Find the volume between the cone $y = sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.












0












$begingroup$



Find the volume between the cone $y = sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.




I know that the volume we're interested in is the volume of the intersection between the sphere of radius $7$ and a an upside down cone in the direction of the $y$-axis, but I have no clue on how to set up the bounds of integration. I'm guessing we are supposed to do this in spherical coordinates, but how would we determine the limits of integration?



Thank you.






multivariable-calculus spherical-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
    $endgroup$
    – G Cab
    Apr 9 '17 at 22:48
















0












$begingroup$



Find the volume between the cone $y = sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.




I know that the volume we're interested in is the volume of the intersection between the sphere of radius $7$ and a an upside down cone in the direction of the $y$-axis, but I have no clue on how to set up the bounds of integration. I'm guessing we are supposed to do this in spherical coordinates, but how would we determine the limits of integration?



Thank you.






multivariable-calculus spherical-coordinates







share|cite|improve this question











$endgroup$












  • $begingroup$
    same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
    $endgroup$
    – G Cab
    Apr 9 '17 at 22:48














0












0








0





$begingroup$



Find the volume between the cone $y = sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.




I know that the volume we're interested in is the volume of the intersection between the sphere of radius $7$ and a an upside down cone in the direction of the $y$-axis, but I have no clue on how to set up the bounds of integration. I'm guessing we are supposed to do this in spherical coordinates, but how would we determine the limits of integration?



Thank you.






multivariable-calculus spherical-coordinates







share|cite|improve this question











$endgroup$





Find the volume between the cone $y = sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.




I know that the volume we're interested in is the volume of the intersection between the sphere of radius $7$ and a an upside down cone in the direction of the $y$-axis, but I have no clue on how to set up the bounds of integration. I'm guessing we are supposed to do this in spherical coordinates, but how would we determine the limits of integration?



Thank you.






multivariable-calculus spherical-coordinates




multivariable-calculus spherical-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 10 '17 at 16:27









Nosrati

26.5k62354




26.5k62354










asked Apr 9 '17 at 22:28









JC1JC1

7127




7127












  • $begingroup$
    same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
    $endgroup$
    – G Cab
    Apr 9 '17 at 22:48


















  • $begingroup$
    same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
    $endgroup$
    – G Cab
    Apr 9 '17 at 22:48
















$begingroup$
same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
$endgroup$
– G Cab
Apr 9 '17 at 22:48




$begingroup$
same as if the cone were along the $z$ axis, so that using spherical coordinates it translates into a limit on just the polar angle. Thus switch $y$ and $z$ in the cone equation.
$endgroup$
– G Cab
Apr 9 '17 at 22:48










2 Answers
2






active

oldest

votes


















0












$begingroup$

If you want to do it in spherical coordinates notice $z=sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $rho=7$.



If $z=r$ then $frac{r}{z}=tan (phi)=1$ so $y=r$ translates to $phi=frac{pi}{4}$.



Hence what we have is,



$$V=int_{0}^{2pi} int_{0}^{frac{pi}{4}} int_{0}^{7} rho^2 sin (phi) drho dphi dtheta$$




Actually because we have $y=sqrt{x^2+z^2}$ I would define a modified version of spherical coordinates. Let $x^2+z^2=r^2$, let $phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 leq phi leq pi$. Let $rho$ be the distance from the origin. Let $theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis.



Then I would compute the Jacobian and it will turn out to be $rho^2 sin (phi)$. The equation $frac{r}{y}=tan (phi)$ will hold and calculations will be pretty much identical.




Think about why $frac{r}{z}=tan (phi)$. Image from Paul's math.



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi, I've previously entered the wrong equation - its now corrected. Apologies
    $endgroup$
    – JC1
    Apr 9 '17 at 22:40










  • $begingroup$
    Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
    $endgroup$
    – JC1
    Apr 9 '17 at 22:45










  • $begingroup$
    I've added a picture.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:51










  • $begingroup$
    Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
    $endgroup$
    – JC1
    Apr 9 '17 at 22:55










  • $begingroup$
    You mean $frac{r}{z}$ right.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:57



















0












$begingroup$

For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:



$$ x= 7/sqrt2 ,y= 7/sqrt2 ,cos t, z= 7/sqrt2 , sin t $$



You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.



$$ k_g= frac{1}{7},, s= 2 pi frac{7}{sqrt2},, int k_g ds = frac{pi}{sqrt2} $$



Integral curvature (solid angle)



$$ int int K dA = 2 pi- frac{ pi}{sqrt2}$$



and Volume is solid angle times $R^3/3$



$$= pi(2 - frac{ 1}{sqrt2}) , frac{7^3}{3}. $$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If you want to do it in spherical coordinates notice $z=sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $rho=7$.



    If $z=r$ then $frac{r}{z}=tan (phi)=1$ so $y=r$ translates to $phi=frac{pi}{4}$.



