Volume between cone and cylinder with a shift












1












$begingroup$


I am trying to solve the following problem:



Compute the volume of the solid bounded by the $ z = 3 sqrt{x^2 + y^2} $ , the plane $ z = 0 $ , and the cylinder $ x^2 +(y−1)^2 = 1 $ .



I know I need to use a triple integral, integrating over $ z, r, $ and $ theta $, but I am not sure how to account for the $ y $ shift. How do I do this?






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
    $endgroup$
    – David H
    May 4 '14 at 3:05










  • $begingroup$
    Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
    $endgroup$
    – user145277
    May 4 '14 at 16:24
















1












$begingroup$


I am trying to solve the following problem:



Compute the volume of the solid bounded by the $ z = 3 sqrt{x^2 + y^2} $ , the plane $ z = 0 $ , and the cylinder $ x^2 +(y−1)^2 = 1 $ .



I know I need to use a triple integral, integrating over $ z, r, $ and $ theta $, but I am not sure how to account for the $ y $ shift. How do I do this?






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
    $endgroup$
    – David H
    May 4 '14 at 3:05










  • $begingroup$
    Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
    $endgroup$
    – user145277
    May 4 '14 at 16:24














1












1








1


1



$begingroup$


I am trying to solve the following problem:



Compute the volume of the solid bounded by the $ z = 3 sqrt{x^2 + y^2} $ , the plane $ z = 0 $ , and the cylinder $ x^2 +(y−1)^2 = 1 $ .



I know I need to use a triple integral, integrating over $ z, r, $ and $ theta $, but I am not sure how to account for the $ y $ shift. How do I do this?






multivariable-calculus







share|cite|improve this question











$endgroup$




I am trying to solve the following problem:



Compute the volume of the solid bounded by the $ z = 3 sqrt{x^2 + y^2} $ , the plane $ z = 0 $ , and the cylinder $ x^2 +(y−1)^2 = 1 $ .



I know I need to use a triple integral, integrating over $ z, r, $ and $ theta $, but I am not sure how to account for the $ y $ shift. How do I do this?






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 4 '14 at 2:08









colormegone

9,71321542




9,71321542










asked May 4 '14 at 1:53









user145277user145277

133




133












  • $begingroup$
    Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
    $endgroup$
    – David H
    May 4 '14 at 3:05










  • $begingroup$
    Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
    $endgroup$
    – user145277
    May 4 '14 at 16:24


















  • $begingroup$
    Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
    $endgroup$
    – David H
    May 4 '14 at 3:05










  • $begingroup$
    Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
    $endgroup$
    – user145277
    May 4 '14 at 16:24
















$begingroup$
Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
$endgroup$
– David H
May 4 '14 at 3:05




$begingroup$
Rewrite the circle equation in polar coordinates and simplify for $r$ as a function of $theta$.
$endgroup$
– David H
May 4 '14 at 3:05












$begingroup$
Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
$endgroup$
– user145277
May 4 '14 at 16:24




$begingroup$
Hi David - that was my original approach, but I cannot figure out how to make it work since the cylinder is shifted, meaning I cannot simply use z = 3r.
$endgroup$
– user145277
May 4 '14 at 16:24










2 Answers
2






active

oldest

votes


















0












$begingroup$

The problem is to compute the volume of the region $R$ bounded by the three surfaces,



$$begin{cases}z=0,\z=3sqrt{x^2+y^2},\x^2+(y-1)^2=1.end{cases}$$



This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(rhocos{phi},rhosin{phi},z)$, the three surface equations become



$$begin{cases}z=0,\z=3sqrt{(rhocos{phi})^2+(rhosin{phi})^2},\(rhocos{phi})^2+(rhosin{phi}-1)^2=1,end{cases}$$



or



$$begin{cases}z=0,\z=3rho,~~~0leqrho\rho=2sin{phi},~~~0leqphi<pi.end{cases}$$



Thus, the volume of $R$ can be found by evaluating the following triple integral:



$$V=iiint_{R}dV\
=int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,rhoint_{0}^{3rho}dz\
=int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,(3rho^2)\
=int_{0}^{pi}dphi,(2sin{phi})^3\
=frac{32}{3}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The answer seems to be not right. It should be $32/3/sqrt 3$.
    $endgroup$
    – duanduan
    Jan 7 '17 at 7:41





















0












$begingroup$

HINT



Convert to cylindrical/polar coordinates



$$ z=3 ,r; z = 6 sin theta; , r = 2 sin theta, $$



the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).






share|cite|improve this answer











$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f780321%2fvolume-between-cone-and-cylinder-with-a-shift%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The problem is to compute the volume of the region $R$ bounded by the three surfaces,



    $$begin{cases}z=0,\z=3sqrt{x^2+y^2},\x^2+(y-1)^2=1.end{cases}$$



    This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(rhocos{phi},rhosin{phi},z)$, the three surface equations become



    $$begin{cases}z=0,\z=3sqrt{(rhocos{phi})^2+(rhosin{phi})^2},\(rhocos{phi})^2+(rhosin{phi}-1)^2=1,end{cases}$$



    or



    $$begin{cases}z=0,\z=3rho,~~~0leqrho\rho=2sin{phi},~~~0leqphi<pi.end{cases}$$



    Thus, the volume of $R$ can be found by evaluating the following triple integral:



    $$V=iiint_{R}dV\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,rhoint_{0}^{3rho}dz\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,(3rho^2)\
    =int_{0}^{pi}dphi,(2sin{phi})^3\
    =frac{32}{3}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The answer seems to be not right. It should be $32/3/sqrt 3$.
      $endgroup$
      – duanduan
      Jan 7 '17 at 7:41


















