Parametric Equation of a sphere

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SUMMARY

The parametric equations for the part of the sphere defined by the equation x² + y² + z² = 9, constrained between the planes y = 1 and y = 2, are established as follows: x = 3sin(φ)cos(θ), y = 3cos(φ), and z = 3sin(φ)sin(θ). The bounds for φ are cos(1/3) - 1 < φ < cos(2/3) - 1, while θ ranges from 0 to 2π. This parametrization effectively utilizes the spherical coordinates to describe the specified section of the 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.


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The Attempt at a Solution



Okay knowing that the p=3 I wrote the parametric equations for a sphere as
x=3sin\phicos\theta y=3sin\phisin\theta
z=3cos\phi now the phi bound is 0<\phi<\pi
but I'm not sure what to write for the \theta 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\phi
and from there plug in the values to find the \phi bound to be cos(1/3)-1<\phi<cos(2/3)-1 and the \theta bound should just be 0< \theta<2\pi
 
anubis01 said:
Oh okay so I can say z=3cos\phi
and from there plug in the values to find the \phi bound to be cos(1/3)-1<\phi<cos(2/3)-1 and the \theta bound should just be 0< \theta<2\pi

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<\phi<\pi and since y=3cos\theta the bounds for \theta are
cos(1/3)-1<\theta<cos(2/3)-1
 
anubis01 said:
oh so then phi bound should be 0<\phi<\pi and since y=3cos\theta the bounds for \theta are
cos(1/3)-1<\theta<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<\theta<2\pi
y=3cos\phi the bounds for phi are then
cos(1/3)-1<\phi<cos(2/3)-1

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

and the parametric equation for x=3sin\phicos\theta
and z=3sin\phisin\theta

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

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