Multiple Integrals

Cover Page

Artifacts

Artifact #1	Change of Variables in Multiple Integrals (HW15 #3)
Artifact #2	Triple Integrals in Cylindrical Coordinates (HW14 #1)

Sub-outcome coverage

Sub Outcome	Artifact #1	Artifact #2
Evaluating by hand: compute double or triple integrals over general regions whose boundaries are described by curves or surfaces.	Yes	Yes
Coordinate systems: draw, set up, and evaluate triple integrals in cylindrical and/or spherical coordinates.		Yes
Change of Variables: construct transformations, compute their Jacobians, and apply the Change of Variables theorem.	Yes

Artifact 1

Problem

Evaluate \[ \iint_{R} \frac{-5x - 6y}{4x + 2y}\, dA, \] where \(R\) is the parallelogram enclosed by the lines \[ -5x - 6y = 0, \qquad -5x - 6y = 9, \qquad 4x + 2y = 1, \qquad 4x + 2y = 8. \]

Solution:

Parallelograms are a pain to integrate over directly, so we switch to a change of variables.

One good choice would be: \[ u = -5x - 6y, \qquad v = 4x + 2y. \] This turns the four boundary lines into the much friendlier bounds: \[ u = 0, \qquad u = 9, \qquad v = 1, \qquad v = 8. \]

Interactive

Drag the sliders to see how a point in the \(uv\)-rectangle maps back to the parallelogram in the \(xy\)-plane.

viewof uval = Inputs.range([0, 9], {step: 0.1, value: 4.5, label: "u = -5x - 6y"})

viewof vval = Inputs.range([1, 8], {step: 0.1, value: 4.5, label: "v = 4x + 2y"})

xpt = (uval + 3 * vval) / 7
ypt = (-4 * uval - 5 * vval) / 14
xyCorners = [[0, 1], [9, 1], [9, 8], [0, 8], [0, 1]].map(([u, v]) => [(u + 3 * v) / 7, (-4 * u - 5 * v) / 14])

html`<div style="display:flex; gap:1rem; flex-wrap:wrap; align-items:flex-start">
  <div style="flex:1 1 260px; min-width:0">
  ${Plot.plot({
    width: 340,
    aspectRatio: 1,
    marginLeft: 45,
    style: {background: "transparent", color: "#334155"},
    x: {label: "x", grid: true},
    y: {label: "y", grid: true},
    marks: [
      Plot.line(xyCorners, {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2}),
      Plot.dot([[xpt, ypt]], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 6})
    ]
  })}
  </div>
  <div style="flex:1 1 260px; min-width:0">
  ${Plot.plot({
    width: 340,
    aspectRatio: 1,
    marginLeft: 45,
    style: {background: "transparent", color: "#334155"},
    x: {label: "u", grid: true, domain: [-1.5, 10.5]},
    y: {label: "v", grid: true, domain: [-0.5, 9]},
    marks: [
      Plot.line([[0, 1], [9, 1], [9, 8], [0, 8], [0, 1]], {x: d => d[0], y: d => d[1], stroke: "#d97706", strokeWidth: 2}),
      Plot.dot([[uval, vval]], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 6})
    ]
  })}
  </div>
</div>`

Recalling the formula for change of variables:

Change of variables

\[ \iint_{R} f(x, y)\, dA = \iint_{R'} f\big(x(u, v), y(u, v)\big) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| du\, dv. \]

To use it we need the Jacobian: \[ \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} -5 & -6 \\ \phantom{-}4 & \phantom{-}2 \end{vmatrix} = -10 + 24 = 14, \] \[ \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{14}, \] \[ dA = \frac{1}{14}\, du\, dv. \]

Next we rewrite the integrand: \[ \frac{-5x - 6y}{4x + 2y} = \frac{u}{v}. \]

Now that we have rewritten the integrand and found the Jacobian, we can finally compute: \[ \begin{aligned} \iint_{R} \frac{-5x - 6y}{4x + 2y}\, dA &= \int_{1}^{8} \int_{0}^{9} \frac{u}{v} \cdot \frac{1}{14}\, du\, dv \\[4pt] &= \int_{1}^{8} \left[ \frac{u^{2}}{28 v} \right]_{0}^{9} dv \\[4pt] &= \int_{1}^{8} \frac{81}{28 v}\, dv \\[4pt] &= \left[ \frac{81}{28} \ln v \right]_{1}^{8} \\[4pt] &= \frac{81}{28} \ln 8 - \frac{81}{28} \ln 1 \\[4pt] &= \frac{81}{28} \ln 8. \end{aligned} \]

Artifact 2

Problem

Evaluate \[ \iiint_{E} e^{z}\, dV, \] where \(E\) is the solid enclosed by the paraboloid \(z = 2 + x^{2} + y^{2}\), the cylinder \(x^{2} + y^{2} = 5\), and the \(xy\)-plane.

