## Evaluate! surface integral over surface

1. The problem statement, all variables and given/known data
Evaluate the surface integral of G over the surface S
S is the parabolic cylinder y=2x^2, 0=< x =<5, 0=< z =<5
G(x,y,z)=6x

Answer is one of the following:
1. (15/8)*(401sqrt(401)-1)
2. (5/8)*(401sqrt(401)-1)
3. (15/8)*(401sqrt(401)+1)
4. (5/8)*(401sqrt(401)+1)

2. Relevant equations

3. The attempt at a solution
Let f(x,y,z)=y-2x^2=0
 Let me try ot reason through what you have. If at the end you were integrating dzdx, I'll assume I should try to paramet(e)rize the surface in x and z. Then r(x,z)=(x,2x^2,z). Then r_x=(1,4x,0), while r_z=(0,0,1). Then cross product is (r_x)x(r_z)=(4x,-1,0). Then the infinitesimal area on the surface is given by dS=sqrt(16x^2+1)dxdz. Now we want to integrate the SCALAR G(x,y,z)=4x against the area. That is, integrate G dS. So $\int_{x=0}^5\int_{z=0}^54x\sqrt{16x^2+1}\ dzdx.$ So my first guess is, you are mixing up VECTOR integrals with SCALAR integrals. You might compare them and their derivations.