{
  "index": "1951-B-7",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "7. Find the volume of the four-dimensional hypersphere \\( x^{2}+y^{2}+z^{2}+ \\) \\( t^{2}=r^{2} \\). and also the hypervolume of its interior \\( x^{2}+y^{2}+z^{2}+t^{2}<r^{2} \\).",
  "solution": "Solution. Let \\( V_{+}(r) \\) be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius \\( r \\). If the hypervolume is \"sliced\" perpendicular to the \\( x \\)-axis, one gets\n\\[\nV_{4}(r)=\\int_{-r}^{r} V_{3}\\left(\\sqrt{r^{2}-x^{2}}\\right) d x\n\\]\nwhere\n\\[\nV_{3}(\\rho)=(4 / 3) \\pi \\rho^{3}\n\\]\nis the ordinary volume of the three-dimensional ball of radius \\( \\rho \\). Hence\n\\[\nV_{4}(r)=(4 / 3) \\pi \\int_{-r}^{r}\\left(r^{2}-x^{2}\\right)^{3 / 2} d x\n\\]\n\nMaking the substitution \\( x=r \\sin \\theta \\)\n\\[\n\\begin{aligned}\nV_{4}(r) & =\\frac{4 \\pi r^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{4} \\theta d \\theta^{\\prime} \\\\\n& =\\frac{\\pi r^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2}\\left(\\frac{3}{2}+2 \\cos 2 \\theta+\\frac{1}{2} \\cos 4 \\theta\\right) d \\theta,\n\\end{aligned}\n\\]\nwhere we have used the double-angle formula \\( 2 \\cos ^{2} \\theta=1+\\cos 2 \\theta \\) twice to simplify the integrand. Hence\n\\[\nV_{4}(r)=\\frac{\\pi r^{4}}{3} \\cdot \\frac{3}{2} \\pi=\\frac{\\pi^{2} r^{4}}{2}\n\\]\nis the hypervolume of the four-dimensional ball.\nLet \\( A_{4}(r) \\) be the three-dimensional volume of the hyperspherical surface of radius \\( r \\). We can relate \\( A_{4} \\) to \\( V_{4} \\) as follows. If we compute \\( V_{4}(r) \\) by considering concentric spherical shells we see that\n\\[\nV_{4}(r)=\\int_{0}^{r} A_{4}(\\rho) d \\rho .\n\\]\n\nHence by the fundamental theorem of calculus\n\\[\n\\frac{d}{d r} V_{4}(r)=A_{4}(r),\n\\]\nso that\n\\[\nA_{4}(r)=2 \\pi^{2} r^{3} .\n\\]\n\nExtension. This method shows that the \\( n \\)-dimensional volume of an \\( \\boldsymbol{n} \\)-dimensional ball of radius \\( r \\) is given by\n\\[\nV_{n}(r)=\\gamma_{n} r^{n}\n\\]\nwhere the \\( \\gamma \\) 's can be calculated recursively by the formula\n\\[\n\\gamma_{n}=\\gamma_{n-1} \\int_{-\\pi / 2}^{j^{\\pi / 2}} \\cos ^{n} \\theta d \\theta\n\\]\n\nThe values of these integrals are obtainable from the reduction formula\n\\[\n\\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{n} \\theta d \\theta=\\frac{n-1}{n} \\int_{-\\pi, 2}^{\\pi / 2} \\cos ^{n-2} d \\theta \\quad \\text { for } n \\geq 2,\n\\]\nwhich leads to the result\n\\[\n\\gamma_{2 n}=\\frac{\\pi^{n}}{n!}, \\quad \\gamma_{2 n+1}=\\frac{2(2 \\pi)^{n}}{1 \\cdot 3 \\cdot 5 \\cdots(2 n+1)} .\n\\]\n\nUsing the \\( \\Gamma \\)-function we can express both of these equations by\n\\[\n\\gamma_{m}=\\frac{\\pi^{m / 2}}{\\Gamma\\left(\\frac{m}{2}+1\\right)}\n\\]\n\nThe \\( (n-1) \\)-dimensional volume of the surface of the \\( n \\)-ball is given by\n\\[\nA_{n}(r)=\\frac{d}{d r}\\left(V_{n}(r)\\right)=n \\gamma_{n} r^{n} \\quad 1=\\frac{2 \\pi^{n / 2} r^{n-1}}{\\Gamma\\left(\\frac{n}{2}\\right)}\n\\]\n\nSee H. P. Evans, American Mathematical Monthly, vol. 54 (1947), pages 592-594.",
  "vars": [
    "x",
    "y",
    "z",
    "t",
    "r",
    "\\\\rho",
    "\\\\theta"
  ],
  "params": [
    "V_+",
    "V_4",
    "V_3",
    "A_4",
    "\\\\gamma_n",
    "\\\\gamma_2n",
    "\\\\gamma_2n+1",
    "n",
    "m",
    "\\\\Gamma"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "coordinatex",
        "y": "coordinatey",
        "z": "coordinatez",
        "t": "coordinatet",
        "r": "radiusvar",
        "\\rho": "radialrho",
        "\\theta": "angletheta",
        "V_+": "hypervolin",
        "V_4": "hypervolfour",
        "V_3": "hypervolthree",
        "A_4": "surfacefour",
        "\\gamma_n": "gammancoeff",
        "\\gamma_2n": "gammaeven",
        "\\gamma_2n+1": "gammaoddn",
        "n": "dimensn",
        "m": "dimensm",
        "\\Gamma": "gammafunc"
      },
      "question": "7. Find the volume of the four-dimensional hypersphere \\( coordinatex^{2}+coordinatey^{2}+coordinatez^{2}+ \\) \\( coordinatet^{2}=radiusvar^{2} \\). and also the hypervolume of its interior \\( coordinatex^{2}+coordinatey^{2}+coordinatez^{2}+coordinatet^{2}<radiusvar^{2} \\).",
