{"id":676,"date":"2025-10-09T15:29:40","date_gmt":"2025-10-09T15:29:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=676"},"modified":"2025-12-02T16:46:37","modified_gmt":"2025-12-02T16:46:37","slug":"double-integrals-over-basic-regions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-basic-regions-learn-it-1\/","title":{"raw":"Double Integrals over Basic Regions: Learn It 1","rendered":"Double Integrals over Basic Regions: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate double integrals when your region is a rectangle<\/li>\r\n \t<li>Find double integrals over more complex regions<\/li>\r\n \t<li>Switch the order of integration to make calculations easier<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Volumes and Double Integrals<\/h2>\r\nIn this section we investigate double integrals and show how we can use them to find the volume of a solid over a rectangular region in the [latex]xy[\/latex]-plane. Many of the properties of double integrals are similar to those we have already discussed for single integrals.\r\n<p class=\"whitespace-normal break-words\">Let's begin by reviewing the definite integral from single-variable calculus. This foundation will help us extend the concept to two dimensions.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Riemann Sums and the Definite Integral<\/strong>\r\n<p class=\"whitespace-normal break-words\">Suppose that [latex]y = f(x)[\/latex] is a continuous function and [latex]f(x) \\ge 0[\/latex] on a closed interval [latex][a,b][\/latex]. The area under the curve and above the [latex]x[\/latex]-axis can be approximated using rectangles.<\/p>\r\n<p class=\"whitespace-normal break-words\">To create this approximation, subdivide the interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals of width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]. If [latex]x_i^\\ast[\/latex] denotes a sample point within the [latex]i^{\\text{th}}[\/latex] subinterval, then [latex]f(x_i^\\ast)[\/latex] gives the height of the [latex]i^{\\text{th}}[\/latex] rectangle. Adding up each of these areas yields an approximation of the total area, called a Riemann sum:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The definite integral of [latex]f(x)[\/latex] on [latex][a,b][\/latex] represents the exact area under the curve [latex]y = f(x)[\/latex], and is defined as the following limit:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int_a^b f(x)dx = \\underset{n\\to \\infty}{\\text{lim}}\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\r\n\r\n<\/section><section id=\"fs-id1167794098425\" data-depth=\"1\">\r\n<p class=\"whitespace-normal break-words\">To extend this concept to two dimensions, we consider the space above a rectangular region [latex]R[\/latex]. Consider a continuous function [latex]f(x,y)\\geq 0[\/latex] of two variables defined on the closed rectangle [latex]R[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = [a,b]\\times[c,d] = {(x,y)\\in {\\mathbb{R}}^2 |a\\leq x\\leq b,c\\leq y\\leq d }[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Here [latex][a,b]\\times[c,d][\/latex] denotes the Cartesian product of the two closed intervals [latex][a,b][\/latex] and [latex][c,d][\/latex]. It consists of rectangular pairs [latex](x, y)[\/latex] such that [latex]a\\leq x\\leq b[\/latex] and [latex]c\\leq y\\leq d[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The graph of [latex]f[\/latex] represents a surface above the [latex]xy[\/latex]-plane with equation [latex]z=f(x, y)[\/latex], where [latex]z[\/latex] is the height of the surface at the point [latex](x, y)[\/latex]. Let [latex]S[\/latex] be the solid that lies above [latex]R[\/latex] and under the graph of [latex]f[\/latex] (Figure 1). The base of the solid is the rectangle [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We want to find the volume [latex]V[\/latex] of the solid [latex]S[\/latex].<\/p>\r\n\r\n<\/section>[caption id=\"attachment_1299\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-1299\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141334\/5-1-1.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d).\" width=\"487\" height=\"434\" \/> Figure 1.\u00a0The graph of\u00a0[latex]f(x,y)[\/latex] over the rectangle\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane is a curved surface.