The Central Idea
Suppose a vector field
\[ \mathbf{F} \]passes through an oriented surface \(S\).
We want to measure the field component that crosses the surface rather than the component that slides along it.
The local quantity is
\[ \boxed{ \mathbf{F}\cdot\mathbf{n} } \]where \(\mathbf{n}\) is the chosen unit normal.
Flux accumulates this normal component across the entire surface.
The Flux Integral
The flux of a vector field \(\mathbf{F}\) through an oriented surface \(S\) is
\[ \boxed{ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS }. \]This is a vector surface integral, but the final result is a scalar.
What the Dot Product Does
Since
\[ \mathbf{F}\cdot\mathbf{n} = \|\mathbf{F}\| \|\mathbf{n}\| \cos\theta \]and
\[ \|\mathbf{n}\|=1, \]we have
\[ \boxed{ \mathbf{F}\cdot\mathbf{n} = \|\mathbf{F}\|\cos\theta }. \]Thus the dot product extracts the component of the vector field normal to the surface.
Maximum Positive Flux Density
If the vector field points in exactly the same direction as the chosen normal, then
\[ \theta=0 \]and
\[ \cos\theta=1. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{n} = \|\mathbf{F}\|. \]The field passes directly through the surface in the positive orientation.
Negative Flux Density
If the field points opposite the chosen normal, then
\[ \mathbf{F}\cdot\mathbf{n}<0. \]This means the field passes through the surface in the negative orientation.
Zero Flux Density
If the vector field is tangent to the surface, then it is perpendicular to the normal:
\[ \theta=\frac{\pi}{2}. \]Therefore,
\[ \boxed{ \mathbf{F}\cdot\mathbf{n}=0 }. \]The field moves along the surface rather than through it.
Orientation Is Essential
Flux requires an oriented surface.
If the normal \(\mathbf{n}\) is replaced by
\[ -\mathbf{n}, \]then
\[ \mathbf{F}\cdot(-\mathbf{n}) = - \mathbf{F}\cdot\mathbf{n}. \]Therefore,
\[ \boxed{ \text{reversing orientation reverses flux sign} }. \]Vector Area Element
Recall the oriented vector area element
\[ d\mathbf{S} = \mathbf{n}\,dS. \]Then the flux integral can be written compactly as
\[ \boxed{ \iint_S \mathbf{F}\cdot d\mathbf{S} }. \]Flux Through a Parameterized Surface
Suppose
\[ \mathbf{r}(u,v) \]parameterizes \(S\).
If
\[ \mathbf{r}_u\times\mathbf{r}_v \]has the desired orientation, then
\[ \boxed{ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u\times\mathbf{r}_v) \,du\,dv }. \]Why We Do Not Normalize the Cross Product
The cross product can be written
\[ \mathbf{r}_u\times\mathbf{r}_v = \mathbf{n} \|\mathbf{r}_u\times\mathbf{r}_v\|. \]Therefore,
\[ (\mathbf{r}_u\times\mathbf{r}_v) \,du\,dv = \mathbf{n}\,dS. \]The cross product already combines orientation and the correct area scale.
Example: Flux Through a Plane Patch
Let
\[ \mathbf{F} = \langle1,2,3\rangle \]and let the surface be
\[ \mathbf{r}(u,v) = \langle u,v,0\rangle, \]over
\[ 0\le u\le2, \qquad 0\le v\le1. \]Then
\[ \mathbf{r}_u = \langle1,0,0\rangle \]and
\[ \mathbf{r}_v = \langle0,1,0\rangle. \]Therefore,
\[ \mathbf{r}_u\times\mathbf{r}_v = \langle0,0,1\rangle. \]The dot product is
\[ \mathbf{F}\cdot (\mathbf{r}_u\times\mathbf{r}_v) = 3. \]Hence,
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \int_0^2\int_0^1 3 \,dv\,du = \boxed{6}. \]Flux Through Graph Surfaces
Suppose the surface is
\[ z=g(x,y). \]A natural parameterization is
\[ \mathbf{r}(x,y) = \langle x,y,g(x,y)\rangle. \]Then
\[ \mathbf{r}_x\times\mathbf{r}_y = \langle-g_x,-g_y,1\rangle. \]This vector points upward because its \(z\)-component is positive.
