Compute the surface integral over the given oriented surface: f=y3i+8j−xk, portion of the plane x+y+z=1 in the octant x, y,z≥0, downward-pointing normal

We are to find the surface integral of

where;

∴

The plane x + y + z = 1

z = 1 - x - y

Since the surface is oriented downward

∴

However,

Flux =

Therefore, we can conclude that the surface integral is

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Parameterize the surface (call it ) by

with and . Take the normal vector to to be

Then the integral of over with the given orientation is

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11