{"id":595,"date":"2025-11-27T09:30:13","date_gmt":"2025-11-27T08:30:13","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=595"},"modified":"2025-11-26T23:11:44","modified_gmt":"2025-11-26T22:11:44","slug":"mathbio-integral-doble","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=595","title":{"rendered":"MathBio: Integral doble"},"content":{"rendered":"<p>As\u00ed como la integral de una funci\u00f3n positiva de una variable se interpreta como el \u00e1rea entre la gr\u00e1fica de la funci\u00f3n y el eje \\({\\displaystyle x}\\), la integral doble de una funci\u00f3n positiva de dos variables representa el volumen de la regi\u00f3n entre la superficie definida por la funci\u00f3n y el plano que contiene el dominio de la funci\u00f3n. <\/p>\n<p>Si definimos la integral doble mediante<\/p>\n<p>\\[\\iint _{R}f(x,y)\\;dxdy=\\lim _{n,m\\to 0}\\sum _{i=0}^{n-1}\\sum _{j=0}^{m-1}f(x_{ij},y_{ij})\\Delta x\\Delta y,\\]<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-medium wp-image-582\" title=\"int2\" src=\"http:\/\/uploads.jesussoto.es\/area770_5b.png\" alt=\"\" width=\"399\" height=\"293\" \/><\/p>\n<p>observamos que esto equivale a determinar el volumen de una funci\u00f3n positiva \\(f (x, y)\\) de dos variables, definida en una regi\u00f3n del plano \\(xy\\),<br \/>\n\\[V=\\iint_R f(x,y)\\, dA\\]<\/p>\n<p>Hacemos que \\(dA=dxdy\\) pero no tiene por qu\u00e9 ser as\u00ed. La integral doble as\u00ed definida cumple: \\[\\iint_R(\\lambda f(x,y)+\\mu g(x,y))\\,dA=\\lambda\\iint_R f(x,y)\\,dA+\\mu\\iint_R g(x,y)\\,dA.\\]<br \/>\nPara calcular la integral doble utilizamos el concepto de integral iterada, de modo que<br \/>\n\\[\\iint_R f(x,y)\\,dA=\\int_a^b\\left[\\int_c^df(x,y)dy\\right]dx,\\] donde \\(R=[a.b]\\times[c,d]\\).<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar el volumen bajo la superficie de \\(f(x,y)=x^2y\\) en el rect\u00e1ngulo \\(R:[1,2]\\times [1,3]\\)<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1a() {\n  var htmlShow1a = document.getElementById(\"html-show1a\");\n  if (htmlShow1a.style.display === \"none\") {\n    htmlShow1a.style.display = \"block\";\n  } else {\n    htmlShow1a.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a\" style=\"display: none;\">\nNo piden la integral \\[\\iint_Rx^2y\\ dA\\] Utilizando la igualdad anterior<br \/>\n\\[\\iint_Rx^2y\\ dA=\\int_1^2\\int_1^3x^2y\\ dy\\ dx=\\int_1^2x^2\\left[\\frac{y^2}{2}\\right|_1^3\\ dx=\\] \\[=\\int_1^2x^2\\left[\\frac{3^2}{2}-\\frac{1^2}{2}\\right]\\ dx=4\\int_1^2x^2\\ dx=4\\left[\\frac{x^3}{3}\\right|_1^2=\\frac{28}{3}\\]\n<\/div>\n<hr \/>\n<p>De este modo, si existe el l\u00edmite existe la integral doble, y esta se puede calcular mediante el teorema de Fubini:<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-full wp-image-583\" title=\"fubini\" src=\"http:\/\/uploads.jesussoto.es\/2011\/05\/fubini.png\" alt=\"\" width=\"625\" height=\"82\" \/><\/p>\n<p>Este teorema es muy \u00fatil porque nos permite decidir qu\u00e9 integramos antes para facilitar el c\u00e1lculo.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar el volumen bajo la superficie de \\(f(x,y)=y\\ e^{xy}\\) en el rect\u00e1ngulo \\(R:[0,1]\\times [1,2]\\)<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1b() {\n  var htmlShow1b = document.getElementById(\"html-show1b\");\n  if (htmlShow1b.style.display === \"none\") {\n    htmlShow1b.style.display = \"block\";\n  } else {\n    htmlShow1b.