{"id":1366,"date":"2025-10-17T14:30:15","date_gmt":"2025-10-17T14:30:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1366"},"modified":"2025-11-06T13:23:19","modified_gmt":"2025-11-06T13:23:19","slug":"conservative-vector-fields-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/conservative-vector-fields-learn-it-2\/","title":{"raw":"Conservative Vector Fields: Learn It 2","rendered":"Conservative Vector Fields: Learn It 2"},"content":{"raw":"

Fundamental Theorem for Line Integrals<\/h2>\r\nNow that we understand some basic curves and regions, we can generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function [latex]f[\/latex] has an antiderivative [latex]F[\/latex], then the integral of [latex]f[\/latex] from [latex]a[\/latex] to [latex]b[\/latex] depends only on the values of [latex]F[\/latex] at [latex]a[\/latex] and at [latex]b[\/latex]:\r\n

[latex]\\displaystyle\\int_a^bf(x) \\ dx =F(b)-F(a)[\/latex].<\/p>\r\n

If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. Let's see how this works with an example.<\/p>\r\n\r\n

Let [latex]{\\bf{F}}(x,y)=\\langle2x,4y\\rangle[\/latex]. Calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex], where [latex]C[\/latex] is the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] (Figure 4).[reveal-answer q=\"243750312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"243750312\"]\r\n

We use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]\u00a0to calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex]. Curve [latex]C[\/latex] can be parameterized by [latex]{\\bf{r}}(t)=\\langle2t,2t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex]. Then, [latex]{\\bf{F}}({\\bf{r}}(t))=\\langle4t,8t\\rangle[\/latex] and [latex]{\\bf{r}}^\\prime(t)=\\langle2,2\\rangle[\/latex], which implies that<\/p>\r\n

[latex]\\begin{aligned}\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1\\langle4t,8t\\rangle\\cdot\\langle2,2\\rangle \\ dt \\\\&=\\displaystyle\\int_0^1(8t+16t)dt=\\displaystyle\\int_0^124t \\ dt \\\\&=[12t^2]_0^1=12.\\end{aligned}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_2031\" align=\"aligncenter\" width=\"455\"]\"A Figure 4. The value of line integral [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex] depends only on the value of the potential function of [latex]\\bf{F}[\/latex] at the endpoints of the curve.[\/caption]\r\n

Notice that [latex]F=\\nabla{f}[\/latex], where [latex]f(x, y)=x^{2}+2y^{2}[\/latex]. If we think of the gradient as a derivative, then [latex]f[\/latex] is an \u201cantiderivative\u201d of\u00a0[latex]\\bf{F}[\/latex]. In the case of single-variable integrals, the integral of derivative [latex]g'(x)[\/latex] is [latex]g(b)-g(a)[\/latex], where\u00a0a<\/em>\u00a0is the start point of the interval of integration and\u00a0b<\/em>\u00a0is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral\u00a0[latex]\\bf{F}[\/latex]\u00a0to be [latex]f(P_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of the curve of integration and [latex]P_0[\/latex] is the start point. Notice that this is the case for this example:<\/p>\r\n

[latex]\\large{\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=12}[\/latex]<\/p>\r\n

and<\/p>\r\n

[latex]\\large{f(2,2)-f(0,0)=4+8-0=12}[\/latex].<\/p>\r\n

In other words, the integral of a \u201cderivative\u201d can be calculated by evaluating an \u201cantiderivative\u201d at the endpoints of the curve and subtracting, just as for single-variable integrals.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field.\r\n\r\n

\r\n

the fundamental theorem for line integrals<\/h3>\r\n

Let [latex]C[\/latex] be a piecewise smooth curve with parameterization [latex]{\\bf{r}}(t)[\/latex], [latex]a\\leq{t}\\leq{b}[\/latex]. Let [latex]f[\/latex] be a function of two or three variables with first-order partial derivatives that exist and are continuous on [latex]C[\/latex]. Then,<\/p>\r\n