      "solution": "Solution. Let \\( hypervolin(radiusvar) \\) be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius \\( radiusvar \\). If the hypervolume is \"sliced\" perpendicular to the \\( coordinatex \\)-axis, one gets\n\\[\nhypervolfour(radiusvar)=\\int_{-radiusvar}^{radiusvar} hypervolthree\\left(\\sqrt{radiusvar^{2}-coordinatex^{2}}\\right) d coordinatex\n\\]\nwhere\n\\[\nhypervolthree(radialrho)=(4 / 3) \\pi radialrho^{3}\n\\]\nis the ordinary volume of the three-dimensional ball of radius \\( radialrho \\). Hence\n\\[\nhypervolfour(radiusvar)=(4 / 3) \\pi \\int_{-radiusvar}^{radiusvar}\\left(radiusvar^{2}-coordinatex^{2}\\right)^{3 / 2} d coordinatex\n\\]\n\nMaking the substitution \\( coordinatex=radiusvar \\sin angletheta \\)\n\\[\n\\begin{aligned}\nhypervolfour(radiusvar) & =\\frac{4 \\pi radiusvar^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{4} angletheta d angletheta^{\\prime} \\\\\n& =\\frac{\\pi radiusvar^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2}\\left(\\frac{3}{2}+2 \\cos 2 angletheta+\\frac{1}{2} \\cos 4 angletheta\\right) d angletheta,\n\\end{aligned}\n\\]\nwhere we have used the double-angle formula \\( 2 \\cos ^{2} angletheta=1+\\cos 2 angletheta \\) twice to simplify the integrand. Hence\n\\[\nhypervolfour(radiusvar)=\\frac{\\pi radiusvar^{4}}{3} \\cdot \\frac{3}{2} \\pi=\\frac{\\pi^{2} radiusvar^{4}}{2}\n\\]\nis the hypervolume of the four-dimensional ball.\nLet \\( surfacefour(radiusvar) \\) be the three-dimensional volume of the hyperspherical surface of radius \\( radiusvar \\). We can relate \\( surfacefour \\) to \\( hypervolfour \\) as follows. If we compute \\( hypervolfour(radiusvar) \\) by considering concentric spherical shells we see that\n\\[\nhypervolfour(radiusvar)=\\int_{0}^{radiusvar} surfacefour(radialrho) d radialrho .\n\\]\n\nHence by the fundamental theorem of calculus\n\\[\n\\frac{d}{d radiusvar} hypervolfour(radiusvar)=surfacefour(radiusvar),\n\\]\nso that\n\\[\nsurfacefour(radiusvar)=2 \\pi^{2} radiusvar^{3} .\n\\]\n\nExtension. This method shows that the \\( dimensn \\)-dimensional volume of an \\( \\boldsymbol{dimensn} \\)-dimensional ball of radius \\( radiusvar \\) is given by\n\\[\nV_{dimensn}(radiusvar)=\\gammancoeff radiusvar^{dimensn}\n\\]\nwhere the \\( \\gamma \\) 's can be calculated recursively by the formula\n\\[\n\\gammancoeff=\\gamma_{dimensn-1} \\int_{-\\pi / 2}^{j^{\\pi / 2}} \\cos ^{dimensn} angletheta d angletheta\n\\]\n\nThe values of these integrals are obtainable from the reduction formula\n\\[\n\\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{dimensn} angletheta d angletheta=\\frac{dimensn-1}{dimensn} \\int_{-\\pi, 2}^{\\pi / 2} \\cos ^{dimensn-2} d angletheta \\quad \\text { for } dimensn \\geq 2,\n\\]\nwhich leads to the result\n\\[\ngammaeven=\\frac{\\pi^{dimensn}}{dimensn!}, \\quad gammaoddn=\\frac{2(2 \\pi)^{dimensn}}{1 \\cdot 3 \\cdot 5 \\cdots(2 dimensn+1)} .\n\\]\n\nUsing the \\( gammafunc \\)-function we can express both of these equations by\n\\[\n\\gamma_{dimensm}=\\frac{\\pi^{dimensm / 2}}{gammafunc\\left(\\frac{dimensm}{2}+1\\right)}\n\\]\n\nThe \\( (dimensn-1) \\)-dimensional volume of the surface of the \\( dimensn \\)-ball is given by\n\\[\nA_{dimensn}(radiusvar)=\\frac{d}{d radiusvar}\\left(V_{dimensn}(radiusvar)\\right)=dimensn \\gammancoeff radiusvar^{dimensn} \\quad 1=\\frac{2 \\pi^{dimensn / 2} radiusvar^{dimensn-1}}{gammafunc\\left(\\frac{dimensn}{2}\\right)}\n\\]\n\nSee H. P. Evans, American Mathematical Monthly, vol. 54 (1947), pages 592-594."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "eucalyptus",
        "y": "blackthorn",
        "z": "marigolds",
        "t": "schoolbus",
        "r": "afterglow",
        "\\rho": "sandstone",
        "\\theta": "arrowhead",
        "V_+": "moonstone",
        "V_4": "courtyard",
        "V_3": "lumberjack",
        "A_4": "blueprint",
        "\\gamma_n": "driftwood",
        "\\gamma_2n": "peppermint",
        "\\gamma_2n+1": "campfires",
        "n": "waterfall",
        "m": "cottonseed",
        "\\Gamma": "thornbush"
      },