[\/caption]\r\n<p class=\"whitespace-normal break-words\">We divide the region [latex]R[\/latex] into small rectangles [latex]R_{ij}[\/latex], each with area [latex]\\Delta{A}[\/latex] and with sides [latex]\\Delta{x}[\/latex] and [latex]\\Delta{y}[\/latex] (Figure 2). We do this by dividing the interval [latex][a,b][\/latex] into [latex]m[\/latex] subintervals and dividing the interval [latex][c,d][\/latex] into [latex]n[\/latex] subintervals. Hence [latex]\\Delta{x}=\\frac{b-a}{m}[\/latex], [latex]\\Delta{y}=\\frac{d-c}{n}[\/latex], and [latex]\\Delta{A}=\\Delta{x}\\Delta{y}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1300\" align=\"aligncenter\" width=\"911\"]<img class=\"size-full wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141459\/5-1-2.jpeg\" alt=\"In the xy plane, there is a rectangle with corners (a, c), (a, d), (b, c), and (b, d). Between a and b on the x axis lines are drawn from a, x1, x2, \u2026, xi, \u2026, b with distance Delta x between each line; between c and d on the y axis lines are drawn from c, y1, y2, \u2026, yj, \u2026, d with distance Delta y between each line. Among the resulting subrectangles, the one in the second column and third row up has a point marked (x*23, y*23). The rectangle Rij is marked with upper right corner (xi, yj). Within this rectangle the point (x*ij, y*ij) is marked.\" width=\"911\" height=\"527\" \/> Figure 2. Rectangle\u00a0[latex]R[\/latex] is divided into small rectangles [latex]R_{ij}[\/latex], each with an area [latex]\\Delta{A}[\/latex].[\/caption]The volume of a thin rectangular box above [latex]{R_{ij}}[\/latex] is [latex]{f{({x{_{ij}^*}},{y{_{ij}^*}})}}{\\Delta{A}}[\/latex], where [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] is an arbitrary sample point in each [latex]{R_{ij}}[\/latex] as shown in the following figure.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1301\" align=\"aligncenter\" width=\"487\"]<img class=\"size-full wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141611\/5-1-3.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d). There are additional squares drawn to correspond to changes of Delta x and Delta y. On the surface, a square is marked and its projection onto the plane is marked as Rij. The average value for this small square is f(x*ij, y*ij).\" width=\"487\" height=\"474\" \/> Figure 3.\u00a0A thin rectangular box above\u00a0[latex]R_{ij}[\/latex] with height\u00a0[latex]f(x^{\\ast}_{ij},y^{\\ast}_{ij})[\/latex].[\/caption]\r\n<p class=\"whitespace-normal break-words\">Using the same idea for all the subrectangles, we obtain an approximate volume of the solid [latex]S[\/latex] as [latex]{V}{\\approx} \\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\ {f{({x{_{ij}^*}},{y{_{ij}^*}})}{\\Delta{A}}}[\/latex]. This sum is known as a <strong>double Riemann sum<\/strong> and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.<\/p>\r\n<p class=\"whitespace-normal break-words\">As we have seen in the single-variable case, we obtain a better approximation to the actual volume if [latex]m[\/latex] and [latex]n[\/latex] become larger. The exact volume is given by:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{{V}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f{({x{_{ij}^*}},{y{_{ij}^*}}){\\Delta{A}}}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">or equivalently,<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\lim_{\\Delta x,\\Delta y\\to0}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_{ij}^*,y_{ij}^*)\\Delta A}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base [latex]R[\/latex]. Now we are ready to define the double integral.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Double Integral<\/h3>\r\nThe\u00a0<span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term206\" data-type=\"term\">double integral<\/span> of the function [latex]f(x, y)[\/latex] over the rectangular region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is defined as:\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\r\n\r\n<ul>\r\n \t<li>If [latex]f(x, y)\\geq 0[\/latex], then the volume [latex]V[\/latex] of the solid [latex]S[\/latex], which lies above [latex]R[\/latex] in the [latex]xy[\/latex]-plane and under the graph of [latex]f[\/latex], is the double integral of the function [latex]f(x, y)[\/latex] over the rectangle [latex]R[\/latex].<\/li>\r\n \t<li>If the function is ever negative, then the double integral can be considered a \u201csigned\u201d volume in a manner similar to the way we defined net signed area in single-variable calculus.