[latex]\\displaystyle\\int_C\\nabla{f}. \\ d{\\bf{r}}=f({\\bf{r}}(b))-f({\\bf{r}}(a))[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

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

\r\n

By\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex],<\/p>\r\n

[latex]\\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt[\/latex].<\/p>\r\n

By the chain rule,<\/p>\r\n

[latex]\\frac{d}{dt}(f({\\bf{r}}(t))=\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t)[\/latex].<\/p>\r\n

Therefore, by the Fundamental Theorem of Calculus,<\/p>\r\n

[latex]\\begin{aligned}\r\n\\displaystyle\\int_C\\nabla{f}\\cdot \\ d{\\bf{r}}&=\\displaystyle\\int_a^b\\nabla{f}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\\r\n&=\\displaystyle\\int_a^b\\frac{d}{dt}(f({\\bf{r}}(t)) \\ dt \\\\\r\n&=[{f}({\\bf{r}}(t))]_{t=a}^{t=b} \\\\\r\n&={f}({\\bf{r}}(b))-{f}({\\bf{r}}(a))\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n

We know that if [latex]\\mathbf{F}[\/latex] is a conservative vector field, there are potential functions [latex]f[\/latex] such that [latex]\\nabla f=\\mathbf{F}[\/latex]. Therefore [latex]\\displaystyle\\int_C\\mathbf{F}\\cdot d\\mathbf{r}=\\displaystyle\\int_C\\nabla f\\cdot d\\mathbf{r}=f(\\mathbf{r}(b))-f(\\mathbf{r}(a))[\/latex].<\/p>\r\n

In other words, just as with the Fundamental Theorem of Calculus, computing the line integral [latex]\\displaystyle\\int_C\\mathbf{F}\\cdot d\\mathbf{r}[\/latex] where [latex]\\mathbf{F}[\/latex] is conservative is a two-step process:<\/p>\r\n\r\n

    \r\n \t
  1. Find a potential function (\"antiderivative\") [latex]f[\/latex] for [latex]\\mathbf{F}[\/latex]<\/li>\r\n \t
  2. Compute the value of [latex]f[\/latex] at the endpoints of [latex]C[\/latex] and calculate their difference [latex]f(\\mathbf{r}(b))-f(\\mathbf{r}(a))[\/latex].<\/li>\r\n<\/ol>\r\n
    There is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A continuous function of one variable must have an antiderivative. However, a vector field, even if it is continuous, does not necessarily have a potential function.<\/section>
    \r\n

    Calculate integral [latex]\\displaystyle\\int_c{\\bf{f}}.d{\\bf{r}}[\/latex], where [latex]{\\bf{F}}(x,y,z)=\\left\\langle2x\\ln{y},\\frac{x^2}y+z^2,2yz\\right\\rangle[\/latex] and [latex]C[\/latex] is a curve with parameterization [latex]{\\bf{r}}(t)=\\langle{t^2},t,t\\rangle[\/latex],\u00a0[latex]1\\leq{t}\\leq1{e}[\/latex]<\/p>\r\n\r\n

      \r\n \t
    1. without using the Fundamental Theorem of Line Integrals and<\/li>\r\n \t
    2. using the Fundamental Theorem of Line Integrals.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"847519662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847519662\"]\r\n
        \r\n \t
      1. \r\n
          \r\n \t
        1. First, let\u2019s calculate the integral without the Fundamental Theorem for Line Integrals and instead use\u00a0[latex]\\displaystyle\\int_{C} {\\bf{F}}\\cdot{ds}=\\displaystyle\\int_{C}{\\bf{F}}\\cdot{\\bf{T}}ds=\\displaystyle\\int_{a}^{b}{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^{\\prime}(t)dt[\/latex]:\r\n<\/span>\r\n
          [latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=\\displaystyle\\int_1^e{\\bf{F}}(r(t)).r^\\prime(t) \\ dt \\\\\r\n&=\\displaystyle\\int_1^e\\left\\langle2t^2\\ln{t},\\frac{t^4}t+t^2,2t^2\\right\\rangle. \\langle2t,1,1\\rangle dt \\\\\r\n&=\\displaystyle\\int_1^e(4t^3\\ln{t}+t^3+3t^2) \\ dt \\\\\r\n&=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt +\\displaystyle\\int_1^e(t^3+3t^2) \\ dt \\\\\r\n&=\\displaystyle\\int_1^e4t^3\\ln{t} \\ dt+\\left[\\frac{t^4}4+t^3\\right]_1^e \\\\\r\n&=2\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54.\r\n\\end{align}[\/latex].<\/div>\r\n\r\n<\/span>Integral [latex]\\displaystyle\\int_1^et^3\\ln{t}[\/latex] requires integration by parts. Let [latex]u=\\ln{t}[\/latex] and [latex]dv=t^3[\/latex]. Then [latex]u=\\ln{t}[\/latex],\u00a0[latex]dv=t^3[\/latex],\r\n<\/span>and\r\n<\/span>\r\n
          \r\n

