Answer to Question #148519 in Calculus for qwerty

Question #148519
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)
1
Expert's answer
2020-12-10T18:47:48-0500

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

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