{"id":25581,"date":"2026-08-25T11:30:00","date_gmt":"2026-08-25T16:30:00","guid":{"rendered":"https:\/\/blog.espol.edu.ec\/algoritmos101\/?p=25581"},"modified":"2026-08-28T21:24:29","modified_gmt":"2026-08-29T02:24:29","slug":"s2eva2026paoi_t1-volumen-sillon-giratorio","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2026paoi_t1-volumen-sillon-giratorio\/","title":{"rendered":"s2Eva2026PAOI_T1 Volumen de sill\u00f3n giratorio"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2026PAOI_T1 Volumen de sill\u00f3n giratorio<\/a><\/p>\n\n\n\n

El volumen para cada segmento se calcula mediante:<\/p>\n\n\n\n V = \\int_a^b \\pi \\left( f(x) \\right)^2 dx <\/span>\n\n\n\n

\n
\n f_1 (x)= 0.05 e^{10.52x} <\/span>\n\n\n\n

0 \u2264 x \u2264 0.2<\/p>\n\n\n\n f_2 (x) = - 0.2 \\sin \\left(\\frac{10}{3} \\pi (x-0.2) \\right)+ 0.4099<\/span>\n\n\n\n

0.2 < x \u2264 0.34<\/p>\n\n\n\n f_3 (x)=0.15 e^{1.95(x-0.34)}+0.06104 <\/span>\n\n\n\n

0.34 < x \u2264 0.64<\/p>\n\n\n\n g(x)=1-e^{-2.01(x-0.44)} <\/span>\n\n\n\n

0.44 < x \u2264 0.64<\/p>\n<\/div>\n\n\n\n

\n
\"Sill\u00f3n<\/figure>\n<\/div>\n<\/div>\n\n\n\n

literal a<\/h2>\n\n\n\n

Integrar con cuadratura de Gauss, intervalo 0 \u2264 x \u2264 0.2<\/p>\n\n\n\n V_1 = \\int_a^b \\pi \\left( 0.05 e^{10.52x} \\right)^2 dx <\/span>\n\n\n\n f_1(x) = \\pi 0.05 e^{2(10.52)x}<\/span>\n\n\n\n

x_a = \\frac{0.2+0}{2} - \\frac{0.2-0}{2}\\left(\\frac{1}{\\sqrt{3}} \\right) = 0.04226<\/span><\/p>\n\n\n\n x_b = \\frac{0.2+0}{2} + \\frac{0.2-0}{2}\\left(\\frac{1}{\\sqrt{3}} \\right) =0.1577<\/span>\n\n\n\n f_1(x_a) = \\pi 0.05 e^{2(10.52)(0.04226)} = 0.01911 <\/span>\n\n\n\n f_1(x_b) = \\pi0.05 e^{2(10.52)(0.1577)} = 0.2169 <\/span>\n\n\n\n I \\cong \\frac{0.2-0}{2}(0.01911 + 0.2169) =0.02360 <\/span>\n\n\n\n

literal b<\/h2>\n\n\n\n

Integrar con Simpson 1\/3, intervalo 0.2 < x \u2264 0.34<\/p>\n\n\n\n V_2 = \\int_a^b \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (x-0.2) \\right)+ 0.4099 \\right)^2 dx <\/span>\n\n\n\n f_2(x) = \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (x-0.2) \\right)+ 0.4099 \\right)^2<\/span>\n\n\n\n

con al menos 3 tramos en el intervalo, pero se necesita que sean m\u00faltiplos de 2. Por lo que se usan 4 tramos.<\/p>\n\n\n\n

h = (0.34 - 0.2)\/4 = 0.035<\/p>\n\n\n\n

xi = [0.2, 0.235, 0.27, 0.305, 0.34]<\/p>\n\n\n\n f_2(0.2) = \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (0.2-0.2) \\right)+ 0.4099 \\right)^2 = 0.5279<\/span>\n\n\n\n f_2(0.235)= \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (0.235-0.2) \\right)+ 0.4099 \\right)^2 =0.3594 <\/span>\n\n\n\n f_2(0.27) = \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (0.27-0.2) \\right)+ 0.4099 \\right)^2=0.2395<\/span>\n\n\n\n f_2(0.305) = \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (0.305-0.2) \\right)+ 0.4099 \\right)^2=0.1687<\/span>\n\n\n\n f_2(0.34) = \\pi \\left( - 0.2 \\sin \\left(\\frac{10}{3} \\pi (0.34-0.2) \\right)+ 0.4099 \\right)^2=0.1399<\/span>\n\n\n\n V_2 = \\frac{0.035}{3} \\left( 0.5279 + 4 (0.3594) + 0.2395 \\right) + <\/span>\n\n\n\n + \\frac{0.035}{3} \\left( 0.2395 + 4 (0.1687) + 0.1399 \\right) = 0.03803 <\/span>\n\n\n\n

