{"id":1349,"date":"2025-10-17T13:48:12","date_gmt":"2025-10-17T13:48:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1349"},"modified":"2025-11-05T20:05:46","modified_gmt":"2025-11-05T20:05:46","slug":"line-integrals-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/line-integrals-learn-it-4\/","title":{"raw":"Line Integrals: Learn It 4","rendered":"Line Integrals: Learn It 4"},"content":{"raw":"

Flux and Circulation<\/h2>\r\n

We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used to calculate fluid flow across a curve, and circulation is important for characterizing conservative gradient fields. Both concepts are used heavily throughout the rest of this course and are especially important for Green's theorem, Stokes' theorem, and the divergence theorem.<\/p>\r\n\r\n

Understanding Flux<\/h3>\r\n

Let [latex]C[\/latex] be a plane curve and let [latex]{\\textbf{F}}[\/latex] be a vector field in the plane. Imagine [latex]C[\/latex] is a membrane across which fluid flows, but [latex]C[\/latex] does not impede the flow of the fluid. In other words, [latex]C[\/latex] is an idealized membrane invisible to the fluid. Suppose [latex]{\\textbf{F}}[\/latex] represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing [latex]C[\/latex]?<\/p>\r\n

Recall that the line integral of [latex]{\\textbf{F}}[\/latex] along [latex]C[\/latex] is [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex]\u2014the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector [latex]{\\textbf{N}}(t)[\/latex] and instead compute integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{N}} ds[\/latex], we determine the flux<\/strong> across [latex]C[\/latex].<\/p>\r\n

The definition of integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{N}} ds[\/latex] is the same as integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex], except the [latex]{\\textbf{T}}[\/latex] in the Riemann sum is replaced with [latex]{\\textbf{N}}[\/latex]. Therefore, the flux across [latex]C[\/latex] is defined as:<\/p>\r\n

[latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds=\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{i=1}^n{\\bf{F}}(P_i^*)\\cdot{\\bf{N}}(P_i^*)\\Delta{s_i}[\/latex],<\/p>\r\n

where [latex]P_i^*[\/latex] and [latex]\\Delta s[\/latex] are defined as they were for integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex]. A flux integral is perpendicular to a vector line integral, because [latex]{\\textbf{N}}[\/latex] and [latex]{\\textbf{T}}[\/latex] are perpendicular vectors.<\/p>\r\n\r\n

If [latex]{\\textbf{F}}[\/latex] is a velocity field of a fluid and [latex]C[\/latex] is a curve that represents a membrane, then the flux of [latex]{\\textbf{F}}[\/latex] across [latex]C[\/latex] is the quantity of fluid flowing across [latex]C[\/latex] per unit time, or the rate of flow.<\/section>\r\n

More formally, let [latex]C[\/latex] be a plane curve parameterized by [latex]{\\textbf{r}}(t)=\\langle x(t),y(t)\\rangle[\/latex], [latex]a\\leq t\\leq b[\/latex]. Let [latex]{\\textbf{n}}(t)=\\langle y'(t),-x'(t)\\rangle[\/latex] be the vector that is normal to [latex]C[\/latex] at the endpoint of [latex]{\\textbf{r}}(t)[\/latex] and points to the right as we traverse [latex]C[\/latex] in the positive direction (Figure 3). Then [latex]{\\textbf{N}}(t)=\\frac{{\\textbf{n}}(t)}{||{\\textbf{n}}(t)||}[\/latex] is the unit normal vector to [latex]C[\/latex] at the endpoint of [latex]{\\textbf{r}}(t)[\/latex] that points to the right as we traverse [latex]C[\/latex].<\/p>\r\n\r\n

\r\n

flux across a curve<\/h3>\r\nThe flux<\/strong> of [latex]{\\textbf{F}}[\/latex] across [latex]C[\/latex] is line integral:\r\n

[latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot\\frac{{\\textbf{n}}(t)}{||{\\textbf{n}}(t)||}ds[\/latex]<\/p>\r\n\r\n<\/section>[caption id=\"\" align=\"aligncenter\" width=\"522\"]\"A Figure 3. The flux of vector field [latex]\\bf{F}[\/latex]\u00a0across curve [latex]C[\/latex] is computed by an integral similar to a vector line integral.[\/caption]\r\n

We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral.<\/p>\r\n\r\n

\r\n

calculating flux across a curve<\/h3>\r\n

Let [latex]{\\bf{F}}[\/latex] be a vector field and let [latex]C[\/latex] be a smooth curve with parameterization [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex]. The flux of [latex]{\\bf{F}}[\/latex] across [latex]C[\/latex] is:<\/p>\r\n