Upward Flux Through a Graph
If
\[ \mathbf{F} = \langle P,Q,R\rangle \]and \(S\) is
\[ z=g(x,y) \]with upward orientation, then
\[ \boxed{ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iint_D \mathbf{F}(x,y,g(x,y)) \cdot \langle-g_x,-g_y,1\rangle \,dx\,dy }. \]Component Formula for Upward Flux
Expanding the dot product gives
\[ \boxed{ \iint_D \left( -Pg_x-Qg_y+R \right) \,dx\,dy } \]where \(P\), \(Q\), and \(R\) are evaluated on the surface
\[ z=g(x,y). \]Downward Flux Through a Graph
For downward orientation, use
\[ \langle g_x,g_y,-1\rangle. \]Therefore, downward flux is the negative of upward flux:
\[ \boxed{ \Phi_{\text{down}} = - \Phi_{\text{up}} }. \]Example: Upward Flux Through \(z=x+y\)
Let
\[ \mathbf{F}(x,y,z) = \langle x,y,1\rangle \]and let
\[ z=x+y \]over
\[ 0\le x\le1, \qquad 0\le y\le1. \]Here
\[ g_x=1, \qquad g_y=1. \]The upward vector area factor is
\[ \langle-1,-1,1\rangle. \]Therefore,
\[ \mathbf{F}\cdot \langle-1,-1,1\rangle = -x-y+1. \]Thus,
\[ \Phi = \int_0^1\int_0^1 (1-x-y) \,dy\,dx. \]Evaluating,
\[ \boxed{ \Phi=0 }. \]Flux Through a Flat Coordinate Plane
Coordinate planes produce especially simple normals.
For the \(xy\)-plane,
\[ \mathbf{n} = \pm\mathbf{k}. \]For the \(yz\)-plane,
\[ \mathbf{n} = \pm\mathbf{i}. \]For the \(xz\)-plane,
\[ \mathbf{n} = \pm\mathbf{j}. \]Example: Flux Through the \(yz\)-Plane
Let
\[ \mathbf{F}(x,y,z) = \langle2x+1,y,z\rangle \]and let \(S\) be the rectangle in the \(yz\)-plane
\[ x=0, \qquad 0\le y\le2, \qquad 0\le z\le3. \]Choose the positive \(x\)-direction normal
\[ \mathbf{n}=\mathbf{i}. \]On the surface,
\[ \mathbf{F}\cdot\mathbf{n} = 1. \]Therefore,
\[ \Phi = (1)(2)(3) = \boxed{6}. \]Flux Through a Cylinder
The lateral surface of a cylinder of radius \(R\) can be parameterized by
\[ \mathbf{r}(\theta,z) = \langle R\cos\theta, R\sin\theta, z \rangle. \]A convenient outward vector area element is
\[ \boxed{ \mathbf{r}_\theta\times\mathbf{r}_z = \langle R\cos\theta, R\sin\theta, 0 \rangle }. \]Example: Radial Flux Through a Cylinder
Let
\[ \mathbf{F}(x,y,z) = \langle x,y,0\rangle \]and consider the lateral surface of the cylinder
\[ x^2+y^2=R^2, \qquad 0\le z\le H. \]On the surface,
\[ \mathbf{F} = \langle R\cos\theta, R\sin\theta, 0 \rangle. \]Dotting with the outward vector area element gives
\[ R^2. \]Therefore,
\[ \Phi = \int_0^{2\pi}\int_0^H R^2 \,dz\,d\theta. \]Hence,
\[ \boxed{ \Phi = 2\pi R^2H }. \]Flux Through a Sphere
For a sphere of radius \(R\) centered at the origin, the outward unit normal is
\[ \boxed{ \mathbf{n} = \frac{\langle x,y,z\rangle}{R} }. \]This radial geometry often makes flux calculations especially simple.
Example: Radial Flux Through a Sphere
Let
\[ \mathbf{F}(x,y,z) = \langle x,y,z\rangle \]on the sphere
\[ x^2+y^2+z^2=R^2. \]Since
\[ \mathbf{n} = \frac{\langle x,y,z\rangle}{R}, \]we have
\[ \mathbf{F}\cdot\mathbf{n} = \frac{x^2+y^2+z^2}{R} = R. \]Therefore,
\[ \Phi = R \iint_S dS. \]Since the sphere has area \(4\pi R^2\),
\[ \boxed{ \Phi = 4\pi R^3 }. \]Closed-Surface Flux
If \(S\) is closed, flux is often written
\[ \boxed{ \oiint_S \mathbf{F}\cdot\mathbf{n}\,dS }. \]The standard orientation is outward.