literal c<\/h2>\n\n\n\n

Integrar con Simpson 3\/8, intervalo 0.34 < x \u2264 0.64<\/p>\n\n\n\n V_3 = \\int_a^b \\pi \\left( 0.15 e^{1.95(x-0.34)}+0.06104\\right)^2 dx <\/span>\n\n\n\n f_3(x) = \\pi \\left( 0.15 e^{1.95(x-0.34)}+0.06104 \\right)^2<\/span>\n\n\n\n

con al menos 3 tramos en el intervalo, si se alcanza para usar un segmento de 3 tramos.<\/p>\n\n\n\n

h = (0.64-0.34)\/3 = 0.1<\/p>\n\n\n\n

xi = [0.34, 0.44, 0.54, 0.64]<\/p>\n\n\n\n f_3(0.34) = \\pi \\left( 0.15 e^{1.95(0.34-0.34)}+0.06104 \\right)^2 =0.1399<\/span>\n\n\n\n f_3(0.44) = \\pi \\left( 0.15 e^{1.95(0.44-0.34)}+0.06104 \\right)^2=0.1860<\/span>\n\n\n\n f_3(0.54) = \\pi \\left( 0.15 e^{1.95(0.54-0.34)}+0.06104 \\right)^2=0.2508<\/span>\n\n\n\n f_3(0.54) = \\pi \\left( 0.15 e^{1.95(0.64-0.34)}+0.06104 \\right)^2=0.3427<\/span>\n\n\n\n V_3 = \\frac{3}{8} (0.1)\\left( 0.1399 + 3 (0.1860) + 3(0.2508)+0.3427 \\right) =0.06724 <\/span>\n\n\n\n

literal d<\/h2>\n\n\n\n

Se puede seleccionar cualquiera de los m\u00e9todos num\u00e9ricos.
intervalo 0.44 < x \u2264 0.64. Por ejemplo, trapecios:<\/p>\n\n\n\n V_3 = \\int_a^b \\pi \\left( 1-e^{-2.01(x-0.44)} \\right)^2 dx <\/span>\n\n\n\n g(x) = \\pi \\left( 1-e^{-2.01(x-0.44)}\\right)^2<\/span>\n\n\n\n

con al menos 2 tramos en el intervalo, se puede usar dos trapecios.<\/p>\n\n\n\n

h = (0.64-0.44)\/2 = 0.1<\/p>\n\n\n\n

xi = [0.44,0.54,0.64]<\/p>\n\n\n\n g(0.44) = \\pi \\left( 1-e^{ -2.01 (0.44-0.44)}\\right) ^2 = 0 <\/span>\n\n\n\n g(0.54) = \\pi \\left( 1-e^{-2.01( 0.54-0.44)}\\right)^2 = 0.1041<\/span>\n\n\n\n g(0.64) = \\pi \\left( 1-e^{-2.01( 0.64-0.44)}\\right)^2 = 0.3442<\/span>\n\n\n\n V_4 = 0.1 \\left( \\frac{0+0.1041}{2}\\right) + 0.1 \\left(\\frac{0.1041+0.3442}{2}\\right)=0.02762<\/span>\n\n\n\n

literal e<\/h2>\n\n\n\n

La cota de error para cada segmento es: <\/p>\n\n\n\n

0 \u2264 x \u2264 0.2 ; Error \u2245 f(4)<\/sup>(x ), Observaciones al realizar la gr\u00e1fica con algoritmo en literal g<\/p>\n\n\n\n

0.2 < x \u2264 0.34 ; Error O(h5<\/sup>\/90)=(0.0355<\/sup>\/90)=5.8357e-10<\/p>\n\n\n\n

0.34 < x \u2264 0.64 ; Error O(3h5<\/sup>\/80)=(3(0.1)5<\/sup>\/80)=3.75e-7<\/p>\n\n\n\n

