Parametric Equation of a sphere

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Homework Help Overview

The discussion revolves around finding parametric equations for a portion of a sphere defined by the equation x² + y² + z² = 9, specifically the section that lies between the planes y = 1 and y = 2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the parametric equations for the sphere and discuss the implications of the bounds for φ and θ. There is a consideration of whether to use y or z in the parametrization and how that affects the bounds.

Discussion Status

Participants are actively refining their understanding of the bounds for φ and θ in the context of the given planes. Some guidance has been offered regarding the correct parametrization and the relationship between the variables.

Contextual Notes

There is some confusion regarding the bounds for θ and φ, with participants questioning how to properly set them based on the constraints of the problem. The original poster's initial assumptions about the parametrization are being revisited.

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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.
 

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