    Hence what we have is,



    $$V=int_{0}^{2pi} int_{0}^{frac{pi}{4}} int_{0}^{7} rho^2 sin (phi) drho dphi dtheta$$




    Actually because we have $y=sqrt{x^2+z^2}$ I would define a modified version of spherical coordinates. Let $x^2+z^2=r^2$, let $phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 leq phi leq pi$. Let $rho$ be the distance from the origin. Let $theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis.



    Then I would compute the Jacobian and it will turn out to be $rho^2 sin (phi)$. The equation $frac{r}{y}=tan (phi)$ will hold and calculations will be pretty much identical.




    Think about why $frac{r}{z}=tan (phi)$. Image from Paul's math.



    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Hi, I've previously entered the wrong equation - its now corrected. Apologies
      $endgroup$
      – JC1
      Apr 9 '17 at 22:40










    • $begingroup$
      Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
      $endgroup$
      – JC1
      Apr 9 '17 at 22:45










    • $begingroup$
      I've added a picture.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:51










    • $begingroup$
      Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
      $endgroup$
      – JC1
      Apr 9 '17 at 22:55










    • $begingroup$
      You mean $frac{r}{z}$ right.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:57
















    0












    $begingroup$

    If you want to do it in spherical coordinates notice $z=sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $rho=7$.



    If $z=r$ then $frac{r}{z}=tan (phi)=1$ so $y=r$ translates to $phi=frac{pi}{4}$.



    Hence what we have is,



    $$V=int_{0}^{2pi} int_{0}^{frac{pi}{4}} int_{0}^{7} rho^2 sin (phi) drho dphi dtheta$$




    Actually because we have $y=sqrt{x^2+z^2}$ I would define a modified version of spherical coordinates. Let $x^2+z^2=r^2$, let $phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 leq phi leq pi$. Let $rho$ be the distance from the origin. Let $theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis.



    Then I would compute the Jacobian and it will turn out to be $rho^2 sin (phi)$. The equation $frac{r}{y}=tan (phi)$ will hold and calculations will be pretty much identical.




    Think about why $frac{r}{z}=tan (phi)$. Image from Paul's math.



    enter image description here






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Hi, I've previously entered the wrong equation - its now corrected. Apologies
      $endgroup$
      – JC1
      Apr 9 '17 at 22:40










    • $begingroup$
      Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
      $endgroup$
      – JC1
      Apr 9 '17 at 22:45










    • $begingroup$
      I've added a picture.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:51










    • $begingroup$
      Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
      $endgroup$
      – JC1
      Apr 9 '17 at 22:55










    • $begingroup$
      You mean $frac{r}{z}$ right.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:57














    0












    0








    0





    $begingroup$

    If you want to do it in spherical coordinates notice $z=sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $rho=7$.



    If $z=r$ then $frac{r}{z}=tan (phi)=1$ so $y=r$ translates to $phi=frac{pi}{4}$.



    Hence what we have is,



    $$V=int_{0}^{2pi} int_{0}^{frac{pi}{4}} int_{0}^{7} rho^2 sin (phi) drho dphi dtheta$$




    Actually because we have $y=sqrt{x^2+z^2}$ I would define a modified version of spherical coordinates. Let $x^2+z^2=r^2$, let $phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 leq phi leq pi$. Let $rho$ be the distance from the origin. Let $theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis.



    Then I would compute the Jacobian and it will turn out to be $rho^2 sin (phi)$. The equation $frac{r}{y}=tan (phi)$ will hold and calculations will be pretty much identical.




    Think about why $frac{r}{z}=tan (phi)$. Image from Paul's math.



    enter image description here






    share|cite|improve this answer











    $endgroup$



    If you want to do it in spherical coordinates notice $z=sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $rho=7$.



    If $z=r$ then $frac{r}{z}=tan (phi)=1$ so $y=r$ translates to $phi=frac{pi}{4}$.



    Hence what we have is,



    $$V=int_{0}^{2pi} int_{0}^{frac{pi}{4}} int_{0}^{7} rho^2 sin (phi) drho dphi dtheta$$




    Actually because we have $y=sqrt{x^2+z^2}$ I would define a modified version of spherical coordinates. Let $x^2+z^2=r^2$, let $phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 leq phi leq pi$. Let $rho$ be the distance from the origin. Let $theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis.



    Then I would compute the Jacobian and it will turn out to be $rho^2 sin (phi)$. The equation $frac{r}{y}=tan (phi)$ will hold and calculations will be pretty much identical.




    Think about why $frac{r}{z}=tan (phi)$. Image from Paul's math.



