The word flux is derived from the Latin word fluere, which means “to flow.”

REMINDER

Formula for a scalar surface integral in terms of an oriented parametrization: \begin{equation*} \begin{array}{l} \iint_{{\mathcal{S}}\,}f(x,y,z)\, dS\\ = \iint f({G}(u,v))\left \| {{\bf{n}}(u,v)} \right \| \,du\,dv \end{array}\tag{1} \end{equation*}

THEOREM 1 Vector Surface Integral

Let \({G}(u,v)\) be an oriented parametrization of an oriented surface \({\mathcal{S}}\) with parameter domain \({\mathcal{D}}\). Assume that \({G}\) is one-to-one and regular, except possibly at points on the boundary of \({\mathcal{D}}\). Then \begin{equation*} \boxed{ \iint_{{\mathcal{S}}\,}{\bf{F}}{\cdot} d{\bf{S}} = \iint_{{\mathcal{D}}\,} {\bf{F}}({G}(u,v)){\cdot} {\bf{n}}(u,v) \,du\,dv}\tag{3} \end{equation*}

If the orientation of \({\mathcal{S}}\) is reversed, the surface integral changes sign.

EXAMPLE 1

Calculate \(\iint_{{\mathcal{S}}\,}{\bf{F}}{\cdot}\,d{\bf{S}}\), where \({\bf{F}} = \left\langle 0,0,x \right\rangle \) and \({\mathcal{S}}\) is the surface with parametrization \({G}(u,v) = (u^2, v, u^3 - v^2)\) for \( 0\le u\le 1,\,\, 0\le v\le 1\) and oriented by upward-pointing normal vectors.

Solution

Step 1. Compute the tangent and normal vectors.

\[ \begin{array}{rl} {\bf{T}}_u &= \big\langle 2u,0,3u^2\big\rangle,\qquad T_v = \left\langle 0,1,-2v\right\rangle\\ {\bf{n}}(u,v) &= {\bf{T}}_u{\times} {\bf{T}}_v = \left| \begin{array}{ccc} {\bf{i}} & {\bf{j}} & {\bf{k}}\\ 2u & 0 & 3u^2\\ 0 & 1 & -2v \end{array} \right| \\ &= -3u^2{\bf{i}} + 4uv{\bf{j}} + 2u{\bf{k}} = \big\langle-3u^2, 4uv, 2u\big\rangle \end{array} \]

The \(z\)-component of \({\bf{n}}\) is positive on the domain \(0\le u\le 1\), so \({\bf{n}}\) is the upward-pointing normal (Figure 16.55).

Figure 16.55: The surface \({G}(u,v) = (u^2, v, u^3 - v^2)\) with an upward-pointing normal. The vector field \({\bf{F}}= \left\langle 0,0,x \right\rangle \) points in the vertical direction.

Step 2. Evaluate \({\bf{F}}{\cdot} {\bf{n}}\).

Write \({\bf{F}}\) in terms of the parameters \(u\) and \(v\). Since \(x= u^2\), \[ {\bf{F}}({G}(u,v)) = \left\langle 0,0,x \right\rangle = \big\langle 0,0,u^2\big\rangle \] and \[ {\bf{F}}({G}(u,v)){\cdot} {\bf{n}}(u,v) = \big\langle 0,0,u^2\big\rangle {\cdot}\big\langle-3u^2, 4uv, 2u\big\rangle = 2u^3 \]

Step 3. Evaluate the surface integral.

The parameter domain is \(0\le u\le1\), \(0\le v\le 1\), so \[ \begin{array}{rl} \iint_{{\mathcal{S}}\,} {\bf{F}}{\cdot} d{\bf{S}} &= \int_{u=0}^1\int_{v=0}^1 {\bf{F}}({G}(u,v)){\cdot}{\bf{n}}(u,v)\,dv\,du \\ &= \int_{u=0}^1\int_{v=0}^1 2u^3\,dv\,du = \int_{u=0}^1 2u^3 \,du =\frac12 \end{array} \]

EXAMPLE 2 Integral over a Hemisphere

Calculate the flux of \({\bf{F}} = \left\langle z, x, 1 \right\rangle \) across the upper hemisphere \({\mathcal{S}}\) of the sphere \(x^2+y^2+z^2=1\), oriented with outward-pointing normal vectors (Figure 16.56).

Figure 16.56: The vector field \({\bf{F}} = \left\langle z, x, 1 \right\rangle\).

Solution Parametrize the hemisphere by spherical coordinates: \[ {G}(\theta,\phi) = (\cos\theta\sin\phi, \sin\theta\sin\phi,\cos\phi),\qquad 0\le\phi\le \frac{\pi}2, \quad 0 \le \theta< 2\pi \]

Step 1. Compute the normal vector.

According to Eq.(2) in Section 16.4, the outward-pointing normal vector is \[ {\bf{n}} = {\bf{T}}_\phi{\times} {\bf{T}}_\theta = \sin\phi\big\langle\cos\theta\sin \phi, \sin\theta\sin \phi,\cos\phi\big\rangle \]

Step 2. Evaluate \({\bf{F}}{\cdot} {\bf{n}}\).

\[ \begin{array}{rcl} {\bf{F}}({G}(\theta,\phi)) &=& \left\langle z, x, 1\right\rangle = \big\langle\cos \phi, \cos\theta\sin \phi,1 \big\rangle\\ {\bf{F}}({G}(\theta,\phi))\cdot {\bf{n}}(\theta,\phi) &=& \big\langle \cos \phi, \cos\theta\sin \phi,1 \big\rangle {\cdot} \big\langle \cos\theta\sin^2\phi, \sin\theta\sin^2\phi,\cos\phi\sin\phi\big\rangle \\ &=&\cos\theta\sin^2\phi\cos \phi+ \cos\theta\sin\theta \sin^3\phi +\cos\phi\sin\phi \end{array} \]

Step 3. Evaluate the surface integral.

\[ \begin{array}{rl} \iint_{{\mathcal{S}}\,}{\bf{F}}{\cdot} d{\bf{S}} &= \int_{\phi=0}^{\pi/2} \int_{\theta=0}^{2\pi} {\bf{F}}({G}(\theta,\phi)){\cdot} {\bf{n}}(\theta,\phi)\,d\theta\,d\phi\\ & = \int_{\phi=0}^{\pi/2} \int_{\theta=0}^{2\pi} ( \underbrace{\cos\theta\sin^2\phi\cos \phi+ \cos\theta\sin\theta \sin^3\phi}_{\textrm{Integral over }\theta \textrm{ is zero}} +\cos\phi\sin\phi)\,d\theta\,d\phi \end{array} \]

The integrals of \(\cos\theta\) and \(\cos\theta\sin\theta\) over \([0, 2\pi]\) are both zero, so we are left with \[ \int_{\phi=0}^{\pi/2} \int_{\theta=0}^{2\pi}\cos\phi\sin\phi \,d\theta\,d\phi = 2\pi \int_{\phi=0}^{\pi/2} \cos\phi\sin\phi \, d\phi = -2\pi \frac{\cos^2\phi}2\bigg|_0^{\pi/2} = \pi \]

EXAMPLE 3 Surface Integral over a Graph

Calculate the flux of \({\bf{F}} = x^2{\bf{j}}\) through the surface \({\mathcal{S}}\) defined by \(y=1+x^2+z^2\) for \(1\le y \le 5\). Orient \({\mathcal{S}}\) with normal pointing in the negative \(y\)-direction.

Solution This surface is the graph of the function \(y=1+x^2+z^2\), where \(x\) and \(z\) are the independent variables (Figure 16.57).

Step 1. Find a parametrization.

It is convenient to use \(x\) and \(z\) because \(y\) is given explicitly as a function of \(x\) and \(z\). Thus we define \[ {G}(x,z)=(x,1+x^2+z^2,z) \]

What is the parameter domain? The condition \(1\le y\le 5\) is equivalent to \(1\le 1+x^2+z^2 \le 5\) or \(0\le x^2+z^2 \le 4\). Therefore, the parameter domain is the disk of radius \(2\) in the \(xz\)-plane—that is, \({\mathcal{D}}=\{(x,z): x^2+z^2\le 4\}\).

Because the parameter domain is a disk, it makes sense to use the polar variables \(r\) and \(\theta\) in the \(xz\)-plane. In other words, we write \(x=r\cos\theta\), \(z=r\sin\theta\). Then \[ y = 1+x^2+z^2=1+r^2 \] \[ {G}(r,\theta)=(r\cos\theta,1+r^2,r\sin\theta),\quad 0\le\theta\le 2\pi,\quad 0\le r\le 2 \]

Step 2. Compute the tangent and normal vectors.

\[ \begin{array}{rl} {\bf{T}}_r &= \left\langle \cos\theta,2r,\sin\theta\right\rangle,\qquad {\bf{T}}_\theta = \left\langle -r\sin\theta, 0, r\cos\theta\right\rangle\\ {\bf{n}} &= {\bf{T}}_r{\times} {\bf{T}}_\theta = \left| \begin{array}{ccc} {\bf{i}} & {\bf{j}} & {\bf{k}}\\ \cos\theta&2r&\sin\theta\\ -r\sin\theta& 0& r\cos\theta \end{array} \right| = 2r^2\cos\theta{\bf{i}}-r{\bf{j}}+2r^2\sin\theta{\bf{k}} \end{array} \]

The coefficient of \({\bf{j}}\) is \(-r\). Because it is negative, \({\bf{n}}\) points in the negative \(y\)-direction, as required.