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793933224\">Consider the function [latex]{z = f(x,y) = 3x^2 - y}[\/latex] over the rectangular region [latex]{R = {[0,2]} \\times {[0,2]}}[\/latex] (Figure 4).<\/p>\r\n\r\n<ol id=\"fs-id1167793247300\" type=\"a\">\r\n \t<li>Set up a double integral for finding the value of the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the upper right corner point of each square [latex](1,1),(2,1),(1,2),[\/latex] and [latex](2,2)[\/latex] (Figure 4) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\r\n \t<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the midpoint of each square: [latex](1\/2,1\/2),(3\/2,1\/2),(1\/2,3\/2),[\/latex] and [latex](3\/2,3\/2)[\/latex] to approximate the signed volume.<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1303\" align=\"aligncenter\" width=\"483\"]<img class=\"size-full wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141819\/5-1-4.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.\" width=\"483\" height=\"252\" \/> Figure 4.\u00a0The function [latex]\\small{z=f(x,y)}[\/latex] graphed over the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].[\/caption][reveal-answer q=\"193760238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"193760238\"]\r\n<div class=\"os-solution-container\">\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li style=\"text-align: center;\">As we can see, the function [latex]z = f(x,y) = 3x^2 - y[\/latex] is above the plane. To find the signed volume of [latex]S[\/latex], we need to divide the region [latex]R[\/latex] into small rectangles [latex]{R}_{ij}[\/latex] each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex], and choose [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] as sample points in each [latex]{R}_{ij}[\/latex]. Hence, a double integral is set up as\r\n[latex]V = \\underset{R}{\\displaystyle\\iint}(3x^2-y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\left[3(x_{ij}^*)^2-y_{ij}^*\\right]\\Delta A[\/latex].<span data-type=\"newline\">\r\n<\/span><\/li>\r\n \t<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex] we have [latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. Also, the sample points are [latex](1, 1), (2, 1), (1, 2),[\/latex] and [latex](2,2)[\/latex] as shown in the following figure.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_15_01_005\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_1305\" align=\"aligncenter\" width=\"376\"]<img class=\"size-full wp-image-1305\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23142153\/5-1-4a.jpeg\" alt=\"On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.\" width=\"376\" height=\"484\" \/> Figure 5. Subrectangles for the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].[\/caption]<\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Hence,<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[latex]\\hspace{3cm}\\begin{align}\r\n\r\nV&amp;=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\\r\n\r\n&amp;=\\displaystyle\\sum_{i=1}^2(f(x_{i1}^*,y_{i1}^*)+ff(x_{i2}^*,y_{i2}^*))\\Delta A \\\\\r\n\r\n&amp;=(f(x_{11}^*,y_{11}^*)\\Delta A+(f(x_{21}^*,y_{21}^*)\\Delta A+(f(x_{12}^*,y_{12}^*)\\Delta A+(f(x_{22}^*,y_{22}^*)\\Delta A \\\\\r\n&amp;=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1) \\\\\r\n&amp;=(3-1)(1)+(12-1)(1)+(3-1)(1)+(12-2)(1) \\\\\r\n&amp;=2+11+1+10=24.\r\n\\end{align}[\/latex]<\/li>\r\n \t<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex], we have[latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. In this case the sample points are [latex](1\/2, 1\/2), (3\/2, 1\/2), (1\/2, 3\/2),[\/latex] and [latex](3\/2, 3\/2)[\/latex].<span data-type=\"newline\">\r\n<\/span>Hence,<\/li>\r\n<\/ol>\r\n[latex]\\hspace{3cm}\\begin{align}\r\nV&amp;=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\&amp;=f(x_{11}^*,y_{11}^*)\\Delta A+f(x_{21}^*,y_{21}^*)\\Delta A+f(x_{12}^*,y_{12}^*)\\Delta A+f(x_{22}^*,y_{22}^*)\\Delta A \\\\\r\n&amp;=f(1\/2,1\/2)(1)+f(3\/2,1\/2)(1)+f(1\/2,3\/2)(1)+f(3\/2,3\/2)(1) \\\\\r\n&amp;=\\left(\\small{\\frac34}-\\small{\\frac14}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac12}\\right)(1)+\\left(\\small{\\frac34}-\\small{\\frac32}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac32}\\right)(1) \\\\\r\n&amp;=\\frac24+\\frac{25}4+\\left(-\\frac34\\right)+\\frac{21}4=\\frac{45}4.