    0












    $begingroup$

    The problem is to compute the volume of the region $R$ bounded by the three surfaces,



    $$begin{cases}z=0,\z=3sqrt{x^2+y^2},\x^2+(y-1)^2=1.end{cases}$$



    This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(rhocos{phi},rhosin{phi},z)$, the three surface equations become



    $$begin{cases}z=0,\z=3sqrt{(rhocos{phi})^2+(rhosin{phi})^2},\(rhocos{phi})^2+(rhosin{phi}-1)^2=1,end{cases}$$



    or



    $$begin{cases}z=0,\z=3rho,~~~0leqrho\rho=2sin{phi},~~~0leqphi<pi.end{cases}$$



    Thus, the volume of $R$ can be found by evaluating the following triple integral:



    $$V=iiint_{R}dV\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,rhoint_{0}^{3rho}dz\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,(3rho^2)\
    =int_{0}^{pi}dphi,(2sin{phi})^3\
    =frac{32}{3}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The answer seems to be not right. It should be $32/3/sqrt 3$.
      $endgroup$
      – duanduan
      Jan 7 '17 at 7:41
















    0












    0








    0





    $begingroup$

    The problem is to compute the volume of the region $R$ bounded by the three surfaces,



    $$begin{cases}z=0,\z=3sqrt{x^2+y^2},\x^2+(y-1)^2=1.end{cases}$$



    This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(rhocos{phi},rhosin{phi},z)$, the three surface equations become



    $$begin{cases}z=0,\z=3sqrt{(rhocos{phi})^2+(rhosin{phi})^2},\(rhocos{phi})^2+(rhosin{phi}-1)^2=1,end{cases}$$



    or



    $$begin{cases}z=0,\z=3rho,~~~0leqrho\rho=2sin{phi},~~~0leqphi<pi.end{cases}$$



    Thus, the volume of $R$ can be found by evaluating the following triple integral:



    $$V=iiint_{R}dV\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,rhoint_{0}^{3rho}dz\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,(3rho^2)\
    =int_{0}^{pi}dphi,(2sin{phi})^3\
    =frac{32}{3}.$$






    share|cite|improve this answer









    $endgroup$



    The problem is to compute the volume of the region $R$ bounded by the three surfaces,



    $$begin{cases}z=0,\z=3sqrt{x^2+y^2},\x^2+(y-1)^2=1.end{cases}$$



    This is clearly a task best accomplished using cylindrical coordinates. In terms of cylindrical coordinates $(x,y,z)=(rhocos{phi},rhosin{phi},z)$, the three surface equations become



    $$begin{cases}z=0,\z=3sqrt{(rhocos{phi})^2+(rhosin{phi})^2},\(rhocos{phi})^2+(rhosin{phi}-1)^2=1,end{cases}$$



    or



    $$begin{cases}z=0,\z=3rho,~~~0leqrho\rho=2sin{phi},~~~0leqphi<pi.end{cases}$$



    Thus, the volume of $R$ can be found by evaluating the following triple integral:



    $$V=iiint_{R}dV\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,rhoint_{0}^{3rho}dz\
    =int_{0}^{pi}dphiint_{0}^{2sin{phi}}drho,(3rho^2)\
    =int_{0}^{pi}dphi,(2sin{phi})^3\
    =frac{32}{3}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 8 '14 at 0:46









    David HDavid H

    21.7k24693




    21.7k24693












    • $begingroup$
      The answer seems to be not right. It should be $32/3/sqrt 3$.
      $endgroup$
      – duanduan
      Jan 7 '17 at 7:41




















    • $begingroup$
      The answer seems to be not right. It should be $32/3/sqrt 3$.
      $endgroup$
      – duanduan
      Jan 7 '17 at 7:41


















    $begingroup$
    The answer seems to be not right. It should be $32/3/sqrt 3$.
    $endgroup$
    – duanduan
    Jan 7 '17 at 7:41






    $begingroup$
    The answer seems to be not right. It should be $32/3/sqrt 3$.
    $endgroup$
    – duanduan
    Jan 7 '17 at 7:41













    0












    $begingroup$

    HINT



    Convert to cylindrical/polar coordinates



    $$ z=3 ,r; z = 6 sin theta; , r = 2 sin theta, $$



    the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      HINT



      Convert to cylindrical/polar coordinates



      $$ z=3 ,r; z = 6 sin theta; , r = 2 sin theta, $$



      the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        HINT



        Convert to cylindrical/polar coordinates



        $$ z=3 ,r; z = 6 sin theta; , r = 2 sin theta, $$



        the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).






        share|cite|improve this answer











        $endgroup$



        HINT



        Convert to cylindrical/polar coordinates



        $$ z=3 ,r; z = 6 sin theta; , r = 2 sin theta, $$



        the cylinder generator goes through $z$ axis. The volume computation is simpler than Viviani ( cylinder and sphere intersection).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 '18 at 19:43

























        answered Jan 8 '18 at 19:18









        NarasimhamNarasimham

        21.2k62258




        21.2k62258






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f780321%2fvolume-between-cone-and-cylinder-with-a-shift%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Redirect URL with Chrome Remote Debugging Android Devices

            Image
            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more