{
  "index": "1940-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "10. A cylindrical hole of radius \\( r \\) is bored through a cylinder of radius \\( R \\) ( \\( r \\leq R \\) ) so that the axes intersect at right angles.\n(i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form\n\\[\nS=8 r^{2} \\int_{0}^{1} \\frac{1-v^{2}}{\\sqrt{\\left(1-v^{2}\\right)\\left(1-m^{2} v^{2}\\right)}} d v \\quad \\text { where } \\quad m=\\frac{r}{R}\n\\]\n(ii) If\n\\[\nK=\\int_{0}^{1} \\frac{d v}{\\sqrt{\\left(1-v^{2}\\right)\\left(1-m^{2} v^{2}\\right)}} \\text { and } E=\\int_{0}^{1} \\sqrt{\\frac{1-m^{2} v^{2}}{1-v^{2}}} d v\n\\]\nshow that\n\\[\nS=8\\left[R^{2} E-\\left(R^{2}-r^{2}\\right) K\\right]\n\\]",
  "solution": "Solution. Consider the sketch.\n\nLet the two cylindrical surfaces be \\( x^{2}+z^{2}=R^{2} \\) and \\( x^{2}+y^{2}=r^{2} \\), where \\( r \\leq R \\). The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is\n\\[\nz=\\sqrt{R^{2}-x^{2}}\n\\]\n\nThe required area is\n\\[\nS=8 \\iint \\sqrt{1+\\left(\\frac{\\partial z}{\\partial y}\\right)^{2}+\\left(\\frac{\\partial z}{\\partial x}\\right)^{2}} d y d x\n\\]\nwhere the double integral is over the region\n\\[\nx^{2}+y^{2} \\leq r^{2}, \\quad x \\geq 0, \\quad y \\geq 0 .\n\\]\n\nConverted to an iterated integral, this becomes\n\\[\n\\begin{aligned}\nS & =8 \\int_{0}^{r}\\left(\\int_{0}^{\\sqrt{r^{2}-x^{2}}} \\frac{R}{\\sqrt{R^{2}-x^{2}}} d y\\right) d x \\\\\n& =8 R \\int_{0}^{r} \\sqrt{\\frac{r^{2}-x^{2}}{R^{2}-x^{2}}} d x .\n\\end{aligned}\n\\]\n\nNow let \\( x / r=v \\) and \\( r / R=m \\) and simplify further to get\n\\[\nS=8 r^{2} \\int_{0}^{1} \\frac{1-v^{2}}{\\sqrt{\\left(1-v^{2}\\right)\\left(1-m^{2} v^{2}\\right)}} d v,\n\\]\nwhich completes part (i).\nTo obtain (ii), write \\( r^{2}\\left(1-v^{2}\\right)=R^{2}\\left(1-m^{2} v^{2}\\right)-\\left(R^{2}-r^{2}\\right) \\) and substitute to get\n\\[\n\\begin{aligned}\nS & =8 \\int_{0}^{1} \\frac{R^{2}\\left(1-m^{2} v^{2}\\right)}{\\sqrt{\\left(1-v^{2}\\right)\\left(1-m^{2} v^{2}\\right)}} d v-8 \\int_{0}^{1} \\frac{\\left(R^{2}-r^{2}\\right)}{\\sqrt{\\left(1-v^{2}\\right)\\left(1-m^{2} v^{2}\\right)}} d v \\\\\n& =8\\left[R^{2} E-\\left(R^{2}-r^{2}\\right) K\\right]\n\\end{aligned}\n\\]\n\nRemark. The integrals \\( K \\) and \\( E \\) are known as the complete elliptic integrals of the first and second kinds, respectively. Their values have been tabulated in terms of the parameter \\( m \\).",
  "vars": [
    "S",
    "x",
    "y",
    "z",
    "v"
  ],
  "params": [
    "r",
    "R",
    "m",
    "K",
    "E"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "arearesult",
        "x": "axisfirst",
        "y": "axissecond",
        "z": "axisthird",
        "v": "scaleratio",
        "r": "smallradius",
        "R": "largeradius",
        "m": "ratparameter",
        "K": "elliptickone",
        "E": "elliptictwo"
      },
      "question": "A cylindrical hole of radius \\( smallradius \\) is bored through a cylinder of radius \\( largeradius \\) ( \\( smallradius \\leq largeradius \\) ) so that the axes intersect at right angles.\n(i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form\n\\[\narearesult=8\\,smallradius^{2}\\int_{0}^{1}\\frac{1-scaleratio^{2}}{\\sqrt{\\left(1-scaleratio^{2}\\right)\\left(1-ratparameter^{2} scaleratio^{2}\\right)}}\\,d scaleratio\\quad\\text{ where }\\quad ratparameter=\\frac{smallradius}{largeradius}\n\\]\n(ii) If\n\\[\nelliptickone=\\int_{0}^{1}\\frac{d scaleratio}{\\sqrt{\\left(1-scaleratio^{2}\\right)\\left(1-ratparameter^{2} scaleratio^{2}\\right)}}\\quad\\text{and}\\quad elliptictwo=\\int_{0}^{1}\\sqrt{\\frac{1-ratparameter^{2} scaleratio^{2}}{1-scaleratio^{2}}}\\,d scaleratio\n\\]\nshow that\n\\[\narearesult=8\\left[largeradius^{2}\\,elliptictwo-\\left(largeradius^{2}-smallradius^{2}\\right)\\,elliptickone\\right]\n\\]",
