The Flux Across a Surface

Section 9.6 The Flux Across a Surface

Imagine that \(\vect f\) models the velocity of a fluid at any given point in space. We want to know how much fluid passes through a surface in a unit time interval. The first thing we must do is to decide in which direction we want to measure the flux. Mathematically this means that we need to give \(S\) an orientation. Hence suppose that \(S\) is an orientable surface, and that \(\vect n(\vect x)\) is the positive unit normal at every point \(\vect x\in S\text{.}\) Assume that \(\vect g\colon D\to S\) is a parametrisation of \(S\text{.}\) Given a small rectangle \(R\subset D\) with edges of length \(\Delta y_1\) and \(\Delta y_2\) we want to estimate the volume of fluid going across \(\vect g(R)\) in a unit time interval. The volume flowing across is approximately the volume of the parallelepiped spanned by \(\vect f(\vect g(\vect y))\) and the vectors \(\vect v_1\) and \(\vect v_2\) given by (9.6). The volume of that parallelepiped equals the product of the area of its base and its perpendicular height. The situation is depicted in Figure 9.29.

Figure 9.29. Flux through a small “surface element” in a unit time interval. The volume of the two parallelepipeds is the same.

We saw that the area of the base involves the Jacobian, and is given in this situation by

\begin{equation*} \area(\vect g(R)) \approxeq\sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g}(\vect y)\bigr)}\,\Delta y_1\Delta y_2\text{.} \end{equation*}

We get the height by taking the projection onto the direction of the unit normal \(\vect n\text{.}\) Hence the height is \(\vect f\cdot\vect n\text{,}\) and the approximate volume of fluid crossing \(\vect g(R)\) in a unit time interval is

\begin{equation*} \vect f(\vect g(\vect y))\cdot\vect n(\vect g(\vect y)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g}(\vect y)\bigr)}\,\Delta y_1\Delta y_2. \end{equation*}

Adding up the above terms over a partition of \(D\) into small rectangles we get a Riemann sum for the function

\begin{equation*} \vect f\cdot\vect n\sqrt{\det\bigl(\bigl(J_{\vect g}\bigr)^T J_{\vect g}\bigr)}\text{.} \end{equation*}

Hence we make the following definition.

Definition 9.30. Flux of a vector field across a surface.

Suppose that \(\vect f\) is a continuous vector field over the smooth orientable surface \(S\) with positive unit normal \(\vect n\text{.}\) Then we say that

\begin{equation} \int_{S}\vect f\cdot\vect n\,dS =\int_D\vect f(\vect g(\vect y))\cdot\vect n(\vect g(\vect y)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}\,d\vect y\tag{9.8} \end{equation}

is the integral of \(\vect f\) over \(S\text{,}\) or the flux of \(\vect f\) across \(S\).

We next derive a formula allowing us to evaluate the flux of a vector field across a surface. We assume that \(S\) is a smooth orientable surface with a regular parametrisation \(\vect g\colon D\to S\) consistent with the orientation. By this we mean that the positive field of normals is given by (9.3). By taking into account (9.3) and Proposition 9.17 we see that

\begin{align*} f(\vect g(\vect y))&\cdot\vect n(\vect g(\vect y)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}\\ &=f(\vect g(\vect y))\cdot \frac{\dfrac{\partial\vect g}{\partial x_1}(\vect x) \times \dfrac{\partial\vect g}{\partial y_2}(\vect y)} {\Bigl\|\dfrac{\partial\vect g}{\partial y_1}(\vect y) \times \dfrac{\partial\vect g}{\partial y_2}(\vect y)\Bigr\|} \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}\\ &=f(\vect g(\vect y))\cdot \frac{\dfrac{\partial\vect g}{\partial y_1}(\vect x) \times \dfrac{\partial\vect g}{\partial y_2}(\vect y)} { \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}} \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}\\ &=f(\vect g(\vect y))\cdot \Bigl(\frac{\partial\vect g}{\partial y_1}(\vect x) \times\frac{\partial\vect g}{\partial y_2}(\vect y)\Bigr)\text{.} \end{align*}

Using (9.4) we finally get

\begin{align*} f(\vect g(\vect y))&\cdot\vect n(\vect g(\vect y)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g} (\vect y)\bigr)}\\ &=f_1(\vect g(\vect y))\frac{\partial(g_2,g_3)}{\partial(y_1,y_2)} +f_2(\vect g(\vect y))\frac{\partial(g_3,g_1)}{\partial(y_1,y_2)} +f_3(\vect g(\vect y))\frac{\partial(g_1,g_2)}{\partial(y_1,y_2)}\text{.} \end{align*}

Hence we get the following proposition.

