{"id":1052,"date":"2018-02-09T06:00:26","date_gmt":"2018-02-09T11:00:26","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=1052"},"modified":"2026-04-05T20:16:09","modified_gmt":"2026-04-06T01:16:09","slug":"s2eva2017tii_t2-volumen-de-isla","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva20\/s2eva2017tii_t2-volumen-de-isla\/","title":{"rendered":"s2Eva2017TII_T2 Volumen de isla"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2017TII_T2 Volumen de isla<\/a><\/p>\n\n\n\n

isla = np.array([[0,1,0,0,0],\n                 [1,3,1,1,0],\n                 [5,4,3,2,0],\n                 [0,0,1,1,0]])\n\nxi = np.array([0,100,200,300,400])\nyi = np.array([0, 50,100,150])<\/code><\/pre>\n\n\n\n
Tama\u00f1o de la matriz: n=4, m=5<\/p>\n\n\n\n
cantidad de elementos por fila impar, aplica Simpson 1\/3\nhx = (200-0)\/2 =100\nfila=0\n    vector = [0,1,0,0,0]\n    deltaA = (100\/3)(0+4(1)+0) = 4(100\/3)\n    deltaA = (100\/3)(0+4(0)+0) = 0\n    area0 = 4(100\/3) + 0 = 4(100\/3)\nfila=1\n    vector = [1,3,1,1,0]\n    deltaA = (100\/3)(1+4(3)+1) = 14(100\/3)\n    deltaA = (100\/3)(1+4(1)+0) = 5(100\/3)\n    area1 = 14(100\/3) + 5(100\/3) = 19(100\/3)\nfila=2\n    vector = [5,4,3,2,0]\n    deltaA = (100\/3)(5+4(4)+3) = 24(100\/3)\n    deltaA = (100\/3)(3+4(2)+0) = 11(100\/3)\n    area2 = 24(100\/3) + 11(100\/3) = 35(100\/3)\nfila=3\n    vector = [0,0,1,1,0]\n    deltaA = (100\/3)(0+4(0)+1) = (100\/3)\n    deltaA = (100\/3)(1+4(1)+0) = 5(100\/3)\n    area3 = (100\/3) + 5(100\/3) = 6(100\/3)\n\nareas = [ 4(100\/3), 19(100\/3), 35(100\/3), 6(100\/3)]\nareas = (100\/3)[ 4, 19, 35, 6 ]\n\nareas tiene cantidad de elementos par, aplica Simpson 3\/8\n    hy = (150-0)\/3 = 50\n    deltaV = (3\/8)(50)(100\/3)(4+3(19) + 3(35)+ 6)\n           = (25*25)(168)\n    Volumen = 107500<\/code><\/pre>\n\n\n\n
\"s2Eva2017TII_T2<\/figure>\n\n\n\n
tramos:  4 5\nareas:  [  133.33333333   633.33333333  1166.66666667    66.66666667]\nVolumen:  107500.0<\/code><\/pre>\n\n\n\n
Instrucciones en Python para encontrar el valor son:<\/p>\n\n\n
\n# 2da Eval II T 2017. Tema 2\n# Formula de simpson\n# Integraci\u00f3n: Regla Simpson 1\/3 y 3\/8\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom mpl_toolkits.mplot3d import axes3d\n\ndef simpson13(xi,yi):\n    '''\n    las muestras deben ser impares\n    '''\n    area = 0\n    muestras = len(xi)\n    impar = muestras%2\n    if impar == 1:\n        for i in range(0,muestras-2,2):\n            h = (xi[i+2] - xi[i])\/2\n            deltaA = (h\/3)*(yi[i]+4*yi[i+1]+yi[i+2])\n            area = area + deltaA\n    return(area)\n\ndef simpson38(xi,yi):\n    '''\n    las muestras deben ser pares\n    '''\n    area = 0\n    muestras = len(xi)\n    impar = muestras%2\n    if impar == 0:\n        for i in range(0,muestras-3,3):\n            h = (xi[i+3] - xi[i])\/3\n            deltaA = (3*h\/8)*(yi[i]+3*yi[i+1]+3*yi[i+2]+yi[i+3])\n            area = area + deltaA\n    return(area)\n\ndef simpson(xi,yi):\n    '''\n    Selecciona el tipo de algoritmo Simpson\n    '''\n    muestras = len(xi)\n    impar = muestras%2\n    if impar == 1:\n        area = simpson13(xi,yi)\n    else:\n        area = simpson38(xi,yi)\n    return(area)\n    \n\n# INGRESO\nisla = np.array([[0,1,0,0,0],\n                 [1,3,1,1,0],\n                 [5,4,3,2,0],\n                 [0,0,1,1,0]])\n\nxi = np.array([0,100,200,300,400])\nyi = np.array([0, 50,100,150])\n\n# PROCEDIMIENTO\ntamano = np.shape(isla)\nn = tamano[0]\nm = tamano[1]\n\nareas = np.zeros(n,dtype = float)\nfor fila in range(0,n,1):\n    unafranja = isla[fila,:]\n    areas[fila] = simpson(xi,unafranja)\nvolumen = simpson(yi,areas)\n\n# SALIDA\nprint('tramos: ', n,m)\nprint('areas: ', areas)\nprint('Volumen: ', volumen)\n\n# Gr\u00e1fica\nX, Y = np.meshgrid(xi, yi)\nfig = plt.figure()\nax = fig.add_subplot(111, projection = '3d')\nax.plot_wireframe(X,Y,isla)\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2017TII_T2 Volumen de isla Tama\u00f1o de la matriz: n=4, m=5 Instrucciones en Python para encontrar el valor son:<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[58,54],"class_list":["post-1052","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1052","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=1052"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1052\/revisions"}],"predecessor-version":[{"id":23869,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1052\/revisions\/23869"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1052"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1052"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1052"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}