          [latex]du=\\frac{1}t dt[\/latex],[latex]v=\\frac{t^4}4[\/latex].<\/p>\r\nTherefore,\r\n<\/span>\r\n

          [latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_1^et^3\\ln{t}dt&=\\left[\\frac{t^4}4\\ln{t}\\right]_1^e-\\frac14\\displaystyle\\int_1^et^3 \\ dt \\\\\r\n&=\\frac{e^4}4-\\frac1r\\left(\\frac{e^4}4-\\frac14\\right).\r\n\\end{align}[\/latex].<\/div>\r\n\r\n<\/span>Thus,\r\n<\/span>\r\n
          [latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=4\\displaystyle\\int_1^et^3\\ln{t} \\ dt+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&=4\\left(\\frac{e^4}4-\\frac14\\left(\\frac{e^4}4-\\frac14\\right)\\right)+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&=e^4-\\frac{e^4}4+\\frac14+\\frac{e^4}4+e^3-\\frac54 \\\\\r\n&=e^4+e^3-1\r\n\\end{align}[\/latex].<\/div>\r\n<\/div><\/li>\r\n \t
        2. Given that [latex]f(x,y,z)=x^2\\ln{y}+yz^2[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex], let\u2019s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that\r\n<\/span>\r\n
          [latex]\\hspace{6cm}\\begin{align}\r\n\\displaystyle\\int_C{\\bf{F}}.d{\\bf{r}}&=\\displaystyle\\int_C\\nabla{f}.d{\\bf{r}} \\\\\r\n&=f({\\bf{r}}(e))-f({\\bf{r}}(1)) \\\\\r\n&=f(e^2,e,e)-f(1,1,1) \\\\\r\n&=e^4+e^3-1\r\n\\end{align}[\/latex].<\/div>\r\n\r\n<\/span>This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

          The Fundamental Theorem for Line Integrals allows us to calculate many vector line integrals more easily. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.<\/p>\r\n\r\n

          Two Important Consequences<\/h3>\r\n

          The Fundamental Theorem for Line Integrals has two important consequences. The first is that if [latex]\\mathbf{F}[\/latex] is conservative and [latex]C[\/latex] is a closed curve, then the circulation of [latex]\\mathbf{F}[\/latex] along [latex]C[\/latex] is zero\u2014that is, [latex]\\displaystyle\\int_C\\mathbf{F}\\cdot d\\mathbf{r}=0[\/latex].<\/p>\r\n

          To see why this is true, let [latex]f[\/latex] be a potential function for [latex]\\mathbf{F}[\/latex]. Since [latex]C[\/latex] is a closed curve, the terminal point [latex]\\mathbf{r}(b)[\/latex] of [latex]C[\/latex] is the same as the initial point [latex]\\mathbf{r}(a)[\/latex] of [latex]C[\/latex]\u2014that is, [latex]\\mathbf{r}(a)=\\mathbf{r}(b)[\/latex]. Therefore, by the Fundamental Theorem for Line Integrals,<\/p>\r\n