      "solution": "Solution. Consider the sketch.\n\nLet the two cylindrical surfaces be \\( axisfirst^{2}+axisthird^{2}=largeradius^{2} \\) and \\( axisfirst^{2}+axissecond^{2}=smallradius^{2} \\), where \\( smallradius \\leq largeradius \\). The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is\n\\[\naxisthird=\\sqrt{largeradius^{2}-axisfirst^{2}}\n\\]\n\nThe required area is\n\\[\narearesult=8\\iint\\sqrt{1+\\left(\\frac{\\partial axisthird}{\\partial axissecond}\\right)^{2}+\\left(\\frac{\\partial axisthird}{\\partial axisfirst}\\right)^{2}}\\,d axissecond\\,d axisfirst\n\\]\nwhere the double integral is over the region\n\\[\naxisfirst^{2}+axissecond^{2}\\leq smallradius^{2},\\quad axisfirst\\geq0,\\quad axissecond\\geq0.\n\\]\n\nConverted to an iterated integral, this becomes\n\\[\n\\begin{aligned}\narearesult&=8\\int_{0}^{smallradius}\\left(\\int_{0}^{\\sqrt{smallradius^{2}-axisfirst^{2}}}\\frac{largeradius}{\\sqrt{largeradius^{2}-axisfirst^{2}}}\\,d axissecond\\right)d axisfirst\\\\\n&=8\\,largeradius\\int_{0}^{smallradius}\\sqrt{\\frac{smallradius^{2}-axisfirst^{2}}{largeradius^{2}-axisfirst^{2}}}\\,d axisfirst.\n\\end{aligned}\n\\]\n\nNow let \\( axisfirst/smallradius=scaleratio \\) and \\( smallradius/largeradius=ratparameter \\) and simplify further to get\n\\[\narearesult=8\\,smallradius^{2}\\int_{0}^{1}\\frac{1-scaleratio^{2}}{\\sqrt{(1-scaleratio^{2})(1-ratparameter^{2} scaleratio^{2})}}\\,d scaleratio,\n\\]\nwhich completes part (i).\n\nTo obtain (ii), write \\( smallradius^{2}(1-scaleratio^{2})=largeradius^{2}(1-ratparameter^{2} scaleratio^{2})-(largeradius^{2}-smallradius^{2}) \\) and substitute to get\n\\[\n\\begin{aligned}\narearesult&=8\\int_{0}^{1}\\frac{largeradius^{2}(1-ratparameter^{2} scaleratio^{2})}{\\sqrt{(1-scaleratio^{2})(1-ratparameter^{2} scaleratio^{2})}}\\,d scaleratio-8\\int_{0}^{1}\\frac{(largeradius^{2}-smallradius^{2})}{\\sqrt{(1-scaleratio^{2})(1-ratparameter^{2} scaleratio^{2})}}\\,d scaleratio\\\\\n&=8\\left[largeradius^{2}\\,elliptictwo-(largeradius^{2}-smallradius^{2})\\,elliptickone\\right].\n\\end{aligned}\n\\]\n\nRemark. The integrals \\( elliptickone \\) and \\( elliptictwo \\) are known as the complete elliptic integrals of the first and second kinds, respectively. Their values have been tabulated in terms of the parameter \\( ratparameter \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "panoramal",
        "x": "meadowway",
        "y": "harborline",
        "z": "glyphtrack",
        "v": "cobblestone",
        "r": "lanternarc",
        "R": "starlingdew",
        "m": "quartzbeam",
        "K": "willowcrest",
        "E": "emberthroat"
      },
