Question

Evaluate the surface integral SSs (x + y + z)ds, s is the parallelogram with parametric equations x = u + v, y = u - v, z = 1 + 2u +v, 0 <= u <= 2, 0 <= v <=1

Answer 1

The surface element is

dS = || ∂r/∂u x ∂r/dv || du dv

where

r (u, v) = x (u, v) i + y (u, v) j + z (u, v) k

r (u, v) = (u + v) i + (u - v) j + (1 + 2u + v) k

is the vector parameterization for the surface S.

The normal vector to the surface is

∂r/∂u x ∂r/dv

… = (i + j + 2 k) x (i - j + k)

… = 3 i + j - 2 k

and has norm

|| ∂r/∂u x ∂r/dv || = √(3² + 1² + (-2)²) = √(14)

Now compute the integral:

∫∫ (x + y + z) dS = √(14) ∫₀¹ ∫₀² ((u + v) + (u - v) + (1 + 2u + v)) du dv

… = √(14) ∫₀¹ ∫₀² (3u + v + 1) du dv

… = √(14) ∫₀¹ (3/2 u² + uv + u) |₀² dv

… = √(14) ∫₀¹ (8 + 2v) dv

… = √(14) (8v + 2v²) |₀¹

… = 10 √(14)