Flux

Recall that a unit normal vector to a surface can be given by 

\[ \textbf{n} = \dfrac{\textbf{r}_u \times \textbf{r}_v}{ \left| \textbf{r}_u \times \textbf{r}_v \right| }\]

There is another choice for the normal vector to the surface, namely the vector in the opposite direction, \(-\textbf{n}\).

By this point, you may have noticed the similarity between the formulas for the unit normal vector and the surface integral. This idea leads us to the definition of the Flux Integral. Consider a fluid flowing through a surface \(S\). The Flux of the fluid across \(S\) measures the amount of fluid passing through the surface per unit time. If the fluid flow is represented by the vector field \(F\), then for a small piece with area \(\Delta S\) of the surface the flux will equal to 

\[\Delta \text{Flux} = F \cdot n\, \Delta S \]

Adding up all these together and taking a limit, we get

Definition: Flux Integral

Let \(F\) be a differentiable vector field on a surface \(S\) oriented by a unit normal vector \(n\). The flux integral of \(F\) across \(n\) is given by 

\[ \iint_{S} (F \cdot n) d\sigma . \]

Note that the method for choosing the value for \(d\sigma\) for flux is identical to doing it for the integrals described above. Also notice that the denominator of \(n\) and the formula for \(dS\) both involve \( |\textbf{r}_u \times \textbf{r}_v|\). Canceling, we get

\[ \textbf{n}\, dS = \textbf{r}_u \times \textbf{r}_v dv\, du\]

for a surface that is defined by the function \(z = g(x,y)\), we get the nice formula

\[ \textbf{n}\, dS = -g_x(x,y) \hat{\textbf{i}} - g_y(x,y)\hat{\textbf{j}} + \hat{\textbf{k}} \text{(oriented upward)}\]

or

\[ \textbf{n}\, dS = g_x(x,y) \hat{\textbf{i}} + g_y(x,y)\hat{\textbf{j}} - \hat{\textbf{k}} \text{(oriented downward)}\]

Example \(\PageIndex{1}\)

Find the flux of \( F = yz\hat{\textbf{j}} +z^{2}\hat{\textbf{k}} \) outward through the surface \(S\) cut from the cylinder \( y^{2} + z^{2} = 1, z \geq 0 \), by the planes \(x = 0\) and \(x = 1\).

Solution

First we calculate the outward normal field on \(S\). This can be calulated by finding the gradient of \( g(x,y,z) = y^{2} + z^{2} \) and dividing by its magnitude.

\[ n = \frac{\nabla g}{| \nabla g |} = \frac{2y\hat{\textbf{j}} + 2z\hat{\textbf{k}}}{\sqrt{4y^{2} + 4z^{2}}} = \frac{2y\hat{\textbf{i}} + 2z\hat{\textbf{k}}}{2\sqrt{1}} = y\hat{\textbf{j}} + z\hat{\textbf{k}} \]

Next, we calculate the value for \(d\sigma = \frac{\nabla g}{| \nabla g \cdot p|} dA \). Note the similarity to the value of \(n\). All we need to do is find the value of \(p\). Because our cylinder sits with its shadow region in the xy plane, the vector normal to that region is in the \(k\) direction. Thus,

\[ d\sigma = \frac{\nabla g}{| \nabla g \cdot k|} dA = \frac{2}{2z} dA = \frac{1}{z} dA \]

Note: we dropped the absolute value bars in the last step there because the problems specifies \(z \geq 0\) on \(S\).

Now find the value of \( F \cdot n \).

\[\begin{align} F \cdot n &= (yz\hat{\textbf{j}} + z^{2}\hat{\textbf{k}}) \cdot (y\hat{\textbf{j}} + z\hat{\textbf{k}}) \\ &= y^{2}z + z^{3} \\ &= z(y^{2} + z^{2}) \\ &= z &\text{because the surface} \\ &&\text{is defined as } y^{2} + z^{2} = 1 \end{align}\]

Once all the preliminary work is done, plug them into the integral:

\[ \iint_{S} F \cdot n \, d\sigma = \iint_{S} \left(z\right) \left(\frac{1}{z} dA \right) = \iint_{R_{xy}} dA = area(R_{xy}) = 2\].

Example \(\PageIndex{2}\)

Find the flux of \(F = y\hat{\textbf{j}} - z\hat{\textbf{k}} \) through the paraboloid \(S = y = x^{2} + z^{2}, y \leq 1 \).

Solution

The flux can be described by \( \iint _{S} F \cdot n \, d\sigma \) with \( n = \frac{2x\hat{\textbf{i}} - \hat{\textbf{j}} + 2z\hat{\textbf{k}}}{\sqrt{1 + 4x^{2} + 4z^{2}}} \).

Take the dot product of \(F\) and \(n\)

\[(0\hat{\textbf{i}}+y\hat{\textbf{j}}- z\hat{\textbf{k}}) \cdot \frac{2x\hat{\textbf{i}}-\hat{\textbf{j}}+2z\hat{\textbf{k}}}{\sqrt{1+4x^{2}+4z^{2}}} = \frac{1}{\sqrt{1+4x^{2}+4z^{2}}}(-y+2z^{2})\]

Substitute \(x^{2} + z^{2} = y \) to simplify \(n\) to \( -1 + \frac{2z^{2}}{y} \).

Thus, the integral for the flux is

\[\begin{align} &\int_{-1}^1 \int_{-1}^1 \left(-1 + \frac{2z^{2}}{y} \right) dz\, dy \\ &= \int_{-1}^{1} -z + \left. \frac{2z^3}{3y} \right |_{-1}^{1} dy \\ &= \int_{-1}^{1} \left (-1 + \frac{2}{3y} \right) - \left (1 + \frac{-2}{3y} \right) dy \\ &= \int_{-1}^{1} -2 dy \\ &= 0 \end{align}\]

The total flux through the surface is 0.

Example \(\PageIndex{3}\)

Find the flux of 

\[ \textbf{F}(x,y,z) = x\hat{\textbf{i}} + 2y\hat{\textbf{j}} + z\hat{\textbf{k}}\]

across the part of the surface

\[ z = x + y2 \]

with upward pointing normal that lies within the box

\( 0 \le x \le 3 \) and \( 2 \le y \le 5\)

Solution

We compute

\[ NdS = -\hat{\textbf{i}} - 2y\hat{\textbf{j}} + \hat{\textbf{k}} dy\, dx\]

and 

\[ \textbf{F} \cdot \textbf{N}\, dS = -x - 4y^2 + x + y^2 = -3y^2 \]

The flux integral is 

\[ \int_0^3\int_2^5 -3y^2\,dy\,dx = -351\]

Contributors

• Michael Rea (UCD)
• Larry Green (Lake Tahoe Community College)