Find the flux of 𝐹⃑ across the positive oriented 𝑆 if 𝐹⃑ = 𝑥𝑖⃑ + 2𝑗⃑ + 𝑧𝑘⃑⃑ and 𝑆 is the boundary of the region enclosed by the cylinder 𝑦^2 + 𝑧^2 = 1 and the planes 𝑥 = 0 and 𝑥 + 𝑧 = 1

Question

Find the flux of 𝐹⃑ across the positive oriented 𝑆 if 𝐹⃑ = 𝑥𝑖⃑ + 2𝑗⃑ + 𝑧𝑘⃑⃑ and 𝑆 is the boundary of the region enclosed by the cylinder 𝑦^2 + 𝑧^2 = 1 and the planes 𝑥 = 0 and 𝑥 + 𝑧 = 1. (NO use of the Divergence Theorem (section 16.9).)

Answer 1

This is an exercise on Green's theorem, where the flux form of double integral of the field's divergence over the region it encloses determines the outward flux of a vector field across a closed curve. Green's Double Integral for flux Φ is written as follows:

Φ (Flux of field F on surface S)  = ∫∫_S F.dS

Note that both F and S are vectors.

The dot product F.dS determines the intensity of flux flow along the normal vector n to the surface S.

Surface S comprises of three surfaces of a bounded cylinder.

Therefore, our task is to determine the following the expressions:

1) the surface unit vectors of dS on the three surfaces.

2) the values of surface elements of dS on the three surfaces.

3) the products F.dS for the thee surfaces.

4) the double integrals S = ∫∫_S F.dS for the three surfaces.

Given:

Fluid vector: F = x i + 2 j + z k ........(1)

Geometrical surfaces S:

C_y : Cylinder: y² + z² = 1 ......(2)

P_o: : End plane:  x = 0 .........(3)

P₁ : End plane:  x + z = 1. ......(4)

 

Unit vectors of surfaces:

Rewrite equations (2), (3), and (4), in vector form, we get:

C_y : n₁ = 0 i + y j + z k = y j + z k .........( lateral walls of cylinder)

P_o: n₂ = - i + 0 j + 0 k = - i ............(negative for outward flow)

P₁: n₃ = i + 0 j + k = i + k ............( positive outward flow on opposite side of cylinder)

Values of surface are differential dS for the three planes:

(1) Cylinder wall C_y:

We need to transform the cartesian coordinates onto polar coordinates such that we could account for the curvature of the cylinder.

Since the cylinder radius, r = 1 = sqrt( y² + z² ), then

y = 1. cos θ

z = 1. sin θ

x = 1 - z = 1 - 1. sin θ

and

dS_c = 1. dx.dθ ............(5)

Equation (5) accounts for the cylindrical shape along the x-axis of a cylinder of radius 1.

(2) Surface area element on P_o:

dS_po = 1. dz.dy

......= - dx.(-sinθ ) dθ

......= sinθ.dx.dθ ............(6)

(3) Surface area element on P₁:

dS_p1 = 1. dz.dy / | n_3.. i |

Where, the dot product | n_1.. j | accounts for the slanted plane (x + z = 1)

| n₃ . i | = ( i + k ). j = 1

i.e., dS_p1 = dz.dy...... ............(7)

Flux differentials:

Substituting by n₁, n₂, n₃, dS_c , dS_po, dS_p1 in integrand of Green's double integral, we get three integrands of the forms:

(1) F .n₁ . dS_c = (x i + 2 j + z k ) . ( y j + z k). dx.dθ ........-(8)

(2) F .n₂ . dS_po = (x i + 2 j + z k ) . ( - i ). sinθ.dx.dθ ........(9)

(3) F .n₃ . dS_p1 = (x i + 2 j + z k ) . ( i + k ). dz.dy.. ........(10)

Equations (8), (9), and (10), are simplified as follows:

(1a) F .n₁ . dS_c = ( 2y + z² ). dx.dθ ........-(8a)

(2a) F .n₂ . dS_po = - x. sinθ.dx.dθ = 0 ......(9a)

( because x = 0 on that plane )

(3a) F .n₃ . dS_p1 = (x + z). dz.dy.. = dz.dy........(10a)

(because x + z = 1 on that plane)

Net fluxes of the three surfaces:

(1) Φ_c = ∫∫_c ( 2y + z² ). dx.dθ ........-(11)

(2) Φ_po = 0 ..............................--(12)

(3) Φ_p1 = ∫∫_p1 dz.dy.. ....................(13)

Substituting by y = cos θ, z = sin θ, and x = 1-sin θ, we get:

(1a) Φ_c = ∫∫_c ( 2 cos θ + sin ² θ ). dx.dθ ........-(11a)

(2a) Φ_po = 0 ..............................--(12a)

(3a) Φ_p1 = ∫∫_p1 (cos θ dθ).(-sin θ d θ) .. ........-(13a)

Green's Double Integration:

Equation (11a) requires serious attention to the limits of x, since

x = 1 -z = x = 1-sin θ

Therefore, we the first integral is bounded by x = 0 and 1-sin θ.

Thus,

Φ_c = ∫∫_c ( 2 cos θ + sin ² θ ). dx.dθ

......= ∫ ( 2 cos θ + sin ² θ ). [ x ] ₀ ^1-sinθ . dθ (integrating from x = 0 to x = 1-sin θ)

......= ∫ ( 2 cos θ + sin ² θ ). (1-sin θ) . dθ

Distributing the terms of the polynomial, we get

.Φ_c.= ∫ ( 2 cos θ + sin ² θ - 2 cos θ sin θ - sin ³ θ ) . dθ ( θ varies from 0 to 2 π)

Applying basic rules of integration, substitution, and trigonometry, we get

.Φ_c.= [ 2 sin θ + cos θ + (1/2) θ - (1/4) sin(2θ) - (1/3) cos³ θ ] ₀ ^2π

......= [ 1+ π- (1/3) ] - [ 1 - (1/3) ]

......= π ..............-(14)

And we already proved that

Φ_po = 0 ..............-(12a)

Finally,

Φ_p1 = ∫∫_p1 (cos θ dθ).(-sin θ d θ)

Final Answer:

Φ_c + Φ_p0 + Φ_p1 = π + 0 + π = 2 π