Surface Integral: Finding K = $\int\int_S z/2 dA$

[squenshl]

Let S be a parametrised surface given by (x, y, z) = R(u, v) := (u², v², u + v), for 0 \leq u \leq 1 and
0 \leq v \leq 1. How do I find the integral K := \int\int_S z/2 dA.

 

[quasar987]

By definition, this integral is

\int_0^1\int_0^1 \frac{u+v}{2}\sqrt{E(u,v)G(u,v)-(F(u,v))^2}dudv

where
E=R_u\cdot R_u
F=R_u\cdot R_v
G=R_v\cdot R_v

And this you know how to do.

 

[squenshl]

That's easy. Cheers.