Decent problem (Calc 3)

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:

Code:

`∫∫s F dS ... F = e^(-y) i - y j + x sin(z) k`

s: the elliptic cylinder r(u,v) = 2 cos(v) i + sin(v) j + u k

... 0 ≤ u ≤ 5, 0 ≤ v ≤ 2π

∫∫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(θ) ; v = r sin(θ) ; z = z

∂(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

Also, this one isn't too bad without the divergence theorem

because obviously there isn't any flux across the ends.

∫∫s F dS

= ∫∫s F(u,v)·<∂r/∂u x ∂r/∂v > du dv

∂r/∂u x ∂r/∂v = <-cos(v), 2 sin(v), 0>

F(u,v) = <e^(-sin(v)),-sin(v),2 cos(v) sin(u)>

= ∫∫ -cos(v) e^(-sin(v)) - 2 sin²(v) du dv

{(u,v) | 0 ≤ u ≤ 5 ; 0 ≤ v ≤ 2π}

= 5 ∫ -cos(v) e^(-sin(v)) - 2 sin²(v) dv

{(v) | 0 ≤ v ≤ 2π}

= -10π

Answer (1): -10π

2. ∫c F·dr .... F = (x + y) i + (xy) j - (z^2) k ; c: from (0,0,0) to (1,3,1) to (2,-1,4)

→ (0,0,0) to (1,3,1)

x = t ; y = 3t ; z = t ; 0 ≤ t ≤ 1

dx = 1 dt; dy = 3 dt; dz = 1 dt

∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz

= ∫ (t + 3t) + 9t² - t² dt [0,1]

= 14/3

→ (1,3,1) to (2,-1,4)

x = t + 1 ; y = 3 - 4t ; z = 1 + 3t

dx = 1 dt; dy = -4 dt; dz = 3 dt

∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz

= ∫ (t + 1 + 3 - 4t) - 4(t + 1)(3 - 4t) - 3(1 + 3t)^2 dt [0,1]

= ∫ -11t^2 - 17t - 11 dt [0,1]

= -139/6

14/3 - 139/6 = -37/2

Answer (2): -37/2