{"id":601,"date":"2025-12-03T10:15:14","date_gmt":"2025-12-03T09:15:14","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=601"},"modified":"2025-11-26T23:16:06","modified_gmt":"2025-11-26T22:16:06","slug":"mathbio-integral-triple","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=601","title":{"rendered":"MathBio: Integral triple"},"content":{"rendered":"<p>La integral triple de una funci\u00f3n, \\(f(x,y,z)\\), en una regi\u00f3n cerrada del espacio, \\(Q\\), con un volumen \\(V\\), no es m\u00e1s que la generalizaci\u00f3n del concepto de integral simple y doble. As\u00ed<br \/>\n\\[\\iiint_{Q} f(x,y,z) dV = \\underset{l,m,n \\to \\infty}{lim}\\sum_{l}^{i=1}\\sum_{m}^{j=1}\\sum_{n}^{k=1}f(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})\\Delta V\\]<\/p>\n<p>Si la funci\u00f3n es \\(f(x,y,z)=1\\), tendremos el volumen del s\u00f3lido encerrado en \\(Q\\), \\[Volumen=\\iiint_Q\\,dV\\]<\/p>\n<p>De este modo, cuando el s\u00f3lido viene dado por un paralep\u00edpedo, \\(Q=\\{[a,a&#8217;]\\times [b,b&#8217;]\\times [c,c&#8217;]\\}\\), calcular el volumen es resolver mediante integrales iteradas:<br \/>\n\\[\\iiint_Qf(x,y,z)\\,dV=\\int_a^{a&#8217;}\\int_b^{b&#8217;}\\int_c^{c&#8217;}f(x,y,z)\\,dz\\,dy\\,dx.\\]<\/p>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Cu\u00e1l es el volumen encerrado en el recinto \\(Q=[0,1]\\times[1,2]\\times[0,2]\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9a() {\n  var htmlShow9a = document.getElementById(\"html-show9a\");\n  if (htmlShow9a.style.display === \"none\") {\n    htmlShow9a.style.display = \"block\";\n  } else {\n    htmlShow9a.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a\" style=\"display: none;\">\nNos piden calcular \\[\\iiint_Q\\ dV=\\int_{0}^{1}\\int_{1}^{2}\\int_{0}^{2} \\ dz\\,dy\\,dx.\\] Aplicando la definici\u00f3n: \\[\\int_{0}^{1}\\int_{1}^{2}\\int_{0}^{2} \\ dz\\,dy\\,dx=\\int_{0}^{1}\\left[\\int_{1}^{2}\\left[z\\right|_{0}^{2}\\,dy\\right]\\,dx=\\] \\[=\\int_{0}^{1}2\\left[y\\right|_{1}^{2}\\,dx=2\\left[x\\right|_{0}^{2}=4\\]<br \/>\n<iframe loading=\"lazy\" title=\"Matem\u00e1tica Aplicada - Integral triple. Ej.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/hu0_6y9baQM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<p>Igual como en el caso de la integral doble se cumple:\\[\\iiint_Q(\\lambda f(x,y,z)+\\mu g(x,y,z))\\,dV=\\lambda\\iiint_Q f(x,y,z)\\,dV+\\mu\\iiint_Q g(x,y,z)\\,dV.\\]<br \/>\n A veces es m\u00e1s pr\u00e1ctico verlo como la integral doble en una regi\u00f3n del plano XY de una funci\u00f3n; es decir,<\/p>\n<p>\\[V=\\iiint_QdV=\\iint_R\\left(\\int_{f(x,y)}^{g(x,y)}dz\\right)dA,\\]<\/p>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Cu\u00e1l es el volumen encerrado en la regi\u00f3n s\u00f3lida \\(Q=\\{(x,y,z);\\ 0\\leq x\\leq 1,\\ 0\\leq y\\leq 1,\\ 0\\leq z\\leq x-y^2\\}\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9b() {\n  var htmlShow9b = document.getElementById(\"html-show9b\");\n  if (htmlShow9b.style.display === \"none\") {\n    htmlShow9b.style.display = \"block\";\n  } else {\n    htmlShow9b.