{
  "index": "1948-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "4. Let \\( D \\) be a plane region bounded by a circle of radius \\( r \\). Let \\( (x, y) \\) be a point of \\( D \\) and consider a circle of radius \\( \\delta \\) and center at \\( (x, y) \\). Denote by \\( l(x, y) \\) the length of that arc of the circle which is outside \\( D \\). Find\n\\[\n\\lim _{d \\rightarrow 0} \\frac{1}{\\delta^{2}} \\iint_{D} l(x, y) d x d y\n\\]",
  "solution": "Solution. First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \\( (x, y) \\) has polar coordinates \\( (\\rho, \\theta) \\), then \\( l(x, y)=L(\\rho) \\), where\n\\[\nL(\\rho)=0 \\quad \\text { if } 0 \\leq \\rho \\leq r-\\delta\n\\]\nand\n\\[\nL(\\rho)=2 \\delta \\phi=2 \\delta \\arccos \\left(\\frac{r^{2}-\\rho^{2}-\\delta^{2}}{2 \\rho \\delta}\\right) \\quad \\text { if } r-\\delta \\leq \\rho \\leq r,\n\\]\nas we see by applying the law of cosines to triangle \\( O P Q \\). Hence we must find\n\\[\n\\begin{aligned}\nA & =\\lim _{\\delta \\rightarrow \\infty} \\frac{1}{\\delta^{2}} \\int_{0}^{2 \\pi} \\int_{0}^{r} L(\\rho) \\rho d \\rho d \\theta \\\\\n& =\\lim _{\\delta \\rightarrow 0} \\frac{1}{\\delta^{2}} \\int_{0}^{2 \\pi} \\int_{r-\\delta}^{r} 2 \\delta \\rho \\arccos \\left(\\frac{r^{2}-\\rho^{2}-\\delta^{2}}{2 \\rho \\delta}\\right) d \\rho d \\theta \\\\\n& =\\lim _{\\delta \\rightarrow 0} \\frac{4 \\pi}{\\delta} \\int_{r-\\delta}^{r} \\rho \\arccos \\left(\\frac{r^{2}-\\rho^{2}-\\delta^{2}}{2 \\rho \\delta}\\right) d \\rho\n\\end{aligned}\n\\]\n\nWe make the substitution \\( \\rho=r-\\delta u \\) and get\n\\[\nA=\\lim _{\\delta \\rightarrow 0} 4 \\pi \\int_{0}^{1}(r-\\delta u) \\arccos \\left(\\frac{2 u r-\\delta\\left(1+u^{2}\\right)}{2(r-\\delta u)}\\right) d u .\n\\]\n\nSince the integrand is a continuous function of \\( u \\) and \\( \\delta \\) for \\( u \\in[0,1] \\) and \\( \\delta \\in\\left[0, \\frac{1}{2} r\\right] \\), and the domain of integration is bounded, we can conclude that the limit \\( A \\) exists and\n\\[\n\\begin{aligned}\nA & =4 \\pi \\int_{0}^{1} \\lim _{\\delta-0}(r-\\delta u) \\arccos \\left(\\frac{2 u r-\\delta\\left(1+u^{2}\\right)}{2(r-\\delta u)}\\right) d u \\\\\n& =4 \\pi \\int_{0}^{1} r \\arccos u d u .\n\\end{aligned}\n\\]\n\nIntegrating by parts, we obtain\n\\[\n\\begin{aligned}\nA & =4 \\pi r[u \\arccos u]_{0}^{1}+4 \\pi r \\int_{0}^{1} \\frac{u d u}{\\sqrt{1-u^{2}}} \\\\\n& =0+4 \\pi r\\left[-\\sqrt{1-u^{2}}\\right]_{0}^{1}=4 \\pi r\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "y",
    "\\\\rho",
    "\\\\theta",
    "u",
    "\\\\phi"
  ],
  "params": [
    "D",
    "r",
    "\\\\delta",
    "l",
    "L",
    "A",
    "O",
    "P",
    "Q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizcoor",
        "y": "verticcoor",
        "\\rho": "radialdist",
        "\\theta": "polarangle",
        "u": "shrinkvar",
        "\\phi": "chordangle",
        "D": "domainregion",
        "r": "baseradius",
        "\\delta": "smallradius",
        "l": "arclengthfunc",
        "L": "arclengthbig",
        "A": "limitvalue",
        "O": "centpoint",
        "P": "pointpvar",
        "Q": "pointqvar"
      },
      "question": "4. Let \\( domainregion \\) be a plane region bounded by a circle of radius \\( baseradius \\). Let \\( (horizcoor, verticcoor) \\) be a point of \\( domainregion \\) and consider a circle of radius \\( smallradius \\) and center at \\( (horizcoor, verticcoor) \\). Denote by \\( arclengthfunc(horizcoor, verticcoor) \\) the length of that arc of the circle which is outside \\( domainregion \\). Find\n\\[\n\\lim _{d \\rightarrow 0} \\frac{1}{smallradius^{2}} \\iint_{domainregion} arclengthfunc(horizcoor, verticcoor) d horizcoor d verticcoor\n\\]",
      "solution": "Solution. First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \\( (horizcoor, verticcoor) \\) has polar coordinates \\( (radialdist, polarangle) \\), then \\( arclengthfunc(horizcoor, verticcoor)=arclengthbig(radialdist) \\), where\n\\[\narclengthbig(radialdist)=0 \\quad \\text { if } 0 \\leq radialdist \\leq baseradius-smallradius\n\\]\nand\n\\[\narclengthbig(radialdist)=2 smallradius chordangle=2 smallradius \\arccos \\left(\\frac{baseradius^{2}-radialdist^{2}-smallradius^{2}}{2 radialdist smallradius}\\right) \\quad \\text { if } baseradius-smallradius \\leq radialdist \\leq baseradius,\n\\]\nas we see by applying the law of cosines to triangle \\( centpoint\\, pointpvar\\, pointqvar \\). Hence we must find\n\\[\n\\begin{aligned}\nlimitvalue & =\\lim _{smallradius \\rightarrow \\infty} \\frac{1}{smallradius^{2}} \\int_{0}^{2 \\pi} \\int_{0}^{baseradius} arclengthbig(radialdist) radialdist \\, d radialdist \\, d polarangle \\\\\n& =\\lim _{smallradius \\rightarrow 0} \\frac{1}{smallradius^{2}} \\int_{0}^{2 \\pi} \\int_{baseradius-smallradius}^{baseradius} 2 smallradius radialdist \\arccos \\left(\\frac{baseradius^{2}-radialdist^{2}-smallradius^{2}}{2 radialdist smallradius}\\right) \\, d radialdist \\, d polarangle \\\\\n& =\\lim _{smallradius \\rightarrow 0} \\frac{4 \\pi}{smallradius} \\int_{baseradius-smallradius}^{baseradius} radialdist \\arccos \\left(\\frac{baseradius^{2}-radialdist^{2}-smallradius^{2}}{2 radialdist smallradius}\\right) \\, d radialdist\n\\end{aligned}\n\\]\n\nWe make the substitution \\( radialdist = baseradius - smallradius \\, shrinkvar \\) and get\n\\[\nlimitvalue = \\lim _{smallradius \\rightarrow 0} 4 \\pi \\int_{0}^{1} (baseradius - smallradius \\, shrinkvar) \\arccos \\left(\\frac{2 \\, shrinkvar \\, baseradius - smallradius\\left(1 + shrinkvar^{2}\\right)}{2(baseradius - smallradius \\, shrinkvar)}\\right) \\, d shrinkvar .