{"id":1232,"date":"2025-10-15T15:42:10","date_gmt":"2025-10-15T15:42:10","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1232"},"modified":"2025-10-29T16:21:38","modified_gmt":"2025-10-29T16:21:38","slug":"double-integrals-in-polar-coordinates-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-in-polar-coordinates-learn-it-2\/","title":{"raw":"Double Integrals in Polar Coordinates: Learn It 2","rendered":"Double Integrals in Polar Coordinates: Learn It 2"},"content":{"raw":"

General Polar Regions of Integration<\/h2>\r\nTo evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider regions analogous to Type I and Type II as discussed earlier for rectangular coordinates. It is more common to write polar equations as [latex]r = f(\\theta)[\/latex] than [latex]\\theta = f(r)[\/latex], so we describe a general polar region as [latex]R = {(r,\\theta)\\mid \\alpha \\leq \\theta \\leq \\beta, h_1(\\theta) \\leq r \\leq h_2(\\theta) }[\/latex] (see the following figure).\r\n\r\n[caption id=\"attachment_1362\" align=\"aligncenter\" width=\"354\"]\"A Figure 5.\u00a0A general polar region between [latex]\\alpha < 0<\\beta[\/latex] and\u00a0[latex]h_{1}(\\theta) < r<h_{2}(\\theta)[\/latex].[\/caption]
\r\n

double integrals over a general polar regions<\/h3>\r\nIf [latex]{f}{(r,{\\theta})}[\/latex] is continuous on a general polar region [latex]D[\/latex] as described above, then:\r\n

[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta[\/latex]<\/p>\r\n\r\n<\/section>

Evaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{r^2}{\\sin\\theta}{r} \\ {dr} \\ {d{\\theta}}[\/latex] where [latex]D[\/latex] is the region bounded by the polar axis and the upper half of the cardioid [latex]{r} = {1} + {\\cos\\theta}[\/latex].[reveal-answer q=\"425517802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425517802\"]We can describe the region [latex]D[\/latex] as [latex]\\{(r,\\theta)\\mid0\\leq\\theta\\leq\\pi,0\\leq{r}\\leq1+\\cos\\theta\\}[\/latex] as shown in the following figure.[caption id=\"attachment_1363\" align=\"aligncenter\" width=\"425\"]\"A Figure 6.\u00a0The region [latex]D[\/latex] is the top half of a cardioid.[\/caption]Hence, we have
\r\n[latex]\r\n\\begin{array}{ccc}\r\n\\underset{D}{\\displaystyle\\iint}r^2\\sin\\theta{r}dr \\ d\\theta \\hfill &= \\hfill \\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}\\displaystyle\\int_{r=0}^{r=1+\\cos\\theta}(r^2\\sin\\theta)r dr d\\theta \\\\\r\n\\hfill &= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}[r^4]_{r=0}^{r=1+\\cos\\theta}\\sin\\theta d\\theta \\\\\r\n\\hfill &= \\hfill \\frac{1}{4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi}(1+\\cos\\theta)^4\\sin\\theta d\\theta \\\\\r\n\\hfill &= \\hfill -\\frac{1}{4}[\\frac{(1+\\cos\\theta)^5}{5}]_0^{\\pi}=\\frac{8}{5}.\r\n\\end{array}\r\n[\/latex]<\/center>[\/hidden-answer]<\/section>","rendered":"

General Polar Regions of Integration<\/h2>\n

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider regions analogous to Type I and Type II as discussed earlier for rectangular coordinates. It is more common to write polar equations as [latex]r = f(\\theta)[\/latex] than [latex]\\theta = f(r)[\/latex], so we describe a general polar region as [latex]R = {(r,\\theta)\\mid \\alpha \\leq \\theta \\leq \\beta, h_1(\\theta) \\leq r \\leq h_2(\\theta) }[\/latex] (see the following figure).<\/p>\n

\"A
Figure 5.\u00a0A general polar region between [latex]\\alpha < 0<\\beta[\/latex] and\u00a0[latex]h_{1}(\\theta) < r<h_{2}(\\theta)[\/latex].<\/figcaption><\/figure>\n
\n

double integrals over a general polar regions<\/h3>\n

If [latex]{f}{(r,{\\theta})}[\/latex] is continuous on a general polar region [latex]D[\/latex] as described above, then:<\/p>\n

[latex]\\underset{R}{\\displaystyle\\iint}f(r,\\theta)r \\ dr \\ d\\theta=\\displaystyle\\int_{\\theta=\\alpha}^{\\theta=\\beta}\\displaystyle\\int_{r=h_1(\\theta)}^{r=h_2(\\theta)}f(r,\\theta)r \\ dr \\ d\\theta[\/latex]<\/p>\n<\/section>\n

Evaluate the integral [latex]\\underset{D}{\\displaystyle\\iint}{r^2}{\\sin\\theta}{r} \\ {dr} \\ {d{\\theta}}[\/latex] where [latex]D[\/latex] is the region bounded by the polar axis and the upper half of the cardioid [latex]{r} = {1} + {\\cos\\theta}[\/latex].<\/p>\n