{"id":686,"date":"2025-10-09T15:30:44","date_gmt":"2025-10-09T15:30:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=686"},"modified":"2026-05-26T15:05:05","modified_gmt":"2026-05-26T15:05:05","slug":"applications-of-double-integrals-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/applications-of-double-integrals-learn-it-1\/","title":{"raw":"Applications of Double Integrals: Learn It 1","rendered":"Applications of Double Integrals: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region<\/li>\r\n \t<li>Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region<\/li>\r\n \t<li>Solve problems involving improper double integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications of Double Integrals<\/h2>\r\n<p class=\"whitespace-normal break-words\">Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function [latex]f(x,y) = 1[\/latex] over the region [latex]R[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>area of the region<\/h3>\r\nThe area of the region [latex]R[\/latex] is given by:\r\n<p style=\"text-align: center;\">[latex]A(R) = \\underset{R}{\\displaystyle\\iint}1dA[\/latex].<\/p>\r\n\r\n<\/section>This definition makes sense because using [latex]f(x,y) = 1[\/latex] and evaluating the integral make it a product of length and width. Let\u2019s check this formula with an example and see how this works.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the area of the region [latex]{R} = {\\left \\{ (x,y) \\mid 0 \\leq x \\leq 3,0 \\leq y \\leq 2 \\right \\}}[\/latex] by using a double integral, that is, by integrating [latex]1[\/latex] over the region [latex]R[\/latex].[reveal-answer q=\"802346642\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"802346642\"]The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:\r\n<p style=\"text-align: center;\">[latex]\\large{A(R)=\\displaystyle\\int_0^2\\displaystyle\\int_0^31dx \\ dy =\\displaystyle\\int_0^2\\left[x\\bigg|_0^3\\right]dy = \\displaystyle\\int_0^23dy=3\\displaystyle\\int_0^2dy=3y\\bigg|_0^2=3(2)=6}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>We have already seen how double integrals can be used to find the volume of a solid bounded above by a function [latex]f(x,y)[\/latex] over a region [latex]R[\/latex] provided [latex]f(x,y)\\geq 0[\/latex] for all [latex](x,y)[\/latex] in [latex]R[\/latex]. Here is another example to illustrate this concept.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume [latex]V[\/latex] of the solid [latex]S[\/latex] that is bounded by the elliptic paraboloid [latex]2x^2+y^2+z=27[\/latex], the planes [latex]x = 3[\/latex] and [latex]y = 3[\/latex], and the three coordinate planes.[reveal-answer q=\"731255809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"731255809\"]First notice the graph of the surface [latex]z=27-2x^2-y^2[\/latex] in\u00a0Figure 1(a) and above the square region [latex]R_1=[-3,3]\\times[-3,3][\/latex]. However, we need the volume of the solid bounded by the elliptic paraboloid [latex]2x^2+y^2+z=27[\/latex]<span style=\"font-size: 0.9em;\">,<\/span>\u00a0the planes [latex]x = 3[\/latex] and [latex]y = 3[\/latex], and the three coordinate planes.[caption id=\"attachment_1311\" align=\"aligncenter\" width=\"867\"]<img class=\"size-full wp-image-1311\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143310\/5-1-7.jpeg\" alt=\"This figure consists of two figures marked a and b. In figure a, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from negative 3 to positive 3. The shape looks like a sheet that has been pinned at the corners and forced up gently in the middle. In figure b, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from 0 to positive 3. The surface is the upper corner of the figure from part a, and below the surface is marked the solid S.\" width=\"867\" height=\"428\" \/> Figure 1. (a) The surface [latex]\\small{z=27-2x^{2}-y^{2}}[\/latex] above the square region [latex]\\small{R_{1}=[-3,3]\\times[-3,3]}[\/latex].\u00a0(b) The solid [latex]\\small{S}[\/latex] lies under the surface [latex]\\small{z=27-2x^{2}-y^{2}}[\/latex] above the square region [latex]\\small{R_{2}=[0,3]\\times[0,3]}[\/latex].