\n\\]\n\nSince the integrand is a continuous function of \\( shrinkvar \\) and \\( smallradius \\) for \\( shrinkvar \\in [0,1] \\) and \\( smallradius \\in \\left[0, \\frac{1}{2} baseradius\\right] \\), and the domain of integration is bounded, we can conclude that the limit \\( limitvalue \\) exists and\n\\[\n\\begin{aligned}\nlimitvalue & = 4 \\pi \\int_{0}^{1} \\lim _{smallradius \\to 0} (baseradius - smallradius \\, shrinkvar) \\arccos \\left(\\frac{2 \\, shrinkvar \\, baseradius - smallradius\\left(1 + shrinkvar^{2}\\right)}{2(baseradius - smallradius \\, shrinkvar)}\\right) \\, d shrinkvar \\\\\n& = 4 \\pi \\int_{0}^{1} baseradius \\arccos shrinkvar \\, d shrinkvar .\n\\end{aligned}\n\\]\n\nIntegrating by parts, we obtain\n\\[\n\\begin{aligned}\nlimitvalue & = 4 \\pi \\, baseradius [shrinkvar \\arccos shrinkvar]_{0}^{1} + 4 \\pi \\, baseradius \\int_{0}^{1} \\frac{shrinkvar \\, d shrinkvar}{\\sqrt{1 - shrinkvar^{2}}} \\\\\n& = 0 + 4 \\pi \\, baseradius \\left[ - \\sqrt{1 - shrinkvar^{2}} \\right]_{0}^{1} = 4 \\pi \\, baseradius\n\\end{aligned}\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pinecones",
        "y": "driftwood",
        "\\rho": "sandcastle",
        "\\theta": "blueberry",
        "u": "paperclips",
        "\\phi": "wildfire",
        "D": "teapotset",
        "r": "moonlight",
        "\\delta": "starlights",
        "l": "foxgloves",
        "L": "oceanbreeze",
        "A": "doorknob",
        "O": "marshland",
        "P": "evergreens",
        "Q": "hummingbrd"
      },
      "question": "4. Let \\( teapotset \\) be a plane region bounded by a circle of radius \\( moonlight \\). Let \\( (pinecones, driftwood) \\) be a point of \\( teapotset \\) and consider a circle of radius \\( starlights \\) and center at \\( (pinecones, driftwood) \\). Denote by \\( foxgloves(pinecones, driftwood) \\) the length of that arc of the circle which is outside \\( teapotset \\). Find\n\\[\n\\lim _{d \\rightarrow 0} \\frac{1}{starlights^{2}} \\iint_{teapotset} foxgloves(pinecones, driftwood) d pinecones d driftwood\n\\]",
      "solution": "Solution. First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \\( (pinecones, driftwood) \\) has polar coordinates \\( (sandcastle, blueberry) \\), then \\( foxgloves(pinecones, driftwood)=oceanbreeze(sandcastle) \\), where\n\\[\noceanbreeze(sandcastle)=0 \\quad \\text { if } 0 \\leq sandcastle \\leq moonlight-starlights\n\\]\nand\n\\[\noceanbreeze(sandcastle)=2 starlights wildfire=2 starlights \\arccos \\left(\\frac{moonlight^{2}-sandcastle^{2}-starlights^{2}}{2 sandcastle starlights}\\right) \\quad \\text { if } moonlight-starlights \\leq sandcastle \\leq moonlight,\n\\]\nas we see by applying the law of cosines to triangle \\( marshland evergreens hummingbrd \\). Hence we must find\n\\[\n\\begin{aligned}\ndoorknob & =\\lim _{starlights \\rightarrow \\infty} \\frac{1}{starlights^{2}} \\int_{0}^{2 \\pi} \\int_{0}^{moonlight} oceanbreeze(sandcastle) sandcastle d sandcastle d blueberry \\\\\n& =\\lim _{starlights \\rightarrow 0} \\frac{1}{starlights^{2}} \\int_{0}^{2 \\pi} \\int_{moonlight-starlights}^{moonlight} 2 starlights sandcastle \\arccos \\left(\\frac{moonlight^{2}-sandcastle^{2}-starlights^{2}}{2 sandcastle starlights}\\right) d sandcastle d blueberry \\\\\n& =\\lim _{starlights \\rightarrow 0} \\frac{4 \\pi}{starlights} \\int_{moonlight-starlights}^{moonlight} sandcastle \\arccos \\left(\\frac{moonlight^{2}-sandcastle^{2}-starlights^{2}}{2 sandcastle starlights}\\right) d sandcastle\n\\end{aligned}\n\\]\n\nWe make the substitution \\( sandcastle=moonlight-starlights paperclips \\) and get\n\\[\ndoorknob=\\lim _{starlights \\rightarrow 0} 4 \\pi \\int_{0}^{1}(moonlight-starlights paperclips) \\arccos \\left(\\frac{2 paperclips moonlight-starlights(1+paperclips^{2})}{2(moonlight-starlights paperclips)}\\right) d paperclips .\n\\]\n\nSince the integrand is a continuous function of \\( paperclips \\) and \\( starlights \\) for \\( paperclips \\in[0,1] \\) and \\( starlights \\in\\left[0, \\frac{1}{2} moonlight\\right] \\), and the domain of integration is bounded, we can conclude that the limit \\( doorknob \\) exists and\n\\[\n\\begin{aligned}\ndoorknob & =4 \\pi \\int_{0}^{1} \\lim _{starlights-0}(moonlight-starlights paperclips) \\arccos \\left(\\frac{2 paperclips moonlight-starlights(1+paperclips^{2})}{2(moonlight-starlights paperclips)}\\right) d paperclips \\\\\n& =4 \\pi \\int_{0}^{1} moonlight \\arccos paperclips d paperclips .\n\\end{aligned}\n\\]\n\nIntegrating by parts, we obtain\n\\[\n\\begin{aligned}\ndoorknob & =4 \\pi moonlight[paperclips \\arccos paperclips]_{0}^{1}+4 \\pi moonlight \\int_{0}^{1} \\frac{paperclips d paperclips}{\\sqrt{1-paperclips^{2}}} \\\\\n& =0+4 \\pi moonlight\\left[-\\sqrt{1-paperclips^{2}}\\right]_{0}^{1}=4 \\pi moonlight\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "\\rho": "tangentialdist",
