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) = xzey i − xzey j + z k s is the part of the plane x + y + z = 3 in the first octant and has upward orientation incorrect: your answer is incorrect. 9/2

I'm pretty sure your answer of 9/2 is correct...

Parameterize by

with and .

In case youre' not sure where this came from: recall the standard parameterization for a line segment connecting two points and ,

with . Then treating as a "point", we can parameterize a plane containing and another point by using

with .

Now, take the normal vector to to be

Then the flux of across is

(since the first two components of cancel in the dot product)

step-by-step explanation:

abc