Verified answer

For S, we have R(u, v) = <u+v, u-v, 1+2u+v>.

R_u = <1, 1, 2>

R_v = <1, -1, 1>.

So, R_u x R_v = <3, 1, -2>.

==> ||R_u x R_v|| = √14.

So, ∫∫s (x+y+z) ds

= ∫(v = 0 to 1) ∫(u = 0 to 2) ((u+v) + (u-v) + (1+2u+v)) * √14 du dv

= √14 ∫(v = 0 to 1) ∫(u = 0 to 2) (1 + 4u + v) du dv

= √14 ∫(v = 0 to 1) (u + 2u^2 + uv) {for u = 0 to 2} dv

= √14 ∫(v = 0 to 1) (10 + 2v) dv

= √14 (10v + v^2) {for v = 0 to 1}

= 11√14.

I hope this helps!

this is a bit late but here goes: first you take the s parallelogram and take the partials of u and v and end up with r_u=<1,1,2> and r_v=<1,-1,1> so then r_u x r_v = <3,-1,-2> (but since the orientation must be positive, you multiply it by -1 so the k component is positive, thus giving you) <-3,1,2>. and that will be the r_u x r_v you will use. next we attack the F(x,y,z). since we are going to take the dot product of the function and the cross product of the partials, F has to be in terms of u and v as well so just substitute from the equations given from the parallelogram. then you will have F in terms of u and v, which should look like: F(u,v)= <(1+2u+v)e^(u^2-v^2),3(1+2u+v)e^(u^2-v^2),u^2-v^2>. from there take the dot product of F(u,v) and (r_u x r_v) and it will beautifully cancel stuff out till you re left with 2(u^2-v^2) which from there you will do the double integral 2(u^2-v^2) using the boundaries previously given by the parallelogram. doesnt matter which one goes first since they are constants. *sorry, can t do everything otherwise there s no learning involved* but so as to check if you re right or not; the answer is 4.