0.44 < x \u2264 0.64 ; Error O(h3<\/sup>\/12)=((0.1)3<\/sup>\/12)=8.33e-5<\/p>\n\n\n\n

literal f<\/h2>\n\n\n\n

Volumen total del sill\u00f3n<\/p>\n\n\n\n V_{sillon} = V_1 + V_2 + V3 - V_4 <\/span>\n\n\n\n = 0.02360 + 0.03803 + 0.06724 - 0.02762 =0.1012 <\/span>\n\n\n\n

literal g<\/h2>\n\n\n\n

Usando el algoritmo de cuadratura de Gauss b\u00e1sico, actualizando la funci\u00f3n, intervalo y tramos a 1:<\/p>\n\n\n\n

Factores Gauss-Legendre puntos: 2\nxgl: [-0.57735  0.57735]\ncgl: [1.0, 1.0]\ntabla por intervalos [a,b]\n[a,b] : [0.  0.2]\nxi : [0.04226 0.15774]\nfi : [0.01911 0.21697]\narea : 0.02360826289510036\nIntegral:  0.02360826289510036<\/code><\/pre>\n\n\n\n
\"Sill\u00f3n<\/figure>\n\n\n\n
Al observar la gr\u00e1fica, se podr\u00eda intentar hacer el integral con mas segmentos usando el algoritmo y mejorar la precisi\u00f3n del resultado.<\/p>\n\n\n\n
Factores Gauss-Legendre puntos: 2\nxgl: [-0.57735  0.57735]\ncgl: [1.0, 1.0]\ntabla por intervalos [a,b]\n[a,b] : [0.  0.1]\nxi : [0.02113 0.07887]\nfi : [0.01225 0.04128]\narea : 0.00267660576897996\n[a,b] : [0.1 0.2]\nxi : [0.12113 0.17887]\nfi : [0.10045 0.33846]\narea : 0.021945223062764193\nIntegral:  0.02462182883174415<\/code><\/pre>\n\n\n\n
\"Sill\u00f3n<\/figure>\n\n\n\n
Algoritmo en Python<\/p>\n\n\n
\n# Integraci\u00f3n: Cuadratura de Gauss de 2 puntos\n# modelo con varios tramos entre [a,b]\nimport numpy as np\n \n# INGRESO\nfx = lambda x: np.pi*(0.05*np.exp(10.52*x))**2\na = 0 # intervalo de integraci\u00f3n\nb = 0.2\ntramos = 1  # subintervalos a integrar\nprecision = 5 # decimales en tabla\n \n# PROCEDIMIENTO\n# cuadratura de 2 puntos\nn_puntos = 2\nxgl = np.array([-1\/np.sqrt(3), 1\/np.sqrt(3)],dtype=float)\ncgl = [1.,1.]\n# cuadratura de n_puntos, f\u00f3rmulas Gauss-Legendre\n#xgl, cgl = np.polynomial.legendre.leggauss(2)\n \nx_h = np.linspace(a,b,tramos+1)\ntabla = {}\nsuma = 0\nfor k in range(0,tramos,1):\n    a = x_h[k]\n    b = x_h[k+1]\n    centro = (a+b)\/2\n    mitad = (b-a)\/2\n    xa = centro + xgl[0]*mitad\n    xb = centro + xgl[1]*mitad\n \n    area = ((b-a)\/2)*(cgl[0]*fx(xa) + cgl[1]*fx(xb))\n    tabla[k]= {'[a,b]': np.array([a,b]),\n               'xi': np.array([xa,xb]),\n               'fi': np.array([fx(xa),fx(xb)]),\n               'area':area\n               }\n    suma = suma + area\n \n# SALIDA\nnp.set_printoptions(precision)\nprint('Factores Gauss-Legendre puntos:',n_puntos)\nprint('xgl:',xgl)\nprint('cgl:',cgl)\nprint('tabla por intervalos [a,b]')\nfor k in range(0,tramos,1):\n    for entrada in tabla[k]:\n        print(entrada,':',tabla[k][entrada])\nprint('Integral: ', suma)\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2026PAOI_T1 Volumen de sill\u00f3n giratorio El volumen para cada segmento se calcula mediante: 0 \u2264 x \u2264 0.2 0.2 < x \u2264 0.34 0.34 < x \u2264 0.64 0.44 < x \u2264 0.64 literal a Integrar con cuadratura de Gauss, intervalo 0 \u2264 x \u2264 0.2 literal b Integrar con Simpson 1\/3, intervalo 0.2 […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[58,54],"class_list":["post-25581","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25581","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=25581"}],"version-history":[{"count":14,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25581\/revisions"}],"predecessor-version":[{"id":25647,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/25581\/revisions\/25647"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=25581"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=25581"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=25581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}