      "question": "10. A cylindrical hole of radius \\( lanternarc \\) is bored through a cylinder of radius \\( starlingdew \\) ( \\( lanternarc \\leq starlingdew \\) ) so that the axes intersect at right angles.\n(i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form\n\\[\npanoramal=8 \\, lanternarc^{2} \\int_{0}^{1} \\frac{1-cobblestone^{2}}{\\sqrt{\\left(1-cobblestone^{2}\\right)\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}} d cobblestone \\quad \\text { where } \\quad quartzbeam=\\frac{lanternarc}{starlingdew}\n\\]\n(ii) If\n\\[\nwillowcrest=\\int_{0}^{1} \\frac{d cobblestone}{\\sqrt{\\left(1-cobblestone^{2}\\right)\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}} \\text { and } emberthroat=\\int_{0}^{1} \\sqrt{\\frac{1-quartzbeam^{2} \\, cobblestone^{2}}{1-cobblestone^{2}}} d cobblestone\n\\]\nshow that\n\\[\npanoramal=8\\left[starlingdew^{2} \\, emberthroat-\\left(starlingdew^{2}-lanternarc^{2}\\right) \\, willowcrest\\right]\n\\]",
      "solution": "Solution. Consider the sketch.\n\nLet the two cylindrical surfaces be \\( meadowway^{2}+glyphtrack^{2}=starlingdew^{2} \\) and \\( meadowway^{2}+harborline^{2}=lanternarc^{2} \\), where \\( lanternarc \\leq starlingdew \\). The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is\n\\[\nglyphtrack=\\sqrt{starlingdew^{2}-meadowway^{2}}\n\\]\n\nThe required area is\n\\[\npanoramal=8 \\iint \\sqrt{1+\\left(\\frac{\\partial glyphtrack}{\\partial harborline}\\right)^{2}+\\left(\\frac{\\partial glyphtrack}{\\partial meadowway}\\right)^{2}} d harborline \\, d meadowway\n\\]\nwhere the double integral is over the region\n\\[\nmeadowway^{2}+harborline^{2} \\leq lanternarc^{2}, \\quad meadowway \\geq 0, \\quad harborline \\geq 0 .\n\\]\n\nConverted to an iterated integral, this becomes\n\\[\n\\begin{aligned}\npanoramal & =8 \\int_{0}^{lanternarc}\\left(\\int_{0}^{\\sqrt{lanternarc^{2}-meadowway^{2}}} \\frac{starlingdew}{\\sqrt{starlingdew^{2}-meadowway^{2}}} d harborline\\right) d meadowway \\\\\n& =8 \\, starlingdew \\int_{0}^{lanternarc} \\sqrt{\\frac{lanternarc^{2}-meadowway^{2}}{starlingdew^{2}-meadowway^{2}}} d meadowway .\n\\end{aligned}\n\\]\n\nNow let \\( meadowway / lanternarc=cobblestone \\) and \\( lanternarc / starlingdew=quartzbeam \\) and simplify further to get\n\\[\npanoramal=8 \\, lanternarc^{2} \\int_{0}^{1} \\frac{1-cobblestone^{2}}{\\sqrt{\\left(1-cobblestone^{2}\\right)\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}} d cobblestone,\n\\]\nwhich completes part (i).\nTo obtain (ii), write \\( lanternarc^{2}\\left(1-cobblestone^{2}\\right)=starlingdew^{2}\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)-\\left(starlingdew^{2}-lanternarc^{2}\\right) \\) and substitute to get\n\\[\n\\begin{aligned}\npanoramal & =8 \\int_{0}^{1} \\frac{starlingdew^{2}\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}{\\sqrt{\\left(1-cobblestone^{2}\\right)\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}} d cobblestone-8 \\int_{0}^{1} \\frac{\\left(starlingdew^{2}-lanternarc^{2}\\right)}{\\sqrt{\\left(1-cobblestone^{2}\\right)\\left(1-quartzbeam^{2} \\, cobblestone^{2}\\right)}} d cobblestone \\\\\n& =8\\left[starlingdew^{2} \\, emberthroat-\\left(starlingdew^{2}-lanternarc^{2}\\right) \\, willowcrest\\right]\n\\end{aligned}\n\\]\n\nRemark. The integrals \\( willowcrest \\) and \\( emberthroat \\) are known as the complete elliptic integrals of the first and second kinds, respectively. Their values have been tabulated in terms of the parameter \\( quartzbeam \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "perimeter",
        "x": "verticalcoord",
        "y": "horizontalcoord",
        "z": "deepcoord",
        "v": "constantval",
        "r": "diameter",
        "R": "circumference",
        "m": "productval",
        "K": "incomplete",
        "E": "hyperbolic"
      },