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b\" style=\"display: none;\">\nNo piden la integral \\[\\iint_Ry\\ e^{xy}\\ dA\\] En este caso, vemos que \\[\\int y\\ e^{xy}\\ dx= e^{xy}+C,\\] luego, utilizando el Teorema de Fubini nos resulta \\[\\iint_Ry\\ e^{xy}\\ dA=\\int_1^2\\left[\\int_1^0y\\ e^{xy}\\ dx\\right]\\ dy=\\int_1^2\\left[e^{y}-e^{0}\\right]\\ dy=\\left[e^{y}-y\\right|_1^2=e^{2}-e-1\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea R=[0,1]x[0,1], \u00bfcu\u00e1l es el valor de la integral dada? \\[\\iint_R e^{x+y}\\ dA\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1bx2() {\n  var htmlShow1bx2 = document.getElementById(\"html-show1bx2\");\n  if (htmlShow1bx2.style.display === \"none\") {\n    htmlShow1bx2.style.display = \"block\";\n  } else {\n    htmlShow1bx2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bx2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bx2\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"Matem\u00e1tica aplicada - Integral doble Ej. 2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/B8566KUpiZo?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<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea R=[0,1]x[0,1], \u00bfcu\u00e1l es el valor de la integral dada? \\[\\iint_Rxe^{xy}\\ dA\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1bx3() {\n  var htmlShow1bx3 = document.getElementById(\"html-show1bx3\");\n  if (htmlShow1bx3.style.display === \"none\") {\n    htmlShow1bx3.style.display = \"block\";\n  } else {\n    htmlShow1bx3.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bx3()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bx3\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"Matem\u00e1tica Aplicada - Integral doble. Ej.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/2j3J0Q32Vzw?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>Nos tenemos por qu\u00e9 restringirnos a una regi\u00f3n cuadrada R, podemos considerar una regi\u00f3n plana D delimitada por curvas, las denominadas regiones de Tipo 1 y II. As\u00ed calculamos el volumen bajo una funci\u00f3n \\(z=f(x,y)\\) y sobre la regi\u00f3n D. Adem\u00e1s, la integral doble nos permite definir el \u00e1rea de esa regi\u00f3n, si queremos calcular el \u00e1rea encerada en D, esta ser\u00e1 \\[A(D)=\\iint_D\\,dA.\\]<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar el \u00e1rea del disco \\({\\displaystyle D=\\{(x,y)\\in \\mathbb {R} ^{2}:2\\leq x\\leq 4\\;;\\;3\\leq y\\leq 6\\}}\\) encerrada por la funci\u00f3n \\(f(x,y)=cy\\) donde \\(c\\) es una constante real<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1bx() {\n  var htmlShow1bx = document.getElementById(\"html-show1bx\");\n  if (htmlShow1bx.style.display === \"none\") {\n    htmlShow1bx.style.display = \"block\";\n  } else {\n    htmlShow1bx.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bx()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bx\" style=\"display: none;\">\n\\[{\\displaystyle {\\begin{aligned}\\iint _{D}c\\;dA&#038;=\\int _{3}^{6}\\int _{2}^{4}c\\;dxdy\\\\&#038;=c\\int _{3}^{6}\\int _{2}^{4}dxdy\\\\&#038;=c\\;A(D)\\\\&#038;=c\\;(3\\times 2)=6c\\end{aligned}}}\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfCu\u00e1l es el \u00e1rea definida mediante la recta \\(y=x-1\\) y la par\u00e1bola \\(y^2=2x+6\\)?<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1bx6() {\n  var htmlShow1bx6 = document.getElementById(\"html-show1bx6\");\n  if (htmlShow1bx6.style.display === \"none\") {\n    htmlShow1bx6.style.display = \"block\";\n  } else {\n    htmlShow1bx6.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bx6()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bx6\" style=\"display: none;\">\nSe dice que una regi\u00f3n plana \\(D\\) es tipo II si yace entre las gr\u00e1ficas de dos funciones continuas de y, es decir,\\[{ D=\\{(x,y)\\in \\mathbb {R} ^{2}:c\\leq y\\leq d\\;;\\;h_1(x)\\leq x\\leq h_2(x)\\}}.