[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n

The proof of calculating flux across a curve is similar to the proof of the scalar line integral calculation. Before deriving the formula, note that [latex]||{\\bf{n}}(t)||=||\\langle y'(t),-x'(t)\\rangle||=\\sqrt{(y^\\prime(t))^2+(x^\\prime(t))^2}=||{\\bf{r}}^\\prime(t)||[\/latex]. Therefore,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}ds \\\\\r\n&=\\displaystyle\\int_a^b{\\bf{F}}\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}||{\\bf{r}}^\\prime(t)||dt \\\\\r\n&=\\displaystyle\\int_a^b{\\bf{F}}[({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt\r\n\\end{aligned}[[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>

Calculate the flux of [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex] across a unit circle oriented counterclockwise (Figure 4).\r\n[caption id=\"attachment_2025\" align=\"aligncenter\" width=\"496\"]\"A Figure 4. A unit circle in vector field [latex]F=\u27e82x,2y\u27e9[\/latex].[\/caption][reveal-answer q=\"589274091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589274091\"]\r\n

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization [latex]{\\bf{r}}(t)\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. The normal vector to a unit circle is [latex]\\langle\\cos{t},\\sin{t}\\rangle[\/latex]. Therefore, the flux is<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\int_0^{2\\pi}\\langle2\\cos{t},2\\sin{t}\\rangle\\cdot\\langle\\cos{t},\\sin{t}\\rangle \\ dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}(2\\cos^2{t},2\\sin^2{t}) \\ dt = 2\\displaystyle\\int_0^{2\\pi}(\\cos^2{t},\\sin^2{t}) \\ dt \\\\\r\n&=2\\displaystyle\\int_0^{2\\pi} \\ dt = 4\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Alternative Notation for Flux<\/strong>\r\n

Let [latex]{\\textbf{F}}(x,y)=\\langle P(x,y),Q(x,y)\\rangle[\/latex] be a two-dimensional vector field. Recall that integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}}ds[\/latex] is sometimes written as [latex]\\displaystyle\\int_CP dx+Q dy[\/latex].<\/p>\r\n

Analogously, flux [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{N}}ds[\/latex] is sometimes written in the notation [latex]\\displaystyle\\int_C-Q dx+P dy[\/latex], because the unit normal vector [latex]{\\textbf{N}}[\/latex] is perpendicular to the unit tangent [latex]{\\textbf{T}}[\/latex]. Rotating the vector [latex]d{\\textbf{r}}=\\langle dx,dy\\rangle[\/latex] by [latex]90\u00b0[\/latex] results in vector [latex]\\langle dy,-dx\\rangle[\/latex].<\/p>\r\n\r\n<\/section>\r\n

Understanding Circulation<\/h3>\r\n

The line integral of vector field [latex]{\\textbf{F}}[\/latex] along an oriented closed curve is called the circulation<\/strong> of [latex]{\\textbf{F}}[\/latex] along [latex]C[\/latex]. Circulation line integrals have their own notation:<\/p>\r\n

[latex]\\displaystyle\\oint_C{\\textbf{F}}\\cdot{\\textbf{T}}ds[\/latex].<\/p>\r\n

The circle on the integral symbol denotes that [latex]C[\/latex] is \"circular\" in that it has no endpoints.<\/p>\r\n

To see where the term circulation comes from and what it measures, let [latex]{\\textbf{v}}[\/latex] represent the velocity field of a fluid and let [latex]C[\/latex] be an oriented closed curve. At a particular point [latex]P[\/latex], the closer the direction of [latex]{\\textbf{v}}(P)[\/latex] is to the direction of [latex]{\\textbf{T}}(P)[\/latex], the larger the value of the dot product [latex]{\\textbf{v}}(P)\\cdot{\\textbf{T}}(P)[\/latex].<\/p>\r\n

The maximum value of [latex]{\\textbf{v}}(P)\\cdot{\\textbf{T}}(P)[\/latex] occurs when the two vectors point in the exact same direction. The minimum value occurs when the two vectors point in opposite directions. Thus, the value of the circulation [latex]\\displaystyle\\oint_C{\\textbf{v}}\\cdot{\\textbf{T}}ds[\/latex] measures the tendency of the fluid to move in the direction of [latex]C[\/latex].<\/p>\r\n\r\n

Let [latex]{\\textbf{F}}=\\langle -y,x\\rangle[\/latex] and let [latex]C[\/latex] represent the unit circle oriented counterclockwise. Calculate the circulation of [latex]{\\textbf{F}}[\/latex] along [latex]C[\/latex].[hidden-answer a=\"039258487\"]\r\n