Positive closed-surface flux means more field leaves the enclosed region than enters it.
Outward Versus Inward Flux
For a closed surface,
\[ \boxed{ \Phi_{\text{inward}} = - \Phi_{\text{outward}} }. \]A problem should specify orientation, or use the standard convention that closed surfaces are oriented outward.
Flux and Fluid Flow
If
\[ \mathbf{v}(x,y,z) \]is a fluid velocity field, then
\[ \mathbf{v}\cdot\mathbf{n} \]measures normal velocity through the surface.
Integrating across \(S\) gives volumetric flow rate when the velocity field and units are appropriate.
Mass Flux
If a fluid has mass density
\[ \rho(x,y,z) \]and velocity field
\[ \mathbf{v}, \]then the mass-flux vector is
\[ \rho\mathbf{v}. \]The rate of mass crossing an oriented surface is
\[ \boxed{ \iint_S \rho\mathbf{v}\cdot\mathbf{n}\,dS }. \]Flux and Electric Fields
Electric flux uses an electric field
\[ \mathbf{E} \]and an oriented surface:
\[ \boxed{ \Phi_E = \iint_S \mathbf{E}\cdot\mathbf{n}\,dS }. \]The geometric meaning is the same: measure the field component passing through the surface.
Flux and Gravitational Fields
Gravitational vector fields can also be integrated through surfaces.
The sign and interpretation depend on the chosen orientation and the fact that idealized gravitational fields generally point toward mass sources.
Flux and Divergence
Divergence measures local source or sink behavior:
\[ \nabla\cdot\mathbf{F}. \]Flux measures net field crossing a finite surface:
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS. \]The Divergence Theorem will connect these two ideas exactly.
Local Flux Density
Divergence can be viewed as the limiting outward flux per unit volume:
\[ \boxed{ \nabla\cdot\mathbf{F} = \lim_{V\to0} \frac{ \text{outward flux through }\partial V }{ \text{volume of }V } }. \]This interpretation explains why positive divergence is source-like and negative divergence is sink-like.
Constant Fields and Closed Surfaces
A constant vector field has zero divergence.
Intuitively, whatever enters one side of a closed surface exits elsewhere.
Therefore, its net outward flux through a closed surface is zero.
Example: Constant Field Through a Closed Box
Let
\[ \mathbf{F} = \langle1,0,0\rangle. \]Through the right face of a rectangular box, outward flux is positive.
Through the left face, outward flux is equally negative.
Through the remaining faces, the field is tangent.
Therefore the total outward flux is
\[ \boxed{0}. \]Flux Through Piecewise Surfaces
If a surface is built from pieces,
\[ S=S_1\cup S_2\cup\cdots\cup S_n, \]then
\[ \boxed{ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \sum_{k=1}^n \iint_{S_k} \mathbf{F}\cdot\mathbf{n}_k\,dS }. \]Each piece must use an orientation consistent with the overall surface.
Example: Closed Cylinder as Three Pieces
A closed cylinder consists of:
- the lateral surface;
- the top disk;
- the bottom disk.
Direct flux calculation requires summing the flux through all three pieces.
The Divergence Theorem will often provide a more efficient alternative.
Surface Symmetry Can Simplify Flux
Before parameterizing, inspect the relationship between the field and the surface.
If
\[ \mathbf{F}\cdot\mathbf{n} \]is constant on the surface, then
\[ \boxed{ \Phi = (\mathbf{F}\cdot\mathbf{n}) \times \text{surface area} }. \]This can eliminate most of the integration.
Radial Symmetry
For a radial field and a sphere centered at the same origin, the field is parallel or antiparallel to the surface normal.
This makes
\[ \mathbf{F}\cdot\mathbf{n} \]especially simple.
Radial symmetry is one of the most useful patterns in flux problems.
Units of Flux
Since flux integrates a field across area,
\[ \boxed{ [\text{flux}] = [\mathbf{F}] \times [\text{area}] }. \]The exact physical unit depends on what the vector field represents.
Common Error: Using \(\|\mathbf{F}\|\) Instead of \(\mathbf{F}\cdot\mathbf{n}\)
Flux measures the normal component of the field, not the full magnitude.
The correct local quantity is
\[ \boxed{ \mathbf{F}\cdot\mathbf{n} }. \]A large vector tangent to the surface contributes zero flux.
Common Error: Forgetting Orientation
Flux is signed.
