# Surface of a Cylinder inside a Sphere

## 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.

andrewkirk
Homework Helper
Gold Member
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##.

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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.

Thank you so much, that was it. The intersection projection was parabolic, not spherical as I assumed.

SammyS