{"id":233356,"date":"2025-06-13T05:27:20","date_gmt":"2025-06-13T05:27:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233356"},"modified":"2025-06-13T05:27:22","modified_gmt":"2025-06-13T05:27:22","slug":"find-the-volume-of-the-ice-cream-cone-d-cut-from-the-solid-sphere-p-1-by-the-cone-10-pts","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/find-the-volume-of-the-ice-cream-cone-d-cut-from-the-solid-sphere-p-1-by-the-cone-10-pts\/","title":{"rendered":"Find the volume of the “ice cream cone” D cut from the solid sphere p < 1 by the cone (10 pts)"},"content":{"rendered":"\n

Find the volume of the “ice cream cone” D cut from the solid sphere p < 1 by the cone (10 pts). 2) Evaluate the integral (10 pts) \u00e2\u02c6\u00ab\u00e2\u02c6\u00ab V(x + y)(y – 2x)^2 dy dx.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Problem 1: Find the volume of the \u201cice cream cone\u201d DDD cut from the solid sphere \u03c1<1\\rho < 1\u03c1<1 by the cone<\/strong><\/h3>\n\n\n\n

We are asked to find the volume of the region inside the sphere<\/strong> \u03c1<1\\rho < 1\u03c1<1 and above the cone<\/strong>. This region looks like an ice cream cone<\/strong>, and the best approach is to use spherical coordinates<\/strong>.<\/p>\n\n\n\n

Given:<\/strong><\/h4>\n\n\n\n
    \n
  • Sphere: \u03c1<1\\rho < 1\u03c1<1<\/li>\n\n\n\n
  • Cone: \u03d5=\u03c04\\phi = \\frac{\\pi}{4}\u03d5=4\u03c0\u200b<\/li>\n<\/ul>\n\n\n\n

    In spherical coordinates:<\/p>\n\n\n\n

      \n
    • \u03c1\\rho\u03c1 is the radial distance<\/li>\n\n\n\n
    • \u03d5\\phi\u03d5 is the angle from the positive z-axis (polar angle)<\/li>\n\n\n\n
    • \u03b8\\theta\u03b8 is the angle in the xy-plane from the positive x-axis (azimuthal angle)<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 1: Set Up the Limits<\/strong><\/h3>\n\n\n\n

      The region lies inside the sphere<\/strong> \u03c1<1\\rho < 1\u03c1<1, but only where \u03d5\u2208[0,\u03c04]\\phi \\in [0, \\frac{\\pi}{4}]\u03d5\u2208[0,4\u03c0\u200b], i.e., above the cone.<\/p>\n\n\n\n

      So, the limits of integration are:<\/p>\n\n\n\n

        \n
      • 0\u2264\u03b8\u22642\u03c00 \\leq \\theta \\leq 2\\pi0\u2264\u03b8\u22642\u03c0<\/li>\n\n\n\n
      • 0\u2264\u03d5\u2264\u03c040 \\leq \\phi \\leq \\frac{\\pi}{4}0\u2264\u03d5\u22644\u03c0\u200b<\/li>\n\n\n\n
      • 0\u2264\u03c1\u226410 \\leq \\rho \\leq 10\u2264\u03c1\u22641<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Step 2: Volume Element in Spherical Coordinates<\/strong><\/h3>\n\n\n\n

        The volume element in spherical coordinates is: dV=\u03c12sin\u2061\u03d5\u2009d\u03c1\u2009d\u03d5\u2009d\u03b8dV = \\rho^2 \\sin\\phi \\, d\\rho \\, d\\phi \\, d\\thetadV=\u03c12sin\u03d5d\u03c1d\u03d5d\u03b8<\/p>\n\n\n\n

        \n\n\n\n

        Step 3: Set Up the Triple Integral<\/strong><\/h3>\n\n\n\n

        V=\u222b02\u03c0\u222b0\u03c04\u222b01\u03c12sin\u2061\u03d5\u2009d\u03c1\u2009d\u03d5\u2009d\u03b8V = \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{4}} \\int_0^1 \\rho^2 \\sin\\phi \\, d\\rho \\, d\\phi \\, d\\thetaV=\u222b02\u03c0\u200b\u222b04\u03c0\u200b\u200b\u222b01\u200b\u03c12sin\u03d5d\u03c1d\u03d5d\u03b8<\/p>\n\n\n\n

        \n\n\n\n

        Step 4: Evaluate the Integral<\/strong><\/h3>\n\n\n\n

        First integrate with respect to \u03c1\\rho\u03c1: \u222b01\u03c12\u2009d\u03c1=[\u03c133]01=13\\int_0^1 \\rho^2 \\, d\\rho = \\left[\\frac{\\rho^3}{3}\\right]_0^1 = \\frac{1}{3}\u222b01\u200b\u03c12d\u03c1=[3\u03c13\u200b]01\u200b=31\u200b<\/p>\n\n\n\n