      "question": "7. Find the volume of the four-dimensional hypersphere \\( eucalyptus^{2}+blackthorn^{2}+marigolds^{2}+ \\) \\( schoolbus^{2}=afterglow^{2} \\) and also the hypervolume of its interior \\( eucalyptus^{2}+blackthorn^{2}+marigolds^{2}+schoolbus^{2}<afterglow^{2} \\).",
      "solution": "Solution. Let \\( moonstone(afterglow) \\) be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius \\( afterglow \\). If the hypervolume is \"sliced\" perpendicular to the \\( eucalyptus \\)-axis, one gets\n\\[\ncourtyard(afterglow)=\\int_{-afterglow}^{afterglow} lumberjack\\left(\\sqrt{afterglow^{2}-eucalyptus^{2}}\\right) d eucalyptus\n\\]\nwhere\n\\[\nlumberjack(sandstone)=(4 / 3) \\pi sandstone^{3}\n\\]\nis the ordinary volume of the three-dimensional ball of radius \\( sandstone \\). Hence\n\\[\ncourtyard(afterglow)=(4 / 3) \\pi \\int_{-afterglow}^{afterglow}\\left(afterglow^{2}-eucalyptus^{2}\\right)^{3 / 2} d eucalyptus\n\\]\n\nMaking the substitution \\( eucalyptus=afterglow \\sin arrowhead \\)\n\\[\n\\begin{aligned}\ncourtyard(afterglow) & =\\frac{4 \\pi afterglow^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{4} arrowhead \\, d arrowhead^{\\prime} \\\\\n& =\\frac{\\pi afterglow^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2}\\left(\\frac{3}{2}+2 \\cos 2 arrowhead+\\frac{1}{2} \\cos 4 arrowhead\\right) d arrowhead,\n\\end{aligned}\n\\]\nwhere we have used the double-angle formula \\( 2 \\cos ^{2} arrowhead=1+\\cos 2 arrowhead \\) twice to simplify the integrand. Hence\n\\[\ncourtyard(afterglow)=\\frac{\\pi afterglow^{4}}{3} \\cdot \\frac{3}{2} \\pi=\\frac{\\pi^{2} afterglow^{4}}{2}\n\\]\nis the hypervolume of the four-dimensional ball.\nLet \\( blueprint(afterglow) \\) be the three-dimensional volume of the hyperspherical surface of radius \\( afterglow \\). We can relate \\( blueprint \\) to \\( courtyard \\) as follows. If we compute \\( courtyard(afterglow) \\) by considering concentric spherical shells we see that\n\\[\ncourtyard(afterglow)=\\int_{0}^{afterglow} blueprint(sandstone) d sandstone .\n\\]\n\nHence by the fundamental theorem of calculus\n\\[\n\\frac{d}{d afterglow} courtyard(afterglow)=blueprint(afterglow),\n\\]\nso that\n\\[\nblueprint(afterglow)=2 \\pi^{2} afterglow^{3} .\n\\]\n\nExtension. This method shows that the \\( waterfall \\)-dimensional volume of an \\( \\boldsymbol{waterfall} \\)-dimensional ball of radius \\( afterglow \\) is given by\n\\[\nV_{waterfall}(afterglow)=driftwood afterglow^{waterfall}\n\\]\nwhere the \\( driftwood \\) 's can be calculated recursively by the formula\n\\[\ndriftwood=\\gamma_{waterfall-1} \\int_{-\\pi / 2}^{j^{\\pi / 2}} \\cos ^{waterfall} arrowhead d arrowhead\n\\]\n\nThe values of these integrals are obtainable from the reduction formula\n\\[\n\\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{waterfall} arrowhead d arrowhead=\\frac{waterfall-1}{waterfall} \\int_{-\\pi, 2}^{\\pi / 2} \\cos ^{waterfall-2} d arrowhead \\quad \\text { for } waterfall \\geq 2,\n\\]\nwhich leads to the result\n\\[\npeppermint=\\frac{\\pi^{waterfall}}{waterfall!}, \\quad campfires=\\frac{2(2 \\pi)^{waterfall}}{1 \\cdot 3 \\cdot 5 \\cdots(2 waterfall+1)} .\n\\]\n\nUsing the \\( thornbush \\)-function we can express both of these equations by\n\\[\n\\gamma_{cottonseed}=\\frac{\\pi^{cottonseed / 2}}{thornbush\\left(\\frac{cottonseed}{2}+1\\right)}\n\\]\n\nThe \\( (waterfall-1) \\)-dimensional volume of the surface of the \\( waterfall \\)-ball is given by\n\\[\nA_{waterfall}(afterglow)=\\frac{d}{d afterglow}\\left(V_{waterfall}(afterglow)\\right)=waterfall driftwood afterglow^{waterfall} \\quad 1=\\frac{2 \\pi^{waterfall / 2} afterglow^{waterfall-1}}{thornbush\\left(\\frac{waterfall}{2}\\right)}\n\\]\n\nSee H. P. Evans, American Mathematical Monthly, vol. 54 (1947), pages 592-594."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "z": "flatdimension",
        "t": "staticmeasure",
        "r": "diameter",
        "\\rho": "centervalue",
        "\\theta": "straightness",
        "V_+": "emptyvolume",
        "V_4": "voidmeasure",
        "V_3": "flatmeasure",
        "A_4": "innerbulk",
        "\\gamma_n": "chaosindex",