The same field and same geometric surface can produce opposite flux values under opposite orientations.
Common Error: Normalizing the Cross Product Unnecessarily
In a parameterized flux integral,
\[ \mathbf{F} \cdot (\mathbf{r}_u\times\mathbf{r}_v) \,du\,dv \]already includes the area scaling.
Normalizing the cross product and forgetting to multiply by \(dS\) loses that scaling factor.
Common Error: Using the Wrong Cross-Product Order
Since
\[ \mathbf{r}_v\times\mathbf{r}_u = - \mathbf{r}_u\times\mathbf{r}_v, \]changing the cross-product order changes the orientation and therefore the flux sign.
Common Error: Forgetting to Evaluate the Field on the Surface
If the surface is parameterized by
\[ \mathbf{r}(u,v), \]then use
\[ \boxed{ \mathbf{F}(\mathbf{r}(u,v)) } \]inside the parameter integral.
Common Error: Confusing Flux with Surface Area
Surface area is
\[ \iint_S 1\,dS. \]Flux is
\[ \iint_S \mathbf{F}\cdot\mathbf{n}\,dS. \]The second integral depends on both the vector field and surface orientation.
A Reliable Procedure for Parameterized Flux Integrals
- Parameterize the surface as \(\mathbf{r}(u,v)\).
- Determine the parameter domain \(D\).
- Compute \(\mathbf{r}_u\) and \(\mathbf{r}_v\).
- Compute \(\mathbf{r}_u\times\mathbf{r}_v\).
- Check whether the cross product has the requested orientation.
- Reverse its sign if necessary.
- Evaluate \(\mathbf{F}\) on the surface.
- Compute the dot product.
- Integrate over \(D\).
- Interpret the sign and units.
A Reliable Procedure for Graph-Surface Flux
- Write the surface as \(z=g(x,y)\) when convenient.
- Identify the projection region \(D\).
- Compute \(g_x\) and \(g_y\).
- For upward orientation, use \(\langle-g_x,-g_y,1\rangle\).
- For downward orientation, reverse the sign.
- Replace \(z\) in the field by \(g(x,y)\).
- Compute the dot product.
- Integrate over \(D\).
A Reliable Procedure for Closed-Surface Flux
- Confirm the surface is closed.
- Use outward orientation unless told otherwise.
- Inspect symmetry before setting up integrals.
- If computing directly, divide the surface into manageable pieces.
- Compute outward flux on every piece.
- Add all contributions.
- Later, compare whether the Divergence Theorem offers a shorter route.
Example: Full Parameterized Flux Workflow
Let
\[ \mathbf{F}(x,y,z) = \langle x,y,z\rangle \]and let \(S\) be the surface
\[ \mathbf{r}(u,v) = \langle u,v,u+v\rangle \]over
\[ 0\le u\le1, \qquad 0\le v\le1, \]oriented upward.
Compute tangent vectors:
\[ \mathbf{r}_u = \langle1,0,1\rangle, \] \[ \mathbf{r}_v = \langle0,1,1\rangle. \]Their cross product is
\[ \mathbf{r}_u\times\mathbf{r}_v = \langle-1,-1,1\rangle. \]Its \(z\)-component is positive, so the orientation is upward.
On the surface,
\[ \mathbf{F} = \langle u,v,u+v\rangle. \]Dot:
\[ \langle u,v,u+v\rangle \cdot \langle-1,-1,1\rangle = -u-v+u+v = 0. \]Therefore,
\[ \boxed{ \Phi=0 }. \]Example: Full Graph-Flux Workflow
Let
\[ \mathbf{F}(x,y,z) = \langle0,0,z\rangle \]and let
\[ z=x^2+y^2 \]above the disk
\[ x^2+y^2\le1, \]oriented upward.
Here
\[ g_x=2x, \qquad g_y=2y. \]The upward vector area factor is
\[ \langle-2x,-2y,1\rangle. \]On the surface,
\[ \mathbf{F} = \langle0,0,x^2+y^2\rangle. \]Therefore,
\[ \mathbf{F}\cdot \langle-2x,-2y,1\rangle = x^2+y^2. \]Thus,
\[ \Phi = \iint_D (x^2+y^2) \,dA. \]Use polar coordinates:
\[ \Phi = \int_0^{2\pi}\int_0^1 r^2r \,dr\,d\theta. \]Hence,
\[ \boxed{ \Phi = \frac{\pi}{2} }. \]Practice 1: Constant Field Through a Plane
Let
\[ \mathbf{F} = \langle2,1,4\rangle \]and let \(S\) be the rectangle
\[ 0\le x\le3, \qquad 0\le y\le2, \qquad z=0, \]oriented upward.