        "\\theta": "straightdistance",
        "u": "constantvalue",
        "\\phi": "linearshift",
        "D": "unboundedregion",
        "r": "diameterlength",
        "\\delta": "largespread",
        "l": "insidearea",
        "L": "insideareafunc",
        "A": "infinitevalue",
        "O": "peripherypoint",
        "P": "outsidepoint",
        "Q": "outlierpoint"
      },
      "question": "4. Let \\( unboundedregion \\) be a plane region bounded by a circle of radius \\( diameterlength \\). Let \\( (verticalaxis, horizontalaxis) \\) be a point of \\( unboundedregion \\) and consider a circle of radius \\( largespread \\) and center at \\( (verticalaxis, horizontalaxis) \\). Denote by \\( insidearea(verticalaxis, horizontalaxis) \\) the length of that arc of the circle which is outside \\( unboundedregion \\). Find\n\\[\n\\lim _{d \\rightarrow 0} \\frac{1}{largespread^{2}} \\iint_{unboundedregion} insidearea(verticalaxis, horizontalaxis) d verticalaxis d horizontalaxis\n\\]",
      "solution": "Solution. First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \\( (verticalaxis, horizontalaxis) \\) has polar coordinates \\( (tangentialdist, straightdistance) \\), then \\( insidearea(verticalaxis, horizontalaxis)=insideareafunc(tangentialdist) \\), where\n\\[\ninsideareafunc(tangentialdist)=0 \\quad \\text { if } 0 \\leq tangentialdist \\leq diameterlength-largespread\n\\]\nand\n\\[\ninsideareafunc(tangentialdist)=2 largespread linearshift=2 largespread \\arccos \\left(\\frac{diameterlength^{2}-tangentialdist^{2}-largespread^{2}}{2 tangentialdist largespread}\\right) \\quad \\text { if } diameterlength-largespread \\leq tangentialdist \\leq diameterlength,\n\\]\nas we see by applying the law of cosines to triangle \\( peripherypoint\\ outsidepoint\\ outlierpoint \\). Hence we must find\n\\[\n\\begin{aligned}\ninfinitevalue & =\\lim _{largespread \\rightarrow \\infty} \\frac{1}{largespread^{2}} \\int_{0}^{2 \\pi} \\int_{0}^{diameterlength} insideareafunc(tangentialdist) tangentialdist d tangentialdist d straightdistance \\\\\n& =\\lim _{largespread \\rightarrow 0} \\frac{1}{largespread^{2}} \\int_{0}^{2 \\pi} \\int_{diameterlength-largespread}^{diameterlength} 2 largespread tangentialdist \\arccos \\left(\\frac{diameterlength^{2}-tangentialdist^{2}-largespread^{2}}{2 tangentialdist largespread}\\right) d tangentialdist d straightdistance \\\\\n& =\\lim _{largespread \\rightarrow 0} \\frac{4 \\pi}{largespread} \\int_{diameterlength-largespread}^{diameterlength} tangentialdist \\arccos \\left(\\frac{diameterlength^{2}-tangentialdist^{2}-largespread^{2}}{2 tangentialdist largespread}\\right) d tangentialdist\n\\end{aligned}\n\\]\n\nWe make the substitution \\( tangentialdist=diameterlength-largespread constantvalue \\) and get\n\\[\ninfinitevalue=\\lim _{largespread \\rightarrow 0} 4 \\pi \\int_{0}^{1}(diameterlength-largespread constantvalue) \\arccos \\left(\\frac{2 constantvalue diameterlength-largespread\\left(1+constantvalue^{2}\\right)}{2(diameterlength-largespread constantvalue)}\\right) d constantvalue .\n\\]\n\nSince the integrand is a continuous function of \\( constantvalue \\) and \\( largespread \\) for \\( constantvalue \\in[0,1] \\) and \\( largespread \\in\\left[0, \\frac{1}{2} diameterlength\\right] \\), and the domain of integration is bounded, we can conclude that the limit \\( infinitevalue \\) exists and\n\\[\n\\begin{aligned}\ninfinitevalue & =4 \\pi \\int_{0}^{1} \\lim _{largespread-0}(diameterlength-largespread constantvalue) \\arccos \\left(\\frac{2 constantvalue diameterlength-largespread\\left(1+constantvalue^{2}\\right)}{2(diameterlength-largespread constantvalue)}\\right) d constantvalue \\\\\n& =4 \\pi \\int_{0}^{1} diameterlength \\arccos constantvalue d constantvalue .\n\\end{aligned}\n\\]\n\nIntegrating by parts, we obtain\n\\[\n\\begin{aligned}\ninfinitevalue & =4 \\pi diameterlength[constantvalue \\arccos constantvalue]_{0}^{1}+4 \\pi diameterlength \\int_{0}^{1} \\frac{constantvalue d constantvalue}{\\sqrt{1-constantvalue^{2}}} \\\\\n& =0+4 \\pi diameterlength\\left[-\\sqrt{1-constantvalue^{2}}\\right]_{0}^{1}=4 \\pi diameterlength\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "\\rho": "bdkspqme",
        "\\theta": "lwhfztro",
        "u": "fvrndklo",
        "\\phi": "ckmzjifa",
        "D": "ugsnwvqd",
        "r": "xrtpoksl",
        "\\delta": "nqphsrae",
        "l": "otqzmpaw",
        "L": "mdbvshqe",
        "A": "sxoqklen",
        "O": "iphgjwra",
        "P": "tcldzxke",
        "Q": "vrsnapui"