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9b\" style=\"display: none;\">\nNos piden calcular \\[\\iiint_Q\\ dV=\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{x-y^2} \\ dz\\,dy\\,dx.\\] Ve\u00e1moslo: \\[\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{x-y^2} \\ dz\\,dy\\,dx=\\int_{0}^{1}\\int_{0}^{1}\\left[\\int_{0}^{x-y^2} \\ dz\\right]\\,dy\\,dx=\\]\\[=\\int_{0}^{1}\\left[\\int_{0}^{1}\\left[z\\right|_{0}^{x-y^2}\\,dy\\right]\\,dx=\\int_{0}^{1}\\left[\\int_{0}^{1}\\left[{x-y^2}\\right]\\,dy\\right]\\,dx=\\] \\[=\\int_{0}^{1}\\left[yx-\\frac{y^3}{3}\\right|_0^1\\,dx=\\int_{0}^{1}\\left[x-\\frac{1}{3}\\right]\\,dx=\\left[\\frac{x^2}{2}-\\frac{x}{3}\\right|_{0}^{1}=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}\\]\n<\/div>\n<hr \/>\n<p>En general, si \\(Q=\\{(x,y,z)|a\\leq x \\leq b,\\, h_1(x)\\leq y \\leq h_2(x),\\,g_1(x,y)\\leq z \\leq g_1(x,y)\\}\\), tendremos<br \/>\n\\[\\iiint_{Q}\\, f(x,y,z) dV = \\int_{a}^{b}\\int_{h_1(x)}^{h_2(x)}\\int_{g_1(x,y)}^{g_2(x,y)}\\, f(x,y,z) dz dy dx\\]<\/p>\n<p>Veamos unos ejemplos:<\/p>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Calcular \\(\\int_{0}^{1}\\int_{0}^{y}\\int_{0}^{x} \\cos(x+y+z)dz\\,dx\\,dy\\), <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9() {\n  var htmlShow9 = document.getElementById(\"html-show9\");\n  if (htmlShow9.style.display === \"none\") {\n    htmlShow9.style.display = \"block\";\n  } else {\n    htmlShow9.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9\" style=\"display: none;\">\nResolvamos primero<br \/>\n\\[ I_1 = \\int_{0}^{x} \\cos(x+y+z)dz= \\left[ \\sin(x+y+z) \\right]_{z=0}^{z=x}=\\sin(2x+y) &#8211; \\sin(x+y) \\]<br \/>\nAhora,<br \/>\n\\[<br \/>\n\\begin{align}<br \/>\nI_2 &#038;= \\int_{0}^{y} \\left( \\sin(2x+y) &#8211; \\sin(x+y) \\right) dx \\\\<br \/>\n&#038;= \\left[ -\\frac{1}{2}\\cos(2x+y) + \\cos(x+y) \\right]_{x=0}^{x=y}\\\\<br \/>\n&#038;=-\\frac{1}{2}\\cos(3y) + \\cos(2y)-\\frac{1}{2}\\cos(y)<br \/>\n\\end{align}<br \/>\n\\]<br \/>\nPor \u00faltimo,<br \/>\n\\[<br \/>\n\\begin{align}<br \/>\nI &#038;= \\int_{0}^{1} \\left( \\cos(2y) &#8211; \\frac{1}{2}\\cos(3y) &#8211; \\frac{1}{2}\\cos(y) \\right) dy\\\\<br \/>\n&#038;= \\left[ \\frac{1}{2}\\sin(2y) &#8211; \\frac{1}{6}\\sin(3y) &#8211; \\frac{1}{2}\\sin(y) \\right]_{0}^{1}\\\\<br \/>\n&#038;= \\frac{1}{2}\\sin(2) &#8211; \\frac{1}{6}\\sin(3) &#8211; \\frac{1}{2}\\sin(1)<br \/>\n\\end{align}<br \/>\n\\]<\/p>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Calcular \\(\\int_{0}^{\\sqrt{\\pi}}\\int_{0}^{x}\\int_{0}^{xz} x^2\\, \\sin(y)dy\\,dz\\,dx\\), <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv8() {\n  var htmlShow8 = document.getElementById(\"html-show8\");\n  if (htmlShow8.style.display === \"none\") {\n    htmlShow8.style.display = \"block\";\n  } else {\n    htmlShow8.