Question

1) Find the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ wit outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi.

2) Find the work done by the force field F(x, y, z) = (x + y) i + (xy) j - (z^2) k on a particle that moves along line segments from (0,0,0) to (1,3,1) to (2,-1,4).

Answer 1

sorry wat 2 high 4 me

Answer 2

\[\int\limits{\int\limits{s F dS}} = ∫∫∫v ∇·F dV\]
\[∇·F = x \cos(z) - 1\]

\[= ∫∫∫v x \cos(z) - 1 dx dy dz\]
\[v: x² + 4y² ≤ 4 ; 0 ≤ z ≤ 5\]
let u = x , v = 2y , z = z
∂(u,v)/∂(x,y) = 2

= 1/2 ∫∫∫v u cos(z) - 1 du dv dz
v: u² + v² ≤ 4 ; 0 ≤ z ≤ 5

let u = r cos(theta) ;
v = r sin(theta)
z = z
\[\frac{∂(u,v)}{∂(r,θ)} = r\]
\[= 1/2 ∫∫∫ (r \cos(θ) \cos(z) - 1) r dr dθ dz\]
\[{(r,θ,z) | 0 ≤ r ≤ 2 ; 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}\]

\[= 1/3 ∫∫ (4 \cos(θ) \cos(z) - 3) dθ dz\]
\[{(θ,z) | 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5}\]

\[= -2π ∫ dz\](z) | 0 ≤ z ≤ 5

\[= -10π\]

Answer 3

oh
2 smaalllllllll

Answer 4

you know the second one

Answer 5

is the first one right? or u dont know tha answer?

Answer 6

im just checking what i got wrong on my hw and yeah the answer is right

Answer 7

oh ok im thinkin

Answer 8

k thanks lana :)

Answer 9

(0,0,0) -->> (1,3,1)
x = t y = 3t z = t
dx =dt........ dy = 3 dt...... dz = dt

work done is integration of F.dr
int{ (x + y)dx + (xy)dy - (z^2) dz

Answer 10

now substitute x=t y=3t and z=t
and the dx dy and dz values in terms of dt

Answer 11

then integrate