      "question": "10. A cylindrical hole of radius \\( diameter \\) is bored through a cylinder of radius \\( circumference \\) ( \\( diameter \\leq circumference \\) ) so that the axes intersect at right angles.\n(i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form\n\\[\nperimeter=8 \\, diameter^{2} \\int_{0}^{1} \\frac{1-constantval^{2}}{\\sqrt{\\left(1-constantval^{2}\\right)\\left(1-productval^{2} \\, constantval^{2}\\right)}} d constantval \\quad \\text { where } \\quad productval=\\frac{diameter}{circumference}\n\\]\n(ii) If\n\\[\nincomplete=\\int_{0}^{1} \\frac{d constantval}{\\sqrt{\\left(1-constantval^{2}\\right)\\left(1-productval^{2} \\, constantval^{2}\\right)}} \\text { and } hyperbolic=\\int_{0}^{1} \\sqrt{\\frac{1-productval^{2} \\, constantval^{2}}{1-constantval^{2}}} d constantval\n\\]\nshow that\n\\[\nperimeter=8\\left[circumference^{2} \\, hyperbolic-\\left(circumference^{2}-diameter^{2}\\right) \\, incomplete\\right]\n\\]",
      "solution": "Solution. Consider the sketch.\n\nLet the two cylindrical surfaces be \\( verticalcoord^{2}+deepcoord^{2}=circumference^{2} \\) and \\( verticalcoord^{2}+horizontalcoord^{2}=diameter^{2} \\), where \\( diameter \\leq circumference \\). The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is\n\\[\ndeepcoord=\\sqrt{circumference^{2}-verticalcoord^{2}}\n\\]\n\nThe required area is\n\\[\nperimeter=8 \\iint \\sqrt{1+\\left(\\frac{\\partial deepcoord}{\\partial horizontalcoord}\\right)^{2}+\\left(\\frac{\\partial deepcoord}{\\partial verticalcoord}\\right)^{2}} d horizontalcoord d verticalcoord\n\\]\nwhere the double integral is over the region\n\\[\nverticalcoord^{2}+horizontalcoord^{2} \\leq diameter^{2}, \\quad verticalcoord \\geq 0, \\quad horizontalcoord \\geq 0 .\n\\]\n\nConverted to an iterated integral, this becomes\n\\[\n\\begin{aligned}\nperimeter & =8 \\int_{0}^{diameter}\\left(\\int_{0}^{\\sqrt{diameter^{2}-verticalcoord^{2}}} \\frac{circumference}{\\sqrt{circumference^{2}-verticalcoord^{2}}} d horizontalcoord\\right) d verticalcoord \\\\\n& =8 \\, circumference \\int_{0}^{diameter} \\sqrt{\\frac{diameter^{2}-verticalcoord^{2}}{circumference^{2}-verticalcoord^{2}}} d verticalcoord .\n\\end{aligned}\n\\]\n\nNow let \\( verticalcoord / diameter=constantval \\) and \\( diameter / circumference=productval \\) and simplify further to get\n\\[\nperimeter=8 \\, diameter^{2} \\int_{0}^{1} \\frac{1-constantval^{2}}{\\sqrt{\\left(1-constantval^{2}\\right)\\left(1-productval^{2} \\, constantval^{2}\\right)}} d constantval,\n\\]\nwhich completes part (i).\nTo obtain (ii), write \\( diameter^{2}\\left(1-constantval^{2}\\right)=circumference^{2}\\left(1-productval^{2} \\, constantval^{2}\\right)-\\left(circumference^{2}-diameter^{2}\\right) \\) and substitute to get\n\\[\n\\begin{aligned}\nperimeter & =8 \\int_{0}^{1} \\frac{circumference^{2}\\left(1-productval^{2} \\, constantval^{2}\\right)}{\\sqrt{\\left(1-constantval^{2}\\right)\\left(1-productval^{2} \\, constantval^{2}\\right)}} d constantval-8 \\int_{0}^{1} \\frac{\\left(circumference^{2}-diameter^{2}\\right)}{\\sqrt{\\left(1-constantval^{2}\\right)\\left(1-productval^{2} \\, constantval^{2}\\right)}} d constantval \\\\\n& =8\\left[circumference^{2} \\, hyperbolic-\\left(circumference^{2}-diameter^{2}\\right) \\, incomplete\\right]\n\\end{aligned}\n\\]\n\nRemark. The integrals \\( incomplete \\) and \\( hyperbolic \\) are known as the complete elliptic integrals of the first and second kinds, respectively. Their values have been tabulated in terms of the parameter \\( productval \\)."