[\/caption]\r\n<p id=\"fs-id1167793640035\">Now let\u2019s look at the graph of the surface in\u00a0Figure 1(b). We determine the volume [latex]V[\/latex] by evaluating the double integral over [latex]R_2[\/latex]:<\/p>\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nV&amp;=\\underset{R}{\\displaystyle\\iint}z \\ dA =\\underset{R}{\\displaystyle\\iint}(27-2x^2-y^2) \\ dA \\\\\r\n\r\n&amp;=\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=3}(27-2x^2-y^2)dx \\ dy &amp;\\quad &amp;\\text{Convert to iterated integral.} \\\\\r\n\r\n&amp;=\\displaystyle\\int_{y=0}^{y=3}\\left[27x-\\frac23x^3-y^2x\\right]\\Bigg|_{x-0}^{x=3}dy &amp;\\quad &amp;\\text{Integrate with respect to }x. \\\\\r\n\r\n&amp;=\\displaystyle\\int_{y=0}^{y=3}(63-3y^2)dy=63y-y^3\\bigg|_{y=0}^{y=3}=162.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]325846[\/ohm_question]<\/section>Similarly, we can define the average value of a function of two variables over a region [latex]R[\/latex]. The main difference is that we divide by an area instead of the width of an interval.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Recall that we defined the average value of a function of one variable on an interval [latex][a,b][\/latex] as:\r\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\dfrac{1}{b-a}}\\displaystyle\\int_a^b{f(x)dx}[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>average value of a function of two variables<\/h3>\r\nThe average value of a function of two variables over a region [latex]R[\/latex] is:\r\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\dfrac{1}{\\text{Area }R}} \\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex]<\/p>\r\n\r\n<\/section>In the next example we find the average value of a function over a rectangular region. This is a good example of obtaining useful information for an integration by making individual measurements over a grid, instead of trying to find an algebraic expression for a function.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793638512\">The weather map in\u00a0Figure 2\u00a0shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4\u20138 inches (100\u2013200 mm) of rain in some parts of the Midwest on September 22\u201323, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1313\" align=\"aligncenter\" width=\"900\"]<img class=\"size-full wp-image-1313\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143550\/5-1-8.jpeg\" alt=\"A map of Wisconsin and Minnesota that shows many cities with numbers affixed to them. The highest numbers come in a narrow band, and the map is colored accordingly. The map effectively looks like a contour map, but instead of elevations, it uses these numbers.\" width=\"900\" height=\"485\" \/> Figure 2. Effects of Hurricane Karl, which dumped 4\u20138 inches (100\u2013200 mm) of rain in some parts of southwest Wisconsin, southern Minnesota, and southeast South Dakota over a span of 300 miles east to west and 250 miles north to south.[\/caption]\r\n\r\n[reveal-answer q=\"705059936\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705059936\"]\r\n<p id=\"fs-id1167794050660\">Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles [latex](m = 2 \\ and\\ n = 3)[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>as shown in\u00a0Figure 3. Assume [latex]f(x,y)[\/latex] denotes the storm rainfall in inches at a point approximately [latex]x[\/latex] miles to the east of the origin and [latex]y[\/latex] miles to the north of the origin. Let [latex]R[\/latex] represent the entire area of [latex]250 \\times 300 = 75,000[\/latex] square miles. Then the area of each subrectangle is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta{A}} = {\\frac{1}{6}} {(75,000)} = {12,500}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793584502\">Assume [latex]({x^*_{ij}},{y^*_{ij}})[\/latex] are approximately the midpoints of each subrectangle [latex]R_{ij}[\/latex]. Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:<\/p>\r\n<p id=\"fs-id1167793584555\">At [latex](x_{11}, y_{11})[\/latex] the rainfall is 0.08.<\/p>\r\n<p id=\"fs-id1167793937944\">At [latex](x_{12}, y_{12})[\/latex] the rainfall is 0.08.<\/p>\r\n<p id=\"fs-id1167793937972\">At [latex](x_{13}, y_{13})[\/latex] the rainfall is 0.01.<\/p>\r\n<p id=\"fs-id1167793495332\">At [latex](x_{21}, y_{21})[\/latex] the rainfall is 1.70.<\/p>\r\n<p id=\"fs-id1167793495360\">At [latex](x_{22}, y_{22})[\/latex] the rainfall is 1.74.