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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 <u> .</u>
Answer:
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 :
=