      },
      "question": "4. Let \\( ugsnwvqd \\) be a plane region bounded by a circle of radius \\( xrtpoksl \\). Let \\( (qzxwvtnp, hjgrksla) \\) be a point of \\( ugsnwvqd \\) and consider a circle of radius \\( nqphsrae \\) and center at \\( (qzxwvtnp, hjgrksla) \\). Denote by \\( otqzmpaw(qzxwvtnp, hjgrksla) \\) the length of that arc of the circle which is outside \\( ugsnwvqd \\). Find\n\\[\n\\lim _{d \\rightarrow 0} \\frac{1}{nqphsrae^{2}} \\iint_{ugsnwvqd} otqzmpaw(qzxwvtnp, hjgrksla) d qzxwvtnp d hjgrksla\n\\]",
      "solution": "Solution. First convert the integral to polar coordinates, taking the origin at the center of the given circle. If a point \\( (qzxwvtnp, hjgrksla) \\) has polar coordinates \\( (bdkspqme, lwhfztro) \\), then \\( otqzmpaw(qzxwvtnp, hjgrksla)=mdbvshqe(bdkspqme) \\), where\n\\[\nmdbvshqe(bdkspqme)=0 \\quad \\text { if } 0 \\leq bdkspqme \\leq xrtpoksl-nqphsrae\n\\]\nand\n\\[\nmdbvshqe(bdkspqme)=2 nqphsrae ckmzjifa=2 nqphsrae \\arccos \\left(\\frac{xrtpoksl^{2}-bdkspqme^{2}-nqphsrae^{2}}{2 bdkspqme nqphsrae}\\right) \\quad \\text { if } xrtpoksl-nqphsrae \\leq bdkspqme \\leq xrtpoksl,\n\\]\nas we see by applying the law of cosines to triangle \\( iphgjwra tcldzxke vrsnapui \\). Hence we must find\n\\[\n\\begin{aligned}\nsxoqklen & =\\lim _{nqphsrae \\rightarrow \\infty} \\frac{1}{nqphsrae^{2}} \\int_{0}^{2 \\pi} \\int_{0}^{xrtpoksl} mdbvshqe(bdkspqme) bdkspqme d bdkspqme d lwhfztro \\\\\n& =\\lim _{nqphsrae \\rightarrow 0} \\frac{1}{nqphsrae^{2}} \\int_{0}^{2 \\pi} \\int_{xrtpoksl-nqphsrae}^{xrtpoksl} 2 nqphsrae bdkspqme \\arccos \\left(\\frac{xrtpoksl^{2}-bdkspqme^{2}-nqphsrae^{2}}{2 bdkspqme nqphsrae}\\right) d bdkspqme d lwhfztro \\\\\n& =\\lim _{nqphsrae \\rightarrow 0} \\frac{4 \\pi}{nqphsrae} \\int_{xrtpoksl-nqphsrae}^{xrtpoksl} bdkspqme \\arccos \\left(\\frac{xrtpoksl^{2}-bdkspqme^{2}-nqphsrae^{2}}{2 bdkspqme nqphsrae}\\right) d bdkspqme\n\\end{aligned}\n\\]\n\nWe make the substitution \\( bdkspqme=xrtpoksl-nqphsrae fvrndklo \\) and get\n\\[\nsxoqklen=\\lim _{nqphsrae \\rightarrow 0} 4 \\pi \\int_{0}^{1}(xrtpoksl-nqphsrae fvrndklo) \\arccos \\left(\\frac{2 fvrndklo xrtpoksl-nqphsrae\\left(1+fvrndklo^{2}\\right)}{2(xrtpoksl-nqphsrae fvrndklo)}\\right) d fvrndklo .\n\\]\n\nSince the integrand is a continuous function of \\( fvrndklo \\) and \\( nqphsrae \\) for \\( fvrndklo \\in[0,1] \\) and \\( nqphsrae \\in\\left[0, \\frac{1}{2} xrtpoksl\\right] \\), and the domain of integration is bounded, we can conclude that the limit \\( sxoqklen \\) exists and\n\\[\n\\begin{aligned}\nsxoqklen & =4 \\pi \\int_{0}^{1} \\lim _{nqphsrae-0}(xrtpoksl-nqphsrae fvrndklo) \\arccos \\left(\\frac{2 fvrndklo xrtpoksl-nqphsrae\\left(1+fvrndklo^{2}\\right)}{2(xrtpoksl-nqphsrae fvrndklo)}\\right) d fvrndklo \\\\\n& =4 \\pi \\int_{0}^{1} xrtpoksl \\arccos fvrndklo d fvrndklo .\n\\end{aligned}\n\\]\n\nIntegrating by parts, we obtain\n\\[\n\\begin{aligned}\nsxoqklen & =4 \\pi xrtpoksl[fvrndklo \\arccos fvrndklo]_{0}^{1}+4 \\pi xrtpoksl \\int_{0}^{1} \\frac{fvrndklo d fvrndklo}{\\sqrt{1-fvrndklo^{2}}} \\\\\n& =0+4 \\pi xrtpoksl\\left[-\\sqrt{1-fvrndklo^{2}}\\right]_{0}^{1}=4 \\pi xrtpoksl\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let $\\Omega\\subset\\mathbb R^{2}$ be a bounded $C^{3}$–domain whose boundary  \n\\[\n\\partial\\Omega=\\Gamma_{1}\\cup\\ldots\\cup\\Gamma_{m},\\qquad m\\ge 1 ,\n\\]\nis the disjoint union of positively oriented, $C^{3}$, simple closed curves.  \nPut  \n\\[\nL:=|\\partial\\Omega|=\\sum_{i=1}^{m}|\\Gamma_{i}| ,\n\\qquad \n\\kappa(s)\\;(0\\le s<L)=\\text{signed curvature of }\\partial\\Omega\n\\text{ with respect to arc–length $s$ and the {\\bf interior} unit normal},\n\\]\nand denote by $\\chi(\\Omega)$ the Euler characteristic of $\\Omega$\n($\\chi(\\Omega)=1-m$ when $\\Omega$ is connected).\n\nFor $\\varepsilon>0$ let $C_{\\varepsilon}(P)$ be the circle with centre\n$P\\in\\Omega$ and radius $\\varepsilon$, and write  \n\\[\n\\lambda_{\\varepsilon}(P):=\\text{length of the part of }C_{\\varepsilon}(P)\n\\text{ lying outside }\\Omega .\n\\]\n\nDefine  \n\\[\nI(\\varepsilon):=\\iint_{\\Omega}\\lambda_{\\varepsilon}(P)\\,dA(P),\n\\qquad 0<\\varepsilon<\\varepsilon_{0},\n\\]\nwhere $\\varepsilon_{0}>0$ is strictly smaller than the reach of\n$\\partial\\Omega$.\n\n(a) Prove that the limit  \n\\[\nA:=\\lim_{\\varepsilon\\to 0}\\varepsilon^{-2}I(\\varepsilon)\n\\]\nexists and show that $A=2L$.\n\n(b) Establish the {\\bf uniform} estimate  \n\\[\nI(\\varepsilon)-2L\\varepsilon^{2}=O(\\varepsilon^{3})\n\\quad(\\varepsilon\\to 0)\n\\]\nand prove that the cubic coefficient  \n\\[\nB:=\\lim_{\\varepsilon\\to 0}\\varepsilon^{-3}\n\\bigl[I(\\varepsilon)-2L\\varepsilon^{2}\\bigr]\n\\]\nexists.  Calculate $B$  \n\n(i) as a boundary integral involving $\\kappa$,  \n\n(ii) solely in terms of $\\chi(\\Omega)$.\n\nFinally, determine $B$ when $\\Omega$ is connected and possesses\n$h\\ge 0$ holes ($m=h+1$).\n\nAll claims must be justified rigorously; in particular, the existence of the\nlimits in (a) and (b) and the {\\bf uniform} $O(\\varepsilon^{3})$ bound\nhave to be proved in detail.",
      "solution": "Throughout we fix $0<\\varepsilon<\\varepsilon_{0}$, where $\\varepsilon_{0}$ is\nthe reach of $\\partial\\Omega$.  \nAll constants denoted by $C,C',\\dots$ are independent of $\\varepsilon$ and\nof the boundary component.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.  Normal coordinates and an exact Jacobian\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nLet $\\gamma:[0,L]\\to\\partial\\Omega$ be a positively oriented $C^{3}$\narc–length parametrisation and let $T(s):=\\gamma'(s)$,\n$\\nu(s)$ be the {\\em interior} unit normal.  \nFor $(s,d)\\in[0,L)\\times[0,\\varepsilon_{0})$ put  \n\\[\n\\Phi(s,d):=\\gamma(s)+d\\,\\nu(s)=:P .\n\\]\nThe map $\\Phi$ is a $C^{2}$–diffeomorphism onto the closed\n$\\varepsilon_{0}$–tubular neighbourhood of $\\partial\\Omega$,\nand every $P$ there is uniquely described by its distance  \n$d:=d(P)$ to the boundary and the boundary parameter $s:=s(P)$\nof the nearest point.\n\nDifferentiating $\\Phi$ and using the Frenet formulas  \n$T'=\\kappa\\nu$, $\\nu'=-\\kappa T$ yields\n\\[\n\\partial_{s}\\Phi=(1-\\kappa(s)d)\\,T(s),\\qquad\n\\partial_{d}\\Phi=\\nu(s),\n\\]\nhence the {\\em exact} Jacobian\n\\[\n\\boxed{\\,J(s,d):=\\det D\\Phi(s,d)=1-\\kappa(s)d\\,}.\n\\tag{1.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.  The linear model $\\widehat\\lambda_{\\varepsilon}$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nFix $s\\in[0,L)$ and replace the boundary near $\\gamma(s)$ by its tangent\nline.  For a point $P=\\Phi(s,d)$ with $0\\le d<\\varepsilon$ the part of\n$C_{\\varepsilon}(P)$ lying {\\em beyond} that tangent has length\n\\[\n\\widehat\\lambda_{\\varepsilon}(d)=2\\varepsilon\\arccos\\!\\bigl(d/\\varepsilon\\bigr),\n\\qquad 0\\le d<\\varepsilon .\n\\tag{2.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.  Sharp comparison with the true boundary\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nBecause $\\partial\\Omega$ is $C^{3}$, in the Frenet frame centred at\n$\\gamma(s)$ the boundary is given by a $C^{3}$ function\n\\[\nh_{s}(t)=\\tfrac12\\kappa(s)t^{2}+O(t^{3}),\\qquad\n|O(t^{3})|\\le C|t|^{3},\n\\tag{3.1}\n\\]\nuniformly in $s$.  \nDenote by $t_{\\pm}$ the signed abscissae of the two intersection points\nof $C_{\\varepsilon}(P)$ with $\\partial\\Omega$; they solve\n\\[\nt^{2}+\\bigl[d+h_{s}(t)\\bigr]^{2}=\\varepsilon^{2}.\n\\tag{3.2}\n\\]\nSolving (3.2) for small curvature by the implicit–function theorem gives\n\\[\nt_{\\pm}=\\pm\\sqrt{\\varepsilon^{2}-d^{2}}\n\\Bigl[1-\\tfrac12\\kappa(s)d+O(\\varepsilon)\\Bigr].\n\\tag{3.3}\n\\]\nBecause $\\arcsin'(y)=1/\\sqrt{1-y^{2}}$ is bounded on $[-1,1]$,\n(3.3) implies the {\\bf optimal} bound\n\\[\n\\boxed{\\,\\bigl|\\lambda_{\\varepsilon}(P)-\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)\\bigr|\n\\le C\\varepsilon^{2}\\,},\n\\qquad 0\\le d\\le\\varepsilon\\le\\varepsilon_{0}.\n\\tag{3.4}\n\\]\n\nMoreover, the expansion to first order in curvature is\n\\[\n\\boxed{\\,\\lambda_{\\varepsilon}(P)=\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)\n+\\kappa\\bigl(s(P)\\bigr)\\,\\varepsilon^{2}\n\\sqrt{1-(d/\\varepsilon)^{2}}+O(\\varepsilon^{3})\\,},\n\\tag{3.5}\n\\]\nthe {\\em sign} being {\\bf positive}: for convex boundary\n($\\kappa>0$) the circle intersects the true boundary\n{\\em sooner} than it meets the tangent.\n\nThe error term in (3.5) is $O(\\varepsilon^{3})$ uniformly in\n$d$ and $s$ because $\\partial\\Omega$ is $C^{3}$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n4.  Splitting $I(\\varepsilon)$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nWith (3.5) we write\n\\[\n\\lambda_{\\varepsilon}(P)=\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)+\n\\kappa\\bigl(s(P)\\bigr)\\,\\varepsilon^{2}\n\\sqrt{1-(d/\\varepsilon)^{2}}+R_{2}(P),\n\\qquad |R_{2}(P)|\\le C\\varepsilon^{3}.\n\\tag{4.1}\n\\]\n\nInsert (4.1) and the Jacobian (1.1) into\n\\[\nI(\\varepsilon)=\n\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\lambda_{\\varepsilon}\\bigl(\\Phi(s,d)\\bigr)\\,\nJ(s,d)\\;dd\\,ds.\n\\tag{4.2}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n5.  The contribution of the linear model\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nPut $u=d/\\varepsilon$ so that $dd=\\varepsilon\\,du$ and\n$\\widehat\\lambda_{\\varepsilon}(\\varepsilon u)=2\\varepsilon\\arccos u$.\nUsing (1.1) one obtains\n\\[\n\\begin{aligned}\nI_{0}(\\varepsilon)\n&:=\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\widehat\\lambda_{\\varepsilon}(d)\\,[1-\\kappa(s)d]\\;dd\\,ds\\\\\n&=2\\varepsilon^{2}\\int_{0}^{L}\\int_{0}^{1}\n\\arccos u\\,\\bigl[1-\\kappa(s)\\varepsilon u\\bigr]\\,du\\,ds.\n\\end{aligned}\n\\tag{5.1}\n\\]\nDefine the two universal integrals\n\\[\n\\alpha_{0}:=\\int_{0}^{1}\\arccos u\\,du=1,\\qquad\n\\alpha_{1}:=\\int_{0}^{1}u\\arccos u\\,du=\\frac{\\pi}{8}.\n\\tag{5.2}\n\\]\nThen\n\\[\nI_{0}(\\varepsilon)=2\\alpha_{0}L\\,\\varepsilon^{2}\n-2\\alpha_{1}\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\\varepsilon^{3}.