<\/p>\r\n<p id=\"fs-id1167793495388\">At [latex](x_{23}, y_{23})[\/latex] the rainfall is 3.00.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1314\" align=\"aligncenter\" width=\"730\"]<img class=\"size-full wp-image-1314\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143727\/5-1-8a.jpeg\" alt=\"Another version of the previous storm map, but this time with lines drawn for x = 100, 200, and 300 and for y = 125 and 250. There is a dot in the center of each of the resulting rectangles.\" width=\"730\" height=\"415\" \/> Figure 3. Storm rainfall with rectangular axes and showing the midpoints of each subrectangle.[\/caption]\r\n<p id=\"fs-id1167793423290\">According to our definition, the average storm rainfall in the entire area during those two days was<\/p>\r\n[latex]\\begin{align}\r\n\r\nf_{\\text{ave}}&amp;=\\frac1{\\text{Area }R}\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy=\\frac1{75,000}\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy \\\\\r\n\r\n&amp;\\cong\\frac1{75,000}\\sum_{i=1}^3\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta{A} \\\\\r\n\r\n&amp;\\cong\\frac1{75,00}\\left[f(x_{11}^*,y_{11}^*)\\Delta{A}+f(x_{12}^*,y_{12}^*)\\Delta{A}+f(x_{13}^*,y_{13}^*)\\Delta{A}+f(x_{21}^*,y_{21}^*)\\Delta{A}+f(x_{22}^*,y_{22}^*)\\Delta{A}+f(x_{23}^*,y_{23}^*)\\Delta{A}\\right] \\\\\r\n\r\n&amp;\\cong\\frac1{75,000}[0.08+0.08+0.01+1.70+1.74+3.00]\\Delta{A} \\\\\r\n\r\n&amp;\\cong\\frac1{75,00}[0.08+0.08+0.01+1.70+1.74+3.00]12,500 \\\\\r\n\r\n&amp;\\cong\\frac5{30}[0.08+0.08+0.01+1.70+1.74+3.00] \\\\\r\n\r\n&amp;\\cong1.10.\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793221644\">During September 22\u201323, 2010 this area had an average storm rainfall of approximately [latex]1.10[\/latex] inches.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region<\/li>\n<li>Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region<\/li>\n<li>Solve problems involving improper double integrals<\/li>\n<\/ul>\n<\/section>\n<h2>Applications of Double Integrals<\/h2>\n<p class=\"whitespace-normal break-words\">Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function [latex]f(x,y) = 1[\/latex] over the region [latex]R[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>area of the region<\/h3>\n<p>The area of the region [latex]R[\/latex] is given by:<\/p>\n<p style=\"text-align: center;\">[latex]A(R) = \\underset{R}{\\displaystyle\\iint}1dA[\/latex].<\/p>\n<\/section>\n<p>This definition makes sense because using [latex]f(x,y) = 1[\/latex] and evaluating the integral make it a product of length and width. Let\u2019s check this formula with an example and see how this works.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the area of the region [latex]{R} = {\\left \\{ (x,y) \\mid 0 \\leq x \\leq 3,0 \\leq y \\leq 2 \\right \\}}[\/latex] by using a double integral, that is, by integrating [latex]1[\/latex] over the region [latex]R[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q802346642\">Show Solution<\/button><\/p>\n<div id=\"q802346642\" class=\"hidden-answer\" style=\"display: none\">The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{A(R)=\\displaystyle\\int_0^2\\displaystyle\\int_0^31dx \\ dy =\\displaystyle\\int_0^2\\left[x\\bigg|_0^3\\right]dy = \\displaystyle\\int_0^23dy=3\\displaystyle\\int_0^2dy=3y\\bigg|_0^2=3(2)=6}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>We have already seen how double integrals can be used to find the volume of a solid bounded above by a function [latex]f(x,y)[\/latex] over a region [latex]R[\/latex] provided [latex]f(x,y)\\geq 0[\/latex] for all [latex](x,y)[\/latex] in [latex]R[\/latex]. Here is another example to illustrate this concept.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the volume [latex]V[\/latex] of the solid [latex]S[\/latex] that is bounded by the elliptic paraboloid [latex]2x^2+y^2+z=27[\/latex], the planes [latex]x = 3[\/latex] and [latex]y = 3[\/latex], and the three coordinate planes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q731255809\">Show Solution<\/button><\/p>\n<div id=\"q731255809\" class=\"hidden-answer\" style=\"display: none\">First notice the graph of the surface [latex]z=27-2x^2-y^2[\/latex] in\u00a0Figure 1(a) and above the square region [latex]R_1=[-3,3]\\times[-3,3][\/latex]. However, we need the volume of the solid bounded by the elliptic paraboloid [latex]2x^2+y^2+z=27[\/latex]<span style=\"font-size: 0.9em;\">,<\/span>\u00a0the planes [latex]x = 3[\/latex] and [latex]y = 3[\/latex], and the three coordinate planes.