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 |
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 set up every later triple integral. |
| Justification | The essay ties the artifact back to the definition of metacognition step by step. |
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.
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 \[ 2x + 3z = 12 \quad \Longrightarrow \quad x = \frac{12 - 3z}{2}. \]
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, because the second pass is where my mathematical intuition grows. This quiz is a clear case. The setup was wrong but the arithmetic was clean, so the only way to learn from it was to come back to the problem on my own and rebuild it from the geometry. I also carried the lesson forward as a habit. Because this failure was partly a quick, careless error rather than a pure gap in understanding, I now double and triple check my work on later quizzes before I commit to an answer.
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 honestly is what pushed me to drill these problems externally on my own, and the improvement is real: I now set up and compute these integrals accurately and without the bound errors that used to trip me up. 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 automatic. Drawing the mistaken region and the correct one together, the prism and the tetrahedron, is exactly the kind of check that this practice taught me, and it is what makes this particular mistake impossible for me to repeat.