    },
    "garbled_string": {
      "map": {
        "S": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "pbvndtqe",
        "z": "lmfqcrua",
        "v": "dsxjplzi",
        "r": "nxpdcove",
        "R": "uwbrhaes",
        "m": "kztqpluw",
        "K": "oblghrse",
        "E": "fsqvlimk"
      },
      "question": "10. A cylindrical hole of radius \\( nxpdcove \\) is bored through a cylinder of radius \\( uwbrhaes \\) ( \\( nxpdcove \\leq uwbrhaes \\) ) so that the axes intersect at right angles.\n(i) Show that the area of the larger cylinder which is inside the smaller can be expressed in the form\n\\[\nqzxwvtnp=8 nxpdcove^{2} \\int_{0}^{1} \\frac{1-dsxjplzi^{2}}{\\sqrt{\\left(1-dsxjplzi^{2}\\right)\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}} d dsxjplzi \\quad \\text { where } \\quad kztqpluw=\\frac{nxpdcove}{uwbrhaes}\n\\]\n(ii) If\n\\[\noblghrse=\\int_{0}^{1} \\frac{d dsxjplzi}{\\sqrt{\\left(1-dsxjplzi^{2}\\right)\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}} \\text { and } fsqvlimk=\\int_{0}^{1} \\sqrt{\\frac{1-kztqpluw^{2} dsxjplzi^{2}}{1-dsxjplzi^{2}}} d dsxjplzi\n\\]\nshow that\n\\[\nqzxwvtnp=8\\left[uwbrhaes^{2} fsqvlimk-\\left(uwbrhaes^{2}-nxpdcove^{2}\\right) oblghrse\\right]\n\\]",
      "solution": "Solution. Consider the sketch.\n\nLet the two cylindrical surfaces be \\( hjgrksla^{2}+lmfqcrua^{2}=uwbrhaes^{2} \\) and \\( hjgrksla^{2}+pbvndtqe^{2}=nxpdcove^{2} \\), where \\( nxpdcove \\leq uwbrhaes \\). The shaded area shown in the diagram is the part of the required area that lies in one octant. The equation of this surface is\n\\[\nlmfqcrua=\\sqrt{uwbrhaes^{2}-hjgrksla^{2}}\n\\]\n\nThe required area is\n\\[\nqzxwvtnp=8 \\iint \\sqrt{1+\\left(\\frac{\\partial lmfqcrua}{\\partial pbvndtqe}\\right)^{2}+\\left(\\frac{\\partial lmfqcrua}{\\partial hjgrksla}\\right)^{2}} d pbvndtqe d hjgrksla\n\\]\nwhere the double integral is over the region\n\\[\nhjgrksla^{2}+pbvndtqe^{2} \\leq nxpdcove^{2}, \\quad hjgrksla \\geq 0, \\quad pbvndtqe \\geq 0 .\n\\]\n\nConverted to an iterated integral, this becomes\n\\[\n\\begin{aligned}\nqzxwvtnp & =8 \\int_{0}^{nxpdcove}\\left(\\int_{0}^{\\sqrt{nxpdcove^{2}-hjgrksla^{2}}} \\frac{uwbrhaes}{\\sqrt{uwbrhaes^{2}-hjgrksla^{2}}} d pbvndtqe\\right) d hjgrksla \\\\\n& =8 uwbrhaes \\int_{0}^{nxpdcove} \\sqrt{\\frac{nxpdcove^{2}-hjgrksla^{2}}{uwbrhaes^{2}-hjgrksla^{2}}} d hjgrksla .\n\\end{aligned}\n\\]\n\nNow let \\( hjgrksla / nxpdcove=dsxjplzi \\) and \\( nxpdcove / uwbrhaes=kztqpluw \\) and simplify further to get\n\\[\nqzxwvtnp=8 nxpdcove^{2} \\int_{0}^{1} \\frac{1-dsxjplzi^{2}}{\\sqrt{\\left(1-dsxjplzi^{2}\\right)\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}} d dsxjplzi,\n\\]\nwhich completes part (i).\nTo obtain (ii), write \\( nxpdcove^{2}\\left(1-dsxjplzi^{2}\\right)=uwbrhaes^{2}\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)-\\left(uwbrhaes^{2}-nxpdcove^{2}\\right) \\) and substitute to get\n\\[\n\\begin{aligned}\nqzxwvtnp & =8 \\int_{0}^{1} \\frac{uwbrhaes^{2}\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}{\\sqrt{\\left(1-dsxjplzi^{2}\\right)\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}} d dsxjplzi-8 \\int_{0}^{1} \\frac{\\left(uwbrhaes^{2}-nxpdcove^{2}\\right)}{\\sqrt{\\left(1-dsxjplzi^{2}\\right)\\left(1-kztqpluw^{2} dsxjplzi^{2}\\right)}} d dsxjplzi \\\\\n& =8\\left[uwbrhaes^{2} fsqvlimk-\\left(uwbrhaes^{2}-nxpdcove^{2}\\right) oblghrse\\right]\n\\end{aligned}\n\\]\n\nRemark. The integrals \\( oblghrse \\) and \\( fsqvlimk \\) are known as the complete elliptic integrals of the first and second kinds, respectively. Their values have been tabulated in terms of the parameter \\( kztqpluw \\)."
