parameterization of a part of a sphere

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I have to parametrize $D=x^2+y^2+z^2le 25,yle -4$.



  • I can see the I have to parametrize 2 surfaces :

    • ($S_1$) the intersection between the plane $z=-4$ and the sphere: ($x^2+z^2le 9$)

    • ($S_2$) The part of the sphere between $-5le y le -4$


$(S_1)$ should be :



$r(u,v)=(ucos(v),-4,usin(v))$



$0le u le 3$



$0le v le 2pi$



$(S_2)$, I need help with this, Can you help me find out $phi,theta$



$r(phi,theta)=(5sinphi costheta,...,5sinphi sintheta)$(x,z might be wrong)



EDIT:



I know that the parametrization of a sphere of radius 5 is : $r(phi,theta)=(5sinphi costheta,5sinphi sintheta,5cosphi)$




multivariable-calculus surfaces parametrization




share|cite|improve this question





















  • If you use sin and cos in your MathJax, your formatting will look a lot better.
    – saulspatz
    Jul 31 at 21:28






  • 1




    $S_1:r(u,v)=(ucos(v),-4,usin(v))$
    – Doug M
    Jul 31 at 21:31










  • As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
    – Doug M
    Jul 31 at 22:10














up vote
0
down vote

favorite












I have to parametrize $D=x^2+y^2+z^2le 25,yle -4$.



  • I can see the I have to parametrize 2 surfaces :

    • ($S_1$) the intersection between the plane $z=-4$ and the sphere: ($x^2+z^2le 9$)

    • ($S_2$) The part of the sphere between $-5le y le -4$


$(S_1)$ should be :



$r(u,v)=(ucos(v),-4,usin(v))$



$0le u le 3$



$0le v le 2pi$



$(S_2)$, I need help with this, Can you help me find out $phi,theta$



$r(phi,theta)=(5sinphi costheta,...,5sinphi sintheta)$(x,z might be wrong)



EDIT:



I know that the parametrization of a sphere of radius 5 is : $r(phi,theta)=(5sinphi costheta,5sinphi sintheta,5cosphi)$




multivariable-calculus surfaces parametrization




share|cite|improve this question





















  • If you use sin and cos in your MathJax, your formatting will look a lot better.
    – saulspatz
    Jul 31 at 21:28






  • 1




    $S_1:r(u,v)=(ucos(v),-4,usin(v))$
    – Doug M
    Jul 31 at 21:31










  • As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
    – Doug M
    Jul 31 at 22:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to parametrize $D=x^2+y^2+z^2le 25,yle -4$.



  • I can see the I have to parametrize 2 surfaces :

    • ($S_1$) the intersection between the plane $z=-4$ and the sphere: ($x^2+z^2le 9$)

    • ($S_2$) The part of the sphere between $-5le y le -4$


$(S_1)$ should be :



$r(u,v)=(ucos(v),-4,usin(v))$



$0le u le 3$



$0le v le 2pi$



$(S_2)$, I need help with this, Can you help me find out $phi,theta$



$r(phi,theta)=(5sinphi costheta,...,5sinphi sintheta)$(x,z might be wrong)



EDIT:



I know that the parametrization of a sphere of radius 5 is : $r(phi,theta)=(5sinphi costheta,5sinphi sintheta,5cosphi)$




multivariable-calculus surfaces parametrization




share|cite|improve this question













I have to parametrize $D=x^2+y^2+z^2le 25,yle -4$.