We use the standard parameterization of the unit circle: [latex]{\\bf{r}}(t)=\\langle\\cos{t},\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex] and [latex]{\\bf{r}}'(t)=\\langle-\\sin{t},\\cos{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}ds&=\\displaystyle\\int_0^{2\\pi}\\langle-\\sin{t},\\cos{t}\\rangle\\cdot\\langle-\\sin{t},\\cos{t}\\rangle \\ dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}(\\sin^2{t}+\\cos^2{t}) \\ dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi} \\ dt = 2\\pi\r\n\\end{aligned}[\/latex].<\/p>\r\n

Notice that the circulation is positive. The reason for this is that the orientation of [latex]C[\/latex] \u201cflows\u201d with the direction of\u00a0[latex]\\bf{F}[\/latex]. At any point along the circle, the tangent vector and the vector from\u00a0[latex]\\bf{F}[\/latex]\u00a0form an angle of less than 90\u00b0, and therefore the corresponding dot product is positive.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

In the above example, what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by [latex]-C[\/latex]. Then:<\/p>\r\n

[latex]\\displaystyle\\oint_{-C}{\\bf{F}}\\cdot{\\bf{T}} \\ ds=-\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}}=-2\\pi[\/latex].<\/p>\r\n

Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of\u00a0[latex]\\bf{F}[\/latex].<\/p>\r\n\r\n

Calculate the work done on a particle that traverses circle [latex]C[\/latex] of radius 2 centered at the origin, oriented counterclockwise, by field [latex]{\\bf{F}}(x,y)=\\langle-2,y\\rangle[\/latex]. Assume the particle starts its movement at [latex](1, 0)[\/latex].[reveal-answer q=\"802581366\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"802581366\"]\r\n

The work done by\u00a0[latex]\\bf{F}[\/latex]\u00a0on the particle is the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex]: [latex]\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds[\/latex]. We use the parameterization [latex]{\\bf{r}}(t)=\\langle2\\cos{t},2\\sin{t}\\rangle[\/latex], [latex]0\\leq{t}\\leq2\\pi[\/latex] for [latex]C[\/latex]. Then, [latex]{\\bf{r}}^\\prime(t)=\\langle-2\\cos{t},2\\sin{t}\\rangle[\/latex] and [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle-2,2\\sin{t}\\rangle[\/latex]. Therefore, the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\oint_C{\\bf{F}}\\cdot{\\bf{T}} \\ ds&=\\displaystyle\\int_0^{2\\pi}\\langle-2,2\\sin{t}\\rangle\\cdot\\langle-2,2\\cos{t}\\rangle \\ dt \\\\\r\n&=\\displaystyle\\int_0^{2\\pi}(4\\sin{t}+4\\sin{t}\\cos{t}) \\ dt \\\\\r\n&=\\left[-4\\cos{t}+4\\sin^2{t}\\right]_0^{2\\pi} \\\\\r\n&=(-4\\cos(2\\pi)+2\\sin^2(2\\pi))-(-4\\cos(0)+4\\sin^2(0)) \\\\\r\n&=-4+4=0\r\n\\end{aligned}[\/latex].<\/p>\r\n

The force field does zero work on the particle.<\/p>\r\n

Notice that the circulation of\u00a0[latex]\\bf{F}[\/latex]\u00a0along [latex]C[\/latex] is zero. Furthermore, notice that since\u00a0[latex]\\bf{F}[\/latex]\u00a0is the gradient of [latex]f(x,y)=-2x+\\frac{y^2}2[\/latex],\u00a0[latex]\\bf{F}[\/latex]\u00a0is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Flux and Circulation<\/h2>\n

We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used to calculate fluid flow across a curve, and circulation is important for characterizing conservative gradient fields. Both concepts are used heavily throughout the rest of this course and are especially important for Green’s theorem, Stokes’ theorem, and the divergence theorem.<\/p>\n

Understanding Flux<\/h3>\n

Let [latex]C[\/latex] be a plane curve and let [latex]{\\textbf{F}}[\/latex] be a vector field in the plane. Imagine [latex]C[\/latex] is a membrane across which fluid flows, but [latex]C[\/latex] does not impede the flow of the fluid. In other words, [latex]C[\/latex] is an idealized membrane invisible to the fluid. Suppose [latex]{\\textbf{F}}[\/latex] represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing [latex]C[\/latex]?<\/p>\n

Recall that the line integral of [latex]{\\textbf{F}}[\/latex] along [latex]C[\/latex] is [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex]\u2014the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector [latex]{\\textbf{N}}(t)[\/latex] and instead compute integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{N}} ds[\/latex], we determine the flux<\/strong> across [latex]C[\/latex].<\/p>\n

The definition of integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{N}} ds[\/latex] is the same as integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex], except the [latex]{\\textbf{T}}[\/latex] in the Riemann sum is replaced with [latex]{\\textbf{N}}[\/latex]. Therefore, the flux across [latex]C[\/latex] is defined as:<\/p>\n

