Let S be the part of the paraboloid x^2 + y^2 = z with 0 <= z <= 1. Set up but do not evaluate the integral ∫∫_S (yz)dS.

Question

Let S be the part of the paraboloid x^2 + y^2 = z with 0 <= z <= 1. Set up but do not evaluate the integral ∫∫_S (yz)dS.

Answer 1

The wording of this question is a bit confusing because the symbol S seems to be used for two things: S is part of the paraboloid ( z between 0 and 1) but also a function of yz. . If it is a function, it is not clear whether a scalar or vector function is involved.

For now I will assume the simplest case - that is, we want the area of the surface for z between 0 and 1

This is given by A = ∫∫ dx dy sqrt( (dz/dx)² + (dz/dy)² +1 )

with z = x² + y²

letting ρ = x² + y² This becomes A = 2π ∫ ρ sqrt( 4ρ² +1) dρ

with the limits 0 and 1