Step 3. Evaluate \({\bf{F}}{\cdot} {\bf{n}}\).

\[ \begin{array}{rl} {\bf{F}}({G}(r,\theta)) &= x^2{\bf{j}} = r^2\cos^2\theta{\bf{j}} = \big\langle 0, r^2\cos^2\theta, 0\big\rangle\\ {\bf{F}}({G}(r,\theta)){\cdot} {\bf{n}} &= \big\langle 0, r^2\cos^2\theta, 0\big\rangle{\cdot} \left\langle 2r^2\cos\theta,-r,2r^2\sin\theta\right\rangle = -r^3 \cos^2\theta\\ \iint_{{\mathcal{S}}\,}{\bf{F}}{\cdot}\, d{\bf{S}} &= \iint_{{\mathcal{D}}\,} {\bf{F}}({G}(r,\theta)){\cdot} {\bf{n}}\,dr\,d\theta = \int_0^{2\pi}\int_0^2\,\,( -r^3 \cos^2\theta)\,dr\,d\theta\\ &= -\left(\int_0^{2\pi}\cos^2\theta\,d\theta\right) \left(\int_0^2\,\, r^3 \,dr\right)\\ &= -(\pi)\left(\frac{2^4}4\right) = -4\pi \end{array} \]

CONCEPTUAL INSIGHT

Since a vector surface integral depends on the orientation of the surface, this integral is defined only for surfaces that have two sides. However, some surfaces, such as the Möbius strip (discovered in 1858 independently by August Möbius and Johann Listing), cannot be oriented because they are one-sided. You can construct a Möbius strip \(M\) with a rectangular strip of paper: Join the two ends of the strip together with a \(180^\circ\) twist. Unlike an ordinary two-sided strip, the Möbius strip \(M\) has only one side, and it is impossible to specify an outward direction in a consistent manner (Figure 16.58). If you choose a unit normal vector at a point \(P\) and carry that unit vector continuously around \(M\), when you return to \(P\), the vector will point in the opposite direction. Therefore, we cannot integrate a vector field over a Möbius strip, and it is not meaningful to speak of the “flux” across \(M\). On the other hand, it is possible to integrate a scalar function. For example, the integral of mass density would equal the total mass of the Möbius strip.

Flow Rate through a Surface

For a fluid with velocity vector field \({\bf{v}}\), \begin{equation*} \boxed{\textrm{Flow rate across the }{\mathcal{S}}\textrm{ (volume per unit time)} = \iint_{{\mathcal{S}}\,} {\bf{v}}{\cdot}\,d{\bf{S}}}\tag{4} \end{equation*}

EXAMPLE 4

Let \({\bf{v}} = \left\langle x^2+y^2,0,z^2\right\rangle\) be the velocity field (in centimeters per second) of a fluid in \({\bf{R}}^3\). Compute the flow rate through the upper hemisphere \({\mathcal{S}}\) of the unit sphere centered at the origin.

Solution We use spherical coordinates: \[ x = \cos\theta\sin\phi,\qquad y = \sin\theta\sin\phi,\qquad z = \cos\phi \]

The upper hemisphere corresponds to the ranges \(0 \le \phi \le \frac{\pi}2\) and \(0 \le \theta \le 2\pi\). By Eq.(2) in Section 16.4, the upward-pointing normal is \[ {\bf{n}} = {\bf{T}}_{\phi} {\times} {\bf{T}}_{\theta} = \sin\phi\big\langle \cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi\big\rangle \]

We have \(x^2+y^2 =\sin^2\phi\), so \[ \begin{array}{rl} {\bf{v}} &= \big\langle x^2+y^2,0,z^2\big\rangle = \big\langle \sin^2\phi, 0, \cos^2\phi\big\rangle\\ {\bf{v}}{\cdot}{\bf{n}} &=\sin\phi\big\langle \sin^2\phi, 0, \cos^2\phi\big\rangle{\cdot}\big\langle \cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi\big\rangle \\ &=\sin^4\phi\cos\theta + \sin\phi \cos^3\phi\\ \iint_{{\mathcal{S}}\,} {\bf{v}} {\cdot} d{\bf{S}} &= \int_{\phi=0}^{\pi/2}\int_{\theta =0}^{2\pi}(\sin^4\phi\cos\theta + \sin\phi \cos^3\phi)\,d\theta\,d\phi \end{array} \]

The integral of \(\sin^4\phi\cos\theta\) with respect to \(\theta\) is zero, so we are left with \[ \begin{array}{rl} \int_{\phi=0}^{\pi/2}\int_{\theta=0}^{2\pi} \sin\phi \cos^3\phi\,d\theta\,d\phi &= 2\pi \int_{\phi=0}^{\pi/2} \cos^3\phi\sin\phi \,d\phi\\ &= 2\pi\left(-\frac{\cos^4\phi}4\right)\bigg|_{\phi=0}^{\pi/2} = \frac{\pi}2~\mathrm{cm^3/s} \end{array} \]

Since \({\bf{n}}\) is an upward-pointing normal, this is the rate at which fluid flows across the hemisphere from below to above.

