find the flux of F=<0,y,z> across the parabolic sheet given by x=2y^2+2z^2+3, with x <= 5
Flux = I = ∫∫F.n dS
for S over the surface, g(x,y,z), where
g(x,y,z) is the part of
x=2y²+2z²+3 where x≤5, which translates to
g(x,y,z)=x-2y²-2z²-3
and the region is the part of the paraboloid where y²+z²≤1 after substitution of x≤5.
This region is circular with radius =1, which facilitates integration later.
Now:
F = <0,y,z>
n=∇g/||g||
so I can be evaluated:
I = ∫∫F.n dS
The surface integral can be evaluated by projecting it to the region R on the y-z plane by multiplying:
dS over S = ||g|| dA over R
So
I = ∫∫F.n dS over S
= ∫∫F.n ||g|| dA over R
where R is the circular region r≤1 in polar coordinates.
The way I have defined g(x,y,z) results in negative flux.
Post if you could use more help.
To find the flux of a vector field across a surface, we can use the surface integral. The formula for the surface integral of a vector field F across a surface S is:
∬S F · dS
where F is the vector field, dS is the differential area vector of the surface S, and the double integral is taken over the surface S.
In this case, the vector field F is given as F = <0, y, z> and the surface S is the parabolic sheet given by x = 2y^2 + 2z^2 + 3, with x ≤ 5.
To evaluate the surface integral, we need to determine an expression for the differential area vector dS.
Since the surface S is defined implicitly by the equation x = 2y^2 + 2z^2 + 3, we can express the differential area vector dS in terms of the partial derivatives of the function that defines the surface.
The general expression for the differential area vector dS of a surface defined implicitly by the equation F(x, y, z) = 0 is given by:
dS = (∂F/∂x) dx + (∂F/∂y) dy + (∂F/∂z) dz
In our case, the equation of the surface is x = 2y^2 + 2z^2 + 3. So, we have:
F(x, y, z) = x - (2y^2 + 2z^2 + 3) = 0
Taking the partial derivatives with respect to x, y, and z, we get:
∂F/∂x = 1
∂F/∂y = -4y
∂F/∂z = -4z
Substituting these partial derivatives into the expression for the differential area vector dS, we have:
dS = dx - 4y dy - 4z dz
Now, we can evaluate the surface integral by substituting the expression for F and dS into the original formula:
∬S F · dS = ∬S (0, y, z) · (dx - 4y dy - 4z dz)
Integrating over the surface S given by x = 2y^2 + 2z^2 + 3 and x ≤ 5 will require parametrization of the surface and the proper bounds of integration. The specific parametrization and bounds for this problem are not provided in the original question.