How could we calculate the mass flux of the fluid across [latex]S[/latex]?

The rate of flow, measured in mass per unit time per unit area, is [latex]\rho \mathbf{v}[/latex]. To calculate the mass flux across [latex]S[/latex], chop [latex]S[/latex] into small pieces [latex]S_{ij}[/latex].

If [latex]S_{ij}[/latex] is small enough, then it can be approximated by a tangent plane at some point [latex]P[/latex] in [latex]S_{ij}[/latex]. Therefore, the unit normal vector at [latex]P[/latex] can be used to approximate [latex]\mathbf{N}(x,y,z)[/latex] across the entire piece [latex]S_{ij}[/latex], because the normal vector to a plane does not change as we move across the plane.

The component of the vector [latex]\rho \mathbf{v}[/latex] at [latex]P[/latex] in the direction of [latex]\mathbf{N}[/latex] is [latex]\rho \mathbf{v} \cdot \mathbf{N}[/latex] at [latex]P[/latex]. Since [latex]S_{ij}[/latex] is small, the dot product [latex]\rho \mathbf{v} \cdot \mathbf{N}[/latex] changes very little as we vary across [latex]S_{ij}[/latex], and therefore [latex]\rho \mathbf{v} \cdot \mathbf{N}[/latex] can be taken as approximately constant across [latex]S_{ij}[/latex].

To approximate the mass of fluid per unit time flowing across [latex]S_{ij}[/latex] (and not just locally at point [latex]P[/latex]), we need to multiply [latex](\rho \mathbf{v} \cdot \mathbf{N})(P)[/latex] by the area of [latex]S_{ij}[/latex]. Therefore, the mass of fluid per unit time flowing across [latex]S_{ij}[/latex] in the direction of [latex]\mathbf{N}[/latex] can be approximated by [latex](\rho \mathbf{v} \cdot \mathbf{N}) \Delta S_{ij}[/latex], where [latex]\mathbf{N}[/latex], [latex]\rho[/latex], and [latex]\mathbf{v}[/latex] are all evaluated at [latex]P[/latex] (Figure 5).

This is analogous to the flux of two-dimensional vector field [latex]\mathbf{F}[/latex] across plane curve [latex]C[/latex], in which we approximated flux across a small piece of [latex]C[/latex] with the expression [latex](\mathbf{F} \cdot \mathbf{N}) \Delta s[/latex].

To approximate the mass flux across [latex]S[/latex], form the sum [latex]\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^n(\rho{\bf{v}}\cdot{\bf{N}})\nabla{S_{ij}}[/latex] As pieces [latex]S_{ij}[/latex] get smaller, this sum gets arbitrarily close to the mass flux. Therefore, the mass flux is:

[latex]\displaystyle\iint_s\rho{\bf{v}}\cdot{\bf{N}}dS=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^n(\rho{\bf{v}}\cdot{\bf{N}})\nabla{S_{ij}}[/latex].

This is a surface integral of a vector field. Letting the vector field [latex]\rho \mathbf{v}[/latex] be an arbitrary vector field [latex]\mathbf{F}[/latex] leads to the following definition.

surface integral of a vector field

Let [latex]\mathbf{F}[/latex] be a continuous vector field with a domain that contains oriented surface [latex]S[/latex] with unit normal vector [latex]\mathbf{N}[/latex]. The surface integral of [latex]\mathbf{F}[/latex] over [latex]S[/latex] is:

[latex]\displaystyle\iint_S{\bf{F}}\cdot{d}S=\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S[/latex].

This integral is also called the flux of [latex]\mathbf{F}[/latex] across [latex]S[/latex].

Notice the parallel between this definition and the definition of vector line integral [latex]\displaystyle\int_C{\bf{F}}\cdot{\bf{N}}{d}s[/latex]. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (two-dimensional) rather than a curve (one-dimensional).

Computing Surface Integrals of Vector Fields

Just as with vector line integrals, surface integral [latex]\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S[/latex] is easier to compute after surface [latex]S[/latex] has been parameterized. Let [latex]\mathbf{r}(u,v)[/latex] be a parameterization of [latex]S[/latex] with parameter domain [latex]D[/latex].

The unit normal vector is given by [latex]\mathbf{N} = \frac{\mathbf{t}_u \times \mathbf{t}_v}{||\mathbf{t}_u \times \mathbf{t}_v||}[/latex]. Using the surface integral formula for scalar-valued functions, we have:

[latex]\begin{aligned} \displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S&=\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S \\ &=\displaystyle\iint_S{\bf{F}}\cdot\frac{{\bf{t}}_u\times{\bf{t}}_v}{||{\bf{t}}_u\times{\bf{t}}_v||}dS \\ &=\displaystyle\iint_D\left({\bf{F}}({\bf{r}}(u,v))\cdot\frac{{\bf{t}}_u\times{\bf{t}}_v}{||{\bf{t}}_u\times{\bf{t}}_v||}\right)||{\bf{t}}_u\times{\bf{t}}_v||dA \\ &=\displaystyle\iint_D({\bf{F}}({\bf{r}}(u,v))\cdot({\bf{t}}_u\times{\bf{t}}_v))dA \end{aligned}[/latex].

Notice that the magnitude terms cancel, simplifying the computation considerably.

computing surface integrals

For a vector field [latex]\mathbf{F}[/latex] and oriented surface [latex]S[/latex] with parameterization [latex]\mathbf{r}(u,v)[/latex] over domain [latex]D[/latex]:

[latex]\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S=\displaystyle\iint_D({\bf{F}}({\bf{r}}(u,v))\cdot({\bf{t}}_u\times{\bf{t}}_v))dA[/latex].

