Let F = <x, z, xz> evaluate ∫∫F⋅dS for the following region:
x2+y2≤z≤1 and x≥0
∫∫∫(∇⋅F)dV = ∫∫F⋅dS
The Attempt at a Solution
This is the graph of the entire function:
Thank you Wolfram Alpha.
But my surface is just the half of this paraboloid where x is positive. So I thought if I looked down the x-axis I would get something like this:
But only the right half of the circle (from -3π/2 to π/2)...
The integral I set up is the following:
∫∫∫xdzdxdy (x is the dot product of ∇ and F)
I converted to polar coordinates
I ended up getting
(4/15)[sin(π/2)-sin(-3π/2)] = 0
Answer should be 4/15 according to the back of the book.