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) = xy i + yz j + zx k s is the part of the paraboloid z = 8 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

The flux of F across S is 8.627.

Step-by-step explanation:

Given,

F(x, y, z) = xy i + yz j + zx k

or F=(xy, yz, zx)

S is the part of the paraboloid above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

By differentiating z with respect to x we get:

fx=-2x

By differentiating z with respect to y we get:

fy=-2y

So, Surface integral is given by:

Integrating with respect y:

After Substituting limits of y, we get:

Integrating with respect x:

After Substituting limits of x, we get:

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Parameterize

with and .

Take the normal vector to to be

Then the flux of across is

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

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step-by-step explanation: