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.
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
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.