Stokes’ Theorem

Cover Page

Artifacts

Artifact #1	Surface Integral by Stokes’ Theorem (HW23 #5)
Artifact #2	Line Integral by Stokes’ Theorem (HW23 #6)

Sub-outcome coverage

Sub Outcome	Artifact #1	Artifact #2
Do a line integral instead: calculate a vector surface integral by using Stokes’ Theorem to evaluate an equivalent line integral.	Yes
Do a surface integral instead: calculate a line integral by using Stokes’ Theorem to evaluate an equivalent surface integral.		Yes

Artifact 1

Problem

Let \(\vec{F} = \left\langle xyz, \; xy, \; x^{2}yz \right\rangle\). Use Stokes’ Theorem to evaluate \[ \iint_{S} \operatorname{curl} \vec{F} \cdot d\vec{S}, \] where \(S\) consists of the top and the four sides (but not the bottom) of the cube with one corner at \((-4, -4, -4)\) and the diagonal corner at \((-3, -3, -3)\), oriented with outward normals.

Solution:

A surface integral of a curl over five faces of a cube is a lot of separate flux computations, so we use Stokes’ Theorem to replace it with a single line integral.

Stokes’ Theorem

For an oriented surface \(S\) with boundary curve \(C\) oriented consistently, \[ \iint_{S} \operatorname{curl} \vec{F} \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r}. \] The flux of a curl depends only on the boundary curve, so any surface with the same boundary gives the same value.

The five faces of \(S\) form a box that is open on the bottom. The one and only edge of that open box is the square where the missing bottom face would attach, the square at \(z = -4\) with \(-4 \le x \le -3\) and \(-4 \le y \le -3\). That square is the boundary curve \(C\).

For the orientation, the side faces carry outward normals, and the right-hand rule then sends \(C\) counterclockwise as seen from above (looking down the \(z\)-axis). We take that direction.

Interactive

Drag to rotate. The blue faces are the surface \(S\) (top and four sides, no bottom). The orange loop at the bottom is the boundary curve \(C\), the only edge of the open box, traversed counterclockwise as seen from above.

Plotly = require("https://cdn.plot.ly/plotly-2.27.0.min.js")

boxView = {
  const a = -4, b = -3;
  const lin = (p, q, n) => Array.from({length: n}, (_, i) => p + (q - p) * i / (n - 1));
  const us = lin(a, b, 12);

  const grid = (fx, fy, fz, uu, vv) => {
    const X = [], Y = [], Z = [];
    for (const s of uu) {
      const xr = [], yr = [], zr = [];
      for (const t of vv) { xr.push(fx(s, t)); yr.push(fy(s, t)); zr.push(fz(s, t)); }
      X.push(xr); Y.push(yr); Z.push(zr);
    }
    return {X, Y, Z};
  };
  const surf = (g) => ({
    type: "surface", x: g.X, y: g.Y, z: g.Z, showscale: false, opacity: 0.45,
    colorscale: [[0, "#2563eb"], [1, "#2563eb"]], hoverinfo: "skip",
    lighting: {ambient: 0.7, diffuse: 0.7}
  });

  const top   = grid((s, t) => s, (s, t) => t, () => b, us, us);
  const xlow  = grid((s, t) => a, (s, t) => s, (s, t) => t, us, us);
  const xhigh = grid((s, t) => b, (s, t) => s, (s, t) => t, us, us);
  const ylow  = grid((s, t) => s, (s, t) => a, (s, t) => t, us, us);
  const yhigh = grid((s, t) => s, (s, t) => b, (s, t) => t, us, us);

  const loop = {
    type: "scatter3d", mode: "lines+markers",
    x: [a, b, b, a, a], y: [a, a, b, b, a], z: [a, a, a, a, a],
    line: {color: "#d97706", width: 6}, marker: {size: 3, color: "#d97706"}, hoverinfo: "skip"
  };

  const div = html`<div></div>`;
  Plotly.newPlot(div, [surf(top), surf(xlow), surf(xhigh), surf(ylow), surf(yhigh), loop], {
    paper_bgcolor: "rgba(0,0,0,0)", plot_bgcolor: "rgba(0,0,0,0)",
    font: {color: "#334155"}, margin: {l: 0, r: 0, t: 0, b: 0}, height: 440,
    scene: {
      xaxis: {title: "x", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      yaxis: {title: "y", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      zaxis: {title: "z", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      aspectmode: "cube",
      camera: {eye: {x: 1.7, y: 1.7, z: 1.2}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

