{"id":995,"date":"2025-06-20T17:27:19","date_gmt":"2025-06-20T17:27:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=995"},"modified":"2025-09-10T18:02:07","modified_gmt":"2025-09-10T18:02:07","slug":"calculus-with-parametric-curves-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-with-parametric-curves-learn-it-3\/","title":{"raw":"Calculus with Parametric Curves: Learn It 3","rendered":"Calculus with Parametric Curves: Learn It 3"},"content":{"raw":"

Integrals Involving Parametric Equations<\/h2>\r\n

Now that we can find derivatives of parametric curves, let's tackle finding the area under these curves. Consider the cycloid defined by [latex]x(t) = t - \\sin t[\/latex] and [latex]y(t) = 1 - \\cos t[\/latex]. How do we find the area of the shaded region below?<\/p>\r\n\r\n

<\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 7. Graph of a cycloid with the arch over [latex]\\left[0,2\\pi \\right][\/latex] highlighted.[\/caption]<\/figure>\r\n

To find the area under a parametric curve defined by [latex]x = x(t)[\/latex] and [latex]y = y(t)[\/latex] for [latex]a \\le t \\le b[\/latex], we'll use the familiar rectangle approximation method.<\/p>\r\n

Start by partitioning the interval [latex][a, b][\/latex] into [latex]n[\/latex] subintervals: [latex]t_0 = a < t_1 < t_2 < \\cdots < t_n = b[\/latex].<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"233\"]\"A Figure 8. Approximating the area under a parametrically defined curve.[\/caption]<\/figure>\r\n

For each rectangle:<\/p>\r\n\r\n

    \r\n \t
  • Height<\/strong>: [latex]y(\\overline{t}_i)[\/latex] for some [latex]\\overline{t}_i[\/latex] in the [latex]i[\/latex]th subinterval<\/li>\r\n \t
  • Width<\/strong>: [latex]x(t_i) - x(t_{i-1})[\/latex]<\/li>\r\n<\/ul>\r\n

    The area of the [latex]i[\/latex]th rectangle becomes:<\/p>\r\n

    [latex]{A}_{i}=y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/p>\r\n

    The total approximate area is:<\/p>\r\n\r\n

    [latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)\\right)[\/latex].<\/div>\r\nTo convert this into a form we can integrate, multiply and divide by [latex]\\Delta t = t_i - t_{i-1}[\/latex]:\r\n
    [latex]{A}_{n}=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{{t}_{i}-{t}_{i - 1}}\\right)\\left({t}_{i}-{t}_{i - 1}\\right)=\\displaystyle\\sum _{i=1}^{n}y\\left(x\\left({\\overline{t}}_{i}\\right)\\right)\\left(\\frac{x\\left({t}_{i}\\right)-x\\left({t}_{i - 1}\\right)}{\\Delta t}\\right)\\Delta t[\/latex].<\/div>\r\n

    Taking the limit as [latex]n[\/latex] approaches infinity gives:<\/p>\r\n\r\n

    [latex]A=\\underset{n\\to \\infty }{\\text{lim}}{A}_{n}={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt[\/latex].<\/div>\r\nAs [latex]n[\/latex] approaches infinity, the change in [latex]x[\/latex] over smaller and smaller time intervals becomes the instantaneous rate of change [latex]x'(t)[\/latex]. This transformation relies on the Mean Value Theorem.\r\n\r\n
    Mean Value Theorem\r\n[latex]\\\\[\/latex]\r\n<\/strong>If [latex]f[\/latex] is continuous on [latex][a,b][\/latex] and differentiable on [latex](a,b)[\/latex], then there exists at least one point [latex]c \\in (a,b)[\/latex] such that:\r\n

    [latex]f'(c) = \\frac{f(b) - f(a)}{b - a}[\/latex]<\/p>\r\n

    This theorem guarantees that the average rate of change equals the instantaneous rate of change at some point in the interval.<\/p>\r\n\r\n<\/section>The preceding result leads to the following theorem.\r\n\r\n

    \r\n

    theorem: area under a parametric curve<\/h3>\r\n

    For a non-self-intersecting plane curve defined by:<\/p>\r\n\r\n

      \r\n \t
    • [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]a \\le t \\le b[\/latex]<\/li>\r\n \t
    • [latex]x(t)[\/latex] is differentiable<\/li>\r\n<\/ul>\r\n

      The area under the curve is:<\/p>\r\n

      [latex]A = \\int_a^b y(t) \\cdot x'(t) dt[\/latex]<\/p>\r\nImportant:<\/strong> This formula assumes the curve doesn't cross itself and that [latex]x(t)[\/latex] increases as [latex]t[\/latex] increases from [latex]a[\/latex] to [latex]b[\/latex].\r\n\r\n<\/section>

      \r\n
      \r\n

      Find the area under the curve of the cycloid defined by the equations<\/p>\r\n\r\n

      [latex]x\\left(t\\right)=t-\\sin{t},y\\left(t\\right)=1-\\cos{t},0\\le t\\le 2\\pi [\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n
      \r\n

      Using the theorem, we have<\/p>\r\n\r\n

      [latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}y\\left(t\\right){x}^{\\prime }\\left(t\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1-\\cos{t}\\right)\\left(1-\\cos{t}\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+{\\cos}^{2}t\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(1 - 2\\cos{t}+\\frac{1+\\cos2t}{2}\\right)dt\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{2\\pi }\\left(\\frac{3}{2}-2\\cos{t}+\\frac{\\cos2t}{2}\\right)dt\\hfill \\\\ & ={\\frac{3t}{2}-2\\sin{t}+\\frac{\\sin2t}{4}|}_{0}^{2\\pi }\\hfill \\\\ & =3\\pi .\\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.