  • I can see the I have to parametrize 2 surfaces :

    • ($S_1$) the intersection between the plane $z=-4$ and the sphere: ($x^2+z^2le 9$)

    • ($S_2$) The part of the sphere between $-5le y le -4$


$(S_1)$ should be :



$r(u,v)=(ucos(v),-4,usin(v))$



$0le u le 3$



$0le v le 2pi$



$(S_2)$, I need help with this, Can you help me find out $phi,theta$



$r(phi,theta)=(5sinphi costheta,...,5sinphi sintheta)$(x,z might be wrong)



EDIT:



I know that the parametrization of a sphere of radius 5 is : $r(phi,theta)=(5sinphi costheta,5sinphi sintheta,5cosphi)$





multivariable-calculus surfaces parametrization





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 22:17
























asked Jul 31 at 21:23









NPLS

1819




1819











  • If you use sin and cos in your MathJax, your formatting will look a lot better.
    – saulspatz
    Jul 31 at 21:28






  • 1




    $S_1:r(u,v)=(ucos(v),-4,usin(v))$
    – Doug M
    Jul 31 at 21:31










  • As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
    – Doug M
    Jul 31 at 22:10
















  • If you use sin and cos in your MathJax, your formatting will look a lot better.
    – saulspatz
    Jul 31 at 21:28






  • 1




    $S_1:r(u,v)=(ucos(v),-4,usin(v))$
    – Doug M
    Jul 31 at 21:31










  • As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
    – Doug M
    Jul 31 at 22:10















If you use sin and cos in your MathJax, your formatting will look a lot better.
– saulspatz
Jul 31 at 21:28




If you use sin and cos in your MathJax, your formatting will look a lot better.
– saulspatz
Jul 31 at 21:28




1




1




$S_1:r(u,v)=(ucos(v),-4,usin(v))$
– Doug M
Jul 31 at 21:31




$S_1:r(u,v)=(ucos(v),-4,usin(v))$
– Doug M
Jul 31 at 21:31












As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
– Doug M
Jul 31 at 22:10




As for your other surface, I am tempted to say plug in $ x = 5 sin phi cos theta, z = 5 sin phi sin theta$ into $x^2 + y^2 + z^2 = 25$ I also wonder how you know that those the correct parameterization for spherical coordinates without knowing the 3rd coordinate. Then I think I should just let you off the hook, $y = cosphi$
– Doug M
Jul 31 at 22:10










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










The way I see it is you need to parametrize the solid bounded by the sphere of radius $5$ and the plane $y=-4$. Here is how I would proceed :



Use spherical coordinates as follows : let $y=rho cos phi$, $z=rho sin phi cos theta$ and $x=rho sin phi sin theta$, such that the sphere has equation
$$
rho =5,
$$
and the plane $y=-4$ has equation
$$
rho cos phi=-4
$$



Now things become easy. The projection of the solid in the $yz$ plane is the domain
$$
D= ; frac4 cos phi le rho le 5 , cos^-1left(frac-45right)le phi le pi
$$
And it follows that
$$
E = (rho,phi,theta );
$$






share|cite|improve this answer





















  • why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
    – NPLS
    Aug 1 at 8:41











  • The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
    – Kuifje
    Aug 1 at 8:48










  • Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
    – NPLS
    Aug 1 at 9:03











  • My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
    – Kuifje
    Aug 1 at 9:10











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The way I see it is you need to parametrize the solid bounded by the sphere of radius $5$ and the plane $y=-4$. Here is how I would proceed :



Use spherical coordinates as follows : let $y=rho cos phi$, $z=rho sin phi cos theta$ and $x=rho sin phi sin theta$, such that the sphere has equation
$$
rho =5,
$$
and the plane $y=-4$ has equation
$$
rho cos phi=-4
$$



Now things become easy. The projection of the solid in the $yz$ plane is the domain
$$
D= ; frac4 cos phi le rho le 5 , cos^-1left(frac-45right)le phi le pi
$$
And it follows that
$$
E = (rho,phi,theta );
$$






share|cite|improve this answer





















  • why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
    – NPLS
    Aug 1 at 8:41











  • The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
    – Kuifje
    Aug 1 at 8:48










  • Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
    – NPLS
    Aug 1 at 9:03











  • My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
    – Kuifje
    Aug 1 at 9:10















up vote
1
down vote



accepted










The way I see it is you need to parametrize the solid bounded by the sphere of radius $5$ and the plane $y=-4$. Here is how I would proceed :