On the square \(C\) the height is fixed at \(z = -4\), so \(dz = 0\) and only the first two components of \(\vec{F}\) contribute. There \(\vec{F} = \langle xyz, \, xy, \, x^{2}yz \rangle\) becomes \[ \vec{F}\big|_{z = -4} = \left\langle -4xy, \; xy, \; -4x^{2}y \right\rangle, \] so \[ \vec{F} \cdot d\vec{r} = -4xy \, dx + xy \, dy. \]

We walk \(C\) counterclockwise from above through the corners \[ (-4, -4) \to (-3, -4) \to (-3, -3) \to (-4, -3) \to (-4, -4). \]

Edge 1: \((-4, -4) \to (-3, -4)\). Here \(y = -4\) and \(dy = 0\), with \(x\) from \(-4\) to \(-3\): \[ \int_{-4}^{-3} -4x(-4) \, dx = \int_{-4}^{-3} 16x \, dx = 16 \left[ \frac{x^{2}}{2} \right]_{-4}^{-3} = 16 \left( \frac{9}{2} - \frac{16}{2} \right) = -56. \]

Edge 2: \((-3, -4) \to (-3, -3)\). Here \(x = -3\) and \(dx = 0\), with \(y\) from \(-4\) to \(-3\): \[ \int_{-4}^{-3} (-3)y \, dy = -3 \left[ \frac{y^{2}}{2} \right]_{-4}^{-3} = -3 \left( \frac{9}{2} - \frac{16}{2} \right) = \frac{21}{2}. \]

Edge 3: \((-3, -3) \to (-4, -3)\). Here \(y = -3\) and \(dy = 0\), with \(x\) from \(-3\) to \(-4\): \[ \int_{-3}^{-4} -4x(-3) \, dx = \int_{-3}^{-4} 12x \, dx = 12 \left[ \frac{x^{2}}{2} \right]_{-3}^{-4} = 12 \left( \frac{16}{2} - \frac{9}{2} \right) = 42. \]

Edge 4: \((-4, -3) \to (-4, -4)\). Here \(x = -4\) and \(dx = 0\), with \(y\) from \(-3\) to \(-4\): \[ \int_{-3}^{-4} (-4)y \, dy = -4 \left[ \frac{y^{2}}{2} \right]_{-3}^{-4} = -4 \left( \frac{16}{2} - \frac{9}{2} \right) = -14. \]

Adding the four edges: \[ \iint_{S} \operatorname{curl} \vec{F} \cdot d\vec{S} = -56 + \frac{21}{2} + 42 - 14 = -28 + \frac{21}{2} = -\frac{35}{2}. \]

Artifact 2

Problem

Let \(\vec{F} = \left\langle 5x + y^{2}, \; 7y + z^{2}, \; 2z + x^{2} \right\rangle\). Use Stokes’ Theorem to evaluate \[ \oint_{C} \vec{F} \cdot d\vec{r}, \] where \(C\) is the triangle with vertices \((9, 0, 0)\), \((0, 9, 0)\), and \((0, 0, 9)\), oriented counterclockwise as viewed from above.

Solution:

The line integral would mean parametrizing three separate edges, so instead we run Stokes’ Theorem the other way and replace it with one surface integral over the flat triangle the curve bounds.

First we compute the curl: \[ \operatorname{curl} \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\[2pt] \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\[6pt] 5x + y^{2} & 7y + z^{2} & 2z + x^{2} \end{vmatrix} = \left\langle -2z, \; -2x, \; -2y \right\rangle. \]

The three vertices satisfy \(x + y + z = 9\), so \(C\) bounds the flat triangle \(S\) lying in that plane. Counterclockwise as viewed from above means we orient \(S\) with an upward normal, so we write the plane as \(z = 9 - x - y\) and project onto the triangle \(T\) in the \(xy\)-plane with vertices \((0, 0)\), \((9, 0)\), and \((0, 9)\).

