\theta<0[/math]) from the positive x-axis to the ray from the pole toward [math]P[/math] if [math]r>0[/math] or to the opposite ray if [math]r<0[/math].[br][br]In the applet below, you can input various values of [math]r[/math] and [math]\theta[/math] to find out the position of the point [math]P[/math]. Moreover, you can verify that the polar coordinates [math](r,\theta+\pi)[/math] and [math](-r,\theta)[/math] represent the same point.[br][br]

If you are given the polar coordinates [math](r,\theta)[/math], you can convert them to the usual Cartesian coordinates using the following well-known formulas:[br][br][math]x=r\cos\theta, \ \ y=r\sin\theta[/math][br][br][br]On the other hand, if you are given the Cartesian cooridnates [math](x,y)[/math], you can find the polar coordinates using the following equations:[br][br][math]r^2=x^2+y^2, \ \ \tan\theta =\frac yx[/math][br]

[b][size=150]Polar equations and polar curves[/size][/b][br][br]A [b]polar equation[/b] is an equation in [math]r[/math] and [math]\theta[/math]. The set of points in 2D space whose polar coordinates satisfy such equation is the [b]polar curve[/b] of the equation. In fact, for some curves, especially the ones that are closed loops, they can be more conveniently described by polar equations.[br][br][u]Basic examples[/u]:[br][br]Circle centered at the origin with radius [math]r_0>0[/math]:[br]Its polar equation is [math]r=r_0[/math].[br][br][br]A line through the origin such that the angle between the line and the x-axis is [math]\theta_0[/math]:[br]Its polar equation is [math]\theta=\theta_0[/math].[br][br][br]The circle [math]x^2+(y-a)^2=a^2[/math], where [math]a>0[/math]:[br][br]We can obtain its polar equation by substitutions [math]x=r\cos\theta[/math] and [math]y=r\sin\theta[/math] as follows:[br][br][math]r^2\cos^2\theta+(r\sin\theta-a)^2=a^2[/math][br][math]\implies r^2-2ar\sin\theta=0\implies r=2a\sin\theta[/math][br][br]([u]Note[/u]: Similarly, the circle [math](x-a)^2+y^2=a^2[/math] has the polar equation [math]r=2a\cos\theta[/math].)[br][br][br][u]More interesting examples[/u]:[br][br][b]Archimedean spiral[/b][br]Polar equation: [math]r=k\theta[/math], where [math]k>0[/math][br][br][br][b]Cardioid[/b] (a heart-shaped curve)[br]Polar equation: [math]r=1+\sin\theta[/math][br][br][br][b]Rose family[/b][br]Polar equation: [math]r=a\cos\left(\frac mn\theta\right)[/math], where [math]m,n[/math] are positive integers[br][br][br]In the applet below, you can plot the above polar curves and trace the curves using the slider.[br][br][br][br]

[b][size=150]Double integrals in polar coordinates[/size][/b][br][br]Given a function of two variables [math]z=f(x,y)[/math] defined on a region [math]S[/math], we already learned the definition of its double integral over [math]S[/math], i.e. [math]\iint_S f(x,y) \ dA[/math]. If [math]S[/math] is the kind of regions like sectors or rotationally symmetric shapes, which can be more easily described using polar coordinates, the double integral would be much easier to compute if we can rewrite it in terms of polar coordinates.[br][br]Suppose we consider [math]S=\left\{(r,\theta) \ | \ a\leq r \leq b, \ c\leq \theta \leq d\right\}[/math], the [b]polar rectangular region[/b], as shown in the applet below. We divide [math][a,b][/math] into [math]n[/math] subintervals of equal length [math]\frac{b-a}n=\Delta r[/math]. Also, we divide [math][c,d][/math] into [math]m[/math] subintervals of equal length [math]\frac{d-c}m=\Delta\theta[/math]. The arcs and rays divide [math]S[/math] into [math]N=mn[/math] polar rectangles. [br][br]Let [math](r_k^*,\theta_k^*)[/math] be the centre of the [math]k^{\text{th}}[/math] polar rectangle, then the volume [math]V[/math] under the the graph of [math]f[/math] can be approximated by [br][br][math]V\approx \sum_{k=1}^N f(r_k^*,\theta_k^*)\Delta A_k[/math][br][br]where [math]f(r,\theta)=f(r\cos\theta,r\sin\theta)[/math] and [math]A_k[/math] is the area of the [math]k^{\text{th}}[/math] polar rectangle.[br][br]To compute [math]\Delta A_k[/math], we use the formula for the area [math]A[/math] of the sector - [math]A=\frac 12 r^2\alpha[/math], where [math]r[/math] and [math]\alpha[/math] are the radius and the angle (in radian) of the sector respectively. Therefore, we have[br][br][math]\Delta A_k=\frac 12 \left(r_k^*+\frac{\Delta r}2\right)^2\Delta\theta-\frac 12 \left(r_k^*-\frac{\Delta r}2\right)^2\Delta\theta=r_k^*\Delta r\Delta\theta[/math][br][br]([u]Note[/u]: Conceptually, [math]dA[/math] can be regarded as [math]r dr \ d\theta[/math].)[br][br]Taking [math]n,m\to \infty[/math], we have[br][br][math]V=\iint_S f(r,\theta) \ dA=\lim_{N\to\infty}\sum_{k=1}^N f(r_k^*,\theta_k^*)r_k^* \Delta r\Delta\theta[/math][br][br]And by Fubini's theorem, we have[br][br][math]\iint_S f(r,\theta) \ dA=\int_c^d\left(\int_a^b f(r,\theta)r \ dr\right) \ d\theta[/math][br][br][br]In the applet below, you can change the range of [math]r[/math] and [math]\theta[/math] to obtain a different polar rectangular region. Observe that the small polar rectangle is getting larger when it is further away from the origin. The factor "[math]r[/math]" in the [math]dA=rdr \ d\theta[/math] corresponds to such scaling.[br][br]