[latex]\\displaystyle\\int_c{\\bf{F}}\\cdot{\\bf{N}} \\ ds=\\displaystyle\\lim_{n\\to\\infty}\\displaystyle\\sum_{i=1}^n{\\bf{F}}(P_i^*)\\cdot{\\bf{N}}(P_i^*)\\Delta{s_i}[\/latex],<\/p>\n

where [latex]P_i^*[\/latex] and [latex]\\Delta s[\/latex] are defined as they were for integral [latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot{\\textbf{T}} ds[\/latex]. A flux integral is perpendicular to a vector line integral, because [latex]{\\textbf{N}}[\/latex] and [latex]{\\textbf{T}}[\/latex] are perpendicular vectors.<\/p>\n

If [latex]{\\textbf{F}}[\/latex] is a velocity field of a fluid and [latex]C[\/latex] is a curve that represents a membrane, then the flux of [latex]{\\textbf{F}}[\/latex] across [latex]C[\/latex] is the quantity of fluid flowing across [latex]C[\/latex] per unit time, or the rate of flow.<\/section>\n

More formally, let [latex]C[\/latex] be a plane curve parameterized by [latex]{\\textbf{r}}(t)=\\langle x(t),y(t)\\rangle[\/latex], [latex]a\\leq t\\leq b[\/latex]. Let [latex]{\\textbf{n}}(t)=\\langle y'(t),-x'(t)\\rangle[\/latex] be the vector that is normal to [latex]C[\/latex] at the endpoint of [latex]{\\textbf{r}}(t)[\/latex] and points to the right as we traverse [latex]C[\/latex] in the positive direction (Figure 3). Then [latex]{\\textbf{N}}(t)=\\frac{{\\textbf{n}}(t)}{||{\\textbf{n}}(t)||}[\/latex] is the unit normal vector to [latex]C[\/latex] at the endpoint of [latex]{\\textbf{r}}(t)[\/latex] that points to the right as we traverse [latex]C[\/latex].<\/p>\n

\n

flux across a curve<\/h3>\n

The flux<\/strong> of [latex]{\\textbf{F}}[\/latex] across [latex]C[\/latex] is line integral:<\/p>\n

[latex]\\displaystyle\\int_C{\\textbf{F}}\\cdot\\frac{{\\textbf{n}}(t)}{||{\\textbf{n}}(t)||}ds[\/latex]<\/p>\n<\/section>\n

\"A
Figure 3. The flux of vector field [latex]\\bf{F}[\/latex]\u00a0across curve [latex]C[\/latex] is computed by an integral similar to a vector line integral.<\/figcaption><\/figure>\n

We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral.<\/p>\n

\n

calculating flux across a curve<\/h3>\n

Let [latex]{\\bf{F}}[\/latex] be a vector field and let [latex]C[\/latex] be a smooth curve with parameterization [latex]{\\bf{r}}(t)=\\langle{x}(t),y(t)\\rangle[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]{\\bf{n}}(t)=\\langle{y}^\\prime(t),-x^\\prime(t)\\rangle[\/latex]. The flux of [latex]{\\bf{F}}[\/latex] across [latex]C[\/latex] is:<\/p>\n

[latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds=\\displaystyle\\int_a^b{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt[\/latex].<\/p>\n<\/section>\n

\n

Proof<\/strong><\/p>\n

\n

The proof of calculating flux across a curve is similar to the proof of the scalar line integral calculation. Before deriving the formula, note that [latex]||{\\bf{n}}(t)||=||\\langle y'(t),-x'(t)\\rangle||=\\sqrt{(y^\\prime(t))^2+(x^\\prime(t))^2}=||{\\bf{r}}^\\prime(t)||[\/latex]. Therefore,<\/p>\n

[latex]\\begin{aligned} \\displaystyle\\int_C{\\bf{F}}\\cdot{\\bf{N}}ds&=\\displaystyle\\int_C{\\bf{F}}\\cdot\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}ds \\\\ &=\\displaystyle\\int_a^b{\\bf{F}}\\frac{{\\bf{n}}}{||{\\bf{n}}(t)||}||{\\bf{r}}^\\prime(t)||dt \\\\ &=\\displaystyle\\int_a^b{\\bf{F}}[({\\bf{r}}(t))\\cdot{\\bf{n}}(t)dt \\end{aligned}[[\/latex].<\/p>\n

[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

Calculate the flux of [latex]{\\bf{F}}=\\langle2x,2y\\rangle[\/latex] across a unit circle oriented counterclockwise (Figure 4).<\/p>\n
\"A
Figure 4. A unit circle in vector field [latex]F=\u27e82x,2y\u27e9[\/latex].<\/figcaption><\/figure>\n