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.
2 answers:
Answer:
Area = 32/3
Step-by-step explanation:
x = u ²
y = uv
z = 12v ²
0 ≤ u ≤ 2
0 ≤ v ≤ 1
Since ru = <2u, v, 0> and rv = <0, u, 24v>, we have
Where 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 rv
ru × rv = 24v²i - 48uv²j + 2u²k
absolute value of ru × rv = 2u² + 24v²
We can now find the area
∫₀² du ∫₀¹ dv (2u² + 24v²) = ∫₀² du [2u²v + 8v³]₀¹ = 32/3
Detailed description can be found in the attachment
You might be interested in