Vector calculus. Surface integrals of the second kind. Flux

What it is

The flux of a vector field through an oriented surface:

Φ = S \iint F \cdot d S = S \iint F \cdot n d S,

where $n$ is the unit normal of the chosen side. Changing the side changes the sign of the flux, so fix the orientation before you start computing, not by adjusting the sign at the end.

The working formula for an explicit surface

If $S$ is given as $z = z (x, y)$ over a domain $D$ and the chosen normal makes an acute angle with the $O z$ axis (that is, $n_{z} > 0$ ), then

Φ = D \iint (- P z_{x} - Q z_{y} + R) d x d y .

The factor $1 + z_{x}^{2} + z_{y}^{2}$ does not appear here: it cancels against the normalisation of the normal. This is the main difference from the integral of the first kind, and it is exactly where most mistakes are made.

The same thing written conveniently: the vector $N = (- z_{x}, - z_{y}, 1)$ is a non-normalised normal with $n_{z} > 0$ , and $Φ = \iint_{D} F \cdot N d x d y$ . For a plane $A x + B y + C z = D$ with $C > 0$ this $N$ is proportional to $(A, B, C)$ .

The parametric form

Φ = D_{uv} \iint F (r (u, v)) \cdot (r_{u} \times r_{v}) d u d v,

and the order of the factors in the cross product fixes the side: $r_{v} \times r_{u}$ gives the opposite one. For the lateral surface of the cylinder $x^{2} + y^{2} = R^{2}$ the outward normal is simply $n = (cos θ, sin θ, 0)$ , and $d S = R d θ d z$ .

Common mistakes

Putting in a spurious factor $1 + z_{x}^{2} + z_{y}^{2}$ — the typical confusion with the integral of the first kind.
Taking the normal of the wrong side. The check: a normal at an acute angle to $O z$ has a positive third component; the outward normal of a closed surface points away from the solid.
Applying the divergence theorem to a surface that is not closed. A piece of the lateral surface of a cylinder bounds no volume, and no divergence will give its flux. The surface can be closed with caps, but then their flux must be subtracted — and if the surface is also cut off by an angle, flat side walls are added and the trick stops saving time. Direct computation is shorter.

Problems

Problem 1 (The flux of a field through a piece of a plane). Find the flux of the vector field $a = 2 x i + y j - 2 z k$ through the part of the plane $P : 2 x + \frac{y}{2} + z = 1$ lying in the first octant, in the direction of the normal that makes an acute angle with the $O z$ axis.

Solution.

The normal. Express $z$ from the equation of the plane:

z = 1 - 2 x - \frac{y}{2}, z_{x} = - 2, z_{y} = - \frac{1}{2} .

The non-normalised normal with a positive third component:

N = (- z_{x}; - z_{y}; 1) = (2; \frac{1}{2}; 1) .

This is exactly the vector of the plane's coefficients, and its third component is positive — the "acute angle with $O z$ " condition holds, and nothing needs to be turned round.

The formula. For an explicitly given surface

Φ = D \iint a \cdot N d x d y .

There is no factor $1 + z_{x}^{2} + z_{y}^{2}$ here — it cancelled against the normalisation of the normal. This is the main difference from the integral of the first kind.

The integrand.

a \cdot N = 2 x \cdot 2 + y \cdot \frac{1}{2} + (- 2 z) \cdot 1 = 4 x + \frac{y}{2} - 2 z .

Substitute $z = 1 - 2 x - y /2$ :

= 4 x + \frac{y}{2} - 2 + 4 x + y = 8 x + \frac{3 y}{2} - 2.

The projection. The first octant means $x ⩾ 0$ , $y ⩾ 0$ , $z ⩾ 0$ . The last condition gives

1 - 2 x - \frac{y}{2} ⩾ 0 ⟹ y ⩽ 2 - 4 x .

The domain $D$ is the triangle with vertices $(0; 0)$ , $(0; 2)$ , $(1/2; 0)$ .

Integration. The inner integral in $y$ from $0$ to $2 - 4 x$ :

[8 x y + \frac{3 y ^{2}}{4} - 2 y]_{0}^{2 - 4 x} = 8 x (2 - 4 x) + \frac{3 ( 2 - 4 x ) ^{2}}{4} - 2 (2 - 4 x) .

= 16 x - 32 x^{2} + 3 (1 - 4 x + 4 x^{2}) - 4 + 8 x = - 20 x^{2} + 12 x - 1.

The outer integral:

Φ = \int_{0}^{1/2} (- 20 x^{2} + 12 x - 1) d x = [- \frac{20 x ^{3}}{3} + 6 x^{2} - x]_{0}^{1/2} .

= - \frac{20}{24} + \frac{6}{4} - \frac{1}{2} = - \frac{5}{6} + \frac{3}{2} - \frac{1}{2} = \frac{1}{6} .

Answer: $Φ = \frac{1}{6}$ .

Problem 2 (The flux of a field through a piece of a plane (variant)). Find the flux of the vector field $a = x i + 2 y j + z k$ through the part of the plane $x + y + 2 z = 2$ lying in the first octant, in the direction of the normal that makes an acute angle with the $O z$ axis.

