Answer:
Result;

Step-by-step explanation:
Where:
F(x, y, z) = 2(x·i +y·j +z·k) and
S: z = 0, z = 4 -x² - y²
For the solid region between the paraboloid
z = 4 - x² - y²
div F
For S: z = 0, z = 4 -x² - y²
We have the equation of a parabola
To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)
We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

Solving we have;

Plugging the values for N₁ and N₂, we have

Where:
F(x, y, z) = 2(xi +yj +zk) we have




= 32π.
Yea its deez, and thats only because im an expert
Answer: 392
Work:
7×8+3×8×14
56+24×14
56+336
392
Sqaure
rhombus
parallelogram
quadrilateral
hope it helps
They are the same because if you add a zero it doesnt change it