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Evaluate the surface integral


S
F · dS
for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = yj − zk,
S consists of the paraboloid
y = x2 + z2, 0 ≤ y ≤ 1,
and the disk
x2 + z2 ≤ 1, y = 1.

Respuesta :

LammettHash

Since S is closed, you can use the divergence theorem, which says

[tex]\displaystyle\iint_S \vec F(x,y,z) \cdot \mathrm d\vec S=\iiint_R \mathrm{div}\vec F(x,y,z)\,\mathrm dV[/tex]

where R is the interior of the surface S.

We have

div F (x, y, z) = ∂/∂x (0) + ∂/∂y (y) + ∂/∂z (-z) = 1 - 1 = 0

so the flux would be 0.

lhmarianateixeira

In this exercise we have to calculate the flux by the divergent theorem:

The flux would be ZERO

Since S is closed, you can use the divergence theorem, which says:

[tex]\int\limits\int\limits_s {F(x,y,z} \, ds= \int\limits \int\limits\int\limits_R {divF(x, y, z)} \, dv[/tex]

where R is the interior of the surface S. We have:

[tex]\nabla F (x, y, z) = 1-1= 0[/tex]

so the flux would be ZERO.

See more about vectorial calculus at : brainly.com/question/6960786

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