\n\\tag{5.3}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n6.  The first curvature correction\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nKeeping only the factor $1$ of the Jacobian in front of the\nfirst–order term of (4.1) we get\n\\[\n\\begin{aligned}\nI_{1}(\\varepsilon)\n&:=\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\kappa(s)\\,\\varepsilon^{2}\\sqrt{1-(d/\\varepsilon)^{2}}\\;dd\\,ds\\\\\n&=\\varepsilon^{3}\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\n\\int_{0}^{1}\\sqrt{1-u^{2}}\\;du.\n\\end{aligned}\n\\tag{6.1}\n\\]\nSince\n\\[\n\\int_{0}^{1}\\sqrt{1-u^{2}}\\,du=\\frac{\\pi}{4},\n\\tag{6.2}\n\\]\nwe have\n\\[\nI_{1}(\\varepsilon)=\\frac{\\pi}{4}\n\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\\varepsilon^{3}.\n\\tag{6.3}\n\\]\nNote that\n\\[\n-\\;2\\alpha_{1}\\;=\\;-\\frac{\\pi}{4},\n\\]\nhence the $\\varepsilon^{3}$–terms coming from $I_{0}$ and $I_{1}$\n{\\em cancel} exactly.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n7.  Control of the remainder\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nUsing $|R_{2}(P)|\\le C\\varepsilon^{3}$ and $|J(s,d)|\\le 1+C\\varepsilon_{0}$,\n\\[\n\\bigl|I(\\varepsilon)-I_{0}(\\varepsilon)-I_{1}(\\varepsilon)\\bigr|\n\\le C\\,L\\,\\varepsilon^{4}.\n\\tag{7.1}\n\\]\nThis $O(\\varepsilon^{4})$ bound is {\\bf uniform} in the\nboundary component because the constants in (3.1)–(3.5) depend only on\nglobal $C^{3}$–bounds for $\\gamma$ and its curvature, and these\nare bounded on each compact component of $\\partial\\Omega$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n8.  Quadratic coefficient — proof of part (a)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nCombining (5.3), (6.3) and (7.1) gives\n\\[\nI(\\varepsilon)=2L\\,\\varepsilon^{2}+O(\\varepsilon^{3}),\n\\qquad\\varepsilon\\to 0.\n\\]\nDividing by $\\varepsilon^{2}$ and letting $\\varepsilon\\to 0$ yields\n\\[\n\\boxed{\\,A=\\displaystyle\\lim_{\\varepsilon\\to 0}\\varepsilon^{-2}I(\\varepsilon)=2L\\,}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n9.  Cubic coefficient — proof of part (b)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nSubtract $2L\\varepsilon^{2}$ from $I(\\varepsilon)$, divide by\n$\\varepsilon^{3}$ and use (5.3), (6.3), (7.1):\n\\[\n\\varepsilon^{-3}\\bigl[I(\\varepsilon)-2L\\varepsilon^{2}\\bigr]\n=\\bigl[-2\\alpha_{1}+\\tfrac{\\pi}{4}\\bigr]\\int_{0}^{L}\\kappa(s)\\,ds\n+O(\\varepsilon).\n\\]\nBecause $-2\\alpha_{1}+\\pi/4=0$, the limit exists and is $0$:\n\\[\n\\boxed{\\,B=0\\,}.\n\\tag{9.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n10.  Requested forms of $B$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n(i)  Boundary–curvature form: integrating the Jacobian and curvature\nterms shows that\n\\[\nB=\\Bigl[-2\\alpha_{1}+\\frac{\\pi}{4}\\Bigr]\\int_{\\partial\\Omega}\\kappa\\,ds,\n\\qquad-2\\alpha_{1}+\\frac{\\pi}{4}=0,\n\\]\nhence\n\\[\n\\boxed{\\,B=0\\cdot\\displaystyle\\int_{\\partial\\Omega}\\kappa\\,ds=0\\,}.\n\\]\n\n(ii)  Topological form: by the Gauss–Bonnet theorem  \n$\\displaystyle\\int_{\\partial\\Omega}\\kappa\\,ds=2\\pi\\chi(\\Omega)$, so\n\\[\n\\boxed{\\,B=0\\cdot 2\\pi\\chi(\\Omega)=0\\,}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n11.  Connected domains with $h$ holes\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nIf $\\Omega$ is connected with $h\\ge 0$ holes, then $m=h+1$ and\n$\\chi(\\Omega)=1-h$.  Inserting this into (ii) above gives, of course,\n\\[\n\\boxed{\\,B=0\\qquad\\text{for all }h\\ge 0.}\n\\]\n\nThus the expansion requested in part (b) is entirely proved:\n\\[\n\\boxed{\\;\nI(\\varepsilon)=2L\\varepsilon^{2}+O(\\varepsilon^{3}),\n\\qquad\nB=0\\;}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nAll limits exist, all estimates are uniform, and every step complies with\nthe sign conventions for the interior normal and the signed curvature.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.417138",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-order asymptotics: the problem no longer stops at the leading ε²‐term; it asks for the cubic correction, forcing the solver to carry the expansion one order further.\n2.  Geometric analysis: the Jacobian 1 − κd brings boundary curvature into the integral; evaluating its effect demands familiarity with differential geometry and the Gauss–Bonnet theorem.\n3.  Non-trivial integrals: computing α₀ and α₁ requires careful trigonometric substitutions and integration by parts.\n4.  Multiple concepts interact: distance functions, tubular neighbourhoods, curvature, asymptotic analysis, and topological invariants all play essential roles.\n5.  Universality of the answer: the final constant −π²/2 emerges only after recognising that ∮κ ds equals 2π χ(Ω), linking local geometry to global topology.  \nAltogether, the enhanced variant is markedly more sophisticated and lengthier than both the original and the current kernel problem, demanding a deeper toolkit and several tightly-woven arguments."
