Divergence Theorem

Cover Page

Artifacts

Artifact #1	Flux Through a Hemispherical Shell (HW24 #5)
Artifact #2	Flux Through Part of a Cylinder’s Surface

Sub-outcome coverage

Sub Outcome	Artifact #1	Artifact #2
Do a volume integral instead: calculate a vector surface integral by using the Divergence Theorem to evaluate an equivalent triple integral.	Yes
Do a surface integral instead: calculate a volume integral by using the Divergence Theorem to evaluate an equivalent vector surface integral.
Multiple parts of a surface: calculate a vector surface integral by identifying the surface as part of the boundary of a solid, then using the Divergence Theorem.		Yes

Artifact 1

Problem

Calculate \[ \iint_{S} \vec{F} \cdot d\vec{S}, \qquad \vec{F} = \left\langle 4x^{3}z, \; 4y^{3}z, \; 3z^{4} \right\rangle, \] where \(S\) is the surface of the solid bounded by the hemispheres \(z = \sqrt{16 - x^{2} - y^{2}}\) and \(z = \sqrt{1 - x^{2} - y^{2}}\) and the plane \(z = 0\).

Solution:

The surface \(S\) is the full closed boundary of the solid (the outer hemisphere of radius \(4\), the inner hemisphere of radius \(1\), and the flat ring between them on the plane \(z = 0\)). A direct surface integral would mean three separate pieces, so we use the Divergence Theorem to turn it into a single triple integral over the solid.

Divergence Theorem

For a solid \(E\) with closed boundary surface \(S\) oriented outward, \[ \iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{E} \operatorname{div} \vec{F} \, dV. \]

The divergence is \[ \operatorname{div} \vec{F} = \frac{\partial}{\partial x}\!\left( 4x^{3}z \right) + \frac{\partial}{\partial y}\!\left( 4y^{3}z \right) + \frac{\partial}{\partial z}\!\left( 3z^{4} \right) = 12x^{2}z + 12y^{2}z + 12z^{3} = 12z \left( x^{2} + y^{2} + z^{2} \right). \]

The solid is the upper half of the region between two spheres, so spherical coordinates fit. With \[ x^{2} + y^{2} + z^{2} = \rho^{2}, \qquad z = \rho \cos\varphi, \qquad dV = \rho^{2} \sin\varphi \, d\rho \, d\varphi \, d\theta, \] the divergence becomes \[ \operatorname{div} \vec{F} = 12 \, (\rho \cos\varphi)\, \rho^{2} = 12 \rho^{3} \cos\varphi. \]

Interactive

Drag to rotate. The blue cap is the outer hemisphere of radius \(4\), the orange cap is the inner hemisphere of radius \(1\), and the gray ring is the floor on \(z = 0\). Together they enclose the solid we integrate over.

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

shellView = {
  const lin = (p, q, n) => Array.from({length: n}, (_, i) => p + (q - p) * i / (n - 1));
  const thetas = lin(0, 2 * Math.PI, 48);
  const phis = lin(0, Math.PI / 2, 24);
  const rs = lin(1, 4, 16);

  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, color, opacity) => ({
    type: "surface", x: g.X, y: g.Y, z: g.Z, showscale: false, opacity,
    colorscale: [[0, color], [1, color]], hoverinfo: "skip",
    lighting: {ambient: 0.65, diffuse: 0.8, roughness: 0.8}
  });

  const hemi = (R) => grid(
    (p, t) => R * Math.sin(p) * Math.cos(t),
    (p, t) => R * Math.sin(p) * Math.sin(t),
    (p) => R * Math.cos(p), phis, thetas);
  const ring = grid((r, t) => r * Math.cos(t), (r, t) => r * Math.sin(t), () => 0, rs, thetas);

  const div = html`<div></div>`;
  Plotly.newPlot(div, [
    surf(ring, "#94a3b8", 0.75),
    surf(hemi(1), "#d97706", 0.95),
    surf(hemi(4), "#2563eb", 0.45)
  ], {
    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: "data",
      camera: {eye: {x: 1.6, y: 1.6, z: 1.0}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

The bounds are \(1 \le \rho \le 4\) (between the two spheres), \(0 \le \varphi \le \tfrac{\pi}{2}\) (the upper half), and \(0 \le \theta \le 2\pi\). The integrand separates, so \[ \begin{aligned} \iint_{S} \vec{F} \cdot d\vec{S} &= \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{1}^{4} 12 \rho^{3} \cos\varphi \cdot \rho^{2} \sin\varphi \, d\rho \, d\varphi \, d\theta \\[4pt] &= 12 \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{\pi/2} \sin\varphi \cos\varphi \, d\varphi \right) \left( \int_{1}^{4} \rho^{5} \, d\rho \right). \end{aligned} \]

Each factor is standard: \[ \int_{0}^{2\pi} d\theta = 2\pi, \qquad \int_{0}^{\pi/2} \sin\varphi \cos\varphi \, d\varphi = \left[ \frac{\sin^{2}\varphi}{2} \right]_{0}^{\pi/2} = \frac{1}{2}, \qquad \int_{1}^{4} \rho^{5} \, d\rho = \left[ \frac{\rho^{6}}{6} \right]_{1}^{4} = \frac{4096 - 1}{6} = \frac{4095}{6}. \]

Multiplying the three factors: \[ \iint_{S} \vec{F} \cdot d\vec{S} = 12 \cdot 2\pi \cdot \frac{1}{2} \cdot \frac{4095}{6} = 2 \cdot 4095 \, \pi = 8190\pi. \]

Artifact 2

Problem

Let \(\vec{F} = \left\langle x, \; y, \; z^{2} \right\rangle\). Calculate the outward flux \[ \iint_{S} \vec{F} \cdot d\vec{S}, \] where \(S\) is the lateral surface and the bottom disk (but not the top) of the solid cylinder \(x^{2} + y^{2} \le 1\), \(0 \le z \le 2\).