\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\n<p id=\"9\" data-type=\"commentary-title\"><strong><span class=\"os-title-label\">Analysis<\/span><\/strong><\/p>\r\n<p id=\"fs-id1167794067077\">Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\"><strong>Notation and Integrability<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Note that we developed the concept of double integral using a rectangular region [latex]R[\/latex]. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into [latex]R[\/latex], particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region [latex]R[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Also, the heights may not be exact if the surface [latex]z = f(x,y)[\/latex] is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid [latex]S[\/latex] approach [latex]0[\/latex] as [latex]m[\/latex] and [latex]n[\/latex] approach infinity. The double integral of the function [latex]z = f(x,y)[\/latex] exists provided that the function [latex]f[\/latex] is not too discontinuous. If the function is bounded and continuous over [latex]R[\/latex] except on a finite number of smooth curves, then the double integral exists and we say that [latex]f[\/latex] is <strong>integrable<\/strong> over [latex]R[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]\\Delta{A}=\\Delta{x}\\Delta{y}=\\Delta y\\Delta x[\/latex], we can express [latex]dA[\/latex] as [latex]dxdy[\/latex] or [latex]dydx[\/latex]. This means that, when we are using rectangular coordinates, the double integral over a region [latex]R[\/latex] denoted by [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex] can be written as<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy[\/latex] or [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dydx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Now let's list some of the properties that can be helpful to compute double integrals.<\/p>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate double integrals when your region is a rectangle<\/li>\n<li>Find double integrals over more complex regions<\/li>\n<li>Switch the order of integration to make calculations easier<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Volumes and Double Integrals<\/h2>\n<p>In this section we investigate double integrals and show how we can use them to find the volume of a solid over a rectangular region in the [latex]xy[\/latex]-plane. Many of the properties of double integrals are similar to those we have already discussed for single integrals.<\/p>\n<p class=\"whitespace-normal break-words\">Let&#8217;s begin by reviewing the definite integral from single-variable calculus. This foundation will help us extend the concept to two dimensions.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Riemann Sums and the Definite Integral<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Suppose that [latex]y = f(x)[\/latex] is a continuous function and [latex]f(x) \\ge 0[\/latex] on a closed interval [latex][a,b][\/latex]. The area under the curve and above the [latex]x[\/latex]-axis can be approximated using rectangles.<\/p>\n<p class=\"whitespace-normal break-words\">To create this approximation, subdivide the interval [latex][a,b][\/latex] into [latex]n[\/latex] subintervals of width [latex]\\Delta x = \\frac{b-a}{n}[\/latex]. If [latex]x_i^\\ast[\/latex] denotes a sample point within the [latex]i^{\\text{th}}[\/latex] subinterval, then [latex]f(x_i^\\ast)[\/latex] gives the height of the [latex]i^{\\text{th}}[\/latex] rectangle. Adding up each of these areas yields an approximation of the total area, called a Riemann sum:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The definite integral of [latex]f(x)[\/latex] on [latex][a,b][\/latex] represents the exact area under the curve [latex]y = f(x)[\/latex], and is defined as the following limit:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int_a^b f(x)dx = \\underset{n\\to \\infty}{\\text{lim}}\\sum_{i=1}^n f(x_i^\\ast)\\Delta x[\/latex]<\/p>\n<\/section>\n<section id=\"fs-id1167794098425\" data-depth=\"1\">\n<p class=\"whitespace-normal break-words\">To extend this concept to two dimensions, we consider the space above a rectangular region [latex]R[\/latex]. Consider a continuous function [latex]f(x,y)\\geq 0[\/latex] of two variables defined on the closed rectangle [latex]R[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R = [a,b]\\times[c,d] = {(x,y)\\in {\\mathbb{R}}^2 |a\\leq x\\leq b,c\\leq y\\leq d }[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Here [latex][a,b]\\times[c,d][\/latex] denotes the Cartesian product of the two closed intervals [latex][a,b][\/latex] and [latex][c,d][\/latex]. It consists of rectangular pairs [latex](x, y)[\/latex] such that [latex]a\\leq x\\leq b[\/latex] and [latex]c\\leq y\\leq d[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The graph of [latex]f[\/latex] represents a surface above the [latex]xy[\/latex]-plane with equation [latex]z=f(x, y)[\/latex], where [latex]z[\/latex] is the height of the surface at the point [latex](x, y)[\/latex]. Let [latex]S[\/latex] be the solid that lies above [latex]R[\/latex] and under the graph of [latex]f[\/latex] (Figure 1). The base of the solid is the rectangle [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We want to find the volume [latex]V[\/latex] of the solid [latex]S[\/latex].<\/p>\n<\/section>\n<figure id=\"attachment_1299\" aria-describedby=\"caption-attachment-1299\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1299\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141334\/5-1-1.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d).\" width=\"487\" height=\"434\" \/><figcaption id=\"caption-attachment-1299\" class=\"wp-caption-text\">Figure 1.\u00a0The graph of\u00a0[latex]f(x,y)[\/latex] over the rectangle\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane is a curved surface.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">We divide the region [latex]R[\/latex] into small rectangles [latex]R_{ij}[\/latex], each with area [latex]\\Delta{A}[\/latex] and with sides [latex]\\Delta{x}[\/latex] and [latex]\\Delta{y}[\/latex] (Figure 2). We do this by dividing the interval [latex][a,b][\/latex] into [latex]m[\/latex] subintervals and dividing the interval [latex][c,d][\/latex] into [latex]n[\/latex] subintervals. Hence [latex]\\Delta{x}=\\frac{b-a}{m}[\/latex], [latex]\\Delta{y}=\\frac{d-c}{n}[\/latex], and [latex]\\Delta{A}=\\Delta{x}\\Delta{y}[\/latex].<\/p>\n<figure id=\"attachment_1300\" aria-describedby=\"caption-attachment-1300\" style=\"width: 911px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141459\/5-1-2.jpeg\" alt=\"In the xy plane, there is a rectangle with corners (a, c), (a, d), (b, c), and (b, d). Between a and b on the x axis lines are drawn from a, x1, x2, \u2026, xi, \u2026, b with distance Delta x between each line; between c and d on the y axis lines are drawn from c, y1, y2, \u2026, yj, \u2026, d with distance Delta y between each line. Among the resulting subrectangles, the one in the second column and third row up has a point marked (x*23, y*23). The rectangle Rij is marked with upper right corner (xi, yj). Within this rectangle the point (x*ij, y*ij) is marked.\" width=\"911\" height=\"527\" \/><figcaption id=\"caption-attachment-1300\" class=\"wp-caption-text\">Figure 2. Rectangle\u00a0[latex]R[\/latex] is divided into small rectangles [latex]R_{ij}[\/latex], each with an area [latex]\\Delta{A}[\/latex].<\/figcaption><\/figure>\n<p>The volume of a thin rectangular box above [latex]{R_{ij}}[\/latex] is [latex]{f{({x{_{ij}^*}},{y{_{ij}^*}})}}{\\Delta{A}}[\/latex], where [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] is an arbitrary sample point in each [latex]{R_{ij}}[\/latex] as shown in the following figure.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1301\" aria-describedby=\"caption-attachment-1301\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141611\/5-1-3.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d). There are additional squares drawn to correspond to changes of Delta x and Delta y. On the surface, a square is marked and its projection onto the plane is marked as Rij. The average value for this small square is f(x*ij, y*ij).\" width=\"487\" height=\"474\" \/><figcaption id=\"caption-attachment-1301\" class=\"wp-caption-text\">Figure 3.\u00a0A thin rectangular box above\u00a0[latex]R_{ij}[\/latex] with height\u00a0[latex]f(x^{\\ast}_{ij},y^{\\ast}_{ij})[\/latex].<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">Using the same idea for all the subrectangles, we obtain an approximate volume of the solid [latex]S[\/latex] as [latex]{V}{\\approx} \\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\ {f{({x{_{ij}^*}},{y{_{ij}^*}})}{\\Delta{A}}}[\/latex]. This sum is known as a <strong>double Riemann sum<\/strong> and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.