(1)<\/strong>\u3000\u66f2\u9762 \\(S\\) \u3068, \\(2\\) \u3064\u306e\u5e73\u9762 \\(x = 1\\) \u304a\u3088\u3073 \\(x = -1\\) \u3067\u56f2\u307e\u308c\u308b\u7acb\u4f53\u306e\u4f53\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000(1)<\/strong> \u306e\u7acb\u4f53\u306e\u5e73\u9762 \\(y = 0\\) \u306b\u3088\u308b\u5207\u308a\u53e3\u3092, \u5e73\u9762 \\(y = 0\\) \u306b\u304a\u3044\u3066\u56f3\u793a\u305b\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int _ 0^1 \\sqrt{t^2+1} \\, dt\\) \u306e\u5024\u3092 \\(t = \\dfrac{e^s -e^{-s}}{2}\\) \u3068\u7f6e\u63db\u3059\u308b\u3053\u3068\u306b\u3088\u3063\u3066\u6c42\u3081\u3088. \u3053\u308c\u3092\u7528\u3044\u3066, (2)<\/strong> \u306e\u5207\u308a\u53e3\u306e\u9762\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u7dda\u5206 PQ \u4e0a\u306e\u70b9\u306f\r\n\\[\r\nt \\overrightarrow{\\text{OP}} +( 1-t ) \\overrightarrow{\\text{OQ}} = \\left( \\begin{array}{c} 2t-1 \\\\ 1-t \\\\ t \\end{array} \\right)\r\n\\]\r\n\u3068\u8868\u305b\u308b.\r\n\\[\\begin{align}\r\n& \\left\\{ \\begin{array}{l} x = 2t-1 \\\\ y = 1-t \\\\ z = t \\end{array} \\right. \\\\\r\n\\text{\u2234} \\quad & \\left\\{ \\begin{array}{l} t = \\dfrac{x+1}{2} = z \\\\ y = \\dfrac{1-x}{2} \\end{array} \\right.\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\u5e73\u9762 \\(x=u\\) \u3068 PQ \u3068\u306e\u4ea4\u70b9\u306f\r\n\\[\r\n\\left( u , \\dfrac{1-u}{2} , \\dfrac{1+u}{2} \\right)\r\n\\]\r\n\u3053\u306e\u70b9\u3068\u70b9 \\(( u , 0 , 0 )\\) \u3068\u306e\u8ddd\u96e2\u3092 \\(l(u)\\) \u3068\u3059\u308b\u3068\r\n\\[\\begin{align}\r\nl^2(u) & = \\left( \\dfrac{1-u}{2} \\right)^2 + \\left( \\dfrac{1+u}{2} \\right)^2 \\\\\r\n& = \\dfrac{u^2+1}{2}\r\n\\end{align}\\]\r\n\u3086\u3048\u306b\u6c42\u3081\u308b\u4f53\u7a4d \\(V\\) \u306f\r\n\\[\\begin{align}\r\nV & = \\pi \\displaystyle\\int _ {-1}^1 l^2(u) \\, du = \\dfrac{\\pi}{2} \\cdot 2 \\int _ 0^1 ( u^2+1 ) \\, du \\\\\r\n& = \\pi \\left[ \\dfrac{u^3}{3} +u \\right] _ 0^1 \\\\\r\n& = \\underline{\\dfrac{4 \\pi}{3}}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \\(y=0\\) \u306b\u304a\u3051\u308b\u5207\u308a\u53e3\u306f\r\n\\[\r\nx = \\pm 1 , \\ z = \\pm l(x) = \\pm \\dfrac{\\sqrt{2}}{2} \\sqrt{x^2+1}\r\n\\]\r\n\u306b\u56f2\u307e\u308c\u305f\u9818\u57df\u306b\u306a\u308b. (3)<\/strong><\/p>\r\n \\(I = \\displaystyle\\int _ 0^1 \\sqrt{t^2+1} \\, dt\\) \u3068\u304a\u304f.
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
\r\n\\(x\\) \u8ef8\u306b\u3064\u3044\u3066\u5bfe\u79f0\u306a\u306e\u3067, \\(y \\geqq 0\\) \u306e\u9818\u57df\u306b\u3064\u3044\u3066\u8003\u3048\u308b.\r\n\\[\\begin{align}\r\nl'(x) & = \\dfrac{\\sqrt{2}}{2} \\cdot \\dfrac{2x}{2 \\sqrt{x^2+1}} \\\\\r\n& = \\dfrac{\\sqrt{2} x}{2 \\sqrt{x^2+1}}\r\n\\end{align}\\]\r\n\\(l'(x) = 0\\) \u3092\u89e3\u304f\u3068, \\(x = 0\\) .
