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

g

step-by-step explanation:

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.