<\/p>\n<p class=\"whitespace-normal break-words\">As we have seen in the single-variable case, we obtain a better approximation to the actual volume if [latex]m[\/latex] and [latex]n[\/latex] become larger. The exact volume is given by:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{{V}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f{({x{_{ij}^*}},{y{_{ij}^*}}){\\Delta{A}}}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">or equivalently,<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\large{V=\\displaystyle\\lim_{\\Delta x,\\Delta y\\to0}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_{ij}^*,y_{ij}^*)\\Delta A}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base [latex]R[\/latex]. Now we are ready to define the double integral.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Double Integral<\/h3>\n<p>The\u00a0<span id=\"d9da6d44-3ba3-4a90-adbc-dec26572e480_term206\" data-type=\"term\">double integral<\/span> of the function [latex]f(x, y)[\/latex] over the rectangular region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is defined as:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}f(x_i^*,y_j^*)\\Delta A}[\/latex].<\/p>\n<ul>\n<li>If [latex]f(x, y)\\geq 0[\/latex], then the volume [latex]V[\/latex] of the solid [latex]S[\/latex], which lies above [latex]R[\/latex] in the [latex]xy[\/latex]-plane and under the graph of [latex]f[\/latex], is the double integral of the function [latex]f(x, y)[\/latex] over the rectangle [latex]R[\/latex].<\/li>\n<li>If the function is ever negative, then the double integral can be considered a \u201csigned\u201d volume in a manner similar to the way we defined net signed area in single-variable calculus.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793933224\">Consider the function [latex]{z = f(x,y) = 3x^2 - y}[\/latex] over the rectangular region [latex]{R = {[0,2]} \\times {[0,2]}}[\/latex] (Figure 4).<\/p>\n<ol id=\"fs-id1167793247300\" type=\"a\">\n<li>Set up a double integral for finding the value of the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\n<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the upper right corner point of each square [latex](1,1),(2,1),(1,2),[\/latex] and [latex](2,2)[\/latex] (Figure 4) to approximate the signed volume of the solid [latex]S[\/latex] that lies above [latex]R[\/latex] and \u201cunder\u201d the graph of [latex]f[\/latex].<\/li>\n<li>Divide [latex]R[\/latex] into four squares with [latex]m = n = 2[\/latex], and choose the sample point as the midpoint of each square: [latex](1\/2,1\/2),(3\/2,1\/2),(1\/2,3\/2),[\/latex] and [latex](3\/2,3\/2)[\/latex] to approximate the signed volume.<\/li>\n<\/ol>\n<figure id=\"attachment_1303\" aria-describedby=\"caption-attachment-1303\" style=\"width: 483px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23141819\/5-1-4.jpeg\" alt=\"In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.\" width=\"483\" height=\"252\" \/><figcaption id=\"caption-attachment-1303\" class=\"wp-caption-text\">Figure 4.\u00a0The function [latex]\\small{z=f(x,y)}[\/latex] graphed over the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q193760238\">Show Solution<\/button><\/p>\n<div id=\"q193760238\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"os-solution-container\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li style=\"text-align: center;\">As we can see, the function [latex]z = f(x,y) = 3x^2 - y[\/latex] is above the plane. To find the signed volume of [latex]S[\/latex], we need to divide the region [latex]R[\/latex] into small rectangles [latex]{R}_{ij}[\/latex] each with area [latex]{\\Delta{A}}[\/latex] and with sides [latex]{\\Delta{x}}[\/latex] and [latex]{\\Delta{y}}[\/latex], and choose [latex]{({x{_{ij}^*}},{y{_{ij}^*}})}[\/latex] as sample points in each [latex]{R}_{ij}[\/latex]. Hence, a double integral is set up as<br \/>\n[latex]V = \\underset{R}{\\displaystyle\\iint}(3x^2-y)dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^{m}\\sum_{j=1}^{n}\\left[3(x_{ij}^*)^2-y_{ij}^*\\right]\\Delta A[\/latex].<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex] we have [latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. Also, the sample points are [latex](1, 1), (2, 1), (1, 2),[\/latex] and [latex](2,2)[\/latex] as shown in the following figure.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_15_01_005\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<figure id=\"attachment_1305\" aria-describedby=\"caption-attachment-1305\" style=\"width: 376px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1305\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/23142153\/5-1-4a.jpeg\" alt=\"On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.