{"id":1029,"date":"2025-06-20T17:29:44","date_gmt":"2025-06-20T17:29:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1029"},"modified":"2025-08-27T13:51:16","modified_gmt":"2025-08-27T13:51:16","slug":"area-and-arc-length-in-polar-coordinates-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/area-and-arc-length-in-polar-coordinates-learn-it-2\/","title":{"raw":"Area and Arc Length in Polar Coordinates: Learn It 2","rendered":"Area and Arc Length in Polar Coordinates: Learn It 2"},"content":{"raw":"

Arc Length in Polar Curves<\/h2>\r\nWhen you need to find the length of a curve defined in polar coordinates, you can adapt the familiar arc length formula from rectangular coordinates.\r\n\r\n
Rectangular Arc Length<\/strong>: For a parameterized curve [latex](x(t), y(t))[\/latex], arc length is [latex]L = \\int_a^b \\sqrt{\\left(\\frac{dx}{dt}\\right)^2 + \\left(\\frac{dy}{dt}\\right)^2} dt[\/latex]<\/section>\r\n

To derive the polar formula, we start with the conversion equations:<\/p>\r\n\r\n

[latex]x=r\\cos\\theta =f\\left(\\theta \\right)\\cos\\theta \\:\\:\\text{and} \\:\\:y=r\\sin\\theta =f\\left(\\theta \\right)\\sin\\theta [\/latex],<\/div>\r\nTaking derivatives with respect to [latex]\\theta[\/latex]:\r\n
[latex]\\begin{array}{c}\\frac{dx}{d\\theta }={f}^{\\prime }\\left(\\theta \\right)\\cos\\theta -f\\left(\\theta \\right)\\sin\\theta \\hfill \\\\ \\frac{dy}{d\\theta }={f}^{\\prime }\\left(\\theta \\right)\\sin\\theta +f\\left(\\theta \\right)\\cos\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\nNow we substitute these into the arc length formula, replacing the parameter [latex]t[\/latex] with [latex]\\theta[\/latex]:\r\n

[latex]\\begin{align}\r\nL &= \\int_{\\alpha}^{\\beta} \\sqrt{\\left(\\frac{dx}{d\\theta}\\right)^2 + \\left(\\frac{dy}{d\\theta}\\right)^2} d\\theta \\\\\r\n&= \\int_{\\alpha}^{\\beta} \\sqrt{\\left(f'(\\theta)\\cos\\theta - f(\\theta)\\sin\\theta\\right)^2 + \\left(f'(\\theta)\\sin\\theta + f(\\theta)\\cos\\theta\\right)^2} d\\theta\r\n\\end{align}[\/latex]<\/p>\r\nExpanding the terms under the square root and using the identity [latex]\\cos^2\\theta + \\sin^2\\theta = 1[\/latex]:\r\n

[latex]\\begin{align}\r\nL &= \\int_{\\alpha}^{\\beta} \\sqrt{[f'(\\theta)]^2(\\cos^2\\theta + \\sin^2\\theta) + [f(\\theta)]^2(\\cos^2\\theta + \\sin^2\\theta)} d\\theta \\\\\r\n&= \\int_{\\alpha}^{\\beta} \\sqrt{[f'(\\theta)]^2 + [f(\\theta)]^2} d\\theta \\\\\r\n&= \\int_{\\alpha}^{\\beta} \\sqrt{r^2 + \\left(\\frac{dr}{d\\theta}\\right)^2} d\\theta\r\n\\end{align}[\/latex]<\/p>\r\n

This gives us the following theorem.<\/p>\r\n\r\n

\r\n

arc length of a curve defined by a polar function<\/h3>\r\n

Let [latex]f[\/latex] be a function whose derivative is continuous on [latex][\\alpha, \\beta][\/latex]. The length of the curve [latex]r = f(\\theta)[\/latex] from [latex]\\theta = \\alpha[\/latex] to [latex]\\theta = \\beta[\/latex] is:<\/p>\r\n\r\n

[latex]L={\\displaystyle\\int }_{\\alpha }^{\\beta }\\sqrt{{\\left[f\\left(\\theta \\right)\\right]}^{2}+{\\left[{f}^{\\prime }\\left(\\theta \\right)\\right]}^{2}}d\\theta ={\\displaystyle\\int }_{\\alpha }^{\\beta }\\sqrt{{r}^{2}+{\\left(\\frac{dr}{d\\theta }\\right)}^{2}}d\\theta [\/latex].<\/div>\r\n<\/section>
Find the arc length of the cardioid [latex]r=2+2\\cos\\theta [\/latex].[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n
\r\n

When [latex]\\theta =0,r=2+2\\cos0=4[\/latex]. Furthermore, as [latex]\\theta [\/latex] goes from [latex]0[\/latex] to [latex]2\\pi[\/latex], the cardioid is traced out exactly once. Therefore these are the limits of integration. Using [latex]f\\left(\\theta \\right)=2+2\\cos\\theta [\/latex], [latex]\\alpha =0[\/latex], and [latex]\\beta =2\\pi[\/latex], the theorem equation becomes<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill L& ={\\displaystyle\\int }_{\\alpha }^{\\beta }\\sqrt{{\\left[f\\left(\\theta \\right)\\right]}^{2}+{\\left[{f}^{\\prime }\\left(\\theta \\right)\\right]}^{2}}d\\theta \\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{{\\left[2+2\\cos\\theta \\right]}^{2}+{\\left[-2\\sin\\theta \\right]}^{2}}d\\theta \\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi}\\sqrt{4+8\\cos\\theta +4{\\cos}^{2}\\theta +4{\\sin}^{2}\\theta }d\\theta \\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{4+8\\cos\\theta +4\\left({\\cos}^{2}\\theta +{\\sin}^{2}\\theta \\right)}d\\theta \\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{8+8\\cos\\theta }d\\theta \\hfill \\\\ & =2{\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{2+2\\cos\\theta }d\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Next, using the identity [latex]\\cos\\left(2\\alpha \\right)=2{\\cos}^{2}\\alpha -1[\/latex], add 1 to both sides and multiply by 2. This gives [latex]2+2\\cos\\left(2\\alpha \\right)=4{\\cos}^{2}\\alpha [\/latex]. Substituting [latex]\\alpha =\\frac{\\theta}{2}[\/latex] gives [latex]2+2\\cos\\theta =4{\\cos}^{2}\\left(\\frac{\\theta}{2}\\right)[\/latex], so the integral becomes<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill L& =2{\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{2+2\\cos\\theta }d\\theta \\hfill \\\\ & =2{\\displaystyle\\int }_{0}^{2\\pi }\\sqrt{4{\\cos}^{2}\\left(\\frac{\\theta }{2}\\right)}d\\theta \\hfill \\\\ & =2{\\displaystyle\\int }_{0}^{2\\pi }2|\\cos\\left(\\frac{\\theta }{2}\\right)|d\\theta .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from [latex]0[\/latex] to [latex]\\pi [\/latex] and double the answer. This strategy works because cosine is positive between [latex]0[\/latex] and [latex]\\frac{\\pi }{2}[\/latex]. Thus,<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill L& =4{\\displaystyle\\int }_{0}^{2\\pi }|\\cos\\left(\\frac{\\theta }{2}\\right)|d\\theta \\hfill \\\\ & =8{\\displaystyle\\int }_{0}^{\\pi }\\cos\\left(\\frac{\\theta }{2}\\right)d\\theta \\hfill \\\\ & =8{\\left(2\\sin\\left(\\frac{\\theta }{2}\\right)\\right)}_{0}^{\\pi }\\hfill \\\\ & =16.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
Watch the following video to see the worked solution to the example above.