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv8()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show8\" style=\"display: none;\">\n<p>\\[<br \/>\n\\begin{align}<br \/>\nI &#038;=\\int_{0}^{\\sqrt{\\pi}}\\int_{0}^{x}\\int_{0}^{xz} x^2\\, \\sin(y)dy\\,dz\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\int_{0}^{x}\\left(\\int_{0}^{xz} x^2\\, \\sin(y)dy\\right)\\,dz\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\int_{0}^{x}\\left(x^2 \\left[ -\\cos(y) \\right]_{y=0}^{y=xz}\\right)\\,dz\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\int_{0}^{x}\\left(x^2 \\left( 1 &#8211; \\cos(xz) \\right)\\right)\\,dz\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\left(\\int_{0}^{x}\\left(x^2 \\left( 1 &#8211; \\cos(xz) \\right)\\right)\\,dz\\right)\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\left(x^2 \\left[ z &#8211; \\frac{1}{x}\\sin(xz) \\right]_{z=0}^{z=x}\\right)\\,dx \\\\<br \/>\n&#038;=\\int_{0}^{\\sqrt{\\pi}}\\left(x^3 -x\\sin(x^2)\\right)\\,dx \\\\<br \/>\n&#038;\\mbox{(hacemos un cambio de variable}\\,u=x^2 \\mbox{para resolver}\\,  \\int \\,x\\sin(x^2)\\,dx\\\\<br \/>\n&#038;=\\left[ \\frac{x^4}{4} \\right]_{0}^{\\sqrt{\\pi}}+ \\frac{1}{2} \\left[ -\\cos(u) \\right]_{0}^{\\pi}\\\\<br \/>\n&#038;=\\frac{\\pi^2}{4} &#8211; 1<br \/>\n\\end{align}<br \/>\n\\]\n<\/p><\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Calcular \\(\\int_{0}^{1}\\int_{0}^{1}\\int_{0}^{\\sqrt{1-z^2}} \\left(\\frac{z}{y+1}\\right)dx\\,dz\\,dy\\), <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv7() {\n  var htmlShow7 = document.getElementById(\"html-show7\");\n  if (htmlShow7.style.display === \"none\") {\n    htmlShow7.style.display = \"block\";\n  } else {\n    htmlShow7.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv7()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show7\" style=\"display: none;\">\nVamos a calcularla paso a paso.<\/p>\n<p><strong>Paso 1<\/strong>: Integral interior (respecto a \\(x\\))<\/p>\n<p>Integramos \\(\\left(\\frac{z}{y+1}\\right)\\) con respecto a \\(x\\):<br \/>\n\\[<br \/>\n\\begin{aligned}<br \/>\nI_1 &#038;= \\int_{0}^{\\sqrt{1-z^2}} \\left(\\frac{z}{y+1}\\right)dx \\\\<br \/>\n&#038;= \\left(\\frac{z}{y+1}\\right) [x]_{x=0}^{x=\\sqrt{1-z^2}} \\\\<br \/>\n&#038;= \\left(\\frac{z}{y+1}\\right) (\\sqrt{1-z^2} &#8211; 0) \\\\<br \/>\n&#038;= \\frac{z\\sqrt{1-z^2}}{y+1}<br \/>\n\\end{aligned}<br \/>\n\\]<\/p>\n<p><strong>Paso 2<\/strong>: Integral media (respecto a \\(z\\))<\/p>\n<p>Sustituimos e integramos con respecto a \\(z\\):<br \/>\n\\[<br \/>\n\\begin{aligned}<br \/>\nI_2 &#038;= \\int_{0}^{1} \\frac{z\\sqrt{1-z^2}}{y+1} dz \\\\<br \/>\n&#038;= \\frac{1}{y+1} \\int_{0}^{1} z\\sqrt{1-z^2} dz<br \/>\n\\end{aligned}<br \/>\n\\]<br \/>\nPara la integral interna \\(\\int_{0}^{1} z\\sqrt{1-z^2} dz\\), usamos la <strong>sustituci\u00f3n<\/strong> \\(u = 1 &#8211; z^2\\), de donde \\(z\\,dz = -\\frac{1}{2}du\\).