[u]Example[/u]:[br][br]Find the volume of a sphere with radius [math]a[/math] using double integral.[br][br][u]Answer[/u]:[br][br]We may assume that the sphere is centered at the origin. Then the equation of the sphere is[br][br][math]x^2+y^2+z^2=a^2[/math][br][br]By symmetry, it suffices to consider the volume enclosed by the upper hemisphere and the xy-plane. Then the upper hemisphere is the graph of the function [math]z=f(x,y)=\sqrt{a^2-(x^2+y^2)}[/math] on the circular disk [math]S[/math], i.e. [math]S=\left\{(r,\theta) \ | \ 0\leq r \leq a, \ 0\leq \theta \leq 2\pi\right\}[/math].[br][br]Using polar coordinates, we have [math]f(r,\theta)=\sqrt{a^2-r^2}[/math] and the volume of the sphere [math]V[/math] can be expressed as the following double integral:[br][br][math]V=2\iint_S \sqrt{a^2-r^2} \ dA=2\int_0^{2\pi}\left(\int_0^a\sqrt{a^2-r^2} \ r \ dr\right) \ d\theta[/math][br][math]=2\int_0^{2\pi}\left[-\frac 13\left(a^2-r^2\right)^{\frac 32}\right]_0^a \ d\theta=\int_0^{2\pi}\frac 23a^3 \ d\theta =\left[\frac 23 a^3\theta\right]_0^{2\pi}=\frac 43\pi a^3[/math][br][br]

