Use a parameterization to find the flux ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S Bold Upper F times n d sigma of Bold Upper F equals z squared Bold i plus xj minus 4 z Bold k in the outward direction (normal away from the x-axis) across the surface cut from the parabolic cylinder zequals1minusysquared by the planes xequals0, xequals1, and zequals0.

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Use a parameterization to find the flux of F = z²i + xj - 2zk in the outward direction

(normal away from the x-axis) across the surface cut from the parabolic cylinder z = 16 -y²  by the planes x = 0 , x = 1, and z = 0 .

The flux is                  .

Step-by-step explanation:

F(x,y,z) = (z² ,x , - 2z)

To find the flux across a surface :

The sphere z is given as =  16 -y²,    above x = 0 , x= 1, and z = 0

When z = 0 ⇒ 16 - y² = 0

⇒ y = ± 4

The parameterization is y=v⇒z= 16 -v²    and x= u

⇒ r = (x,y, z) = ( u,v,16 - v²)

with 0 ≤ u ≤ 1      and     -4 ≤  v ≤  4

However the normal vector can now be determined as :

=

i pretty sure it is

a .4y4

hope this !