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Evaluate! surface integral over surface

by ride4life
Tags: evaluate, integral, surface
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ride4life
#1
May29-12, 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)=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
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algebrat
#2
May29-12, 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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