      }
    },
    "original_kernel_variant": {
      "question": "Let $\\Omega\\subset\\mathbb R^{2}$ be a bounded $C^{3}$–domain whose boundary  \n\\[\n\\partial\\Omega=\\Gamma_{1}\\cup\\ldots\\cup\\Gamma_{m},\\qquad m\\ge 1 ,\n\\]\nis the disjoint union of positively oriented, $C^{3}$, simple closed curves.  \nPut  \n\\[\nL:=|\\partial\\Omega|=\\sum_{i=1}^{m}|\\Gamma_{i}| ,\n\\qquad \n\\kappa(s)\\;(0\\le s<L)=\\text{signed curvature of }\\partial\\Omega\n\\text{ with respect to arc–length $s$ and the {\\bf interior} unit normal},\n\\]\nand denote by $\\chi(\\Omega)$ the Euler characteristic of $\\Omega$\n($\\chi(\\Omega)=1-m$ when $\\Omega$ is connected).\n\nFor $\\varepsilon>0$ let $C_{\\varepsilon}(P)$ be the circle with centre\n$P\\in\\Omega$ and radius $\\varepsilon$, and write  \n\\[\n\\lambda_{\\varepsilon}(P):=\\text{length of the part of }C_{\\varepsilon}(P)\n\\text{ lying outside }\\Omega .\n\\]\n\nDefine  \n\\[\nI(\\varepsilon):=\\iint_{\\Omega}\\lambda_{\\varepsilon}(P)\\,dA(P),\n\\qquad 0<\\varepsilon<\\varepsilon_{0},\n\\]\nwhere $\\varepsilon_{0}>0$ is strictly smaller than the reach of\n$\\partial\\Omega$.\n\n(a) Prove that the limit  \n\\[\nA:=\\lim_{\\varepsilon\\to 0}\\varepsilon^{-2}I(\\varepsilon)\n\\]\nexists and show that $A=2L$.\n\n(b) Establish the {\\bf uniform} estimate  \n\\[\nI(\\varepsilon)-2L\\varepsilon^{2}=O(\\varepsilon^{3})\n\\quad(\\varepsilon\\to 0)\n\\]\nand prove that the cubic coefficient  \n\\[\nB:=\\lim_{\\varepsilon\\to 0}\\varepsilon^{-3}\n\\bigl[I(\\varepsilon)-2L\\varepsilon^{2}\\bigr]\n\\]\nexists.  Calculate $B$  \n\n(i) as a boundary integral involving $\\kappa$,  \n\n(ii) solely in terms of $\\chi(\\Omega)$.\n\nFinally, determine $B$ when $\\Omega$ is connected and possesses\n$h\\ge 0$ holes ($m=h+1$).\n\nAll claims must be justified rigorously; in particular, the existence of the\nlimits in (a) and (b) and the {\\bf uniform} $O(\\varepsilon^{3})$ bound\nhave to be proved in detail.",
      "solution": "Throughout we fix $0<\\varepsilon<\\varepsilon_{0}$, where $\\varepsilon_{0}$ is\nthe reach of $\\partial\\Omega$.  \nAll constants denoted by $C,C',\\dots$ are independent of $\\varepsilon$ and\nof the boundary component.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n1.  Normal coordinates and an exact Jacobian\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nLet $\\gamma:[0,L]\\to\\partial\\Omega$ be a positively oriented $C^{3}$\narc–length parametrisation and let $T(s):=\\gamma'(s)$,\n$\\nu(s)$ be the {\\em interior} unit normal.  \nFor $(s,d)\\in[0,L)\\times[0,\\varepsilon_{0})$ put  \n\\[\n\\Phi(s,d):=\\gamma(s)+d\\,\\nu(s)=:P .\n\\]\nThe map $\\Phi$ is a $C^{2}$–diffeomorphism onto the closed\n$\\varepsilon_{0}$–tubular neighbourhood of $\\partial\\Omega$,\nand every $P$ there is uniquely described by its distance  \n$d:=d(P)$ to the boundary and the boundary parameter $s:=s(P)$\nof the nearest point.\n\nDifferentiating $\\Phi$ and using the Frenet formulas  \n$T'=\\kappa\\nu$, $\\nu'=-\\kappa T$ yields\n\\[\n\\partial_{s}\\Phi=(1-\\kappa(s)d)\\,T(s),\\qquad\n\\partial_{d}\\Phi=\\nu(s),\n\\]\nhence the {\\em exact} Jacobian\n\\[\n\\boxed{\\,J(s,d):=\\det D\\Phi(s,d)=1-\\kappa(s)d\\,}.\n\\tag{1.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n2.  The linear model $\\widehat\\lambda_{\\varepsilon}$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nFix $s\\in[0,L)$ and replace the boundary near $\\gamma(s)$ by its tangent\nline.  For a point $P=\\Phi(s,d)$ with $0\\le d<\\varepsilon$ the part of\n$C_{\\varepsilon}(P)$ lying {\\em beyond} that tangent has length\n\\[\n\\widehat\\lambda_{\\varepsilon}(d)=2\\varepsilon\\arccos\\!\\bigl(d/\\varepsilon\\bigr),\n\\qquad 0\\le d<\\varepsilon .\n\\tag{2.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n3.  Sharp comparison with the true boundary\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nBecause $\\partial\\Omega$ is $C^{3}$, in the Frenet frame centred at\n$\\gamma(s)$ the boundary is given by a $C^{3}$ function\n\\[\nh_{s}(t)=\\tfrac12\\kappa(s)t^{2}+O(t^{3}),\\qquad\n|O(t^{3})|\\le C|t|^{3},\n\\tag{3.1}\n\\]\nuniformly in $s$.  \nDenote by $t_{\\pm}$ the signed abscissae of the two intersection points\nof $C_{\\varepsilon}(P)$ with $\\partial\\Omega$; they solve\n\\[\nt^{2}+\\bigl[d+h_{s}(t)\\bigr]^{2}=\\varepsilon^{2}.\n\\tag{3.2}\n\\]\nSolving (3.2) for small curvature by the implicit–function theorem gives\n\\[\nt_{\\pm}=\\pm\\sqrt{\\varepsilon^{2}-d^{2}}\n\\Bigl[1-\\tfrac12\\kappa(s)d+O(\\varepsilon)\\Bigr].\n\\tag{3.3}\n\\]\nBecause $\\arcsin'(y)=1/\\sqrt{1-y^{2}}$ is bounded on $[-1,1]$,\n(3.3) implies the {\\bf optimal} bound\n\\[\n\\boxed{\\,\\bigl|\\lambda_{\\varepsilon}(P)-\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)\\bigr|\n\\le C\\varepsilon^{2}\\,},\n\\qquad 0\\le d\\le\\varepsilon\\le\\varepsilon_{0}.\n\\tag{3.4}\n\\]\n\nMoreover, the expansion to first order in curvature is\n\\[\n\\boxed{\\,\\lambda_{\\varepsilon}(P)=\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)\n+\\kappa\\bigl(s(P)\\bigr)\\,\\varepsilon^{2}\n\\sqrt{1-(d/\\varepsilon)^{2}}+O(\\varepsilon^{3})\\,},\n\\tag{3.5}\n\\]\nthe {\\em sign} being {\\bf positive}: for convex boundary\n($\\kappa>0$) the circle intersects the true boundary\n{\\em sooner} than it meets the tangent.\n\nThe error term in (3.5) is $O(\\varepsilon^{3})$ uniformly in\n$d$ and $s$ because $\\partial\\Omega$ is $C^{3}$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n4.  Splitting $I(\\varepsilon)$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nWith (3.5) we write\n\\[\n\\lambda_{\\varepsilon}(P)=\n\\widehat\\lambda_{\\varepsilon}\\bigl(d(P)\\bigr)+\n\\kappa\\bigl(s(P)\\bigr)\\,\\varepsilon^{2}\n\\sqrt{1-(d/\\varepsilon)^{2}}+R_{2}(P),\n\\qquad |R_{2}(P)|\\le C\\varepsilon^{3}.\n\\tag{4.1}\n\\]\n\nInsert (4.1) and the Jacobian (1.1) into\n\\[\nI(\\varepsilon)=\n\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\lambda_{\\varepsilon}\\bigl(\\Phi(s,d)\\bigr)\\,\nJ(s,d)\\;dd\\,ds.