Solution.

The normal. From the equation $z = \frac{2 - x - y}{2}$ , whence $z_{x} = z_{y} = - \frac{1}{2}$ and

N = (\frac{1}{2}; \frac{1}{2}; 1) .

The third component is positive — the orientation is right. The proportional vector $(1; 1; 2)$ could have been taken as well, but then the formula $Φ = \iint_{D} a \cdot N d x d y$ would stop working: in it the normal must have third component equal to one.

The integrand.

a \cdot N = \frac{x}{2} + y + z = \frac{x}{2} + y + \frac{2 - x - y}{2} = 1 + \frac{y}{2} .

The projection. The condition $z ⩾ 0$ gives $x + y ⩽ 2$ , so $D$ is the triangle $0 ⩽ x ⩽ 2$ , $0 ⩽ y ⩽ 2 - x$ .

Integration.

Φ = \int_{0}^{2} \int_{0}^{2 - x} (1 + \frac{y}{2}) d y d x .

The inner integral:

[y + \frac{y ^{2}}{4}]_{0}^{2 - x} = (2 - x) + \frac{( 2 - x ) ^{2}}{4} .

The outer one:

Φ = \int_{0}^{2} (2 - x) d x + \frac{1}{4} \int_{0}^{2} (2 - x)^{2} d x = 2 + \frac{1}{4} \cdot \frac{8}{3} = 2 + \frac{2}{3} = \frac{8}{3} .

Answer: $Φ = \frac{8}{3}$ .

Problem 3 (The flux through part of the lateral surface of a cylinder). Compute the flux of the vector field

F = \frac{1}{π} (2 x i - y j)

through the part of the lateral surface of the cylinder $x^{2} + y^{2} = 4$ where $x ⩾ 0$ , $y ⩾ 0$ and $0 ⩽ z ⩽ 4$ , in the direction of the outward normal.

Solution.

The divergence theorem cannot be used here. The surface is not closed: it is a quarter of the lateral wall and bounds no volume. It can be completed to a closed one (two quarters of the bases plus two flat walls in the planes $x = 0$ and $y = 0$ ), but then the flux through the four added parts has to be computed and subtracted — more work than the direct computation.

Parametrisation. The lateral surface of a cylinder of radius $R = 2$ : $x = 2 cos θ$ , $y = 2 sin θ$ , $z = z$ , $θ \in [0; π /2]$ , $z \in [0; 4]$ .

The normal and the area element. The outward normal to the cylinder is horizontal and points away from the axis: $n = (cos θ; sin θ; 0)$ , $d S = R d θ d z = 2 d θ d z$ .

The integrand.

F \cdot n = \frac{1}{π} (2 x cos θ - y sin θ) = \frac{1}{π} (4 cos^{2} θ - 2 sin^{2} θ) .

Integration. The variables separate, and the integral in $z$ simply gives the height $4$ :

Φ = \frac{2}{π} \int_{0}^{4} d z \int_{0}^{π /2} (4 cos^{2} θ - 2 sin^{2} θ) d θ = \frac{8}{π} \int_{0}^{π /2} (4 cos^{2} θ - 2 sin^{2} θ) d θ .

On $[0; π /2]$ both integrals equal $π /4$ :

\int_{0}^{π /2} cos^{2} θ d θ = \int_{0}^{π /2} sin^{2} θ d θ = \frac{π}{4} .

Therefore

Φ = \frac{8}{π} (4 \cdot \frac{π}{4} - 2 \cdot \frac{π}{4}) = \frac{8}{π} \cdot \frac{π}{2} = 4.

Answer: $Φ = 4$ .

Problem 4 (The flux through part of the lateral surface of a cylinder (variant)). Compute the flux of the vector field

F = \frac{1}{π} (x i + 3 y j)

through the part of the lateral surface of the cylinder $x^{2} + y^{2} = 1$ where $x ⩾ 0$ , $y ⩾ 0$ and $0 ⩽ z ⩽ 2$ , in the direction of the outward normal.

Solution.

Parametrisation. The radius is now $R = 1$ : $x = cos θ$ , $y = sin θ$ , $θ \in [0; π /2]$ , $z \in [0; 2]$ , $n = (cos θ; sin θ; 0)$ , $d S = 1 \cdot d θ d z$ .

The integrand.

F \cdot n = \frac{1}{π} (x cos θ + 3 y sin θ) = \frac{1}{π} (cos^{2} θ + 3 sin^{2} θ) .

Integration.

Φ = \frac{1}{π} \int_{0}^{2} d z \int_{0}^{π /2} (cos^{2} θ + 3 sin^{2} θ) d θ = \frac{2}{π} (\frac{π}{4} + 3 \cdot \frac{π}{4}) = \frac{2}{π} \cdot π = 2.

Here both components of the field point away from the axis, so the flux is positive — the sign can be predicted before computing.

Answer: $Φ = 2$ .

What it is

The working formula for an explicit surface

The parametric form

Common mistakes

Problems

flux.ipynb