        Now the \u03d5\\phi\u03d5-integral: \u222b0\u03c04sin\u2061\u03d5\u2009d\u03d5=[\u2212cos\u2061\u03d5]0\u03c04=\u2212cos\u2061(\u03c04)+cos\u2061(0)=\u221222+1\\int_0^{\\frac{\\pi}{4}} \\sin\\phi \\, d\\phi = \\left[ -\\cos\\phi \\right]_0^{\\frac{\\pi}{4}} = -\\cos\\left(\\frac{\\pi}{4}\\right) + \\cos(0) = -\\frac{\\sqrt{2}}{2} + 1\u222b04\u03c0\u200b\u200bsin\u03d5d\u03d5=[\u2212cos\u03d5]04\u03c0\u200b\u200b=\u2212cos(4\u03c0\u200b)+cos(0)=\u221222\u200b\u200b+1<\/p>\n\n\n\n

        Now the \u03b8\\theta\u03b8-integral: \u222b02\u03c0d\u03b8=2\u03c0\\int_0^{2\\pi} d\\theta = 2\\pi\u222b02\u03c0\u200bd\u03b8=2\u03c0<\/p>\n\n\n\n

        Combine all parts: V=13(1\u221222)\u22c52\u03c0=2\u03c03(1\u221222)V = \\frac{1}{3}(1 – \\frac{\\sqrt{2}}{2}) \\cdot 2\\pi = \\frac{2\\pi}{3} \\left(1 – \\frac{\\sqrt{2}}{2} \\right)V=31\u200b(1\u221222\u200b\u200b)\u22c52\u03c0=32\u03c0\u200b(1\u221222\u200b\u200b)<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Final Answer for Problem 1:<\/strong><\/h3>\n\n\n\n

        V=2\u03c03(1\u221222)\\boxed{V = \\frac{2\\pi}{3} \\left(1 – \\frac{\\sqrt{2}}{2} \\right)}V=32\u03c0\u200b(1\u221222\u200b\u200b)\u200b<\/p>\n\n\n\n

        \n\n\n\n

        Problem 2: Evaluate the integral<\/strong><\/h3>\n\n\n\n

        \u222cV(x+y)(y\u22122x)2\u2009dy\u2009dx\\iint_V (x + y)(y – 2x)^2 \\, dy \\, dx\u222cV\u200b(x+y)(y\u22122x)2dydx<\/p>\n\n\n\n

        We are not given the region of integration, so we assume<\/strong> it is over a simple rectangular region, say: 0\u2264x\u22641,0\u2264y\u226410 \\le x \\le 1, \\quad 0 \\le y \\le 10\u2264x\u22641,0\u2264y\u22641<\/p>\n\n\n\n

        \n\n\n\n

        Step 1: Expand the Integrand<\/strong><\/h3>\n\n\n\n

        Let\u2019s expand (x+y)(y\u22122x)2(x + y)(y – 2x)^2(x+y)(y\u22122x)2<\/p>\n\n\n\n

        First expand (y\u22122x)2(y – 2x)^2(y\u22122x)2: (y\u22122x)2=y2\u22124xy+4×2(y – 2x)^2 = y^2 – 4xy + 4x^2(y\u22122x)2=y2\u22124xy+4×2<\/p>\n\n\n\n

        Now multiply: (x+y)(y2\u22124xy+4×2)=x(y2\u22124xy+4×2)+y(y2\u22124xy+4×2)(x + y)(y^2 – 4xy + 4x^2) = x(y^2 – 4xy + 4x^2) + y(y^2 – 4xy + 4x^2)(x+y)(y2\u22124xy+4×2)=x(y2\u22124xy+4×2)+y(y2\u22124xy+4×2)<\/p>\n\n\n\n

        Compute each term:<\/p>\n\n\n\n

          \n
        • x(y2\u22124xy+4×2)=xy2\u22124x2y+4x3x(y^2 – 4xy + 4x^2) = xy^2 – 4x^2y + 4x^3x(y2\u22124xy+4×2)=xy2\u22124x2y+4×3<\/li>\n\n\n\n
        • y(y2\u22124xy+4×2)=y3\u22124xy2+4x2yy(y^2 – 4xy + 4x^2) = y^3 – 4xy^2 + 4x^2yy(y2\u22124xy+4×2)=y3\u22124xy2+4x2y<\/li>\n<\/ul>\n\n\n\n

          Now add: xy2\u22124x2y+4×3+y3\u22124xy2+4x2yxy^2 – 4x^2y + 4x^3 + y^3 – 4xy^2 + 4x^2yxy2\u22124x2y+4×3+y3\u22124xy2+4x2y<\/p>\n\n\n\n

