{"id":1668,"date":"2025-07-30T13:52:07","date_gmt":"2025-07-30T13:52:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1668"},"modified":"2025-08-20T16:56:12","modified_gmt":"2025-08-20T16:56:12","slug":"calculus-with-parametric-curves-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-with-parametric-curves-learn-it-5\/","title":{"raw":"Calculus with Parametric Curves: Learn It 5","rendered":"Calculus with Parametric Curves: Learn It 5"},"content":{"raw":"

Surface Area Generated by a Parametric Curve<\/h2>\r\n

Recall that when we revolve a function [latex]y = f(x)[\/latex] around the [latex]x[\/latex]-axis from [latex]x = a[\/latex] to [latex]x = b[\/latex], the surface area is:<\/p>\r\n\r\n

[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}f\\left(x\\right)\\sqrt{1+{\\left({f}^{\\prime }\\left(x\\right)\\right)}^{2}}dx[\/latex].<\/div>\r\nNow let's find the surface area when we revolve a parametric curve around the [latex]x[\/latex]-axis, as shown in the following figure.\r\n
<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"341\"]\"A Figure 11. A surface of revolution generated by a parametrically defined curve.[\/caption]<\/figure>\r\n

For a parametric curve [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] where [latex]a \\le t \\le b[\/latex], the analogous formula is:<\/p>\r\n\r\n

[latex]S=2\\pi {\\displaystyle\\int }_{a}^{b}y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime }\\left(t\\right)\\right)}^{2}}dt[\/latex]<\/div>\r\nThis formula requires that [latex]y(t) \\ge 0[\/latex] on [latex][a,b][\/latex]\u2014the curve must lie on or above the [latex]x[\/latex]-axis. Notice that the expression [latex]\\sqrt{[x'(t)]^2 + [y'(t)]^2}[\/latex] is the arc length element we just learned, so we're essentially multiplying the circumference [latex]2\\pi y(t)[\/latex] by the arc length differential.\r\n\r\n
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surface area of revolution (parametric form)<\/h3>\r\n

When revolving a parametric curve [latex]x = x(t)[\/latex], [latex]y = y(t)[\/latex] around the [latex]x[\/latex]-axis for [latex]a \\le t \\le b[\/latex]:<\/p>\r\n

[latex]S = 2\\pi \\int_a^b y(t) \\sqrt{[x'(t)]^2 + [y'(t)]^2} dt[\/latex]<\/p>\r\n

Requirement: [latex]y(t) \\ge 0[\/latex] on [latex][a,b][\/latex]<\/p>\r\n\r\n<\/section>

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Find the surface area of a sphere of radius [latex]r[\/latex] centered at the origin.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558829\"]\r\n

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We start with the curve defined by the equations<\/p>\r\n\r\n

[latex]x\\left(t\\right)=r\\cos{t},y\\left(t\\right)=r\\sin{t},0\\le t\\le \\pi [\/latex].<\/div>\r\n \r\n

This generates an upper semicircle of radius r<\/em> centered at the origin as shown in the following graph.<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"272\"]\"A Figure 12. A semicircle generated by parametric equations.[\/caption]<\/figure>\r\n

When this curve is revolved around the\u00a0[latex]x[\/latex]-axis, it generates a sphere of radius [latex]r[\/latex]. To calculate the surface area of the sphere, we use the above equation:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill S&=2\\pi {\\displaystyle\\int_{a}^{b}} y\\left(t\\right)\\sqrt{{\\left({x}^{\\prime }\\left(t\\right)\\right)}^{2}+{\\left({y}^{\\prime}\\left(t\\right)\\right)}^{2}}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int_{0}^{\\pi}} r\\sin{t}\\sqrt{{\\left(-r\\sin{t}\\right)}^{2}+{\\left(r\\cos{t}\\right)}^{2}}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}{\\sin}^{2}t+{r}^{2}{\\cos}^{2}t}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int }_{0}^{\\pi }r\\sin{t}\\sqrt{{r}^{2}\\left({\\sin}^{2}t+{\\cos}^{2}t\\right)}dt\\hfill \\\\ &=2\\pi {\\displaystyle\\int_{0}^{\\pi}{r}^{2}\\sin{t}dt}\\hfill \\\\ &=2\\pi{r}^{2}\\left(-\\cos{t}|_{0}^{\\pi}\\right)\\hfill \\\\ & =2\\pi {r}^{2}\\left(-\\cos\\pi +\\cos0\\right)\\hfill \\\\ & =4\\pi {r}^{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

This is, in fact, the formula for the surface area of a sphere.<\/p>\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.