{"id":1249,"date":"2025-10-15T15:50:40","date_gmt":"2025-10-15T15:50:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1249"},"modified":"2025-10-31T15:26:23","modified_gmt":"2025-10-31T15:26:23","slug":"triple-integrals-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/triple-integrals-learn-it-4\/","title":{"raw":"Triple Integrals: Learn It 4","rendered":"Triple Integrals: Learn It 4"},"content":{"raw":"<h2>Average Value of a Function of Three Variables<\/h2>\r\nRecall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>average value of a function of three variables<\/h3>\r\n<p id=\"fs-id1167793311703\">If [latex]f(x, y, z)[\/latex] is integrable over a solid bounded region [latex]E[\/latex] with positive volume [latex]V(E)[\/latex], then the average value of the function is:<\/p>\r\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\frac{1}{V(E)}}\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex].<\/p>\r\n<p id=\"fs-id1167793422958\"><em>Note that the volume is <\/em>[latex]{V(E)} = \\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">The temperature at a point [latex](x, y, z)[\/latex] of a solid [latex]E[\/latex] bounded by the coordinate planes and the plane [latex]x+y+z=1[\/latex] is [latex]{T}{(x,y,z)} = {(xy+8z+20)}{^\\circ}{\\text{C}}[\/latex]. Find the average temperature over the solid.[reveal-answer q=\"067483257\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"067483257\"]\r\n<p id=\"fs-id1167793473246\">Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane [latex]x+y+z=1[\/latex] has intercepts\u00a0[latex](1, 0, 0)[\/latex], [latex](0, 1, 0)[\/latex], and [latex](0, 0, 1)[\/latex]. The region [latex]E[\/latex] looks like<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {{1-x},{0}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1-x-y} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793630666\">Hence the triple integral of the temperature is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=1-x}_{y=0}} \\ {\\displaystyle\\int^{z=1-x-y}_{z=0}}(xy+8z+20){dz}{dy}{dx} = {\\frac{147}{40}}[\/latex].<\/p>\r\nThe volume evaluation is\r\n<p style=\"text-align: center;\">[latex]{V}{(E)} = \\underset{E}{\\displaystyle\\iiint}{1dV} = {\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=1-x}_{y=0}} \\ {\\displaystyle\\int^{z=1-x-y}_{z=0}}{1dzdydx} = {\\frac{1}{6}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794160233\">Hence the average value is [latex]{{T}_{\\text{ave}}} = {\\frac{147\/40}{1\/6}} = {\\frac{6(147)}{40}} = {\\frac{441}{20}}[\/latex] degrees Celsius.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2>Average Value of a Function of Three Variables<\/h2>\n<p>Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>average value of a function of three variables<\/h3>\n<p id=\"fs-id1167793311703\">If [latex]f(x, y, z)[\/latex] is integrable over a solid bounded region [latex]E[\/latex] with positive volume [latex]V(E)[\/latex], then the average value of the function is:<\/p>\n<p style=\"text-align: center;\">[latex]{f_{\\text{ave}}} = {\\frac{1}{V(E)}}\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex].<\/p>\n<p id=\"fs-id1167793422958\"><em>Note that the volume is <\/em>[latex]{V(E)} = \\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The temperature at a point [latex](x, y, z)[\/latex] of a solid [latex]E[\/latex] bounded by the coordinate planes and the plane [latex]x+y+z=1[\/latex] is [latex]{T}{(x,y,z)} = {(xy+8z+20)}{^\\circ}{\\text{C}}[\/latex]. Find the average temperature over the solid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q067483257\">Show Solution<\/button><\/p>\n<div id=\"q067483257\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793473246\">Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane [latex]x+y+z=1[\/latex] has intercepts\u00a0[latex](1, 0, 0)[\/latex], [latex](0, 1, 0)[\/latex], and [latex](0, 0, 1)[\/latex]. The region [latex]E[\/latex] looks like<\/p>\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {{1-x},{0}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1-x-y} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167793630666\">Hence the triple integral of the temperature is<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=1-x}_{y=0}} \\ {\\displaystyle\\int^{z=1-x-y}_{z=0}}(xy+8z+20){dz}{dy}{dx} = {\\frac{147}{40}}[\/latex].<\/p>\n<p>The volume evaluation is<\/p>\n<p style=\"text-align: center;\">[latex]{V}{(E)} = \\underset{E}{\\displaystyle\\iiint}{1dV} = {\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=1-x}_{y=0}} \\ {\\displaystyle\\int^{z=1-x-y}_{z=0}}{1dzdydx} = {\\frac{1}{6}}[\/latex].<\/p>\n<p id=\"fs-id1167794160233\">Hence the average value is [latex]{{T}_{\\text{ave}}} = {\\frac{147\/40}{1\/6}} = {\\frac{6(147)}{40}} = {\\frac{441}{20}}[\/latex] degrees Celsius.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":8,"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-1249","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"- Select Header -","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\/1249","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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1249\/revisions"}],"predecessor-version":[{"id":1913,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1249\/revisions\/1913"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1249\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1249"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1249"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1249"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1249"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}