Question

F(x, y) represents a velocity field of a fluid over a surface S defined by z = 6 − 3x − 2y. If the magnitude of the velocity in the direction of the unit normal vector, n̂, on S is 3z⁄√14, compute the flux of F(x, y) over the surface S in the first octant oriented upward, using the projection of S on the xy - plane

Answer 1

http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx

boundary in first quadrant means x from 0 to 2 and y from 0 to 3
(sketch that thing)

Answer 2

n = 3 i +2 j + k

we want F dot n
= |F| |n| cos angle
but we are given that |F| cos angle is 3 z/sqrt14
so what is |n|?
sqrt (9+4+1) = sqrt 14
ha, that helps :)

so F dot n = 3 z
so
do the integral of
3 z
over the surface

Answer 3

thanks for your help. do you mean F.n=3z/sqrt14??

Answer 4

Thanks sir Damon.. Understood with ur explanation..

Answer 5

No, because multiply by sqrt 14 which is |n|

Answer 6

Advanced Engineering Mathematics
By Dennis G. Zill, Michael R. Cullen

you can get it from preview mode on google books page 529 and 530 it got the similar question... be sure to check it out. The answer is 18