Fiding area of a surface

let S be the parallelogram given by
r(u, v) = (u + v) i + (u − v) j + 2u k; 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.

find the area of S

The area of S is represented by the integral

∫[v = 0 to 1] ∫[u = 0 to 1] |r_u x r_v| du dv.

We compute

r_u = i + j + 2k

r_v = i - j + 0k,

r_u x r_v = 2i +2k - 2k,

|r_u x r_v| = √(4 + 4 + 4) = 2√3,

so then

∫[v = 0 to 1] ∫[u = 0 to 1] 2√3 du dv = 2√3.

r(u,v)=u(1,1,2) + v(1,-1,0)
so the two vectors forming the sides of S are (1,1,2) and (1,-1,0).
Call these a and b respectively.

We have a.b (vector dot product) = 1-1+0=0, so a and b are perpendicular,
so the figure is actually a square. Hence the area is ab, where a and b
are the lengths of the 2 vectors.

a=sqrt(1^2+1^2+2^2)=sqrt6
b=sqrt(1^2+1^2)=sqrt2

So Area = ab=sqrt12 = 2sqrt3