        "\\gamma_2n": "evenchaos",
        "\\gamma_2n+1": "oddchaos",
        "n": "microscale",
        "m": "megascale",
        "\\Gamma": "deltafunction"
      },
      "question": "7. Find the volume of the four-dimensional hypersphere \\( verticalaxis^{2}+horizontalaxis^{2}+flatdimension^{2}+ \\) \\( staticmeasure^{2}=diameter^{2} \\). and also the hypervolume of its interior \\( verticalaxis^{2}+horizontalaxis^{2}+flatdimension^{2}+staticmeasure^{2}<diameter^{2} \\).",
      "solution": "Solution. Let \\( emptyvolume(diameter) \\) be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius \\( diameter \\). If the hypervolume is \"sliced\" perpendicular to the \\( verticalaxis \\)-axis, one gets\n\\[\nvoidmeasure(diameter)=\\int_{-diameter}^{diameter} flatmeasure\\left(\\sqrt{diameter^{2}-verticalaxis^{2}}\\right) d verticalaxis\n\\]\nwhere\n\\[\nflatmeasure(centervalue)=(4 / 3) \\pi centervalue^{3}\n\\]\nis the ordinary volume of the three-dimensional ball of radius \\( centervalue \\). Hence\n\\[\nvoidmeasure(diameter)=(4 / 3) \\pi \\int_{-diameter}^{diameter}\\left(diameter^{2}-verticalaxis^{2}\\right)^{3 / 2} d verticalaxis\n\\]\n\nMaking the substitution \\( verticalaxis=diameter \\sin straightness \\)\n\\[\n\\begin{aligned}\nvoidmeasure(diameter) & =\\frac{4 \\pi diameter^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{4} straightness d straightness^{\\prime} \\\\\n& =\\frac{\\pi diameter^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2}\\left(\\frac{3}{2}+2 \\cos 2 straightness+\\frac{1}{2} \\cos 4 straightness\\right) d straightness,\n\\end{aligned}\n\\]\nwhere we have used the double-angle formula \\( 2 \\cos ^{2} straightness=1+\\cos 2 straightness \\) twice to simplify the integrand. Hence\n\\[\nvoidmeasure(diameter)=\\frac{\\pi diameter^{4}}{3} \\cdot \\frac{3}{2} \\pi=\\frac{\\pi^{2} diameter^{4}}{2}\n\\]\nis the hypervolume of the four-dimensional ball.\nLet \\( innerbulk(diameter) \\) be the three-dimensional volume of the hyperspherical surface of radius \\( diameter \\). We can relate \\( innerbulk \\) to \\( voidmeasure \\) as follows. If we compute \\( voidmeasure(diameter) \\) by considering concentric spherical shells we see that\n\\[\nvoidmeasure(diameter)=\\int_{0}^{diameter} innerbulk(centervalue) d centervalue .\n\\]\n\nHence by the fundamental theorem of calculus\n\\[\n\\frac{d}{d diameter} voidmeasure(diameter)=innerbulk(diameter),\n\\]\nso that\n\\[\ninnerbulk(diameter)=2 \\pi^{2} diameter^{3} .\n\\]\n\nExtension. This method shows that the \\( microscale \\)-dimensional volume of an \\( \\boldsymbol{microscale} \\)-dimensional ball of radius \\( diameter \\) is given by\n\\[\nV_{microscale}(diameter)=chaosindex diameter^{microscale}\n\\]\nwhere the \\( chaosindex \\) 's can be calculated recursively by the formula\n\\[\nchaosindex=\\gamma_{microscale-1} \\int_{-\\pi / 2}^{j^{\\pi / 2}} \\cos ^{microscale} straightness d straightness\n\\]\n\nThe values of these integrals are obtainable from the reduction formula\n\\[\n\\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{microscale} straightness d straightness=\\frac{microscale-1}{microscale} \\int_{-\\pi, 2}^{\\pi / 2} \\cos ^{microscale-2} d straightness \\quad \\text { for } microscale \\geq 2,\n\\]\nwhich leads to the result\n\\[\nevenchaos=\\frac{\\pi^{microscale/2}}{(microscale/2)!}, \\quad oddchaos=\\frac{2(2 \\pi)^{(microscale-1)/2}}{1 \\cdot 3 \\cdot 5 \\cdots(2 microscale+1)} .\n\\]\n\nUsing the \\( deltafunction \\)-function we can express both of these equations by\n\\[\nchaosindex=\\frac{\\pi^{microscale / 2}}{deltafunction\\left(\\frac{microscale}{2}+1\\right)}\n\\]\n\nThe \\( (microscale-1) \\)-dimensional volume of the surface of the \\( microscale \\)-ball is given by\n\\[\nA_{microscale}(diameter)=\\frac{d}{d diameter}\\left(V_{microscale}(diameter)\\right)=microscale chaosindex diameter^{microscale} \\quad 1=\\frac{2 \\pi^{microscale / 2} diameter^{microscale-1}}{deltafunction\\left(\\frac{microscale}{2}\\right)}\n\\]\n\nSee H. P. Evans, American Mathematical Monthly, vol. 54 (1947), pages 592-594."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mfplqvde",
