Metacognition

I am aware of and reflect on my own learning process. I reflect on what I learn by recognizing when I do not understand, when I have gained understanding, and when I have demonstrated mastery. I reflect on how I learn by recognizing the specific activities that helped me understand.

Cover Page

Artifact

Artifact 1

Quiz 5, Question 1, rewritten from my own quiz work with the correction shown

Criteria

Criterion	Where it appears
Metacognition	I notice the exact moment my understanding was wrong, then describe the corrected understanding.
Productive failure	The wrong bounds gave a clean but incorrect answer, and fixing them changed how I check later triple integral setups.
Justification	The essay ties the artifact back to the definition of metacognition step by step, using this specific bounds mistake as evidence.

Original Solution

Problem

Let \(S\) be the solid in the first octant, meaning \(x \ge 0\), \(y \ge 0\), and \(z \ge 0\), bounded by the coordinate planes and the plane \[ 2x + y + 3z = 12. \] Use a triple integral to find the exact volume of \(S\).

Initial Attempt

I started from the plane and tried to read the bounds straight off it. I set the outer variable \(z\) from \(0\) to \(4\), since \(3z = 12\) gives \(z = 4\) when \(x = y = 0\). Then I reused the trace of the plane in the \(xy\)-plane, the line \(2x + y = 12\), for the inner two bounds, as if that triangle were the base at every height. At the time this felt reasonable because each individual bound came from the plane, but I had not checked whether the cross sections changed as \(z\) changed.

Original (incorrect) work

I used the bounds \[ 0 \le z \le 4, \qquad 0 \le x \le 6, \qquad 0 \le y \le 12 - 2x, \] so my integral was \[ V = \int_{0}^{4} \int_{0}^{6} \int_{0}^{12 - 2x} 1 \, dy \, dx \, dz. \]

Integrating in \(y\): \[ \int_{0}^{12 - 2x} 1 \, dy = \left[ y \right]_{0}^{12 - 2x} = 12 - 2x. \]

Integrating in \(x\): \[ \int_{0}^{6} (12 - 2x) \, dx = \left[ 12x - x^{2} \right]_{0}^{6} = 72 - 36 = 36. \]

Integrating in \(z\): \[ \int_{0}^{4} 36 \, dz = \left[ 36 z \right]_{0}^{4} = 144. \]

So my original answer was \(\boxed{144}\).

Correction

Where it went wrong:

The line \(y = 12 - 2x\) is only the edge of the solid where the plane meets the floor, that is, where \(z = 0\). By using it at every height I was treating the base triangle as if it were the same for all \(z\), which builds a prism, not the tetrahedron the plane cuts off. The upper surface tilts, so the cross section has to shrink as \(z\) increases. Solving the plane for \(y\) shows the dependence I had dropped: \[ 2x + y + 3z = 12 \quad \Longrightarrow \quad y = 12 - 2x - 3z. \] The top of \(y\) must depend on both \(x\) and \(z\). For the same reason, at a fixed height \(z\) the largest \(x\) happens when \(y = 0\), which gives \[ 2x + 3z = 12 \quad \Longrightarrow \quad x = \frac{12 - 3z}{2}. \] This correction is what made the failure productive: I could see that the error was not in integration, it was in matching the bounds to the geometry.

Interactive

The top shows the prism my mistaken bounds measured (\(V = 144\)). The bottom shows the tetrahedron the plane cuts off (\(V = 48\)). Drag to rotate.

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

compareView = {
  const layoutBase = (title) => ({
    title: {text: title, font: {size: 13, color: "#334155"}, x: 0.5, xanchor: "center"},
    paper_bgcolor: "rgba(0,0,0,0)",
    plot_bgcolor: "rgba(0,0,0,0)",
    font: {color: "#334155"},
    height: 340,
    margin: {l: 0, r: 0, t: 34, b: 0},
    scene: {
      xaxis: {title: "x", range: [0, 6], color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      yaxis: {title: "y", range: [0, 12], color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      zaxis: {title: "z", range: [0, 4], color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      aspectmode: "cube",
      camera: {eye: {x: 1.7, y: 1.7, z: 1.1}}
    }
  });

  // Mistake: triangular prism, floor triangle extruded straight up to z = 4 (V = 144)
  const prism = {
    type: "mesh3d",
    x: [0, 6, 0, 0, 6, 0],
    y: [0, 0, 12, 0, 0, 12],
    z: [0, 0, 0, 4, 4, 4],
    i: [0, 3, 0, 0, 1, 1, 2, 2],
    j: [1, 4, 1, 4, 2, 5, 0, 3],
    k: [2, 5, 4, 3, 5, 4, 3, 5],
    color: "#c0504d", opacity: 0.5, flatshading: true, hoverinfo: "skip", showscale: false
  };