The tesla (T) is the SI unit of magnetic field strength. A one-coulomb point charge passing through a magnetic field of 1 T at 1 m/s experiences a force of 1 newton.

The electric field \({\bf{E}}\) is conservative when the charges are stationary or, more generally, when the magnetic field \({\bf{B}}\) is constant. When \({\bf{B}}\) varies in time, the integral on the right in Eq. (5) is nonzero for some surface, and hence the circulation of \({\bf{E}}\) around the boundary curve \({\mathcal{C}}\) is also nonzero. This shows that \({\bf{E}}\) is not conservative when \({\bf{B}}\) varies in time.

EXAMPLE 5

A varying current of magnitude (\(t\) in seconds) \[ i=28\cos\left(400t\right) \textrm{amperes} \] flows through a straight wire [Figure 16.65]. A rectangular wire loop \({\mathcal{C}}\) of length \(L=1.2~\mathrm{m}\) and width \(H=0.7~\mathrm{m}\) is located a distance \(d=0.1~\mathrm{m}\) from the wire as in the figure. The loop encloses a rectangular surface \({\mathcal{R}}\), which is oriented by normal vectors pointing out of the page.

• (a) Calculate the flux \(\Phi(t)\) of \({\bf{B}}\) through \({\mathcal{R}}\).
• (b) Use Faraday’s Law to determine the voltage drop (in volts) around the loop~\({\mathcal{C}}\).

Solution We choose coordinates \((x,y)\) on rectangle \({\mathcal{R}}\) as in Figure 16.65, so that \(y\) is the distance from the wire and \({\mathcal{R}}\) is the region \[ 0\le x \le L,\qquad d \le y \le H + d \]

Our parametrization of \({\mathcal{R}}\) is simply \(G(x,y)=(x,y)\), for which the normal vector \({\bf{n}}\) is the unit vector perpendicular to \({\mathcal{R}}\), pointing out of the page. The magnetic field \({\bf{B}}\) at \(P=(x,y)\) has magnitude \(\frac{\mu_0|i|}{2\pi y}\). It points out of the page in the direction of \({\bf{n}}\) when \(i\) is positive and into the page when \(i\) is negative. Thus, \[ {\bf{B}} = \frac{\mu_0i}{2\pi y}{\bf{n}} \qquad\textrm{and}\qquad {\bf{B}}{\cdot}{\bf{n}} = \frac{\mu_0i}{2\pi y} \]

• (a) The flux \(\Phi(t)\) of \({\bf{B}}\) through \({\mathcal{R}}\) at time \(t\) is \[ \begin{array}{rl} \Phi(t) &= \iint_{{\mathcal{R}}\,} {\bf{B}}\cdot d{\bf{S}}= \int_{x=0}^L\int_{y=d}^{H+d} {\bf{B}}\cdot {\bf{n}}\,dy\,dx\\ &=\int_{x=0}^L\int_{y=d}^{H+d} \frac{\mu_0i}{2\pi y}\,dy\,dx = \frac{\mu_0Li}{2\pi}\int_{y=d}^{H+d} \frac{dy}{y} \\ &= \frac{\mu_0L}{2\pi}\left(\ln\frac{H+d}{d}\right) i \\ &= \frac{\mu_0(1.2)}{2\pi} \left(\ln\frac{0.8}{0.1}\right) 28 \cos\left(400t\right) \end{array} \] With \(\mu_0 = 4\pi\cdot 10^{-7}\), we obtain \[ \Phi(t) \approx 1.4 \times 10^{-5}\cos\left(400t\right)~\mathrm{T\hbox{-}m^2} \]
• (b) By Faraday’s Law [Eq. (5)], the voltage drop around the rectangular loop \({\mathcal{C}}\), oriented in the counterclockwise direction, is \[ \int_{{\mathcal{C}}} {\bf{E}}{\cdot}\,d{\bf{s}} = -\frac{d \Phi}{d t} \approx -(1.4 \times 10^{-5}) \left(400\right)\sin \left(400 t\right) = -0.0056\sin\left(400t\right)~\mathrm{V} \]