Question

Polar Coordinates: Solve the problems by testing the symmetry, plotting and tracing the curves, and
computing the area, with limits from 0 to 2π, of the following equations

1. r = a cos 2θ
2. r = a sin 3θ
3. r = a (1 + sin θ)
4. r^2 = a^2(cos 2θ)

Answer 1

$\displaystyle \underline{\textsf{Test for symmetry}}\\ (1).\\ r = a\cos(2\theta) \\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (r, -\theta)\\ r=a\cos(-2\theta), r=a\cos(2\theta) \\ \textsf{Since the equation remains the same,}\\ \textsf{the curve is symmetrical about the polar axis.}\\ (x-\textsf{axis})\\ (2).\\ r = a\sin(3\theta) \\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (-r, -\theta)\\ -r=a\sin(-3\theta), r=a\sin(3\theta) \\ \textsf{Since the equation remains the same,}\\ \textsf{the curve is symmetrical about the line}\, \theta = \frac{\pi}{2}\\ (3).\\ r = a(1 + \sin{\theta})\\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (-r, \theta)\\ r = a(1 + \sin{\theta}), -r = a(1 + \sin{\theta})\\ \textsf{The equation does not remains the same.}\\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (-r, -\theta)\\ r = a(1 + \sin{\theta}), -r = a(1 + \sin(-\theta))\\ \textsf{The equation does not remains the same.}\\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (r, -\theta)\\ r = a(1 + \sin{\theta}), r = a(1 + \sin(-\theta))\\ \textsf{The equation does not remains the same.}\\ \textsf{All the replacements change the equation}\\ \textsf{and fails the test. Therefore, the graph}\\ \textsf{may or may not be symmetric with respect}\\ \textsf{to the pole.}\\ (4).\\ r^2 = a^2(\cos(2\theta))\\ \textsf{Replace}\, (r, \theta)\, \textsf{with}\, (r, -\theta)\\ r^2 = a^2(\cos(2\theta)), r^2 = a^2(\cos(-2\theta))\\ \textsf{Since the equation remains the same,}\\ \textsf{the curve is symmetrical about the origin}\\ \underline{\textsf{Compuatations of the areas}}\\ (2).\\ r = a\sin(3\theta), \textsf{at}\, r = 0, \sin(3\theta) = 0\\ \theta = 0, \frac{\pi}{3}\\ (3).\\ a(1 + \sin{\theta}) = 0, \sin{\theta} = -1, \\ \theta = -\frac{\pi}{2}.\\ \textsf{Area of the first half of the curve is from}\\ -\frac{\pi}{2}\,\, \textsf{to}\,\,\frac{\pi}{2}.\\ (4).\\ r^2 = a^2\cos(2\theta) = 0\\ \cos(2\theta) = 0, \cos(2\theta) = \cos\left(\frac{\pi}{2}\right)\\ \theta = \frac{\pi}{4}\\ \cos(2\theta) = 0, \cos(2\theta) = \cos\left(\frac{-\pi}{2}\right)\\ \theta = \frac{-\pi}{4}\\ \textsf{Area of one loop is from}\\ \theta = \frac{-\pi}{4}\,\,\textsf{to}\,\,\frac{\pi}{4}\\ \displaystyle \textsf{The following results are}\\ \textsf{to be applied}\\ \begin{aligned} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2\theta) \, \mathrm{d}\theta &= 0\,\, \textsf{Odd integral} \end{aligned}\\ \begin{aligned} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(2\theta) \, \mathrm{d}\theta &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 - \cos(4\theta)}{2} \, \mathrm{d}\theta \\&= \frac{4\theta - \sin(4\theta)}{8}\biggr\vert_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{4\times \frac{\pi}{2}\times 2}{8} = \frac{\pi}{2} \end{aligned}\\ \begin{aligned} \int_0^{\frac{\pi}{3}} \sin^2(3\theta) \, \mathrm{d}\theta &= \int_0^{\frac{\pi}{3}} \frac{1 - \cos(6\theta)}{2} \, \mathrm{d}\theta \\&= \frac{6\theta - \sin(6\theta)}{12}\biggr\vert_0^{\frac{\pi}{3}} = \frac{\pi}{6} \end{aligned}\\ (1).\\ \begin{aligned} A &= \frac{1}{2}\int_0^{2\pi} r^2 \, \mathrm{d}\theta\\ &= \frac{1}{2}\int_0^{2\pi} a^2\cos^2(2\theta)\, \mathrm{d}\theta = \frac{\pi a^2}{2} \end{aligned}\\ (2).\\ \begin{aligned} A &= \frac{1}{2}\int_0^{\frac{\pi}{3}} r^2 \, \mathrm{d}\theta\\ &= \frac{1}{2}\int_0^{\frac{\pi}{3}} a^2\sin^2(3\theta)\, \mathrm{d}\theta = \frac{a^2\pi}{12} \end{aligned}\\ \textsf{Area of three loops}\, = 3 \times \frac{a^2\pi}{12} = \frac{a^2\pi}{4} \\ (3).\\ \begin{aligned} A &= \frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} r^2 \, \mathrm{d}\theta\\ &= \frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} a^2(1 + 2\sin\theta + \sin^2(2\theta))\, \mathrm{d}\theta \\ &= \frac{a^2\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right) + \frac{\pi}{2}\right)}{2} = \frac{3\pi a^2}{4} \end{aligned}\\ \textsf{Area of the whole curve}\, = 2\times \frac{3\pi a^2}{4} = \frac{3\pi a^2}{2}\\ (4).\\ \begin{aligned} A &= \frac{1}{2}\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} r^2 \, \mathrm{d}\theta\\ &= \frac{1}{2}\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} a^2\cos(2\theta)\, \mathrm{d}\theta = \frac{a^2}{2}\sin(2\theta)\vert_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\&= \frac{a^2}{2}\times 2 = a^2 \end{aligned}\\ \textsf{Area of two loops}\, = 2a^2\\ \textsf{The graphs below are polar plots for}\, a = 3.$1.

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