Question

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 xequals​0, xequals​1, and zequals0.

Answer 1

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Use a parameterization to find the flux $\int\limits \int\limits_s \ F. \ n \ d \sigma$ 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:

$-\frac{512}{3}$

Step-by-step explanation:

F(x,y,z) = (z² ,x , - 2z)

To find the flux across a surface : $\int\limits \int\limits_s \ F. \ n \ dS = \int\limits \int\limits_D \ F. (r_u * r_v) \ dA$

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 :

$r_u = (1,0,0) , \ \ \ r_v = (0,1, - 2v)$

$n = r_u*r_v = \left[\begin{array}{ccc}i&j&k\\1&0&0\\0&1&-2v\end{array}\right] = (0,2v,1)$

$\int\limits \int\limits_s \ F. \ n \ dS = \int\limits \int\limits_D \ F. (r(u,v)) .(r_u*r_v)dA$

$=\int\limits^1_0 \int\limits^4_{-4} [(16-v^2)^2 , u , -2 (16-v^2)](0,2v,1)dvdu$

$=\int\limits^1_0 \int\limits^4_{-4} [2uv-2(16-v^2))dvdu$

$=\int\limits^1_0 (uv^2 -32v+ \frac{2v^3}{3})^ ^4}__{-4}} du$

$=\int\limits^1_0 (u(16-16)-32(8)+\frac{2}{3}(128))du$

$= \int\limits^1_0 (-\frac{512}{3})du$

$= - \frac{512}{3}(u)^ ^1}__0$

= $-\frac{512}{3}$