Calculate the surface integral [latex]\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S[/latex], where [latex]{\bf{F}}=\langle-y,x,0\rangle[/latex] and [latex]S[/latex] is the surface with parameterization [latex]{\bf{r}}(u,v)=\langle{u},v^2-u,u+v\rangle[/latex], [latex]0\leq{u}\leq3, \ 0\leq{v}\leq4[/latex].

Show Solution

The tangent vectors are [latex]{\bf{t}}_u=\langle1,-1,1\rangle[/latex] and [latex]{\bf{t}}_v=\langle0,2v,1\rangle[/latex]. Therefore,

[latex]{\bf{t}}_u\times{\bf{t}}_v=\langle-1-2v,-1,2v\rangle[/latex].

By [latex]\displaystyle\iint_S{\bf{F}}\cdot{\bf{N}}{d}S=\displaystyle\iint_D({\bf{F}}({\bf{r}}(u,v))\cdot({\bf{t}}_u\times{\bf{t}}_v))dA[/latex],

[latex]\begin{aligned} \displaystyle\iint_S{\bf{F}}\cdot{d}S&=\displaystyle\int_0^4\displaystyle\int_0^3{\bf{F}}({\bf{r}}(u,v))\cdot{\bf{t}}_u\times{\bf{t}}_v)dudv \\ &=\displaystyle\int_0^4\displaystyle\int_0^3\langle{u}-v^2,u,0\rangle\cdot\langle-1-2v,-1,2v\rangle{d}udv \\ &=\displaystyle\int_0^4\displaystyle\int_0^3[(u-v^2)(-1-2v)-u]dudv \\ &=\displaystyle\int_0^4\displaystyle\int_0^3(2v^3+v^2-2uv-2u]dudv \\ &=\displaystyle\int_0^4[2v^3u+v^2u-vu^2-u^2]_0^3dv \\ &=\displaystyle\int_0^4(6v^3+3v^2-9v-9)dv \\ &=\left[\frac{3v^2}2+v^3-\frac{9v^2}2-9v\right]_0^4 \\ &=340 \end{aligned}[/latex].

Therefore, the flux of [latex]{\bf{F}}[/latex] across [latex]S[/latex] is [latex]340[/latex].

Let [latex]{\bf{v}}(x,y,z)=\langle2x,2y,z\rangle[/latex] represent a velocity field (with units of meters per second) of a fluid with constant density [latex]80[/latex] kg/m³. Let [latex]S[/latex] be hemisphere [latex]x^{2}+y^{2}+z^{2}=9[/latex] with [latex]z\geq 0[/latex] such that [latex]S[/latex] is oriented outward. Find the mass flow rate of the fluid across [latex]S[/latex].

Show Solution

A parameterization of the surface is:

[latex]{\bf{r}}(\phi,\theta)=\langle3\cos\theta\sin\phi,3\sin\theta\sin\phi,3\cos\phi\rangle, \ 0\leq\theta\leq2\pi, \ 0\leq\phi\leq\pi/2[/latex].

The tangent vectors are:

[latex]{\bf{t}}_\theta\langle-3\sin\theta\sin\phi,3\cos\theta\sin\phi,0\rangle\text{ and }{\bf{t}}_\phi\langle3\cos\theta\cos\phi,3\sin\theta\cos\phi,-3\sin\phi\rangle[/latex],

and their cross product is:

[latex]{\bf{t}}_\phi\times{\bf{t}}_\theta=\langle9\cos\theta\sin^2\phi,9\sin\theta\sin^2\phi,9\sin\phi\cos\phi\rangle d\phid\theta[/latex].

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation [latex]{\bf{N}}=\langle9\cos\theta\sin^2\phi,9\sin\theta\sin^2\phi,9\sin\phi\cos\phi\rangle[/latex] for the sphere.

By the surface integral equation,

[latex]\begin{aligned} \displaystyle\iint_S\rho{\bf{v}}\cdot{d}{\bf{S}}&=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2}{\bf{v}}({\bf{r}}(\phi,\theta))\cdot({\bf{t}}_\phi\times{\bf{t}}_\theta)d\phi{d}\theta \\ &=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2} \\ &=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2}\langle6\cos\theta\sin\phi,6\sin\theta\sin\phi,3\cos\phi\rangle\cdot\langle9\cos\theta\sin^2\phi,9\sin\theta\sin^2\phi,9\sin\phi\cos\phi\rangle \\ &=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2}54\sin^2\phi+27\cos^2\phi\sin\phi{d}\phi{d}\theta \\ &=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2}54(1-\cos^2\phi)\sin\phi+27\cos^2\phi\sin\phi{d}\phi{d}\theta \\ &=80\displaystyle\int_0^{2\pi}\displaystyle\int_0^{\pi/2}54\sin\phi-27\cos^2\phi\sin\phi{d}\phi{d}\theta \\ &=80\displaystyle\int_0^{2\pi}[-54\cos\phi+9\cos^3\phi]_{\phi=0}^{\phi=2\pi}d\theta \\ &=80\displaystyle\int_0^{2\pi}45d\theta=7,200\pi \end{aligned}[/latex].

Therefore, the mass flow rate is [latex]7,200\pi\text{ kg/sec/m^2}[/latex].