Surface of a Cylinder inside a Sphere

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Homework Statement


This is a problem from Boas, Mathematical Methods of the Physical Sciences chapter 5, section 5, number 6.

Find the area of the cylinder x^2+y^2-y=0 inside the sphere x^2+y^2+z^2=1.

Homework Equations


This section deals with projecting curved areas onto a coordinate plane and evaluating using double integrals. The projected surface is related to the curved surface by

dxdy=dA cos γ
or
dA=sec γ dxdy

where γ is measured off the 3rd axis (the one not being projected onto). Also,

sec γ = grad φ / (dφ/dz)

where φ is the equation for the curved surface. So, the differential area can be evaluated through a double integral after choosing the right bounds.

The Attempt at a Solution


In this problem, the cylinder axis is parallel to the z axis, so a projection onto the xy plane will not work. Instead, we can cycle through the axes and modify the equations like:

dA=sec γ dydz
sec γ = grad φ / (dφ/dx)

Where the latter becomes, for our cylinder,
sec γ = 1/(y-y^2)^(1/2)

The equation for the sphere gives the bounds for the double integral, y going from -(1-z^2)^(1/2) and +(1-z^2)^(1/2) while z is just from 0 to 1. I can evaluate one, but not both of the integrals with what I know. I have also tried changing coordinate systems to polar but there was not a simplification.

The answer is supposedly 4, so I feel like I'm missing something obvious.
 

Answers and Replies

  • #2
andrewkirk
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The cylinder is the set of all points whose x,y coordinates lie on the circle in the xy plane with centre C=(0,1/2,0) and radius r=1/2.
We can parametrise that circle by ##\theta##, the angle between the line y=1/2 and the segment CP, where P is the point with parameter ##\theta##.
Express the z coordinate of the intersection between the cylinder above point P in the xyz plane and the upper hemisphere as a function ##g## of ##\theta##, then calculate the surface area of the cylinder above the xy plane as the integral of ##g(\theta)## from 0 to ##2\pi##.
 
  • #3
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Thanks, but I failed to mention I was interested in solutions using the provided equations and not parametrization (which has not been introduced in the text). I realize it is much easier when using that method.
 
  • #4
SammyS
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Homework Statement


This is a problem from Boas, Mathematical Methods of the Physical Sciences chapter 5, section 5, number 6.

Find the area of the cylinder x^2+y^2-y=0 inside the sphere x^2+y^2+z^2=1.

Homework Equations


This section deals with projecting curved areas onto a coordinate plane and evaluating using double integrals. The projected surface is related to the curved surface by

dxdy=dA cos γ
or
dA=sec γ dxdy

where γ is measured off the 3rd axis (the one not being projected onto). Also,

sec γ = grad φ / (dφ/dz)

where φ is the equation for the curved surface. So, the differential area can be evaluated through a double integral after choosing the right bounds.

The Attempt at a Solution


In this problem, the cylinder axis is parallel to the z axis, so a projection onto the xy plane will not work. Instead, we can cycle through the axes and modify the equations like:

dA=sec γ dydz
sec γ = grad φ / (dφ/dx)

Where the latter becomes, for our cylinder,
sec γ = 1/(y-y^2)^(1/2)

The equation for the sphere gives the bounds for the double integral, y going from -(1-z^2)^(1/2) and +(1-z^2)^(1/2) while z is just from 0 to 1. I can evaluate one, but not both of the integrals with what I know. I have also tried changing coordinate systems to polar but there was not a simplification.

The answer is supposedly 4, so I feel like I'm missing something obvious.
I think there is a problem with the integration limits you have.

It looked like determining them might involve a lot of algebra, but your decision to project onto the yz plane makes it very straight forward.

Cylinder: ##\ x^2 + y^2 -y =0 \,. \ ## This gives ##\ x^2 + y^2 = y \,.##

Plug that into:
Sphere: ##\ x^2 + y^2 + z^2 =1 ##

Therefore, ##\ y + z^2 = 1 ##

etc.
 
  • #5
532
31
Thank you so much, that was it. The intersection projection was parabolic, not spherical as I assumed.
 
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