    },
    "kernel_variant": {
      "question": "Let 0<r≤R and let the two axes form an acute angle 0<φ<π/2.\n\nLarge (host) cylinder  \n C_R : y²+z² = R²   (axis Ox)\n\nObliquely–drilled cylinder  \n C_{r,φ} : y² + (z cos φ − x sin φ)² = r²,\n\nwhose axis  \n ℓ : t ↦ (t cos φ, 0, t sin φ)\n\nlies in the xz–plane and meets the x–axis at the angle φ.\n\nDenote by S(φ) the total area of that part of the lateral surface of C_R which is contained in the solid C_{r,φ}.  (No half–space restriction – both z≥0 and z≤0 are included.)\n\nIntroduce the dimensionless ratio n=r/R∈(0,1].\n\n(a)  Prove the representation  \n  S(φ)= 8 r² csc φ ∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.    (1)\n\n(b)  Using the complete elliptic integrals  \n  K(n)=∫₀¹ dw/√[(1−w²)(1−n²w²)],   \n  E(n)=∫₀¹ √[(1−n²w²)/(1−w²)] dw,  \n\n show the closed form  \n  S(φ)= 8 csc φ [ R²E(n) − (R²−r²)K(n) ].           (2)\n\n(c)  Verify the limiting values  \n  lim_{φ→0⁺} S(φ)=+∞ ,  lim_{φ→π/2} S(φ)=8[ R²E(n) − (R²−r²)K(n) ] ,\n\nand explain geometrically why the second limit agrees with the classical right-angle drilling result.",
      "solution": "Throughout put  \n n=r/R  (0<n≤1),  σ=sin φ>0, τ=cos φ>0.\n\nStep 1 – Parameterising the host cylinder.  \nWith the standard angular parameter θ∈[0,2π)\n\n X(x,θ)=(x, R cos θ, R sin θ)                                    (1)\n\nevery fixed θ traces a generator parallel to the x–axis.  \nBecause  \n |∂X/∂x × ∂X/∂θ| = R,  \nthe surface element is\n\n dS = R dx dθ.                                                  (2)\n\nStep 2 – The x–interval cut out by the bore.  \nInsert (1) into the inequality describing the interior of C_{r,φ}:\n\n R² cos²θ + (R sin θ τ − x σ)² ≤ r².                           (3)\n\nFor a fixed θ put\n\n Δ(θ)=√(r² − R² cos²θ),  defined iff |cos θ|≤n.              (4)\n\nWhen |cos θ|>n the inequality has no solution; otherwise (3) becomes\n\n | x − (R τ/σ) sin θ | ≤ Δ(θ)/σ.                                (5)\n\nHence, for every θ with |cos θ|≤n the admissible x–values form the interval\n\n I_θ = [ (R τ sin θ − Δ)/σ , (R τ sin θ + Δ)/σ ],              (6)\n\nwhose length is\n\n L(θ)= 2Δ(θ)/σ = 2R σ^{-1}√(n² − cos²θ).                       (7)\n\nStep 3 – First area integral (full cylinder, not only z≥0).  \nThe set {|cos θ|≤n} consists of the two symmetric arcs  \n [θ₀, π−θ₀] ∪ [π+θ₀, 2π−θ₀], θ₀ := arccos n ∈ (0,π/2].  \nUsing (2) and (7),\n\n S(φ)= R∫_{|cos θ|≤n} L(θ) dθ\n     = 2R∫_{θ₀}^{π−θ₀} L(θ) dθ  (symmetry θ↦θ+π)\n     = 4R² σ^{-1}∫_{θ₀}^{π−θ₀} √(n² − cos²θ) dθ.                (8)\n\nStep 4 – Reduction to a standard integral.  \nPut cos θ = n w (so w∈[−1,1]).  Because\n\n dθ = −n dw /√(1−n²w²) ,\n\nand the integrand is an even function of w, (8) becomes\n\n S(φ)= 8R²n² σ^{-1} ∫₀¹ √(1−w²)/√(1−n²w²) dw.                 (9)\n\nWrite √(1−w²) = (1−w²)/√(1−w²) to get\n\n S(φ)= 8R²n² σ^{-1} ∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.        (10)\n\nSince r²=n²R² and csc φ=σ^{-1}, (10) is exactly formula (1).\n\nStep 5 – Expressing the integral through K and E.  \nDefine  \n\n I₁(n)=∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.                     (11)\n\nWrite 1−w² = 1−n²w² + (n²−1)w² and split I₁:\n\n I₁(n)= ∫₀¹ √{(1−n²w²)/(1−w²)} dw\n        −(1−n²)∫₀¹ w²/√[(1−w²)(1−n²w²)] dw.                  (12)\n\nThe first integral is E(n).  \nThe classical identity\n\n K(n)−E(n)= n²∫₀¹ w²/√[(1−w²)(1−n²w²)] dw                  (13)\n\nimplies  \n\n ∫ w²/√[...] dw = [ K(n)−E(n) ] / n².\n\nSubstituting in (12) gives\n\n I₁(n)= [ E(n) − (1−n²)K(n) ] / n².                          (14)\n\nStep 6 – Closed form for S(φ).  \nInsert (14) in (1):\n\n S(φ)= 8r² csc φ · I₁(n)\n      = 8n²R² csc φ · [E(n) − (1−n²)K(n)]/n²\n      = 8 csc φ [ R²E(n) − (R²−r²)K(n) ],                     (15)\n\nwhich is formula (2).\n\nStep 7 – The requested limits.  \n\n(i)  As φ→0⁺, sin φ≈φ, hence S(φ)∼(8/φ)[R²E(n) − (R²−r²)K(n)] → +∞.  \n  Geometrically, when the two axes nearly coincide the small bore runs\n  almost tangentially to the host surface, producing an arbitrarily\n  long (and therefore unbounded) strip of intersection.