    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 9 '17 at 23:24

























    answered Apr 9 '17 at 22:36









    Ahmed S. AttaallaAhmed S. Attaalla

    14.8k12050




    14.8k12050












    • $begingroup$
      Hi, I've previously entered the wrong equation - its now corrected. Apologies
      $endgroup$
      – JC1
      Apr 9 '17 at 22:40










    • $begingroup$
      Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
      $endgroup$
      – JC1
      Apr 9 '17 at 22:45










    • $begingroup$
      I've added a picture.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:51










    • $begingroup$
      Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
      $endgroup$
      – JC1
      Apr 9 '17 at 22:55










    • $begingroup$
      You mean $frac{r}{z}$ right.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:57


















    • $begingroup$
      Hi, I've previously entered the wrong equation - its now corrected. Apologies
      $endgroup$
      – JC1
      Apr 9 '17 at 22:40










    • $begingroup$
      Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
      $endgroup$
      – JC1
      Apr 9 '17 at 22:45










    • $begingroup$
      I've added a picture.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:51










    • $begingroup$
      Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
      $endgroup$
      – JC1
      Apr 9 '17 at 22:55










    • $begingroup$
      You mean $frac{r}{z}$ right.
      $endgroup$
      – Ahmed S. Attaalla
      Apr 9 '17 at 22:57
















    $begingroup$
    Hi, I've previously entered the wrong equation - its now corrected. Apologies
    $endgroup$
    – JC1
    Apr 9 '17 at 22:40




    $begingroup$
    Hi, I've previously entered the wrong equation - its now corrected. Apologies
    $endgroup$
    – JC1
    Apr 9 '17 at 22:40












    $begingroup$
    Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
    $endgroup$
    – JC1
    Apr 9 '17 at 22:45




    $begingroup$
    Ah, thank you - did not think about that. Also, how did you tell that $r/z = tan(phi)$
    $endgroup$
    – JC1
    Apr 9 '17 at 22:45












    $begingroup$
    I've added a picture.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:51




    $begingroup$
    I've added a picture.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:51












    $begingroup$
    Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
    $endgroup$
    – JC1
    Apr 9 '17 at 22:55




    $begingroup$
    Oh ok. Since $ z = r$, we can say that the angle between of $phi$ will need to be $tan(phi) = z/r $ using SOHCAHTOA . Thank you
    $endgroup$
    – JC1
    Apr 9 '17 at 22:55












    $begingroup$
    You mean $frac{r}{z}$ right.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:57




    $begingroup$
    You mean $frac{r}{z}$ right.
    $endgroup$
    – Ahmed S. Attaalla
    Apr 9 '17 at 22:57











    0












    $begingroup$

    For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:



    $$ x= 7/sqrt2 ,y= 7/sqrt2 ,cos t, z= 7/sqrt2 , sin t $$



    You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.



    $$ k_g= frac{1}{7},, s= 2 pi frac{7}{sqrt2},, int k_g ds = frac{pi}{sqrt2} $$



    Integral curvature (solid angle)



    $$ int int K dA = 2 pi- frac{ pi}{sqrt2}$$



    and Volume is solid angle times $R^3/3$



    $$= pi(2 - frac{ 1}{sqrt2}) , frac{7^3}{3}. $$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:



      $$ x= 7/sqrt2 ,y= 7/sqrt2 ,cos t, z= 7/sqrt2 , sin t $$



      You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.



      $$ k_g= frac{1}{7},, s= 2 pi frac{7}{sqrt2},, int k_g ds = frac{pi}{sqrt2} $$



      Integral curvature (solid angle)



      $$ int int K dA = 2 pi- frac{ pi}{sqrt2}$$



      and Volume is solid angle times $R^3/3$



      $$= pi(2 - frac{ 1}{sqrt2}) , frac{7^3}{3}. $$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:



        $$ x= 7/sqrt2 ,y= 7/sqrt2 ,cos t, z= 7/sqrt2 , sin t $$



        You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.



        $$ k_g= frac{1}{7},, s= 2 pi frac{7}{sqrt2},, int k_g ds = frac{pi}{sqrt2} $$



        Integral curvature (solid angle)



        $$ int int K dA = 2 pi- frac{ pi}{sqrt2}$$



        and Volume is solid angle times $R^3/3$



        $$= pi(2 - frac{ 1}{sqrt2}) , frac{7^3}{3}. $$






        share|cite|improve this answer











        $endgroup$



        For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:



        $$ x= 7/sqrt2 ,y= 7/sqrt2 ,cos t, z= 7/sqrt2 , sin t $$



        You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.



        $$ k_g= frac{1}{7},, s= 2 pi frac{7}{sqrt2},, int k_g ds = frac{pi}{sqrt2} $$



        Integral curvature (solid angle)



        $$ int int K dA = 2 pi- frac{ pi}{sqrt2}$$



        and Volume is solid angle times $R^3/3$



        $$= pi(2 - frac{ 1}{sqrt2}) , frac{7^3}{3}. $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 10 '17 at 13:19

























        answered Apr 9 '17 at 23:16









        NarasimhamNarasimham

        20.8k52158




        20.8k52158






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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