          [latex]\\begin{aligned} \\displaystyle\\oint_C\\mathbf{F}\\cdot d\\mathbf{r}&=\\displaystyle\\oint_C\\nabla f\\cdot d\\mathbf{r} \\ &=f(\\mathbf{r}(b))-f(\\mathbf{r}(a)) \\ &=f(\\mathbf{r}(b))-f(\\mathbf{r}(b)) \\ &=0 \\end{aligned}[\/latex].<\/p>\r\n\r\n

          Recall that the reason a conservative vector field [latex]\\mathbf{F}[\/latex] is called \"conservative\" is because such vector fields model forces in which energy is conserved. Gravity is an example of such a force. If we think of vector field [latex]\\mathbf{F}[\/latex] in integral [latex]\\displaystyle\\oint_C\\mathbf{F}\\cdot d\\mathbf{r}[\/latex] as a gravitational field, then the equation [latex]\\displaystyle\\oint_C\\mathbf{F}\\cdot d\\mathbf{r}=0[\/latex] makes sense. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.<\/section>\r\n

          The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path<\/strong>\u2014meaning they depend only on the endpoints of the given curve and not on the path between the endpoints.<\/p>\r\n\r\n

          \r\n

          path independence<\/h3>\r\nLet [latex]\\mathbf{F}[\/latex] be a vector field with domain [latex]D[\/latex]. The vector field [latex]\\mathbf{F}[\/latex] is independent of path<\/strong> (or path independent<\/strong>) if [latex]\\displaystyle\\int_{C_1}\\mathbf{F}\\cdot d\\mathbf{r}=\\displaystyle\\int_{C_2}\\mathbf{F}\\cdot d\\mathbf{r}[\/latex] for any paths [latex]C_1[\/latex] and [latex]C_2[\/latex] in [latex]D[\/latex] with the same initial and terminal points.\r\n\r\n<\/section>\r\n

          The second consequence is stated formally in the following theorem.<\/p>\r\n\r\n

          \r\n

          path independence of conservative fields<\/h3>\r\nIf [latex]\\mathbf{F}[\/latex] is a conservative vector field, then [latex]\\mathbf{F}[\/latex] is independent of path.\r\n\r\n<\/section>
          \r\n

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

          \r\n

          Let [latex]D[\/latex] denote the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0and let [latex]C_1[\/latex] and [latex]C_2[\/latex] be two paths in [latex]D[\/latex] with the same initial and terminal points (Figure 5). Call the initial point [latex]P_1[\/latex] and the terminal point [latex]P_2[\/latex]. Since\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative, there is a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. By the Fundamental Theorem for Line Integrals,<\/p>\r\n

          [latex]\\large{\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=f({\\bf{P}}_2)-f({\\bf{P}}_1)=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n

          Therefore, [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}=\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n

          To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep. Hiker 3 starts on the steep route but halfway to the top decides it's too difficult, returns to camp, and takes the non-steep path to the top.<\/p>\r\n

          All three hikers are traveling in a gravitational field. Since gravity is a conservative force, the gravitational field is conservative. By independence of path, the total work done by gravity on each hiker is the same because they all started and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total energy each expended is not the same\u2014but the net energy expended against gravity is the same for all three.<\/p>\r\n\r\n[caption id=\"attachment_1381\" align=\"aligncenter\" width=\"655\"]\"A Figure 5. The vector field is conservative, and therefore independent of path.[\/caption]We have shown that if [latex]\\mathbf{F}[\/latex] is conservative, then [latex]\\mathbf{F}[\/latex] is independent of path. It turns out that if the domain of [latex]\\mathbf{F}[\/latex] is open and connected, then the converse is also true. That is, if [latex]\\mathbf{F}[\/latex] is independent of path and the domain of [latex]\\mathbf{F}[\/latex] is open and connected, then [latex]\\mathbf{F}[\/latex] is conservative.\r\n\r\n