        "t": "skjdnfgh",
        "r": "vbchtrms",
        "\\\\rho": "tmnlkqsv",
        "\\\\theta": "ksdhfrwe",
        "V_+": "lkjhgfds",
        "V_4": "qwertyui",
        "V_3": "asdfghjk",
        "A_4": "zxcvbnml",
        "\\\\gamma_n": "poiuytre",
        "\\\\gamma_2n": "mnbvcxza",
        "\\\\gamma_2n+1": "qazwsxed",
        "n": "plokmijn",
        "m": "ujnhbgvf",
        "\\\\Gamma": "ytrewqas"
      },
      "question": "7. Find the volume of the four-dimensional hypersphere \\( qzxwvtnp^{2}+hjgrksla^{2}+mfplqvde^{2}+ \\) \\( skjdnfgh^{2}=vbchtrms^{2} \\). and also the hypervolume of its interior \\( qzxwvtnp^{2}+hjgrksla^{2}+mfplqvde^{2}+skjdnfgh^{2}<vbchtrms^{2} \\).",
      "solution": "Solution. Let \\( lkjhgfds(vbchtrms) \\) be the hypervolume of the interior of the hypersphere, i.e., the four-dimensional ball of radius \\( vbchtrms \\). If the hypervolume is \"sliced\" perpendicular to the \\( qzxwvtnp \\)-axis, one gets\n\\[\nqwertyui(vbchtrms)=\\int_{-vbchtrms}^{vbchtrms} asdfghjk\\left(\\sqrt{vbchtrms^{2}-qzxwvtnp^{2}}\\right) d qzxwvtnp\n\\]\nwhere\n\\[\nasdfghjk(tmnlkqsv)=(4 / 3) \\pi tmnlkqsv^{3}\n\\]\nis the ordinary volume of the three-dimensional ball of radius \\( tmnlkqsv \\). Hence\n\\[\nqwertyui(vbchtrms)=(4 / 3) \\pi \\int_{-vbchtrms}^{vbchtrms}\\left(vbchtrms^{2}-qzxwvtnp^{2}\\right)^{3 / 2} d qzxwvtnp\n\\]\n\nMaking the substitution \\( qzxwvtnp=vbchtrms \\sin ksdhfrwe \\)\n\\[\n\\begin{aligned}\nqwertyui(vbchtrms) & =\\frac{4 \\pi vbchtrms^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{4} ksdhfrwe d ksdhfrwe^{\\prime} \\\\\n& =\\frac{\\pi vbchtrms^{4}}{3} \\int_{-\\pi / 2}^{\\pi / 2}\\left(\\frac{3}{2}+2 \\cos 2 ksdhfrwe+\\frac{1}{2} \\cos 4 ksdhfrwe\\right) d ksdhfrwe,\n\\end{aligned}\n\\]\nwhere we have used the double-angle formula \\( 2 \\cos ^{2} ksdhfrwe=1+\\cos 2 ksdhfrwe \\) twice to simplify the integrand. Hence\n\\[\nqwertyui(vbchtrms)=\\frac{\\pi vbchtrms^{4}}{3} \\cdot \\frac{3}{2} \\pi=\\frac{\\pi^{2} vbchtrms^{4}}{2}\n\\]\nis the hypervolume of the four-dimensional ball.\nLet \\( zxcvbnml(vbchtrms) \\) be the three-dimensional volume of the hyperspherical surface of radius \\( vbchtrms \\). We can relate \\( zxcvbnml \\) to \\( qwertyui \\) as follows. If we compute \\( qwertyui(vbchtrms) \\) by considering concentric spherical shells we see that\n\\[\nqwertyui(vbchtrms)=\\int_{0}^{vbchtrms} zxcvbnml(tmnlkqsv) d tmnlkqsv .\n\\]\n\nHence by the fundamental theorem of calculus\n\\[\n\\frac{d}{d vbchtrms} qwertyui(vbchtrms)=zxcvbnml(vbchtrms),\n\\]\nso that\n\\[\nzxcvbnml(vbchtrms)=2 \\pi^{2} vbchtrms^{3} .\n\\]\n\nExtension. This method shows that the \\( plokmijn \\)-dimensional volume of an \\( \\boldsymbol{plokmijn} \\)-dimensional ball of radius \\( vbchtrms \\) is given by\n\\[\nV_{plokmijn}(vbchtrms)=poiuytre_{plokmijn} vbchtrms^{plokmijn}\n\\]\nwhere the \\( poiuytre \\)'s can be calculated recursively by the formula\n\\[\npoiuytre_{plokmijn}=poiuytre_{plokmijn-1} \\int_{-\\pi / 2}^{j^{\\pi / 2}} \\cos ^{plokmijn} ksdhfrwe d ksdhfrwe\n\\]\n\nThe values of these integrals are obtainable from the reduction formula\n\\[\n\\int_{-\\pi / 2}^{\\pi / 2} \\cos ^{plokmijn} ksdhfrwe d ksdhfrwe=\\frac{plokmijn-1}{plokmijn} \\int_{-\\pi, 2}^{\\pi / 2} \\cos ^{plokmijn-2} d ksdhfrwe \\quad \\text { for } plokmijn \\geq 2,\n\\]\nwhich leads to the result\n\\[\nmnbvcxza_{2 plokmijn}=\\frac{\\pi^{plokmijn}}{plokmijn!}, \\quad qazwsxed_{2 plokmijn+1}=\\frac{2(2 \\pi)^{plokmijn}}{1 \\cdot 3 \\cdot 5 \\cdots(2 plokmijn+1)} .\n\\]\n\nUsing the \\( ytrewqas \\)-function we can express both of these equations by\n\\[\npoiuytre_{plokmijn}=\\frac{\\pi^{plokmijn / 2}}{ytrewqas\\left(\\frac{plokmijn}{2}+1\\right)}\n\\]\n\nThe \\( (plokmijn-1) \\)-dimensional volume of the surface of the \\( plokmijn \\)-ball is given by\n\\[\nA_{plokmijn}(vbchtrms)=\\frac{d}{d vbchtrms}\\left(V_{plokmijn}(vbchtrms)\\right)=plokmijn poiuytre_{plokmijn} vbchtrms^{plokmijn} \\quad 1=\\frac{2 \\pi^{plokmijn / 2} vbchtrms^{plokmijn-1}}{ytrewqas\\left(\\frac{plokmijn}{2}\\right)}\n\\]\n\nSee H. P. Evans, American Mathematical Monthly, vol. 54 (1947), pages 592-594."