          Group like terms: (xy2\u22124xy2)+(\u22124x2y+4x2y)+4×3+y3=\u22123xy2+4×3+y3(xy^2 – 4xy^2) + (-4x^2y + 4x^2y) + 4x^3 + y^3 = -3xy^2 + 4x^3 + y^3(xy2\u22124xy2)+(\u22124x2y+4x2y)+4×3+y3=\u22123xy2+4×3+y3<\/p>\n\n\n\n

          So the integrand becomes: \u22123xy2+4×3+y3-3xy^2 + 4x^3 + y^3\u22123xy2+4×3+y3<\/p>\n\n\n\n

          \n\n\n\n

          Step 2: Integrate<\/strong><\/h3>\n\n\n\n

          \u222b01\u222b01(\u22123xy2+4×3+y3)\u2009dy\u2009dx\\int_0^1 \\int_0^1 (-3xy^2 + 4x^3 + y^3) \\, dy \\, dx\u222b01\u200b\u222b01\u200b(\u22123xy2+4×3+y3)dydx<\/p>\n\n\n\n

          Integrate w.r.t. yyy: \u222b01(\u22123xy2+4×3+y3)\u2009dy=\u22123x\u222b01y2\u2009dy+4×3\u222b01dy+\u222b01y3\u2009dy\\int_0^1 (-3x y^2 + 4x^3 + y^3) \\, dy = -3x \\int_0^1 y^2 \\, dy + 4x^3 \\int_0^1 dy + \\int_0^1 y^3 \\, dy\u222b01\u200b(\u22123xy2+4×3+y3)dy=\u22123x\u222b01\u200by2dy+4×3\u222b01\u200bdy+\u222b01\u200by3dy<\/p>\n\n\n\n

          Compute each:<\/p>\n\n\n\n

            \n
          • \u222b01y2\u2009dy=13\\int_0^1 y^2 \\, dy = \\frac{1}{3}\u222b01\u200by2dy=31\u200b<\/li>\n\n\n\n
          • \u222b01dy=1\\int_0^1 dy = 1\u222b01\u200bdy=1<\/li>\n\n\n\n
          • \u222b01y3\u2009dy=14\\int_0^1 y^3 \\, dy = \\frac{1}{4}\u222b01\u200by3dy=41\u200b<\/li>\n<\/ul>\n\n\n\n

            So: =\u22123x\u22c513+4×3\u22c51+14=\u2212x+4×3+14= -3x \\cdot \\frac{1}{3} + 4x^3 \\cdot 1 + \\frac{1}{4} = -x + 4x^3 + \\frac{1}{4}=\u22123x\u22c531\u200b+4×3\u22c51+41\u200b=\u2212x+4×3+41\u200b<\/p>\n\n\n\n

            Now integrate w.r.t. xxx: \u222b01(\u2212x+4×3+14)\u2009dx=\u222b01\u2212x\u2009dx+\u222b014×3\u2009dx+\u222b0114\u2009dx\\int_0^1 (-x + 4x^3 + \\frac{1}{4}) \\, dx = \\int_0^1 -x \\, dx + \\int_0^1 4x^3 \\, dx + \\int_0^1 \\frac{1}{4} \\, dx\u222b01\u200b(\u2212x+4×3+41\u200b)dx=\u222b01\u200b\u2212xdx+\u222b01\u200b4x3dx+\u222b01\u200b41\u200bdx<\/p>\n\n\n\n

            Compute:<\/p>\n\n\n\n

              \n
            • \u222b01\u2212x\u2009dx=\u221212\\int_0^1 -x \\, dx = -\\frac{1}{2}\u222b01\u200b\u2212xdx=\u221221\u200b<\/li>\n\n\n\n
            • \u222b014×3\u2009dx=4\u22c514=1\\int_0^1 4x^3 \\, dx = 4 \\cdot \\frac{1}{4} = 1\u222b01\u200b4x3dx=4\u22c541\u200b=1<\/li>\n\n\n\n
            • \u222b0114\u2009dx=14\\int_0^1 \\frac{1}{4} \\, dx = \\frac{1}{4}\u222b01\u200b41\u200bdx=41\u200b<\/li>\n<\/ul>\n\n\n\n

              Add: \u221212+1+14=34- \\frac{1}{2} + 1 + \\frac{1}{4} = \\frac{3}{4}\u221221\u200b+1+41\u200b=43\u200b<\/p>\n\n\n\n

              \n\n\n\n

              \u2705 Final Answer for Problem 2:<\/strong><\/h3>\n\n\n\n

              34\\boxed{\\frac{3}{4}}43\u200b\u200b<\/p>\n\n\n\n

              \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

              Find the volume of the “ice cream cone” D cut from the solid sphere p < 1 by the cone (10 pts). 2) Evaluate the integral (10 pts) \u00e2\u02c6\u00ab\u00e2\u02c6\u00ab V(x + y)(y – 2x)^2 dy dx. The Correct Answer and Explanation is: Problem 1: Find the volume of the \u201cice cream cone\u201d DDD cut from […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233356","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233356","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233356"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233356\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233356"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233356"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233356"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}