[u]Double integrals over general polar regions[/u][br][br]Given two polar curves with equations [math]r=g(\theta)[/math] and [math]r=h(\theta)[/math] such that [math]0\leq g(\theta)\leq h(\theta)[/math] for [math]c\leq \theta \leq d[/math]. The [b]polar region[/b] [math]\Omega[/math] is defined as follows:[br][br][math]\Omega=\left\{(r,\theta) \ | \ c\leq\theta\leq d, \ 0\leq g(\theta) \leq r \leq h(\theta)\right\}[/math][br][br]In other words, [math]\Omega[/math] is the region bounded by the polar curve of [math]r=g(\theta)[/math] as the [b]inner boundary[/b] and [math]r=h(\theta)[/math] as the [b]outer boundary[/b], where the angle [math]\theta[/math] sweeps from [math]c[/math] to [math]d[/math]. Then we can express the double integral of the function over [math]\Omega[/math] as the following iterated integral:[br][br][math]\iint_{\Omega}f(r,\theta) \ dA=\int_c^d\left(\int_{g(\theta)}^{h(\theta)} f(r,\theta)r \ dr\right) \ d\theta[/math][br][br][br][u]Example[/u]: Let [math]f(x,y)=\frac{y}{\sqrt{x^2+y^2}}[/math] on [math]\Omega[/math], the region in the first quadrant bounded by the circle [math]r=2[/math] and the cardioid [math]r=2(1+\cos\theta)[/math]. Find the value of [math]\iint_{\Omega} f(x,y) \ dA[/math].[br][br][u]Answer[/u]:[br][br]Using polar coordinates, we have [math]f(r,\theta)=f(r\cos\theta,r\sin\theta)=\sin\theta[/math].[br][br]Since [math]2\leq 2(1+\cos\theta)[/math] for [math]0\leq \theta \leq \frac{\pi}2[/math] (in the first quadrant), the circle is the inner boundary and the cardioid is the outer boundary.[br][br]Therefore, [math]\Omega=\left\{(r,\theta) \ | \ 0\leq \theta\leq \frac{\pi}2, \ 2\leq r \leq 2(1+\cos\theta)\right\}[/math].[br][br]The double integral in polar coordinates is as follows:[br][br][math]\iint_{\Omega}\sin\theta \ dA=\int_0^{\frac{\pi}2}\left(\int_2^{2(1+\cos\theta)}\sin\theta \cdot r \ dr\right) \ d\theta[/math][br][math]=\int_0^{\frac{\pi}2}\left[\frac{r^2}2\sin\theta\right]_2^{2(1+\cos\theta)} \ d\theta[/math][br][math]=\int_0^{\frac{\pi}2}\left(2(1+\cos\theta)^2\sin\theta-2\sin\theta\right) \ d\theta[/math][br][math]=\int_0^{\frac{\pi}2}\left(4\sin\theta\cos\theta+2\sin\theta\cos^2\theta\right) \ d\theta[/math][br][math]=\left[2\sin^2\theta-\frac 23\cos^3\theta\right]_0^{\frac{\pi}2}=2-\left(-\frac 23\right)=\frac 83[/math][br][br][br]In the applet below, you can see the shape of the region [math]\Omega[/math] and the solid under the graph over the region, whose volume is the value of the double integral. In the left panel, the red line segment is the "slice" of the region corresponding to a fixed value of [math]\theta[/math]. The cross section of the solid under the graph along the red line segment corresponds to the integral [math]\int_{g(\theta)}^{h(\theta)}f(r,\theta)r \ dr[/math].[br][br][br]

[u]An application[/u][br][br]The following is the [b]Gaussian integral[/b], which is a very important integral in statistics:[br][br][math]\int_{-\infty}^{\infty} e^{-x^2} \ dx[/math][br][br]There is no elementary way to calculate the indefinite integral [math]\int e^{-x^2} \ dx[/math]. But the above improper integral can be evaluated using the famous "polar coordinates trick".[br][br]First of all, [math]e^{-x^2}[/math] is an even function, we have[br][br][math]\int_{-\infty}^{\infty} e^{-x^2} \ dx=2\int_0^{\infty} e^{-x^2} \ dx[/math][br][br]i.e. It suffices to compute [math]\int_0^{\infty} e^{-x^2} \ dx[/math].[br][br]Let [math]f(r,\theta)=e^{-r^2}[/math] and [math]R=\left\{(r,\theta) \ | \ r\geq 0, \ 0\leq \theta \leq \frac{\pi}2\right\}[/math], i.e., the first quadrant. Consider the following double integral:[br][br][math]\iint_R f(r,\theta) \ dA[/math][br][br]We can compute it as follows:[br][br][math]\iint_R f(r,\theta) \ dA=\int_0^{\frac{\pi}2}\left(\int_0^{\infty}e^{-r^2}r \ dr\right) \ d\theta[/math][br][math]=\int_0^{\frac{\pi}2}\left[-\frac{e^{-r^2}}2\right]_0^{\infty} \ d\theta=\int_0^{\frac{\pi}2}\frac 12 \ d\theta=\frac{\pi}4[/math][br][br]Observe that the above double integral can also be computed using Cartesian coordinates:[br][br][math]\iint_R f(x,y) \ dA=\int_0^{\infty}\left(\int_0^{\infty}e^{-(x^2+y^2)} \ dx\right) \ dy[/math][br][math]=\int_0^{\infty}\left(\int_0^{\infty}e^{-x^2}e^{-y^2} \ dx\right) \ dy[/math][br][math]=\int_0^{\infty}e^{-y^2}\left(\int_0^{\infty}e^{-x^2} \ dx\right) \ dy[/math][br][math]=\left(\int_0^{\infty}e^{-y^2} \ dy\right)\left(\int_0^{\infty}e^{-x^2} \ dx\right)[/math][br][math]=\left(\int_0^{\infty}e^{-x^2} \ dx\right)^2[/math][br][br]Therefore, we have[br][br][math]\left(\int_0^{\infty}e^{-x^2} \ dx\right)^2=\frac{\pi}4[/math][br][math]\implies \int_0^{\infty}e^{-x^2} \ dx=\frac{\sqrt{\pi}}2[/math][br][br]Hence, the Gaussian integral [math]\int_{-\infty}^{\infty} e^{-x^2} \ dx=\sqrt{\pi}[/math].[br][br]