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











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    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






























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