    },
    "kernel_variant": {
      "question": "Let m ≥ 3 be an integer and, for a fixed radius R > 0, split the Euclidean space ℝ^{2m} into two m–tuples of coordinates\nx = (x₁,…,x_m , x_{m+1},…,x_{2m}) ≡ (x′,x″) with ‖x′‖₂²=Σ_{i=1}^{m}x_i² and ‖x″‖₂²=Σ_{i=m+1}^{2m}x_i².  \nDefine the full 2m–ball and the “diagonal‐half–ball’’\n\n B_{R}^{(2m)} = {x ∈ ℝ^{2m} : ‖x‖₂ < R}, \n Ω_{m}(R)     = {x ∈ B_{R}^{(2m)} : ‖x′‖₂ < ‖x″‖₂ }.\n\n(A)   Prove that Vol_{2m}(Ω_{m}(R)) = ½ Vol_{2m}(B_{R}^{(2m)}) and give the value in closed Γ–function form.\n\n(B)   Let S_{R}^{(2m−1)} = ∂B_{R}^{(2m)} be the (2m−1)–sphere of radius R.  \n      Compute the (2m−1)-dimensional surface measure of  \n      Σ_{m}(R) = {x ∈ S_{R}^{(2m−1)} : ‖x′‖₂ < ‖x″‖₂ }.\n\n(C)   Choose a point uniformly at random in B_{1}^{(2m)}.  \n      For U = ‖x′‖₂² / ‖x‖₂² show that U ~ Beta(m/2, m/2), and use this fact to re-derive Part (A) by evaluating P(U < ½).\n\n(D)   In the eight–dimensional case m = 4 let  \n  Q(x) = (‖x′‖₂² − ‖x″‖₂²)².  \n      Compute the average value of Q over Ω₄(R) and the integral  \n      I(R) = ∫_{Ω₄(R)} Q(x) dx.",
      "solution": "Throughout we write n = 2m.  The n–ball volume and surface area are\n\n V_n(R) = π^{n/2}R^{n}/Γ(n/2+1), \n A_n(R) = dV_n/dR = nπ^{n/2}R^{n−1}/Γ(n/2+1).                                (1)\n\n(A)  Volume of Ω_m(R).\n\nBecause the inequality ‖x′‖₂ ≷ ‖x″‖₂ is reversed by the coordinate permutation  \n(x₁,…,x_m , x_{m+1},…,x_{2m}) ↦ (x_{m+1},…,x_{2m},x₁,…,x_m),  \nΩ_m(R) and its complement inside B_{R}^{(n)} are congruent.  Their common boundary  \n‖x′‖₂ = ‖x″‖₂ has Lebesgue measure zero in ℝ^{n}, so the two regions have equal volume:\n\n Vol_n(Ω_m(R)) = ½ Vol_n(B_{R}^{(n)}) = ½ V_n(R).                         (2)\n\nUsing (1) with n = 2m,\n\n Vol_{2m}(Ω_m(R)) = ½·π^{m}R^{2m}/Γ(m+1).                               (3)\n\n(B)  Surface measure of Σ_{m}(R).\n\nThe same symmetry shows that Σ_m(R) occupies one half of the sphere, hence\n\n Vol_{2m−1}(Σ_m(R)) = ½ A_n(R)  \n  = ½·nπ^{n/2}R^{n−1}/Γ(n/2+1)  \n  = m π^{m}R^{2m−1}/Γ(m+1).                                            (4)\n\n(C)  Distribution of U and probabilistic proof of (2).\n\nFor a uniformly random point in the unit n–ball, the radius r = ‖x‖₂ and the direction ξ = x/r are independent.  The squared coordinates of ξ form a Dirichlet(½,…,½) random vector (2m parameters, each ½).  Hence\n\n U = Σ_{i=1}^{m} ξ_i² ~ Beta(a,b) with a = m/2, b = m/2.               (5)\n\nThe Beta density is symmetric about ½, so P(U < ½) = ½.  Because r and U are independent and r < 1 by construction,\n\n P(‖x′‖₂ < ‖x″‖₂) = P(U < ½) = ½.                                     (6)\n\nMultiplying this probability by the total volume V_n(1) gives Vol(Ω_m(1)); scaling by R^{n} yields (3), recovering the geometric proof.\n\n(D)  Quartic moment in the eight–dimensional case (m=4, n=8).\n\nStep 1.  Radial moments inside the 8–ball.  \nFor uniform x ∈ B_{R}^{(8)}, the density of r=‖x‖₂ is\n\n f_r(r) = n r^{n−1}/R^{n}, 0≤r≤R, n=8.                               (7)\n\nHence  \n E[r^4] = ∫_{0}^{R} r^4 f_r(r) dr  \n    = 8∫_{0}^{R} r^{11}/R^{8} dr = 8R^4/(12) = 2R^4/3.               (8)\n\nStep 2.  Angular factor.  \nWith m=4, U ~ Beta(2,2).  For any Borel set A⊂[0,1] we have by symmetry\n\n E[(2U−1)^2 1_{A}] = E[(2U−1)^2 1_{1−A}].                              (9)\n\nThus, with A=[0,½),\n\n E[(2U−1)^2 | U<½] = E[(2U−1)^2].                                    (10)\n\nCompute the unconditional expectation:\n\n Var(U) = ab/[(a+b)^2(a+b+1)] = 4/(16·5)=1/20,  \n E[(2U−1)^2] = 4 Var(U) = 1/5.                                       (11)\n\nHence also E[(2U−1)^2 | U<½] = 1/5.\n\nStep 3.  Average of Q on Ω₄(R).  \nWrite S₁ = ‖x′‖₂² = r²U, S₂ = ‖x″‖₂² = r²(1−U).  Over Ω₄(R) (i.e. U<½),\n\n Q(x) = (S₁−S₂)² = r⁴(2U−1)².\n\nBy independence of r and U,\n\n E_{Ω}[Q] = E[r⁴] · E[(2U−1)² | U<½] = (2R⁴/3)·(1/5)=2R⁴/15.        (12)\n\nStep 4.  Integral of Q.  \nUsing Vol_{8}(Ω₄(R)) = ½ V₈(R) = ½·π⁴R⁸/Γ(5)=π⁴R⁸/48,\n\n I(R) = Vol(Ω₄(R))·E_{Ω}[Q] = (π⁴R⁸/48)·(2R⁴/15) = π⁴R^{12}/360.     (13)\n\nSo\n\n  E_{Ω₄(R)}[Q] = 2R⁴/15,  I(R) = π⁴R^{12}/360.                     (14)",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.442292",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension & more variables: the problem lives in 2m dimensions (≥6, concretely carried out in 8) instead of the original 4- or 6-dimensional setting.  \n2. Additional constraints: the region Ω_m(R) is carved out of the ball by the nonlinear inequality ‖x′‖₂ < ‖x″‖₂, producing a non-separable domain whose boundary is neither spherical nor flat.  \n3. Sophisticated structures: solving Parts (A)–(C) requires recognising Dirichlet and Beta distributions on the sphere, using symmetry arguments, and manipulating Γ-functions; Part (D) needs mixing radial moments with conditional Beta expectations to evaluate a quartic integral.  \n4. Deeper theory: probability on high-dimensional spheres, independence of radial and angular parts, and higher moments of Beta laws must all be deployed.  \n5. Multiple interacting concepts: Euclidean geometry, special functions, probabilistic representation, and higher-order moment calculations intertwine, making the variant substantially harder than simply differentiating or scaling the classical n-ball formulas."
      }
    },
    "original_kernel_variant": {
      "question": "Let m ≥ 3 be an integer and, for a fixed radius R > 0, split the Euclidean space ℝ^{2m} into two m–tuples of coordinates\nx = (x₁,…,x_m , x_{m+1},…,x_{2m}) ≡ (x′,x″) with ‖x′‖₂²=Σ_{i=1}^{m}x_i² and ‖x″‖₂²=Σ_{i=m+1}^{2m}x_i².  \nDefine the full 2m–ball and the “diagonal‐half–ball’’\n\n B_{R}^{(2m)} = {x ∈ ℝ^{2m} : ‖x‖₂ < R}, \n Ω_{m}(R)     = {x ∈ B_{R}^{(2m)} : ‖x′‖₂ < ‖x″‖₂ }.\n\n(A)   Prove that Vol_{2m}(Ω_{m}(R)) = ½ Vol_{2m}(B_{R}^{(2m)}) and give the value in closed Γ–function form.\n\n(B)   Let S_{R}^{(2m−1)} = ∂B_{R}^{(2m)} be the (2m−1)–sphere of radius R.  \n      Compute the (2m−1)-dimensional surface measure of  \n      Σ_{m}(R) = {x ∈ S_{R}^{(2m−1)} : ‖x′‖₂ < ‖x″‖₂ }.\n\n(C)   Choose a point uniformly at random in B_{1}^{(2m)}.  \n      For U = ‖x′‖₂² / ‖x‖₂² show that U ~ Beta(m/2, m/2), and use this fact to re-derive Part (A) by evaluating P(U < ½).\n\n(D)   In the eight–dimensional case m = 4 let  \n  Q(x) = (‖x′‖₂² − ‖x″‖₂²)².  \n      Compute the average value of Q over Ω₄(R) and the integral  \n      I(R) = ∫_{Ω₄(R)} Q(x) dx.",
      "solution": "Throughout we write n = 2m.  The n–ball volume and surface area are\n\n V_n(R) = π^{n/2}R^{n}/Γ(n/2+1), \n A_n(R) = dV_n/dR = nπ^{n/2}R^{n−1}/Γ(n/2+1).                                (1)\n\n(A)  Volume of Ω_m(R).\n\nBecause the inequality ‖x′‖₂ ≷ ‖x″‖₂ is reversed by the coordinate permutation  \n(x₁,…,x_m , x_{m+1},…,x_{2m}) ↦ (x_{m+1},…,x_{2m},x₁,…,x_m),  \nΩ_m(R) and its complement inside B_{R}^{(n)} are congruent.  Their common boundary  \n‖x′‖₂ = ‖x″‖₂ has Lebesgue measure zero in ℝ^{n}, so the two regions have equal volume:\n\n Vol_n(Ω_m(R)) = ½ Vol_n(B_{R}^{(n)}) = ½ V_n(R).                         (2)\n\nUsing (1) with n = 2m,\n\n Vol_{2m}(Ω_m(R)) = ½·π^{m}R^{2m}/Γ(m+1).                               (3)\n\n(B)  Surface measure of Σ_{m}(R).\n\nThe same symmetry shows that Σ_m(R) occupies one half of the sphere, hence\n\n Vol_{2m−1}(Σ_m(R)) = ½ A_n(R)  \n  = ½·nπ^{n/2}R^{n−1}/Γ(n/2+1)  \n  = m π^{m}R^{2m−1}/Γ(m+1).                                            (4)\n\n(C)  Distribution of U and probabilistic proof of (2).\n\nFor a uniformly random point in the unit n–ball, the radius r = ‖x‖₂ and the direction ξ = x/r are independent.  The squared coordinates of ξ form a Dirichlet(½,…,½) random vector (2m parameters, each ½).  Hence\n\n U = Σ_{i=1}^{m} ξ_i² ~ Beta(a,b) with a = m/2, b = m/2.               (5)\n\nThe Beta density is symmetric about ½, so P(U < ½) = ½.  Because r and U are independent and r < 1 by construction,\n\n P(‖x′‖₂ < ‖x″‖₂) = P(U < ½) = ½.                                     (6)\n\nMultiplying this probability by the total volume V_n(1) gives Vol(Ω_m(1)); scaling by R^{n} yields (3), recovering the geometric proof.\n\n(D)  Quartic moment in the eight–dimensional case (m=4, n=8).\n\nStep 1.  Radial moments inside the 8–ball.  \nFor uniform x ∈ B_{R}^{(8)}, the density of r=‖x‖₂ is\n\n f_r(r) = n r^{n−1}/R^{n}, 0≤r≤R, n=8.                               (7)\n\nHence  \n E[r^4] = ∫_{0}^{R} r^4 f_r(r) dr  \n    = 8∫_{0}^{R} r^{11}/R^{8} dr = 8R^4/(12) = 2R^4/3.               (8)\n\nStep 2.  Angular factor.  \nWith m=4, U ~ Beta(2,2).  For any Borel set A⊂[0,1] we have by symmetry\n\n E[(2U−1)^2 1_{A}] = E[(2U−1)^2 1_{1−A}].                              (9)\n\nThus, with A=[0,½),\n\n E[(2U−1)^2 | U<½] = E[(2U−1)^2].                                    (10)\n\nCompute the unconditional expectation:\n\n Var(U) = ab/[(a+b)^2(a+b+1)] = 4/(16·5)=1/20,  \n E[(2U−1)^2] = 4 Var(U) = 1/5.                                       (11)\n\nHence also E[(2U−1)^2 | U<½] = 1/5.\n\nStep 3.  Average of Q on Ω₄(R).  \nWrite S₁ = ‖x′‖₂² = r²U, S₂ = ‖x″‖₂² = r²(1−U).  Over Ω₄(R) (i.e. U<½),\n\n Q(x) = (S₁−S₂)² = r⁴(2U−1)².\n\nBy independence of r and U,\n\n E_{Ω}[Q] = E[r⁴] · E[(2U−1)² | U<½] = (2R⁴/3)·(1/5)=2R⁴/15.        (12)\n\nStep 4.  Integral of Q.  \nUsing Vol_{8}(Ω₄(R)) = ½ V₈(R) = ½·π⁴R⁸/Γ(5)=π⁴R⁸/48,\n\n I(R) = Vol(Ω₄(R))·E_{Ω}[Q] = (π⁴R⁸/48)·(2R⁴/15) = π⁴R^{12}/360.     (13)\n\nSo\n\n  E_{Ω₄(R)}[Q] = 2R⁴/15,  I(R) = π⁴R^{12}/360.                     (14)",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.382119",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension & more variables: the problem lives in 2m dimensions (≥6, concretely carried out in 8) instead of the original 4- or 6-dimensional setting.  \n2. Additional constraints: the region Ω_m(R) is carved out of the ball by the nonlinear inequality ‖x′‖₂ < ‖x″‖₂, producing a non-separable domain whose boundary is neither spherical nor flat.  \n3. Sophisticated structures: solving Parts (A)–(C) requires recognising Dirichlet and Beta distributions on the sphere, using symmetry arguments, and manipulating Γ-functions; Part (D) needs mixing radial moments with conditional Beta expectations to evaluate a quartic integral.  \n4. Deeper theory: probability on high-dimensional spheres, independence of radial and angular parts, and higher moments of Beta laws must all be deployed.  \n5. Multiple interacting concepts: Euclidean geometry, special functions, probabilistic representation, and higher-order moment calculations intertwine, making the variant substantially harder than simply differentiating or scaling the classical n-ball formulas."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}