MATH SOLVE

3 months ago

Q:
# Find the area of the surface. The surface with parametric equations x = u2, y = uv, z = 1 2 v2, 0 ≤ u ≤ 2, 0 ≤ v ≤ 1.

Accepted Solution

A:

Answer:Area = 32/3Step-by-step explanation:x = u
²y = uvz = 12v
²0 ≤ u ≤ 20 ≤ v ≤ 1Since ru = <2u, v, 0> and rv = <0, u, 24v>, we haveWhere ru is the differentiation of x, y, z with respect to u and rv is the differentiation of x, y, z wit respect to v.we find the cross product of ru and rvru × rv = 24v²i - 48uv²j + 2u²kabsolute value of ru × rv = 2u² + 24v²We can now find the area∫₀² du ∫₀¹ dv (2u² + 24v²) = ∫₀² du [2u²v + 8v³]₀¹ = 32/3Detailed description can be found in the attachment