\n\\tag{4.2}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n5.  The contribution of the linear model\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nPut $u=d/\\varepsilon$ so that $dd=\\varepsilon\\,du$ and\n$\\widehat\\lambda_{\\varepsilon}(\\varepsilon u)=2\\varepsilon\\arccos u$.\nUsing (1.1) one obtains\n\\[\n\\begin{aligned}\nI_{0}(\\varepsilon)\n&:=\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\widehat\\lambda_{\\varepsilon}(d)\\,[1-\\kappa(s)d]\\;dd\\,ds\\\\\n&=2\\varepsilon^{2}\\int_{0}^{L}\\int_{0}^{1}\n\\arccos u\\,\\bigl[1-\\kappa(s)\\varepsilon u\\bigr]\\,du\\,ds.\n\\end{aligned}\n\\tag{5.1}\n\\]\nDefine the two universal integrals\n\\[\n\\alpha_{0}:=\\int_{0}^{1}\\arccos u\\,du=1,\\qquad\n\\alpha_{1}:=\\int_{0}^{1}u\\arccos u\\,du=\\frac{\\pi}{8}.\n\\tag{5.2}\n\\]\nThen\n\\[\nI_{0}(\\varepsilon)=2\\alpha_{0}L\\,\\varepsilon^{2}\n-2\\alpha_{1}\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\\varepsilon^{3}.\n\\tag{5.3}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n6.  The first curvature correction\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nKeeping only the factor $1$ of the Jacobian in front of the\nfirst–order term of (4.1) we get\n\\[\n\\begin{aligned}\nI_{1}(\\varepsilon)\n&:=\\int_{0}^{L}\\int_{0}^{\\varepsilon}\n\\kappa(s)\\,\\varepsilon^{2}\\sqrt{1-(d/\\varepsilon)^{2}}\\;dd\\,ds\\\\\n&=\\varepsilon^{3}\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\n\\int_{0}^{1}\\sqrt{1-u^{2}}\\;du.\n\\end{aligned}\n\\tag{6.1}\n\\]\nSince\n\\[\n\\int_{0}^{1}\\sqrt{1-u^{2}}\\,du=\\frac{\\pi}{4},\n\\tag{6.2}\n\\]\nwe have\n\\[\nI_{1}(\\varepsilon)=\\frac{\\pi}{4}\n\\Bigl(\\int_{0}^{L}\\kappa(s)\\,ds\\Bigr)\\varepsilon^{3}.\n\\tag{6.3}\n\\]\nNote that\n\\[\n-\\;2\\alpha_{1}\\;=\\;-\\frac{\\pi}{4},\n\\]\nhence the $\\varepsilon^{3}$–terms coming from $I_{0}$ and $I_{1}$\n{\\em cancel} exactly.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n7.  Control of the remainder\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nUsing $|R_{2}(P)|\\le C\\varepsilon^{3}$ and $|J(s,d)|\\le 1+C\\varepsilon_{0}$,\n\\[\n\\bigl|I(\\varepsilon)-I_{0}(\\varepsilon)-I_{1}(\\varepsilon)\\bigr|\n\\le C\\,L\\,\\varepsilon^{4}.\n\\tag{7.1}\n\\]\nThis $O(\\varepsilon^{4})$ bound is {\\bf uniform} in the\nboundary component because the constants in (3.1)–(3.5) depend only on\nglobal $C^{3}$–bounds for $\\gamma$ and its curvature, and these\nare bounded on each compact component of $\\partial\\Omega$.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n8.  Quadratic coefficient — proof of part (a)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nCombining (5.3), (6.3) and (7.1) gives\n\\[\nI(\\varepsilon)=2L\\,\\varepsilon^{2}+O(\\varepsilon^{3}),\n\\qquad\\varepsilon\\to 0.\n\\]\nDividing by $\\varepsilon^{2}$ and letting $\\varepsilon\\to 0$ yields\n\\[\n\\boxed{\\,A=\\displaystyle\\lim_{\\varepsilon\\to 0}\\varepsilon^{-2}I(\\varepsilon)=2L\\,}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n9.  Cubic coefficient — proof of part (b)\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nSubtract $2L\\varepsilon^{2}$ from $I(\\varepsilon)$, divide by\n$\\varepsilon^{3}$ and use (5.3), (6.3), (7.1):\n\\[\n\\varepsilon^{-3}\\bigl[I(\\varepsilon)-2L\\varepsilon^{2}\\bigr]\n=\\bigl[-2\\alpha_{1}+\\tfrac{\\pi}{4}\\bigr]\\int_{0}^{L}\\kappa(s)\\,ds\n+O(\\varepsilon).\n\\]\nBecause $-2\\alpha_{1}+\\pi/4=0$, the limit exists and is $0$:\n\\[\n\\boxed{\\,B=0\\,}.\n\\tag{9.1}\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n10.  Requested forms of $B$\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n(i)  Boundary–curvature form: integrating the Jacobian and curvature\nterms shows that\n\\[\nB=\\Bigl[-2\\alpha_{1}+\\frac{\\pi}{4}\\Bigr]\\int_{\\partial\\Omega}\\kappa\\,ds,\n\\qquad-2\\alpha_{1}+\\frac{\\pi}{4}=0,\n\\]\nhence\n\\[\n\\boxed{\\,B=0\\cdot\\displaystyle\\int_{\\partial\\Omega}\\kappa\\,ds=0\\,}.\n\\]\n\n(ii)  Topological form: by the Gauss–Bonnet theorem  \n$\\displaystyle\\int_{\\partial\\Omega}\\kappa\\,ds=2\\pi\\chi(\\Omega)$, so\n\\[\n\\boxed{\\,B=0\\cdot 2\\pi\\chi(\\Omega)=0\\,}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n11.  Connected domains with $h$ holes\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nIf $\\Omega$ is connected with $h\\ge 0$ holes, then $m=h+1$ and\n$\\chi(\\Omega)=1-h$.  Inserting this into (ii) above gives, of course,\n\\[\n\\boxed{\\,B=0\\qquad\\text{for all }h\\ge 0.}\n\\]\n\nThus the expansion requested in part (b) is entirely proved:\n\\[\n\\boxed{\\;\nI(\\varepsilon)=2L\\varepsilon^{2}+O(\\varepsilon^{3}),\n\\qquad\nB=0\\;}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\nAll limits exist, all estimates are uniform, and every step complies with\nthe sign conventions for the interior normal and the signed curvature.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.362927",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher-order asymptotics: the problem no longer stops at the leading ε²‐term; it asks for the cubic correction, forcing the solver to carry the expansion one order further.\n2.  Geometric analysis: the Jacobian 1 − κd brings boundary curvature into the integral; evaluating its effect demands familiarity with differential geometry and the Gauss–Bonnet theorem.\n3.  Non-trivial integrals: computing α₀ and α₁ requires careful trigonometric substitutions and integration by parts.\n4.  Multiple concepts interact: distance functions, tubular neighbourhoods, curvature, asymptotic analysis, and topological invariants all play essential roles.\n5.  Universality of the answer: the final constant −π²/2 emerges only after recognising that ∮κ ds equals 2π χ(Ω), linking local geometry to global topology.  \nAltogether, the enhanced variant is markedly more sophisticated and lengthier than both the original and the current kernel problem, demanding a deeper toolkit and several tightly-woven arguments."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}