Solution:

Since both surfaces depend only on \(x^{2} + y^{2}\), cylindrical coordinates are the natural choice, with \(x^{2} + y^{2} = r^{2}\).

In cylindrical coordinates the pieces become:

Cylindrical coordinates

\[ \begin{aligned} &\text{paraboloid: } z = 2 + r^{2}, \qquad \text{cylinder: } r = \sqrt{5}, \qquad \text{floor: } z = 0, \\[4pt] &dV = r\, dz\, dr\, d\theta. \end{aligned} \]

So the bounds are: \[ 0 \le \theta \le 2\pi, \qquad 0 \le r \le \sqrt{5}, \qquad 0 \le z \le 2 + r^{2}. \]

Interactive

Drag to rotate the solid. The blue bowl is the paraboloid top, the orange wall is the cylinder, and the gray disk is the floor.

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

solidView = {
  const R = Math.sqrt(5);
  const lin = (a, b, n) => Array.from({length: n}, (_, i) => a + (b - a) * i / (n - 1));
  const thetas = lin(0, 2 * Math.PI, 48);
  const wallThetas = lin(0.55 * Math.PI, 2 * Math.PI, 42);
  const rs = lin(0, R, 24);
  const zs = lin(0, 7, 24);

  const grid = (fx, fy, fz, us, vs) => {
    const X = [], Y = [], Z = [];
    for (const a of us) {
      const xr = [], yr = [], zr = [];
      for (const b of vs) { xr.push(fx(a, b)); yr.push(fy(a, b)); zr.push(fz(a, b)); }
      X.push(xr); Y.push(yr); Z.push(zr);
    }
    return {X, Y, Z};
  };

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

  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 div = html`<div></div>`;
  Plotly.newPlot(div, [
    surf(wall, "#d97706", 0.38),
    surf(floor, "#94a3b8", 0.72),
    surf(top, "#2563eb", 0.86)
  ], {
    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)"},
      camera: {eye: {x: 1.65, y: 1.65, z: 1.15}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

Now we can compute. Integrating in \(z\) first, with \(r\) coming along for the ride: \[ \int_{0}^{2 + r^{2}} e^{z}\, r\, dz = r \left( e^{2 + r^{2}} - 1 \right). \]

Then in \(r\), we split into two integrals: \[ \int_{0}^{\sqrt{5}} r \left( e^{2 + r^{2}} - 1 \right) dr = \int_{0}^{\sqrt{5}} r\, e^{2 + r^{2}}\, dr - \int_{0}^{\sqrt{5}} r\, dr. \]

For the first integral, we substitute: \[ u = 2 + r^{2}, \qquad du = 2r\, dr, \qquad r\, dr = \frac{1}{2}\, du, \] and the limits change from \(r = 0\) to \(u = 2\), and from \(r = \sqrt{5}\) to \(u = 7\): \[ \begin{aligned} \int_{0}^{\sqrt{5}} r\, e^{2 + r^{2}}\, dr &= \int_{2}^{7} \frac{1}{2} e^{u}\, du \\[4pt] &= \frac{1}{2} \left[ e^{u} \right]_{2}^{7} \\[4pt] &= \frac{1}{2} \left( e^{7} - e^{2} \right). \end{aligned} \]

The second integral is: \[ \int_{0}^{\sqrt{5}} r\, dr = \left[ \frac{r^{2}}{2} \right]_{0}^{\sqrt{5}} = \frac{5}{2}. \]

Putting the two pieces together: \[ \int_{0}^{\sqrt{5}} r \left( e^{2 + r^{2}} - 1 \right) dr = \frac{1}{2} \left( e^{7} - e^{2} \right) - \frac{5}{2}. \]

Finally, integrating in \(\theta\): \[ \begin{aligned} \iiint_{E} e^{z}\, dV &= \int_{0}^{2\pi} \left[ \frac{1}{2} \left( e^{7} - e^{2} \right) - \frac{5}{2} \right] d\theta \\[4pt] &= 2\pi \left[ \frac{1}{2} \left( e^{7} - e^{2} \right) - \frac{5}{2} \right] \\[4pt] &= \pi \left( e^{7} - e^{2} - 5 \right). \end{aligned} \]

Essay

My approach to a multiple integral starts with the geometry: I decide what the region or solid looks like, because the boundaries set both the coordinate system and the limits. The parallel lines in the first problem pointed to a change of variables, and the \(x^{2} + y^{2}\) in the second pointed to cylindrical coordinates. My strength is this setup work, choosing the substitution or coordinate system, computing the Jacobian, and reading the limits off the geometry, and the part I want to keep sharpening is the mechanical follow through, especially keeping the Jacobian as the reciprocal determinant and tracking limits through a substitution. The two artifacts show both sides: the reciprocal step that gives \(dA = \tfrac{1}{14}\, du\, dv\) and the substitution limits that change from \(r\) to \(u\) are exactly the steps where a small slip would change the answer.