\" width=\"376\" height=\"484\" \/><figcaption id=\"caption-attachment-1305\" class=\"wp-caption-text\">Figure 5. Subrectangles for the rectangular region [latex]\\small{R=[0,2]\\times[0,2]}[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Hence,<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>[latex]\\hspace{3cm}\\begin{align}    V&=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\    &=\\displaystyle\\sum_{i=1}^2(f(x_{i1}^*,y_{i1}^*)+ff(x_{i2}^*,y_{i2}^*))\\Delta A \\\\    &=(f(x_{11}^*,y_{11}^*)\\Delta A+(f(x_{21}^*,y_{21}^*)\\Delta A+(f(x_{12}^*,y_{12}^*)\\Delta A+(f(x_{22}^*,y_{22}^*)\\Delta A \\\\  &=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1) \\\\  &=(3-1)(1)+(12-1)(1)+(3-1)(1)+(12-2)(1) \\\\  &=2+11+1+10=24.  \\end{align}[\/latex]<\/li>\n<li>Approximating the signed volume using a Riemann sum with [latex]m = n = 2[\/latex], we have[latex]{\\Delta{A}} = {\\Delta{x}}{\\Delta{y}} = {1 {\\times} 1} = 1[\/latex]. In this case the sample points are [latex](1\/2, 1\/2), (3\/2, 1\/2), (1\/2, 3\/2),[\/latex] and [latex](3\/2, 3\/2)[\/latex].<span data-type=\"newline\"><br \/>\n<\/span>Hence,<\/li>\n<\/ol>\n<p>[latex]\\hspace{3cm}\\begin{align}  V&=\\displaystyle\\sum_{i=1}^2\\sum_{j=1}^2f(x_{ij}^*,y_{ij}^*)\\Delta A \\\\&=f(x_{11}^*,y_{11}^*)\\Delta A+f(x_{21}^*,y_{21}^*)\\Delta A+f(x_{12}^*,y_{12}^*)\\Delta A+f(x_{22}^*,y_{22}^*)\\Delta A \\\\  &=f(1\/2,1\/2)(1)+f(3\/2,1\/2)(1)+f(1\/2,3\/2)(1)+f(3\/2,3\/2)(1) \\\\  &=\\left(\\small{\\frac34}-\\small{\\frac14}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac12}\\right)(1)+\\left(\\small{\\frac34}-\\small{\\frac32}\\right)(1)+\\left(\\small{\\frac{27}4}-\\small{\\frac32}\\right)(1) \\\\  &=\\frac24+\\frac{25}4+\\left(-\\frac34\\right)+\\frac{21}4=\\frac{45}4.  \\end{align}[\/latex]<\/p>\n<\/div>\n<p id=\"9\" data-type=\"commentary-title\"><strong><span class=\"os-title-label\">Analysis<\/span><\/strong><\/p>\n<p id=\"fs-id1167794067077\">Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p class=\"whitespace-normal break-words\"><strong>Notation and Integrability<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Note that we developed the concept of double integral using a rectangular region [latex]R[\/latex]. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into [latex]R[\/latex], particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region [latex]R[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Also, the heights may not be exact if the surface [latex]z = f(x,y)[\/latex] is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid [latex]S[\/latex] approach [latex]0[\/latex] as [latex]m[\/latex] and [latex]n[\/latex] approach infinity. The double integral of the function [latex]z = f(x,y)[\/latex] exists provided that the function [latex]f[\/latex] is not too discontinuous. If the function is bounded and continuous over [latex]R[\/latex] except on a finite number of smooth curves, then the double integral exists and we say that [latex]f[\/latex] is <strong>integrable<\/strong> over [latex]R[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]\\Delta{A}=\\Delta{x}\\Delta{y}=\\Delta y\\Delta x[\/latex], we can express [latex]dA[\/latex] as [latex]dxdy[\/latex] or [latex]dydx[\/latex]. This means that, when we are using rectangular coordinates, the double integral over a region [latex]R[\/latex] denoted by [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dA[\/latex] can be written as<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dxdy[\/latex] or [latex]\\underset{R}{\\displaystyle\\iint}f(x,y)dydx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Now let&#8217;s list some of the properties that can be helpful to compute double integrals.<\/p>\n","protected":false},"author":15,"menu_order":6,"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-676","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\/676","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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/676\/revisions"}],"predecessor-version":[{"id":2747,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/676\/revisions\/2747"}],"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\/676\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=676"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=676"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=676"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=676"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}