<\/p>\n<p>\\[<br \/>\n\\begin{aligned}<br \/>\n\\int_{0}^{1} z\\sqrt{1-z^2} dz &#038;= \\int_{1}^{0} \\sqrt{u} \\left(-\\frac{1}{2}du\\right) \\\\<br \/>\n&#038;= \\frac{1}{2} \\int_{0}^{1} u^{1\/2} du \\\\<br \/>\n&#038;= \\frac{1}{2} \\left[ \\frac{u^{3\/2}}{3\/2} \\right]_{0}^{1} \\\\<br \/>\n&#038;= \\frac{1}{3} (1^{3\/2} &#8211; 0^{3\/2}) = \\frac{1}{3}<br \/>\n\\end{aligned}<br \/>\n\\]<br \/>\nSustituyendo el resultado en \\(I_2\\):<br \/>\n\\[<br \/>\nI_2 = \\frac{1}{y+1} \\cdot \\frac{1}{3} = \\frac{1}{3(y+1)}<br \/>\n\\]<\/p>\n<p><strong>Paso 3<\/strong>: Integral exterior (respecto a \\(y\\))<\/p>\n<p>Resolvemos la \u00faltima integral con respecto a \\(y\\):<br \/>\n\\[<br \/>\n\\begin{aligned}<br \/>\nI &#038;= \\int_{0}^{1} \\frac{1}{3(y+1)} dy \\\\<br \/>\n&#038;= \\frac{1}{3} \\int_{0}^{1} \\frac{1}{y+1} dy \\\\<br \/>\n&#038;= \\frac{1}{3} \\left[ \\ln|y+1| \\right]_{0}^{1} \\\\<br \/>\n&#038;= \\frac{1}{3} \\left( \\ln(2) &#8211; \\ln(1) \\right) \\\\<br \/>\n&#038;= \\frac{1}{3} \\ln(2)<br \/>\n\\end{aligned}<br \/>\n\\]\n<\/p><\/div>\n<hr \/>\n<h3>Bibliograf\u00eda<\/h3>\n<ul>\n<li>Cap\u00edtulo 15 del libro <em>C\u00e1lculo de varias variables<\/em>, de James Stewart.<\/li>\n<\/ul>\n<hr \/>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\">\n<p><strong>Ejercicio:<\/strong> \u00bfCu\u00e1l es el volumen del s\u00f3lido coloreado en la imagen? <\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-medium wp-image-582\" title=\"volumen\" src=\"http:\/\/uploads.jesussoto.es\/area780_4.png\" alt=\"\" width=\"222\" height=\"204\" \/><\/p>\n<div id=\"menu-a\">\n<ul>\n<li>1\/12<\/li>\n<li>4\/13<\/li>\n<li>1<\/li>\n<li>Ninguno de ellos<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv() {\n  var htmlShow = document.getElementById(\"html-show\");\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p>\n<button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>A.)<\/strong><\/p>\n<p><iframe loading=\"lazy\" title=\"Matem\u00e1tica Aplicada - Integral triple. Ej.4 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/Ngk04r0yhQs?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>La integral triple de una funci\u00f3n, \\(f(x,y,z)\\), en una regi\u00f3n cerrada del espacio, \\(Q\\), con un volumen \\(V\\), no es m\u00e1s que la generalizaci\u00f3n del concepto de integral simple y doble. As\u00ed&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-601","post","type-post","status-publish","format-standard","hentry","category-mathbio"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/601","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=601"}],"version-history":[{"count":10,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/601\/revisions"}],"predecessor-version":[{"id":612,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/601\/revisions\/612"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=601"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=601"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=601"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}