Find the flux.
Practice 2: Reverse Orientation
Using Practice 1, find the flux if the same rectangle is oriented downward.
Practice 3: Tangent Field
Let
\[ \mathbf{F} = \langle1,2,0\rangle \]through a region in the \(xy\)-plane oriented upward.
What is the flux?
Practice 4: Parameterized Surface
Let
\[ \mathbf{r}(u,v) = \langle u,v,u-v\rangle. \]Find
\[ \mathbf{r}_u\times\mathbf{r}_v. \]Practice 5: Graph Normal
For
\[ z=2x-y, \]give an upward-pointing vector area factor.
Practice 6: Upward Flux
Let
\[ \mathbf{F}(x,y,z) = \langle0,0,2\rangle \]through
\[ z=2x-y \]above the unit square
\[ 0\le x\le1, \qquad 0\le y\le1, \]oriented upward.
Find the flux.
Practice 7: Radial Sphere Flux
Let
\[ \mathbf{F} = \langle x,y,z\rangle \]on the unit sphere oriented outward.
Find the flux.
Practice 8: Inward Sphere Flux
Using Practice 7, find the flux if the sphere is oriented inward.
Practice 9: Closed Constant Field
What is the total outward flux of a constant vector field through any closed rectangular box?
Practice 10: Interpret the Sign
Suppose the outward flux through a closed surface is
\[ -12. \]What does the negative sign indicate about net field flow relative to the enclosed region?
Practice Answers
Practice 1
The upward unit normal is
\[ \mathbf{n}=\mathbf{k}. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{n}=4. \]The rectangle has area
\[ 3\cdot2=6. \]Hence,
\[ \boxed{ \Phi=24 }. \]Practice 2
\[ \boxed{ \Phi=-24 }. \]Practice 3
The field is tangent to the \(xy\)-plane, so
\[ \mathbf{F}\cdot\mathbf{k}=0. \]Therefore,
\[ \boxed{ \Phi=0 }. \]Practice 4
\[ \mathbf{r}_u = \langle1,0,1\rangle \]and
\[ \mathbf{r}_v = \langle0,1,-1\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}_u\times\mathbf{r}_v = \langle-1,1,1\rangle }. \]Practice 5
Since
\[ g_x=2, \qquad g_y=-1, \]the upward vector area factor is
\[ \boxed{ \langle-2,1,1\rangle }. \]Practice 6
Dot with the upward vector area factor:
\[ \langle0,0,2\rangle \cdot \langle-2,1,1\rangle = 2. \]The projection region has area \(1\), so
\[ \boxed{ \Phi=2 }. \]Practice 7
On the unit sphere,
\[ \mathbf{n} = \langle x,y,z\rangle \]and
\[ \mathbf{F}\cdot\mathbf{n} = x^2+y^2+z^2 = 1. \]Therefore,
\[ \Phi = \text{area of unit sphere} = \boxed{4\pi}. \]Practice 8
\[ \boxed{ -4\pi }. \]Practice 9
A constant field has zero divergence and produces equal entering and leaving contributions on a closed box.
Therefore,
\[ \boxed{ 0 }. \]Practice 10
Negative outward flux means there is more net field entering the enclosed region than leaving it.
What to Remember
Flux through an oriented surface is
\[ \boxed{ \Phi = \iint_S \mathbf{F}\cdot\mathbf{n}\,dS }. \]For a parameterized surface,
\[ \boxed{ \Phi = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u\times\mathbf{r}_v) \,du\,dv }. \]- Flux measures the normal component of a vector field across a surface.
- Positive flux follows the chosen normal orientation.
- Negative flux goes opposite the chosen normal orientation.
- A tangent field contributes zero local flux.
- Reversing orientation reverses flux sign.
- Closed surfaces are usually oriented outward.
- Symmetry can simplify flux calculations substantially.
- Divergence measures local source-sink behavior, while flux measures total crossing through a finite surface.
Where We Go Next
Flux measures how a vector field crosses a surface. The next lesson turns back to vector fields that come from scalar potentials and develops the special consequences of path independence.
The next lesson, Conservative Fields and Potentials, will connect gradient fields, potential functions, path-independent line integrals, closed-loop circulation, and practical tests for conservativeness.