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.

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:

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

so the flux would be ZERO.

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

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.

g(x) = x² + 2

step-by-step explanation:

it is given that the graphs are the same shape.

the only difference is that the red graph is 2 units higher than the blue graph.

that is to say that the red graph is simply the blue graph that has been moved vertically by 2 units in the positive y direction.

hence

g(x) = f(x) + 2

or

g(x) = x² + 2