          \r\n

          the path independence test for conservative fields<\/h3>\r\nIf [latex]\\mathbf{F}[\/latex] is a continuous vector field that is independent of path and the domain [latex]D[\/latex] of [latex]\\mathbf{F}[\/latex] is open and connected, then [latex]\\mathbf{F}[\/latex] is conservative.\r\n\r\n<\/section>
          \r\n

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

          \r\n

          We prove the theorem for vector fields in [latex]\\mathbb{R}^2[\/latex]. The proof for vector fields in [latex]\\mathbb{R}^3[\/latex] is similar. To show that [latex]{\\bf{F}}=\\langle{P},Q\\rangle[\/latex] is conservative, we must find a potential function [latex]f[\/latex] for\u00a0[latex]{\\bf{F}}[\/latex]. To that end, let [latex]X[\/latex] be a fixed point in [latex]D[\/latex]. For any point [latex](x, y)[\/latex] in [latex]D[\/latex], let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex]. Define [latex]f(x, y)[\/latex] by [latex]f(x,y)=\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex]. (Note that this definition of [latex]f[\/latex] makes sense only because\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is independent of path. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0was not independent of path, then it might be possible to find another path [latex]C'[\/latex] from [latex]X[\/latex] to [latex](x, y)[\/latex] such that [latex]\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_C{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], and in such a case [latex]f(x, y)[\/latex] would not be a function.) We want to show that [latex]f[\/latex] has the property [latex]\\nabla{f}={\\bf{F}}[\/latex].<\/p>\r\n

          Since domain [latex]D[\/latex] is open, it is possible to find a disk centered at [latex](x, y)[\/latex] such that the disk is contained entirely inside [latex]D[\/latex]. Let [latex](a, y)[\/latex] with [latex]a<x[\/latex] be a point in that disk. Let [latex]C[\/latex] be a path from [latex]X[\/latex] to [latex](x, y)[\/latex] that consists of two pieces: [latex]C_1[\/latex] and [latex]C_2[\/latex]. The first piece, [latex]C_1[\/latex], is any path from [latex]X[\/latex] to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex]; [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x ,y)[\/latex] (Figure 6). Then<\/p>\r\n

          [latex]\\large{f(x,y)=\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}+\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n

          The first integral does not depend on [latex]x[\/latex], so<\/p>\r\n

          [latex]\\large{f_x=\\frac{\\partial}{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}}[\/latex].<\/p>\r\n

          If we parameterize [latex]C_2[\/latex] by [latex]{\\bf{r}}(t)=\\langle{t},y\\rangle[\/latex], [latex]a\\leq{t}\\leq{x}[\/latex], then<\/p>\r\n

          [latex]\\begin{aligned}\r\nf_x&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}} \\\\\r\n&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot{\\bf{r}}^\\prime(t) \\ dt \\\\\r\n&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\frac{d}{dt}(\\langle{t},y\\rangle \\ dt \\\\\r\n&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^x{\\bf{F}}({\\bf{r}}(t))\\cdot\\langle1,0\\rangle \\ dt \\\\\r\n&=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt\r\n\\end{aligned}[\/latex].<\/p>\r\n

          By the Fundamental Theorem of Calculus (part 1),<\/p>\r\n

          [latex]\\large{f_x=\\frac\\partial{\\partial{x}}\\displaystyle\\int_a^xP(t,y) \\ dt=P(x,y)}[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1382\" align=\"aligncenter\" width=\"428\"]\"A Figure 6. Here, [latex]C_1[\/latex] is any path from [latex]X[\/latex]; to [latex](a, y)[\/latex] that stays inside [latex]D[\/latex], and [latex]C_2[\/latex] is the horizontal line segment from [latex](a, y)[\/latex] to [latex](x, y)[\/latex].[\/caption]\r\n

          A similar argument using a vertical line segment rather than a horizontal line segment shows that [latex]f_y= Q(x, y)[\/latex].<\/p>\r\n

