Multiple Integrals

Cover Page

Artifacts

Artifact #1	Change of Variables in Multiple Integrals (HW15 #3), chosen from homework I did well on
Artifact #2	Triple Integrals in Cylindrical Coordinates (HW14 #1), chosen from homework I did well on

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. The boundary equations already suggest the new variables, which is why this artifact fits the change of variables sub outcome.

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. This is the part of the process that feels most natural to me because it turns the transformation into the correct area-scaling factor: \[ \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} -5 & -6 \\ \phantom{-}4 & \phantom{-}2 \end{vmatrix} = -10 + 24 = 14, \] \[ \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{14}, \] \[ dA = \frac{1}{14}\, du\, dv. \]

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 \(x^{2} + y^{2} = r^{2}\). Recognizing that circular structure is the part of this artifact that feels strongest to me.

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.

PlotlyMultipleIntegrals = 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>`;
  PlotlyMultipleIntegrals.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;
}

So the triple integral becomes: \[ \iiint_{E} e^{z}\, dV = \int_{0}^{2\pi} \int_{0}^{\sqrt{5}} \int_{0}^{2 + r^{2}} e^{z}\, r\, dz\, dr\, d\theta. \]

Integrating in \(z\) first: \[ \int_{0}^{2 + r^{2}} e^{z}\, r\, dz = r \left[ e^{z} \right]_{0}^{2 + r^{2}} = r \left( e^{2 + r^{2}} - 1 \right). \]

Next we integrate in \(r\). The substitution \(u = 2 + r^{2}\) gives \(r\, dr = \tfrac{1}{2}\, du\), so the antiderivative of \(r\, e^{2 + r^{2}}\) is \(\tfrac{1}{2} e^{2 + r^{2}}\), and the antiderivative of \(r\) is \(\tfrac{r^{2}}{2}\): \[ \begin{aligned} \int_{0}^{\sqrt{5}} r \left( e^{2 + r^{2}} - 1 \right) dr &= \left[ \frac{1}{2} e^{2 + r^{2}} - \frac{r^{2}}{2} \right]_{0}^{\sqrt{5}} \\[4pt] &= \left( \frac{1}{2} e^{7} - \frac{5}{2} \right) - \frac{1}{2} e^{2} \\[4pt] &= \frac{1}{2} \left( e^{7} - e^{2} \right) - \frac{5}{2}. \end{aligned} \]

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

Your general approach to problems like this:

My general approach to problems like this is to start with the geometry first. That is because if it is necessary to do a change of coordinate systems, that is the first decision I make before proceeding forward. Looking at the geometry first also has the potential to reveal geometric tricks or shortcuts the problem may present, such as symmetry.

An evaluation of your strengths and areas for improvement in this section:

I would say I am conceptually strong at writing double integrals, especially when the Jacobian is involved. In Artifact 1, setting up \(u = -5x - 6y\) and \(v = 4x + 2y\) made the region easier to understand, and computing the Jacobian felt like a clear step. The part I could use some improvement on is the computation and remembering every factor in the coordinate system. In cylindrical coordinates, I have to be careful not to forget the extra \(r\), and in spherical coordinates I have to be careful with the extra integration factors. In general in math, I sometimes make basic algebra mistakes when I am moving too fast, and there are many of those when computing an integral. Because of the nature of the computation, a slightly incorrect answer to one numeric step could produce a drastically different answer at the end.

Describe how the artifacts you provided support your self-evaluation:

My self-evaluation is strongly supported by my artifacts. Artifact 1 shows that I can connect the boundary lines to a useful transformation and use the Jacobian correctly. Artifact 2 shows that I can recognize when cylindrical coordinates fit the geometry and then set up the bounds with the correct \(r\) factor in \(dV\). Because we had not done double and triple integrals in a while, the setup is where my conceptual understanding came back quickly, while the long computations are exactly where a small slip would have changed the result.

What are the critical pieces of a problem of this type?

I would list in the following order of importance: choosing the right coordinate system, setting proper bounds, and performing accurate computations.