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Use a parameterization to find the flux [tex]\int\limits \int\limits_s \ F. \ n \ d \sigma[/tex] 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 .
Answer:
[tex]-\frac{512}{3}[/tex]
Step-by-step explanation:
F(x,y,z) = (z² ,x , - 2z)
To find the flux across a surface : [tex]\int\limits \int\limits_s \ F. \ n \ dS = \int\limits \int\limits_D \ F. (r_u * r_v) \ dA[/tex]
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 :
[tex]r_u = (1,0,0) , \ \ \ r_v = (0,1, - 2v)[/tex]
[tex]n = r_u*r_v = \left[\begin{array}{ccc}i&j&k\\1&0&0\\0&1&-2v\end{array}\right] = (0,2v,1)[/tex]
[tex]\int\limits \int\limits_s \ F. \ n \ dS = \int\limits \int\limits_D \ F. (r(u,v)) .(r_u*r_v)dA[/tex]
[tex]=\int\limits^1_0 \int\limits^4_{-4} [(16-v^2)^2 , u , -2 (16-v^2)](0,2v,1)dvdu[/tex]
[tex]=\int\limits^1_0 \int\limits^4_{-4} [2uv-2(16-v^2))dvdu[/tex]
[tex]=\int\limits^1_0 (uv^2 -32v+ \frac{2v^3}{3})^ ^4}__{-4}} du[/tex]
[tex]=\int\limits^1_0 (u(16-16)-32(8)+\frac{2}{3}(128))du[/tex]
[tex]= \int\limits^1_0 (-\frac{512}{3})du[/tex]
[tex]= - \frac{512}{3}(u)^ ^1}__0[/tex]
= [tex]-\frac{512}{3}[/tex]