          Therefore [latex]\\nabla{f}={\\bf{F}}[\/latex] and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>We can summarize these results simply: a vector field [latex]\\mathbf{F}[\/latex] on an open and connected domain is conservative if and only if it is independent of path. This is important because conservative vector fields are extremely useful in applications, and these theorems give us a different way of viewing what it means to be conservative.\r\n\r\n

          Use path independence to show that vector field [latex]{\\bf{F}}(x,y)=\\langle{x^2}y,y+5\\rangle[\/latex] is not conservative.\r\n\r\n[reveal-answer q=\"452908\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"452908\"]\r\n

          We can indicate that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative by showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not path independent. We do so by giving two different paths, [latex]C_1[\/latex] and [latex]C_2[\/latex], that both start at [latex](0, 0)[\/latex] and end at [latex](1, 1)[\/latex], and yet [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex].<\/p>\r\n

          Let [latex]C_1[\/latex] be the curve with parameterization [latex]r_1(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] and let [latex]C_2[\/latex] be the curve with parameterization [latex]r_2(t)=\\langle{t},t\\rangle[\/latex], [latex]0\\leq{t}\\leq1[\/latex] (Figure 7). Then<\/p>\r\n

          [latex]\\begin{aligned}\r\n\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1{\\bf{F}}(r_1(t))\\cdot{r}_1^\\prime(t) \\ dt \\\\\r\n&=\\displaystyle\\int_0^1\\langle{t^3},t+5\\rangle\\cdot\\langle1,1\\rangle \\ dt =\\displaystyle\\int_0^1(t^3+t+5) \\ dt \\\\\r\n&=\\left[\\frac{t^4}4+\\frac{t^2}2+5t\\right]_0^1=\\frac{23}4\r\n\\end{aligned}[\/latex]<\/p>\r\n

          and<\/p>\r\n

          [latex]\\begin{aligned}\r\n\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}&=\\displaystyle\\int_0^1{\\bf{F}}(r_2(t))\\cdot{r}_2^\\prime(t) \\ dt \\\\\r\n&=\\displaystyle\\int_0^1\\langle{t^4},t^2+5\\rangle\\cdot\\langle1,2t\\rangle \\ dt =\\displaystyle\\int_0^1(t^4+2t^2+10t) \\ dt \\\\\r\n&=\\left[\\frac{t^5}5+\\frac{t^4}2+5t^2\\right]_0^1=\\frac{57}{10}\r\n\\end{aligned}[\/latex]<\/p>\r\n

          Since [latex]\\displaystyle\\int_{C_1}{\\bf{F}}\\cdot{d}{\\bf{r}}\\ne\\displaystyle\\int_{C_2}{\\bf{F}}\\cdot{d}{\\bf{r}}[\/latex], the value of a line integral of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0depends on the path between two given points. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not independent of path, and\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not conservative.<\/p>\r\n\r\n[caption id=\"attachment_1383\" align=\"aligncenter\" width=\"340\"]\"A Figure 7. Curves [latex]C_1[\/latex] and [latex]C_2[\/latex] are both oriented from left to right.[\/caption][\/hidden-answer]<\/section>

          <\/section>","rendered":"

          Fundamental Theorem for Line Integrals<\/h2>\n

          Now that we understand some basic curves and regions, we can generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function [latex]f[\/latex] has an antiderivative [latex]F[\/latex], then the integral of [latex]f[\/latex] from [latex]a[\/latex] to [latex]b[\/latex] depends only on the values of [latex]F[\/latex] at [latex]a[\/latex] and at [latex]b[\/latex]:<\/p>\n

          [latex]\\displaystyle\\int_a^bf(x) \\ dx =F(b)-F(a)[\/latex].<\/p>\n

          If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. Let’s see how this works with an example.<\/p>\n

          Let [latex]{\\bf{F}}(x,y)=\\langle2x,4y\\rangle[\/latex]. Calculate [latex]\\displaystyle\\int_C{\\bf{F}}. \\ d{\\bf{r}}[\/latex], where [latex]C[\/latex] is the line segment from [latex](0, 0)[\/latex] to [latex](2, 2)[\/latex] (Figure 4).<\/p>\n