Evaluate! surface integral over surface

by ride4life
Tags: evaluate, integral, surface
 P: 33 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 n=gradf/||gradf||=(-4xi+1j+0k)/sqrt(16x^2+1) n*=j G.n=-24x^2/sqrt(16x^2+1) n.n*=1/sqrt(16x^2+1) therefore the double integral over S = SS (G.n/n.n*) dzdx solving the double integral gets -5000
 P: 428 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.

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