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Evaluate! surface integral over surface 
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#1
May2912, 12:31 AM

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)=y2x^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 


#2
May2912, 01:04 AM

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 [itex]\int_{x=0}^5\int_{z=0}^54x\sqrt{16x^2+1}\ dzdx.[/itex] 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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