Parametric Equation of a sphere

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Homework Statement


Find parametric equations for the part of sphere x2+y2+z2=9 that lies between the planes y=1 and y=2.


Homework Equations





The Attempt at a Solution



Okay knowing that the p=3 I wrote the parametric equations for a sphere as
x=3sin[tex]\phi[/tex]cos[tex]\theta[/tex] y=3sin[tex]\phi[/tex]sin[tex]\theta[/tex]
z=3cos[tex]\phi[/tex] now the phi bound is 0<[tex]\phi[/tex]<[tex]\pi[/tex]
but I'm not sure what to write for the [tex]\theta[/tex] bound.
 
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It would be easier if they said between z=2 and z=1, right? Then you would just have to restrict phi. There's no reason why you can't interchange say, y and z in your parametrization.
 
Oh okay so I can say z=3cos[tex]\phi[/tex]
and from there plug in the values to find the [tex]\phi[/tex] bound to be cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1 and the [tex]\theta[/tex] bound should just be 0< [tex]\theta[/tex]<2[tex]\pi[/tex]
 
anubis01 said:
Oh okay so I can say z=3cos[tex]\phi[/tex]
and from there plug in the values to find the [tex]\phi[/tex] bound to be cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1 and the [tex]\theta[/tex] bound should just be 0< [tex]\theta[/tex]<2[tex]\pi[/tex]

Right. But they did say y=1 and y=2. You'll also have to tweak your parametrization so y=3*cos(theta) instead of z=3*cos(theta).
 
oh so then phi bound should be 0<[tex]\phi[/tex]<[tex]\pi[/tex] and since y=3cos[tex]\theta[/tex] the bounds for [tex]\theta[/tex] are
cos(1/3)-1<[tex]\theta[/tex]<cos(2/3)-1
 
anubis01 said:
oh so then phi bound should be 0<[tex]\phi[/tex]<[tex]\pi[/tex] and since y=3cos[tex]\theta[/tex] the bounds for [tex]\theta[/tex] are
cos(1/3)-1<[tex]\theta[/tex]<cos(2/3)-1

Sorry, sorry! I meant to say your parametrization has z=3*cos(phi) and you want to change it so that y=3*cos(phi). Leave theta as it is.
 
alright i think i got it now. So bound for theta 0<[tex]\theta[/tex]<2[tex]\pi[/tex]
y=3cos[tex]\phi[/tex] the bounds for phi are then
cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1

and the parametric equation for x=3sin[tex]\phi[/tex]cos[tex]\theta[/tex]
and z=3sin[tex]\phi[/tex]sin[tex]\theta[/tex]
 
anubis01 said:
alright i think i got it now. So bound for theta 0<[tex]\theta[/tex]<2[tex]\pi[/tex]
y=3cos[tex]\phi[/tex] the bounds for phi are then
cos(1/3)-1<[tex]\phi[/tex]<cos(2/3)-1

and the parametric equation for x=3sin[tex]\phi[/tex]cos[tex]\theta[/tex]
and z=3sin[tex]\phi[/tex]sin[tex]\theta[/tex]

That looks good to me.
 
alright thank you for all the help, its much appreciated.