\r\n\u3057\u305f\u304c\u3063\u3066, \u5897\u6e1b\u8868\u306f\u4e0b\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccccc} x & -1 & \\cdots & 0 & \\cdots & 1\\\\ \\hline l'(x) & & - & 0 & + & \\\\ \\hline l(x) & 1 & \\searrow & \\frac{\\sqrt{2}}{2} & \\nearrow & 1 \\\\ \\end{array}\r\n\\]\r\n\u3086\u3048\u306b\u5207\u308a\u53e3\u306f\u4e0b\u56f3\u659c\u7dda\u90e8.<\/p>\r\n![\"waseda2010_04_01\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/waseda2010_04_01.png\")
\r\n
\r\n\\(t = \\dfrac{e^s -e^{-s}}{2}\\) \u3088\u308a\r\n\\[\r\ndt = \\dfrac{e^s +e^{-s}}{2} ds\r\n\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\n\\dfrac{e^s -e^{-s}}{2} & = 0 \\\\\r\ne^{2s} -1 = \\left( e^s -1 \\right) & \\left( e^s +1 \\right) = 0 \\\\\r\n\\text{\u2234} \\quad s & = 0 , \\\\\r\n\\dfrac{e^s - e^{-s}}{2} & = 1 \\\\\r\ne^{2s} -2e^s -1 & = 0 \\\\\r\ne^s & = 1+\\sqrt{2} \\\\\r\n\\text{\u2234} \\quad s & = \\log \\left( 1+\\sqrt{2} \\right)\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\r\n\\begin{array}{c|ccc} t & 0 & \\rightarrow & 1 \\\\ \\hline s & 0 & \\rightarrow & \\log \\left( 1+\\sqrt{2} \\right) \\end{array}\r\n\\]\r\n\u3086\u3048\u306b\r\n\\[\\begin{align}\r\nI & = \\displaystyle\\int _ 0^{\\log \\left( 1+\\sqrt{2} \\right)} \\sqrt{\\dfrac{\\left( e^s - e^{-s} \\right)^2}{4} +1} \\cdot \\dfrac{e^s + e^{-s}}{2} \\, ds \\\\\r\n& = \\dfrac{1}{4} \\int _ 0^{\\log \\left( 1+\\sqrt{2} \\right)} \\left( e^s + e^{-s} \\right)^2 \\, ds \\\\\r\n& = \\dfrac{1}{4} \\left[ \\dfrac{e^{2s}}{2} +2s -\\dfrac{e^{-2s}}{2} \\right] _ 0^{\\log \\left( 1+\\sqrt{2} \\right)} \\\\\r\n& = \\dfrac{1}{4} \\left\\{ \\dfrac{\\left( 1 +\\sqrt{2} \\right)^2}{2} +2\\log \\left( 1+\\sqrt{2} \\right) -\\dfrac{1}{2 \\left( 1 +\\sqrt{2} \\right)^2} \\right\\} \\\\\r\n& \\qquad -\\dfrac{1}{4} \\left( \\dfrac{1}{2} +0 -\\dfrac{1}{2} \\right) \\\\\r\n& = \\dfrac{1}{8} \\left\\{ \\left( \\sqrt{2} +1 \\right)^2 -\\left( \\sqrt{2} -1 \\right)^2 \\right\\} + \\dfrac{1}{2} \\log \\left( 1+\\sqrt{2} \\right) \\\\\r\n& = \\dfrac{1}{8} \\cdot 4 \\sqrt{2} + \\dfrac{1}{2} \\log \\left( 1+\\sqrt{2} \\right) \\\\\r\n& = \\dfrac{\\sqrt{2} + \\log \\left( 1+\\sqrt{2} \\right) }{2}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u9762\u7a4d \\(S\\) \u306f\r\n\\[\\begin{align}\r\nS & = 4 \\cdot \\dfrac{\\sqrt{2}}{2} \\int _ 0^1 \\sqrt{x^2+1} \\, dx \\\\\r\n& = 2 \\sqrt{2} I = \\underline{2 +\\sqrt{2} \\log \\left( 1+\\sqrt{2} \\right)}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(xyz\\) \u7a7a\u9593\u306b\u304a\u3044\u3066, \\(2\\) \u70b9 P \\((1, 0, 1)\\) , Q \\((-1, 1, 0)\\) \u3092\u8003\u3048\u308b. \u7dda\u5206 PQ \u3092 \\(x\\) \u8ef8\u306e\u5468\u308a\u306b \\(1\\) \u56de\u8ee2\u3057\u3066\u5f97\u3089\u308c\u308b\u66f2\u9762\u3092 \\(S\\) \u3068 … \u7d9a\u304d\u3092\u8aad\u3080