Find the surface area of the part of the sphere x2 + y2 + z2 = 4 that lies betwe

Question

Find the surface area of the part of the sphere x2 + y2 + z2 = 4 that lies between the planes z = -1 and z = 1.

Explanation / Answer

Spherical coordinates are defined as x = ?·sin(f)·cos(?) y = ?·sin(f)·sin(?) z = ?·cos(f) The limits describing the whole sphere or Radius R are: 0 = ? = R 0 = f = p 0 = ? = 2p Describing a spherical surface just fix ? to the radius of the sphere. ? = R = 4^(1/3) = 4^(1/3) The limits for z lead to limits for the azimuthal angle f: -1 = z = 1 -1 = R·cos(f) = 1 -1/R = cos(f) = 1/R => arccos(1/R) = f = arccos(-1/R) For points on the surface the polar angle ? make take any value: 0 = ? = 2p To find the surface integral integrate with dS = R²·sin(f) dfd? and the limits derived above. I'm not sure, what integral you want to calculate, because you question is a bit unclear at this point. e.g the surface of these object is ? dS = 2p arccos(-1/R) ? ? R²·sin(f) d?df = 0 arccos(1/R) arccos(-1/R) ? 2·p·R²·sin(f) df = arccos(1/R) -2·p·R²·(cos(arccos(-1/R) - cos(arccos(+1/R) = 8·p·R = 32·p·1^(1/3)

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