Question:
Find the surface area of the portion of the sphere x^2 + y^2 +z^2 = 4 that lies inside the cylinder x^2 + y^2 = 2y
My work:
Let f(x,y) = sqrt(4- x^2 - y^2)
F(x,y,z) = f(x,y) - z
Required surface area can be calculated using
S = integrate over theta and r (2/(sqrt(4 - r^2))) rdr d(theta), where limit of r, theta as follows:
r: 0 - rsin(theta)
theta: 0 - 2pi
Is this correct?
Your setup is correct, but you need to adjust the limits of integration to reflect the fact that the surface lies inside the cylinder.
First, convert the equation of the cylinder to polar coordinates:
x^2 + y^2 = 2y
r^2 = 2r sin(theta)
r = 2sin(theta)
Thus, the limits of integration for r become 0 to 2sin(theta) since the surface lies inside the cylinder.
So the setup for the surface area becomes:
S = ∬2/(sqrt(4 - r^2)) r dr dθ
Limits:
θ: 0 to π
r: 0 to 2sin(theta)
Now you can evaluate the integral to find the surface area.