Use spherical coordinates as follows : let $y=rho cos phi$, $z=rho sin phi cos theta$ and $x=rho sin phi sin theta$, such that the sphere has equation
$$
rho =5,
$$
and the plane $y=-4$ has equation
$$
rho cos phi=-4
$$



Now things become easy. The projection of the solid in the $yz$ plane is the domain
$$
D= ; frac4 cos phi le rho le 5 , cos^-1left(frac-45right)le phi le pi
$$
And it follows that
$$
E = (rho,phi,theta );
$$






share|cite|improve this answer





















  • why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
    – NPLS
    Aug 1 at 8:41











  • The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
    – Kuifje
    Aug 1 at 8:48










  • Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
    – NPLS
    Aug 1 at 9:03











  • My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
    – Kuifje
    Aug 1 at 9:10













up vote
1
down vote



accepted







up vote
1
down vote



accepted






The way I see it is you need to parametrize the solid bounded by the sphere of radius $5$ and the plane $y=-4$. Here is how I would proceed :



Use spherical coordinates as follows : let $y=rho cos phi$, $z=rho sin phi cos theta$ and $x=rho sin phi sin theta$, such that the sphere has equation
$$
rho =5,
$$
and the plane $y=-4$ has equation
$$
rho cos phi=-4
$$



Now things become easy. The projection of the solid in the $yz$ plane is the domain
$$
D= ; frac4 cos phi le rho le 5 , cos^-1left(frac-45right)le phi le pi
$$
And it follows that
$$
E = (rho,phi,theta );
$$






share|cite|improve this answer













The way I see it is you need to parametrize the solid bounded by the sphere of radius $5$ and the plane $y=-4$. Here is how I would proceed :



Use spherical coordinates as follows : let $y=rho cos phi$, $z=rho sin phi cos theta$ and $x=rho sin phi sin theta$, such that the sphere has equation
$$
rho =5,
$$
and the plane $y=-4$ has equation
$$
rho cos phi=-4
$$



Now things become easy. The projection of the solid in the $yz$ plane is the domain
$$
D= ; frac4 cos phi le rho le 5 , cos^-1left(frac-45right)le phi le pi
$$
And it follows that
$$
E = (rho,phi,theta );
$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 1 at 8:32









Kuifje

6,8112523




6,8112523











  • why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
    – NPLS
    Aug 1 at 8:41











  • The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
    – Kuifje
    Aug 1 at 8:48










  • Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
    – NPLS
    Aug 1 at 9:03











  • My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
    – Kuifje
    Aug 1 at 9:10

















  • why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
    – NPLS
    Aug 1 at 8:41











  • The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
    – Kuifje
    Aug 1 at 8:48










  • Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
    – NPLS
    Aug 1 at 9:03











  • My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
    – Kuifje
    Aug 1 at 9:10
















why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
– NPLS
Aug 1 at 8:41





why is $rho$ between $4/ cosphi$ and $5$ ? Isn't the radius of the sphere constant ?
– NPLS
Aug 1 at 8:41













The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
– Kuifje
Aug 1 at 8:48




The radius of the sphere is constant indeed ($=5$). But you want the inside of the sphere too, between the plane and the sphere.
– Kuifje
Aug 1 at 8:48












Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
– NPLS
Aug 1 at 9:03





Sorry I don't understand, your parametrization is both $S_1$ and $S_2$ ? so it includes the plane and the sphere? becouse I want to calculate the flux, so I think I should have two surfaces because they have different normals
– NPLS
Aug 1 at 9:03













My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
– Kuifje
Aug 1 at 9:10





My parametrization is the solid bounded by $S_1$ and $S_2$, so yes it does include the plane and the sphere (which are the borders of this solid). My suggestion : post a new question with the entire question, not just the parametrization part. To compute the flux there are different ways, but selecting the best way requires the entire question and the context.
– Kuifje
Aug 1 at 9:10













 

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