  // Correct: tetrahedron cut by 2x + y + 3z = 12 (V = 48)
  const tetra = {
    type: "mesh3d",
    x: [0, 6, 0, 0],
    y: [0, 0, 12, 0],
    z: [0, 0, 0, 4],
    i: [0, 0, 0, 1],
    j: [1, 1, 2, 2],
    k: [2, 3, 3, 3],
    color: "#2563eb", opacity: 0.6, flatshading: true, hoverinfo: "skip", showscale: false
  };

  const left = html`<div></div>`;
  const right = html`<div></div>`;
  PlotlyMetacognition.newPlot(left, [prism], layoutBase("Mistake: prism, V = 144"), {displayModeBar: false, responsive: true});
  PlotlyMetacognition.newPlot(right, [tetra], layoutBase("Correct: tetrahedron, V = 48"), {displayModeBar: false, responsive: true});

  return html`<div style="display:flex; flex-direction:column; gap:0.75rem; align-items:stretch">
    <div style="width:100%">${left}</div>
    <div style="width:100%">${right}</div>
  </div>`;
}

Corrected work

The correct bounds are \[ 0 \le z \le 4, \qquad 0 \le x \le \frac{12 - 3z}{2}, \qquad 0 \le y \le 12 - 2x - 3z, \] so \[ V = \int_{0}^{4} \int_{0}^{\frac{12 - 3z}{2}} \int_{0}^{12 - 2x - 3z} 1 \, dy \, dx \, dz. \]

Integrating in \(y\): \[ \int_{0}^{12 - 2x - 3z} 1 \, dy = 12 - 2x - 3z. \]

Integrating in \(x\): \[ \begin{aligned} \int_{0}^{\frac{12 - 3z}{2}} (12 - 2x - 3z) \, dx &= \left[ (12 - 3z) x - x^{2} \right]_{0}^{\frac{12 - 3z}{2}} \\[4pt] &= (12 - 3z)\!\left( \frac{12 - 3z}{2} \right) - \left( \frac{12 - 3z}{2} \right)^{2} \\[4pt] &= \frac{(12 - 3z)^{2}}{2} - \frac{(12 - 3z)^{2}}{4} \\[4pt] &= \frac{(12 - 3z)^{2}}{4}. \end{aligned} \]

So \[ V = \frac{1}{4} \int_{0}^{4} (12 - 3z)^{2} \, dz. \]

Substitute \(u = 12 - 3z\), so \(du = -3 \, dz\) and \(dz = -\tfrac{1}{3} \, du\). The limits change from \(z = 0\) to \(u = 12\), and from \(z = 4\) to \(u = 0\): \[ V = \frac{1}{4} \int_{12}^{0} u^{2} \left( -\frac{1}{3} \right) du = \frac{1}{12} \int_{0}^{12} u^{2} \, du. \]

Integrating in \(u\): \[ V = \frac{1}{12} \left[ \frac{u^{3}}{3} \right]_{0}^{12} = \frac{1}{12} \cdot \frac{12^{3}}{3} = \frac{1}{12} \cdot 576 = 48. \]

So the correct volume is \(\boxed{48}\).

Essay

This artifact demonstrates metacognition as it shows:

I am aware of and reflect on my own learning process. As a graduating mathematics B.S. student, I have learned from experience that it is worth redoing and retrying test problems even when no corrections are offered for the benefit of growing mathematical intuition that can help in the future. For this quiz, the setup was wrong but the arithmetic was accurate, so the only way to learn from it was to come back to the problem on my own and rebuild it from the geometry. Because this failure was partly a quick, careless error rather than a gap in conceptual understanding, I now double and triple check my work on later quizzes before I commit to an answer, especially the way inner bounds depend on the outer variables.

I reflect on what I learn by recognizing when I do not understand, when I have gained understanding, and when I have demonstrated mastery. I had taken this course once before at an earlier date and had to drop it for personal reasons, and at that time double and triple integrals were the topic I struggled with most. Recognizing that weakness is what pushed me to drill these problems externally on my own, and I have seen substantial improvement since that point. This quiz marks the line between not understanding, where I froze the cross section by accident, and understanding, where I can state plainly that the inner limits have to depend on the outer variables.

I reflect on how I learn by recognizing the specific activities that helped me understand. What moved me forward here was not a lecture but deliberate practice: reworking the missed problem from scratch, checking the answer against the shape of the solid, and repeating similar problems until the correct setup became more automatic. The useful habit I took from this artifact is to pause before integrating and ask whether each bound still describes the same solid once the outer variables have been fixed.