Solution:

Here \(S\) is only part of the cylinder’s boundary, the side wall and the bottom, with the top disk left off, so \(S\) is not closed and the Divergence Theorem does not apply to it directly. The fix is to add the missing top disk \(D\) (at \(z = 2\), with outward normal pointing up) to close the surface. The closed surface \(S \cup D\) bounds the full solid cylinder \(E\), so \[ \iint_{S} \vec{F} \cdot d\vec{S} + \iint_{D} \vec{F} \cdot d\vec{S} = \iiint_{E} \operatorname{div} \vec{F} \, dV, \] which rearranges to \[ \iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{E} \operatorname{div} \vec{F} \, dV - \iint_{D} \vec{F} \cdot d\vec{S}. \]

Interactive

Drag to rotate. The blue side wall and gray bottom disk make up the surface \(S\) we want. The dashed orange top disk is the piece we add to close the solid so the Divergence Theorem applies, and its flux is subtracted at the end.

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

cylView = {
  const lin = (p, q, n) => Array.from({length: n}, (_, i) => p + (q - p) * i / (n - 1));
  const thetas = lin(0, 2 * Math.PI, 48);
  const zs = lin(0, 2, 16);
  const rs = lin(0, 1, 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, color, opacity) => ({
    type: "surface", x: g.X, y: g.Y, z: g.Z, showscale: false, opacity,
    colorscale: [[0, color], [1, color]], hoverinfo: "skip",
    lighting: {ambient: 0.7, diffuse: 0.7}
  });

  const wall = grid((z, t) => Math.cos(t), (z, t) => Math.sin(t), (z) => z, zs, thetas);
  const bottom = grid((r, t) => r * Math.cos(t), (r, t) => r * Math.sin(t), () => 0, rs, thetas);
  const top = grid((r, t) => r * Math.cos(t), (r, t) => r * Math.sin(t), () => 2, rs, thetas);

  const topRing = {
    type: "scatter3d", mode: "lines",
    x: thetas.map(t => Math.cos(t)), y: thetas.map(t => Math.sin(t)), z: thetas.map(() => 2),
    line: {color: "#d97706", width: 5, dash: "dash"}, hoverinfo: "skip"
  };

  const div = html`<div></div>`;
  Plotly.newPlot(div, [
    surf(wall, "#2563eb", 0.55),
    surf(bottom, "#94a3b8", 0.85),
    surf(top, "#d97706", 0.18),
    topRing
  ], {
    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: "data",
      camera: {eye: {x: 1.7, y: 1.7, z: 1.1}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

The triple integral. The divergence is \[ \operatorname{div} \vec{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}\!\left( z^{2} \right) = 1 + 1 + 2z = 2 + 2z. \]

In cylindrical coordinates the solid is \(0 \le \theta \le 2\pi\), \(0 \le r \le 1\), \(0 \le z \le 2\), with \(dV = r \, dz \, dr \, d\theta\): \[ \iiint_{E} (2 + 2z) \, dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{2} (2 + 2z) \, r \, dz \, dr \, d\theta. \]

Integrate in \(z\): \[ \int_{0}^{2} (2 + 2z) \, dz = \left[ 2z + z^{2} \right]_{0}^{2} = 4 + 4 = 8. \]

Integrate in \(r\), then \(\theta\): \[ \int_{0}^{2\pi} \int_{0}^{1} 8 r \, dr \, d\theta = \int_{0}^{2\pi} 8 \left[ \frac{r^{2}}{2} \right]_{0}^{1} d\theta = \int_{0}^{2\pi} 4 \, d\theta = 8\pi. \]

The added top disk. On \(D\) the height is \(z = 2\) and the outward normal is \(\vec{k}\), so \(\vec{F} \cdot \vec{k} = z^{2} = 4\) over a disk of radius \(1\): \[ \iint_{D} \vec{F} \cdot d\vec{S} = \iint_{x^{2} + y^{2} \le 1} 4 \, dA = 4 \cdot \pi (1)^{2} = 4\pi. \]

Combine. Subtracting the added piece, \[ \iint_{S} \vec{F} \cdot d\vec{S} = 8\pi - 4\pi = 4\pi. \]

Essay

My approach to a Divergence Theorem problem is to compare the work on each side before computing, since the theorem lets me replace a flux through a surface with a triple integral of the divergence over the solid it bounds. In the first artifact the surface came in three pieces, an outer hemisphere, an inner hemisphere, and a flat ring, but the divergence simplified to \(12z(x^{2} + y^{2} + z^{2})\), which is clean in spherical coordinates, so one separable triple integral replaced three surface integrals. The second artifact shows the move I rely on when a surface is not closed: the side and bottom of the cylinder are only part of its boundary, so I added the missing top disk to close the solid, applied the theorem, and then subtracted the flux through the disk I added. My strength is recognizing that closing-the-surface step and choosing the coordinate system the divergence asks for, spherical for the shell and cylindrical for the cylinder. The area I want to keep improving is staying careful with the outward normal on every separate piece, since the whole second artifact depends on subtracting the top disk with the correct sign, and that bookkeeping is where these problems are easiest to get wrong.