<\/p>\n<figure id=\"attachment_1311\" aria-describedby=\"caption-attachment-1311\" style=\"width: 867px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1311\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143310\/5-1-7.jpeg\" alt=\"This figure consists of two figures marked a and b. In figure a, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from negative 3 to positive 3. The shape looks like a sheet that has been pinned at the corners and forced up gently in the middle. In figure b, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from 0 to positive 3. The surface is the upper corner of the figure from part a, and below the surface is marked the solid S.\" width=\"867\" height=\"428\" \/><figcaption id=\"caption-attachment-1311\" class=\"wp-caption-text\">Figure 1. (a) The surface [latex]\\small{z=27-2x^{2}-y^{2}}[\/latex] above the square region [latex]\\small{R_{1}=[-3,3]\\times[-3,3]}[\/latex].\u00a0(b) The solid [latex]\\small{S}[\/latex] lies under the surface [latex]\\small{z=27-2x^{2}-y^{2}}[\/latex] above the square region [latex]\\small{R_{2}=[0,3]\\times[0,3]}[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1167793640035\">Now let\u2019s look at the graph of the surface in\u00a0Figure 1(b). We determine the volume [latex]V[\/latex] by evaluating the double integral over [latex]R_2[\/latex]:<\/p>\n<p>[latex]\\hspace{5cm}\\begin{align}    V&=\\underset{R}{\\displaystyle\\iint}z \\ dA =\\underset{R}{\\displaystyle\\iint}(27-2x^2-y^2) \\ dA \\\\    &=\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=3}(27-2x^2-y^2)dx \\ dy &\\quad &\\text{Convert to iterated integral.} \\\\    &=\\displaystyle\\int_{y=0}^{y=3}\\left[27x-\\frac23x^3-y^2x\\right]\\Bigg|_{x-0}^{x=3}dy &\\quad &\\text{Integrate with respect to }x. \\\\    &=\\displaystyle\\int_{y=0}^{y=3}(63-3y^2)dy=63y-y^3\\bigg|_{y=0}^{y=3}=162.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm325846\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325846&theme=lumen&iframe_resize_id=ohm325846&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<p>Similarly, we can define the average value of a function of two variables over a region [latex]R[\/latex]. The main difference is that we divide by an area instead of the width of an interval.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">Recall that we defined the average value of a function of one variable on an interval [latex][a,b][\/latex] as:<\/p>\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\dfrac{1}{b-a}}\\displaystyle\\int_a^b{f(x)dx}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>average value of a function of two variables<\/h3>\n<p>The average value of a function of two variables over a region [latex]R[\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\dfrac{1}{\\text{Area }R}} \\underset{R}{\\displaystyle\\iint}{f(x,y)dA}[\/latex]<\/p>\n<\/section>\n<p>In the next example we find the average value of a function over a rectangular region. This is a good example of obtaining useful information for an integration by making individual measurements over a grid, instead of trying to find an algebraic expression for a function.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793638512\">The weather map in\u00a0Figure 2\u00a0shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4\u20138 inches (100\u2013200 mm) of rain in some parts of the Midwest on September 22\u201323, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.<\/p>\n<figure id=\"attachment_1313\" aria-describedby=\"caption-attachment-1313\" style=\"width: 900px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1313\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143550\/5-1-8.jpeg\" alt=\"A map of Wisconsin and Minnesota that shows many cities with numbers affixed to them. The highest numbers come in a narrow band, and the map is colored accordingly. The map effectively looks like a contour map, but instead of elevations, it uses these numbers.