Consider the surface S shown on the graph, whose parameterization is given by: r(u, v) = (v², uv, 4-²) where ≤ ≤2, 0≤ ≤2 z The area differential, ds, is given by: A) ds = √2u²-4uv-2² dA B) dS=√16u¹ +4u²y² + 4v¹ dA C) ds = √4u¹ +4u²v² + 16vª dA == D) dS=√4u + 16u²v² + 4v² da

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A: Let's find.

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