\\]<br \/>\nEn nuestro caso \\[{ D=\\{(x,y)\\in \\mathbb {R} ^{2}:-2\\leq y\\leq 4\\;;\\;\\frac{1}{2}y^2-3\\leq x\\leq y+1\\}}.\\]<br \/>\nEsto se debe a que \\[\\begin{align*}<br \/>\ny=x-1 &#038;\\to x=y+1 \\\\<br \/>\ny^2=2x+6 &#038;\\to x=\\frac{1}{2}y^2-3<br \/>\n\\end{align*}<br \/>\n\\]<br \/>\nEs decir, \\(\\frac{1}{2}y^2-y-4=0\\to \\frac{1}{2} (y + 2) (y &#8211; 4)\\). As\u00ed pues, tendremos \\[\\iint_D\\,dA=\\int_{-2}^4\\int_{\\frac{1}{2}y^2-3}^{y+1} dx=\\int_{-2}^4(-\\frac{1}{2}y^2+y+4)dy=18\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfCu\u00e1l es el valor del \u00e1rea coloreada?<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-medium wp-image-582\" title=\"int2\" src=\"http:\/\/uploads.jesussoto.es\/area770_5.png\" alt=\"\" width=\"336\" height=\"276\" \/>\n<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1bx7() {\n  var htmlShow1bx7 = document.getElementById(\"html-show1bx7\");\n  if (htmlShow1bx7.style.display === \"none\") {\n    htmlShow1bx7.style.display = \"block\";\n  } else {\n    htmlShow1bx7.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bx7()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bx7\" style=\"display: none;\">\nObservando la gr\u00e1fica, nuestro disco ser\u00e1 \\[{ D=\\{(x,y)\\in \\mathbb {R}^{2}:|-1\\leq y\\leq 1;\\  y^2-2\\leq x\\leq e^y\\}}.\\]<br \/>\nLuego \\[\\iint_D dA= \\int_{-1}^1\\left (\\int_{ y^2-2}^{e^y} dx \\right ) dy= \\int_{-1}^1\\left (e^y-y^2+2 \\right ) dy\\]<br \/>\n\\[= \\frac{10}{3} &#8211; \\frac{1}{e}+ e\\approx  5.6837\\]\n<\/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 \u00e1rea del disco \\(D=\\{(x,y);0\\leq x\\leq 2,x^2\\leq y\\leq 2x\\}\\)?<\/p>\n<div id=\"menu-a\">\n<ul>\n<li>3\/4<\/li>\n<li>2\/3<\/li>\n<li>4\/3<\/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>C.)<\/strong><\/p>\n<p>Antes de integrar, comprobamos que efectivamente la recta \\( y = 2x \\) est\u00e1 por encima de la par\u00e1bola \\( y = x^2 \\) en el intervalo \\([0,2]\\):<br \/>\n\\[<br \/>\nx^2 \\leq 2x \\;\\Rightarrow\\; x^2 &#8211; 2x \\leq 0 \\;\\Rightarrow\\; x(x-2) \\leq 0<br \/>\n\\]<br \/>\nEsto se cumple para \\( 0 \\leq x \\leq 2 \\). <\/p>\n<p>As\u00ed pues, el \u00e1rea se calcula como:<br \/>\n\\[<br \/>\nA = \\int_{0}^{2} \\left(\\int_{x^2}^{2x} 1 \\, dy\\right) \\, dx<br \/>\n\\]<\/p>\n<p>Calculemos la integral:<br \/>\n\\[<br \/>\n\\int_{y=x^2}^{2x} 1 \\, dy = [y]_{x^2}^{2x} = 2x &#8211; x^2<br \/>\n\\]<br \/>\nLuego<br \/>\n\\[<br \/>\nA = \\int_{0}^{2} (2x &#8211; x^2) \\, dx=\\left[ x^2 &#8211; \\frac{x^3}{3} \\right]_{0}^{2} = \\left( 4 &#8211; \\frac{8}{3} \\right) &#8211; 0 = \\frac{12}{3} &#8211; \\frac{8}{3} = \\frac{4}{3}<br \/>\n\\]<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>As\u00ed como la integral de una funci\u00f3n positiva de una variable se interpreta como el \u00e1rea entre la gr\u00e1fica de la funci\u00f3n y el eje \\({\\displaystyle x}\\), la integral doble de una&#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-595","post","type-post","status-publish","format-standard","hentry","category-mathbio"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/595","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=595"}],"version-history":[{"count":5,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/595\/revisions"}],"predecessor-version":[{"id":600,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/595\/revisions\/600"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=595"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=595"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=595"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}