\" width=\"900\" height=\"485\" \/><figcaption id=\"caption-attachment-1313\" class=\"wp-caption-text\">Figure 2. Effects of Hurricane Karl, which dumped 4\u20138 inches (100\u2013200 mm) of rain in some parts of southwest Wisconsin, southern Minnesota, and southeast South Dakota over a span of 300 miles east to west and 250 miles north to south.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q705059936\">Show Solution<\/button><\/p>\n<div id=\"q705059936\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794050660\">Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles [latex](m = 2 \\ and\\ n = 3)[\/latex]<span style=\"font-size: 0.9em;\">,\u00a0<\/span>as shown in\u00a0Figure 3. Assume [latex]f(x,y)[\/latex] denotes the storm rainfall in inches at a point approximately [latex]x[\/latex] miles to the east of the origin and [latex]y[\/latex] miles to the north of the origin. Let [latex]R[\/latex] represent the entire area of [latex]250 \\times 300 = 75,000[\/latex] square miles. Then the area of each subrectangle is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta{A}} = {\\frac{1}{6}} {(75,000)} = {12,500}}[\/latex].<\/p>\n<p id=\"fs-id1167793584502\">Assume [latex]({x^*_{ij}},{y^*_{ij}})[\/latex] are approximately the midpoints of each subrectangle [latex]R_{ij}[\/latex]. Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:<\/p>\n<p id=\"fs-id1167793584555\">At [latex](x_{11}, y_{11})[\/latex] the rainfall is 0.08.<\/p>\n<p id=\"fs-id1167793937944\">At [latex](x_{12}, y_{12})[\/latex] the rainfall is 0.08.<\/p>\n<p id=\"fs-id1167793937972\">At [latex](x_{13}, y_{13})[\/latex] the rainfall is 0.01.<\/p>\n<p id=\"fs-id1167793495332\">At [latex](x_{21}, y_{21})[\/latex] the rainfall is 1.70.<\/p>\n<p id=\"fs-id1167793495360\">At [latex](x_{22}, y_{22})[\/latex] the rainfall is 1.74.<\/p>\n<p id=\"fs-id1167793495388\">At [latex](x_{23}, y_{23})[\/latex] the rainfall is 3.00.<\/p>\n<figure id=\"attachment_1314\" aria-describedby=\"caption-attachment-1314\" style=\"width: 730px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1314\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23143727\/5-1-8a.jpeg\" alt=\"Another version of the previous storm map, but this time with lines drawn for x = 100, 200, and 300 and for y = 125 and 250. There is a dot in the center of each of the resulting rectangles.\" width=\"730\" height=\"415\" \/><figcaption id=\"caption-attachment-1314\" class=\"wp-caption-text\">Figure 3. Storm rainfall with rectangular axes and showing the midpoints of each subrectangle.<\/figcaption><\/figure>\n<p id=\"fs-id1167793423290\">According to our definition, the average storm rainfall in the entire area during those two days was<\/p>\n<p>[latex]\\begin{align}    f_{\\text{ave}}&=\\frac1{\\text{Area }R}\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy=\\frac1{75,000}\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy \\\\    &\\cong\\frac1{75,000}\\sum_{i=1}^3\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta{A} \\\\    &\\cong\\frac1{75,00}\\left[f(x_{11}^*,y_{11}^*)\\Delta{A}+f(x_{12}^*,y_{12}^*)\\Delta{A}+f(x_{13}^*,y_{13}^*)\\Delta{A}+f(x_{21}^*,y_{21}^*)\\Delta{A}+f(x_{22}^*,y_{22}^*)\\Delta{A}+f(x_{23}^*,y_{23}^*)\\Delta{A}\\right] \\\\    &\\cong\\frac1{75,000}[0.08+0.08+0.01+1.70+1.74+3.00]\\Delta{A} \\\\    &\\cong\\frac1{75,00}[0.08+0.08+0.01+1.70+1.74+3.00]12,500 \\\\    &\\cong\\frac5{30}[0.08+0.08+0.01+1.70+1.74+3.00] \\\\    &\\cong1.10.    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793221644\">During September 22\u201323, 2010 this area had an average storm rainfall of approximately [latex]1.10[\/latex] inches.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-686","chapter","type-chapter","status-publish","hentry"],"part":432,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/686","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/686\/revisions"}],"predecessor-version":[{"id":2889,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/686\/revisions\/2889"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/432"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/686\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=686"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=686"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=686"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}