\n\n(ii)  For φ→π/2 we have sin φ→1, csc φ→1, and (15) yields\n\n  lim_{φ→π/2} S(φ) = 8[ R²E(n) − (R²−r²)K(n) ].              (16)\n\n  This matches the classical right-angle drilling problem: when φ=π/2\n  the two axes are perpendicular, the solid C_{r,φ} possesses the\n  mirror symmetry z↦−z, and the complete intersection with the entire\n  lateral surface of C_R is exactly twice the area that appears when\n  only the half-space z≥0 is considered in elementary treatments.\n\nThus parts (a)–(c) are proved.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.374395",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional geometric parameter: the cylinders now intersect at an arbitrary angle φ (0<φ<π/2); the orthogonal case is only a limit. All formulas therefore depend simultaneously on n=r/R and on φ, making the geometry genuinely 2-parametric.\n\n2. Coordinate‐free setup: solving the problem required an orthogonal change of variables (u,v,y) adapted to the skew axis ℓ, not merely the standard Cartesian projection used in the original problem.\n\n3. Constraint by a half-space: retaining only H₊ complicates the bounds, forcing a careful treatment of the allowed θ-interval and doubling issues.\n\n4. Integral manipulation: the raw integral that appears is ∫√(1−w²)/√(1−n²w²), not one of the tabulated elliptic forms. Turning it into a combination of E(n) and K(n) needs the non-trivial identity K−E = n²∫w²/…, a step absent from the original exercise.\n\n5. Asymptotic analysis: part (c) asks for limiting behaviour as φ approaches 0 or π/2, introducing an extra layer of calculus and geometric interpretation.\n\nAltogether the problem demands mastery of geometry in oblique coordinates, advanced integral reduction techniques, classical relations between elliptic integrals, and asymptotic reasoning—substantially more technical and conceptual effort than the right-angle, single-parameter case."
      }
    },
    "original_kernel_variant": {
      "question": "Let 0<r≤R and let the two axes form an acute angle 0<φ<π/2.\n\nLarge (host) cylinder  \n C_R : y²+z² = R²   (axis Ox)\n\nObliquely–drilled cylinder  \n C_{r,φ} : y² + (z cos φ − x sin φ)² = r²,\n\nwhose axis  \n ℓ : t ↦ (t cos φ, 0, t sin φ)\n\nlies in the xz–plane and meets the x–axis at the angle φ.\n\nDenote by S(φ) the total area of that part of the lateral surface of C_R which is contained in the solid C_{r,φ}.  (No half–space restriction – both z≥0 and z≤0 are included.)\n\nIntroduce the dimensionless ratio n=r/R∈(0,1].\n\n(a)  Prove the representation  \n  S(φ)= 8 r² csc φ ∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.    (1)\n\n(b)  Using the complete elliptic integrals  \n  K(n)=∫₀¹ dw/√[(1−w²)(1−n²w²)],   \n  E(n)=∫₀¹ √[(1−n²w²)/(1−w²)] dw,  \n\n show the closed form  \n  S(φ)= 8 csc φ [ R²E(n) − (R²−r²)K(n) ].           (2)\n\n(c)  Verify the limiting values  \n  lim_{φ→0⁺} S(φ)=+∞ ,  lim_{φ→π/2} S(φ)=8[ R²E(n) − (R²−r²)K(n) ] ,\n\nand explain geometrically why the second limit agrees with the classical right-angle drilling result.",
      "solution": "Throughout put  \n n=r/R  (0<n≤1),  σ=sin φ>0, τ=cos φ>0.\n\nStep 1 – Parameterising the host cylinder.  \nWith the standard angular parameter θ∈[0,2π)\n\n X(x,θ)=(x, R cos θ, R sin θ)                                    (1)\n\nevery fixed θ traces a generator parallel to the x–axis.  \nBecause  \n |∂X/∂x × ∂X/∂θ| = R,  \nthe surface element is\n\n dS = R dx dθ.                                                  (2)\n\nStep 2 – The x–interval cut out by the bore.  \nInsert (1) into the inequality describing the interior of C_{r,φ}:\n\n R² cos²θ + (R sin θ τ − x σ)² ≤ r².                           (3)\n\nFor a fixed θ put\n\n Δ(θ)=√(r² − R² cos²θ),  defined iff |cos θ|≤n.              (4)\n\nWhen |cos θ|>n the inequality has no solution; otherwise (3) becomes\n\n | x − (R τ/σ) sin θ | ≤ Δ(θ)/σ.                                (5)\n\nHence, for every θ with |cos θ|≤n the admissible x–values form the interval\n\n I_θ = [ (R τ sin θ − Δ)/σ , (R τ sin θ + Δ)/σ ],              (6)\n\nwhose length is\n\n L(θ)= 2Δ(θ)/σ = 2R σ^{-1}√(n² − cos²θ).                       (7)\n\nStep 3 – First area integral (full cylinder, not only z≥0).  \nThe set {|cos θ|≤n} consists of the two symmetric arcs  \n [θ₀, π−θ₀] ∪ [π+θ₀, 2π−θ₀], θ₀ := arccos n ∈ (0,π/2].  \nUsing (2) and (7),\n\n S(φ)= R∫_{|cos θ|≤n} L(θ) dθ\n     = 2R∫_{θ₀}^{π−θ₀} L(θ) dθ  (symmetry θ↦θ+π)\n     = 4R² σ^{-1}∫_{θ₀}^{π−θ₀} √(n² − cos²θ) dθ.                (8)\n\nStep 4 – Reduction to a standard integral.  \nPut cos θ = n w (so w∈[−1,1]).  Because\n\n dθ = −n dw /√(1−n²w²) ,\n\nand the integrand is an even function of w, (8) becomes\n\n S(φ)= 8R²n² σ^{-1} ∫₀¹ √(1−w²)/√(1−n²w²) dw.                 (9)\n\nWrite √(1−w²) = (1−w²)/√(1−w²) to get\n\n S(φ)= 8R²n² σ^{-1} ∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.        (10)\n\nSince r²=n²R² and csc φ=σ^{-1}, (10) is exactly formula (1).\n\nStep 5 – Expressing the integral through K and E.  \nDefine  \n\n I₁(n)=∫₀¹ (1−w²)/√[(1−w²)(1−n²w²)] dw.                     (11)\n\nWrite 1−w² = 1−n²w² + (n²−1)w² and split I₁:\n\n I₁(n)= ∫₀¹ √{(1−n²w²)/(1−w²)} dw\n        −(1−n²)∫₀¹ w²/√[(1−w²)(1−n²w²)] dw.                  (12)\n\nThe first integral is E(n).  \nThe classical identity\n\n K(n)−E(n)= n²∫₀¹ w²/√[(1−w²)(1−n²w²)] dw                  (13)\n\nimplies  \n\n ∫ w²/√[...] dw = [ K(n)−E(n) ] / n².\n\nSubstituting in (12) gives\n\n I₁(n)= [ E(n) − (1−n²)K(n) ] / n².                          (14)\n\nStep 6 – Closed form for S(φ).  \nInsert (14) in (1):\n\n S(φ)= 8r² csc φ · I₁(n)\n      = 8n²R² csc φ · [E(n) − (1−n²)K(n)]/n²\n      = 8 csc φ [ R²E(n) − (R²−r²)K(n) ],                     (15)\n\nwhich is formula (2).\n\nStep 7 – The requested limits.  \n\n(i)  As φ→0⁺, sin φ≈φ, hence S(φ)∼(8/φ)[R²E(n) − (R²−r²)K(n)] → +∞.  \n  Geometrically, when the two axes nearly coincide the small bore runs\n  almost tangentially to the host surface, producing an arbitrarily\n  long (and therefore unbounded) strip of intersection.\n\n(ii)  For φ→π/2 we have sin φ→1, csc φ→1, and (15) yields\n\n  lim_{φ→π/2} S(φ) = 8[ R²E(n) − (R²−r²)K(n) ].              (16)\n\n  This matches the classical right-angle drilling problem: when φ=π/2\n  the two axes are perpendicular, the solid C_{r,φ} possesses the\n  mirror symmetry z↦−z, and the complete intersection with the entire\n  lateral surface of C_R is exactly twice the area that appears when\n  only the half-space z≥0 is considered in elementary treatments.\n\nThus parts (a)–(c) are proved.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.323329",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional geometric parameter: the cylinders now intersect at an arbitrary angle φ (0<φ<π/2); the orthogonal case is only a limit. All formulas therefore depend simultaneously on n=r/R and on φ, making the geometry genuinely 2-parametric.\n\n2. Coordinate‐free setup: solving the problem required an orthogonal change of variables (u,v,y) adapted to the skew axis ℓ, not merely the standard Cartesian projection used in the original problem.\n\n3. Constraint by a half-space: retaining only H₊ complicates the bounds, forcing a careful treatment of the allowed θ-interval and doubling issues.\n\n4. Integral manipulation: the raw integral that appears is ∫√(1−w²)/√(1−n²w²), not one of the tabulated elliptic forms. Turning it into a combination of E(n) and K(n) needs the non-trivial identity K−E = n²∫w²/…, a step absent from the original exercise.\n\n5. Asymptotic analysis: part (c) asks for limiting behaviour as φ approaches 0 or π/2, introducing an extra layer of calculus and geometric interpretation.\n\nAltogether the problem demands mastery of geometry in oblique coordinates, advanced integral reduction techniques, classical relations between elliptic integrals, and asymptotic reasoning—substantially more technical and conceptual effort than the right-angle, single-parameter case."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}