Flux through a graph \(z = g(x, y)\), upward normal

\[ \iint_{S} \vec{G} \cdot d\vec{S} = \iint_{T} \left( -G_{1} \frac{\partial g}{\partial x} - G_{2} \frac{\partial g}{\partial y} + G_{3} \right) dA. \]

Interactive

Drag to rotate. The orange triangle is the surface \(S\) in the plane \(x + y + z = 9\), and the blue loop is the boundary curve \(C\) through the three axis points. Stokes’ Theorem replaces the loop integral with the flux of \(\operatorname{curl}\vec{F}\) through the triangle.

Plotly = require("https://cdn.plot.ly/plotly-2.27.0.min.js")

triView = {
  const tri = {
    type: "mesh3d",
    x: [9, 0, 0], y: [0, 9, 0], z: [0, 0, 9],
    i: [0], j: [1], k: [2],
    color: "#d97706", opacity: 0.55, flatshading: true, hoverinfo: "skip", showscale: false
  };
  const loop = {
    type: "scatter3d", mode: "lines+markers",
    x: [9, 0, 0, 9], y: [0, 9, 0, 0], z: [0, 0, 9, 0],
    line: {color: "#2563eb", width: 6}, marker: {size: 4, color: "#2563eb"}, hoverinfo: "skip"
  };
  const div = html`<div></div>`;
  Plotly.newPlot(div, [tri, loop], {
    paper_bgcolor: "rgba(0,0,0,0)", plot_bgcolor: "rgba(0,0,0,0)",
    font: {color: "#334155"}, margin: {l: 0, r: 0, t: 0, b: 0}, height: 440,
    scene: {
      xaxis: {title: "x", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      yaxis: {title: "y", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      zaxis: {title: "z", color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      aspectmode: "cube",
      camera: {eye: {x: 1.6, y: 1.6, z: 1.3}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

Here \(\vec{G} = \operatorname{curl} \vec{F} = \langle -2z, -2x, -2y \rangle\) and \(g = 9 - x - y\), so \(\dfrac{\partial g}{\partial x} = -1\) and \(\dfrac{\partial g}{\partial y} = -1\). The integrand becomes \[ \begin{aligned} -G_{1} \frac{\partial g}{\partial x} - G_{2} \frac{\partial g}{\partial y} + G_{3} &= -(-2z)(-1) - (-2x)(-1) + (-2y) \\[4pt] &= -2z - 2x - 2y \\[4pt] &= -2(x + y + z). \end{aligned} \]

On the surface \(x + y + z = 9\) at every point, so the integrand is the constant \(-2(9) = -18\). Therefore \[ \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{T} -18 \, dA = -18 \cdot \operatorname{Area}(T). \]

The triangle \(T\) has legs of length \(9\) along the \(x\)- and \(y\)-axes, so \[ \operatorname{Area}(T) = \frac{1}{2}(9)(9) = \frac{81}{2}, \] and \[ \oint_{C} \vec{F} \cdot d\vec{r} = -18 \cdot \frac{81}{2} = -729. \]

Essay

My approach to a Stokes’ Theorem problem is to look at which side of the theorem is set up for me and which side is easier to finish, because the equality lets me swap a surface integral of a curl for a boundary line integral or the reverse. In the first artifact the surface was five faces of a cube, which would be five flux integrals, but the open box has a single boundary edge, so the line integral around that one square was far shorter, and the fact that the flux of a curl depends only on the boundary is what made the swap legal. In the second artifact the trade ran the other way: a triangle in space gives three edges to parametrize, but its boundary bounds one flat surface, and the curl dotted with the upward normal collapsed to the constant \(-18\), leaving only an area to multiply. My strength is spotting that choice and getting the orientation right, since both problems hinge on matching the boundary direction to the normal by the right-hand rule. The area I want to keep improving is the setup of the flux integral over a graph, especially the signs in the \(-G_{1} g_{x} - G_{2} g_{y} + G_{3}\) formula, because that is the step where the whole computation can quietly go wrong.