Proposition 9.31. Flux across parametric surface.

Suppose that \(S\) is an orientable smooth surface, and that \(\vect g\colon D\to S\) is a regular parametrisation consistent with the orientation of \(S\text{.}\) Then

\begin{align*} \int_S\vect f&\cdot\vect n\,dS\\ &=\iint_D f_1(\vect g(\vect y)) \frac{\partial(g_2,g_3)}{\partial(y_1,y_2)} +f_2(\vect g(\vect y))\frac{\partial(g_3,g_1)}{\partial(y_1,y_2)}\\ &\qquad+f_3(\vect g(\vect y))\frac{\partial(g_1,g_2)}{\partial(y_1,y_2)}\,dy_1\,dy_2\text{.} \end{align*}

Let us now look at the special case of an explicitly given surface, that is, \(S\) is the graph of a function over \(h\colon D\to\mathbb R\text{.}\) By convention we orient a graph upwards. From (9.2) and Proposition 9.27 we see that

\begin{equation*} f(\vect g(\vect x))\cdot\vect n(\vect g(\vect x)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect x)\bigr)^T J_{\vect g} (\vect x)\bigr)} =\vect f(\vect x,h(\vect x))\cdot (-\grad h(\vect x),1)\text{.} \end{equation*}

Hence we have the following result.

Proposition 9.32. Flux across a graph.

Suppose that \(S\) is the graph of the smooth function \(h\colon D\to\mathbb R\text{.}\) Then the flux of the continuous vector field \(\vect f\) across \(S\) in the upward direction is

\begin{align*} \int_S\vect f\cdot\vect n\,dS =\iint_D \Bigl(&-f_1\bigl(\vect x,h(\vect x)\bigr) \frac{\partial h}{\partial x_1}(\vect x)\\ &-f_2\bigl(\vect x,h(\vect x)\bigr) \frac{\partial h}{\partial x_2}(\vect x) +f_3\bigl(\vect x,h(\vect x)\bigr)\Bigr) \,dx_1\,dx_2\text{.} \end{align*}

Example 9.33. Flux across a paraboloid.

Calculate the flux of \(\vect f(x,y,z)=(2x,2y,3)\) across the part of the paraboloid \(z=4-x^2-y^2\) lying above the \(xy\)-plane. By convention we orient the surface upwards.

Answer.

The flux is \(44\pi\)

Solution.

By assumption, \(z\geq 0\) so the surface is the graph of \(4-x^2-y^2\) over the disc, \(D\text{,}\) given by \(x^2+y^2\leq 4\text{.}\) According to the above proposition, the flux across the surface is given by

\begin{align*} \int_S\vect f\cdot\vect n\,dS &=\int_D-2x(-2x)-2y(-2y)+3\,dx\,dy\\ &=\int_D4x^2+4y^2+3\,dx\,dy\\ &=\int_0^{2\pi}\int_0^2(4r^2+3)r\,dr\,d\varphi\\ &=2\pi\Bigl[r^2+\frac{3}{2}r^2\Bigr]_0^2\\ &=2\pi(16+3\cdot 2-0)=44\pi\text{.} \end{align*}

Example 9.34. Flux across a sphere.

Calculate the flux of

\begin{equation*} \vect f(x,y,z) :=\Bigl(\frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}},\frac{z}{(x^2+y^2+z^2)^{3/2}} \Bigr) \end{equation*}

out of the sphere with radius \(R\) centred at the origin.

Solution.

The outward pointing unit normal to a sphere of radius \(R\) centred at the origin is given by

\begin{equation*} \vect n(x,y,z)=\frac{1}{R}(x,y,z)\text{.} \end{equation*}

Moreover, for points \((x,y,z)\) on the sphere we have \(x^2+y^2+z^2=R^2\) and so

\begin{equation*} \vect f(x,y,z) = \frac{1}{R^3}(x,y,z)\text{.} \end{equation*}

Using the two facts we get

\begin{equation*} \vect f\cdot\vect n =\frac{1}{R}\frac{1}{R^3}(x,y,z)\cdot(x,y,z) =\frac{1}{R^4}(x^2+y^2+z^2)=\frac{1}{R^2}.\text{.} \end{equation*}

Hence using spherical coordinates we have

\begin{equation*} \int_S\vect f\cdot\vect n\,dS =\frac{1}{R^2}\int_0^{2\pi}\int_0^\pi R^2\sin\theta\,d\theta\,d\varphi =2\pi[-\cos\theta]_0^\pi=4\pi\text{.} \end{equation*}

The flux across the sphere is therefore \(4\pi\text{.}\) Note that it is independent of \(R\text{.}\)

Vector Calculus: An Introduction