Critical Thinking

I evaluate, analyze, and synthesize information and ideas to reach justified conclusions. When given a problem I choose appropriate models and methods, I recognize when more than one method is needed, and I am able to synthesize them.

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Artifact

Artifact 1

HW14, Question 2

Criteria

Criterion	Where it appears
Evaluate and analyze	I read the shape and its symmetry before setting up any integral.
Choose appropriate methods	Cylindrical coordinates for the circular symmetry, a symmetry argument for two of the three coordinates.
Synthesize multiple methods	A geometric symmetry argument and a triple integral computation are combined so that only one integral has to be done.

Artifact

Problem

Find the center of mass of the solid \(S\) bounded by the paraboloid \(z = 2x^{2} + 2y^{2}\) and the plane \(z = 8\). Assume the density is constant.

Solution:

Step 1: Read the shape before computing.

The paraboloid \(z = 2x^{2} + 2y^{2}\) opens upward from the origin, and the plane \(z = 8\) caps it. The paraboloid’s equation depends on \(x\) and \(y\) only as \(x^{2} + y^{2}\), so the solid is a bowl with a flat upper bound, symmetric about the \(z\)-axis.

Center of mass, constant density

The \(x\) moment, \(y\) moment, and \(z\) moment are \[ M_{x} = \iiint_{S} x \, dV, \qquad M_{y} = \iiint_{S} y \, dV, \qquad M_{z} = \iiint_{S} z \, dV, \] and the denominator \(M\) is the volume \[ M = \iiint_{S} dV. \] Each coordinate of the center of mass is a moment divided by \(M\): \[ \bar{x} = \frac{M_{x}}{M}, \qquad \bar{y} = \frac{M_{y}}{M}, \qquad \bar{z} = \frac{M_{z}}{M}. \tag{1} \]

Step 2: Use symmetry for \(\bar{x}\) and \(\bar{y}\).

The solid is symmetric about the \(z\)-axis, since the paraboloid depends on \(x\) and \(y\) only as \(x^{2} + y^{2}\) and \((-x)^{2} = x^{2}\), \((-y)^{2} = y^{2}\): if the point \((x, y, z)\) is in \(S\), then so is \((-x, -y, z)\), and with constant density the two points carry equal mass. Their contributions to the \(x\) moment and the \(y\) moment cancel in pairs. So the two numerator integrals are zero: \[ M_{x} = \iiint_{S} x \, dV = 0, \qquad M_{y} = \iiint_{S} y \, dV = 0, \] which gives \[ \bar{x} = 0, \qquad \bar{y} = 0. \]

Step 3: Convert to cylindrical coordinates.

Because the boundaries depend on \(x^{2} + y^{2}\), cylindrical coordinates fit, with \(x^{2} + y^{2} = r^{2}\).

Cylindrical coordinates

\[ \text{paraboloid: } z = 2r^{2}, \qquad \text{upper bound: } z = 8, \qquad dV = r \, dz \, dr \, d\theta. \]

The paraboloid meets the upper bound where \(2r^{2} = 8\), that is \(r^{2} = 4\), so \(r = 2\). For a fixed radius \(r\), the solid runs vertically from the bowl up to the upper bound. The bounds are \[ 0 \le \theta \le 2\pi, \qquad 0 \le r \le 2, \qquad 2r^{2} \le z \le 8. \]

Step 4: Compute the volume (the denominator).

\[ \iiint_{S} dV = \int_{0}^{2\pi} \int_{0}^{2} \int_{2r^{2}}^{8} r \, dz \, dr \, d\theta. \]

Integrate in \(z\): \[ \int_{2r^{2}}^{8} r \, dz = r \left[ z \right]_{2r^{2}}^{8} = r(8 - 2r^{2}) = 8r - 2r^{3}. \]

Integrate in \(r\): \[ \int_{0}^{2} (8r - 2r^{3}) \, dr = \left[ 4r^{2} - \frac{r^{4}}{2} \right]_{0}^{2} = 16 - 8 = 8. \]

Integrate in \(\theta\): \[ M = \iiint_{S} dV = \int_{0}^{2\pi} 8 \, d\theta = 16\pi. \]

Step 5: Compute the \(z\) moment (the numerator).

\[ M_{z} = \iiint_{S} z \, dV = \int_{0}^{2\pi} \int_{0}^{2} \int_{2r^{2}}^{8} z \, r \, dz \, dr \, d\theta. \]

Integrate in \(z\): \[ \int_{2r^{2}}^{8} z \, r \, dz = r \left[ \frac{z^{2}}{2} \right]_{2r^{2}}^{8} = r \left( \frac{64}{2} - \frac{(2r^{2})^{2}}{2} \right) = r(32 - 2r^{4}) = 32r - 2r^{5}. \]

Integrate in \(r\): \[ \int_{0}^{2} (32r - 2r^{5}) \, dr = \left[ 16r^{2} - \frac{r^{6}}{3} \right]_{0}^{2} = 64 - \frac{64}{3} = \frac{128}{3}. \]

Integrate in \(\theta\): \[ M_{z} = \iiint_{S} z \, dV = \int_{0}^{2\pi} \frac{128}{3} \, d\theta = \frac{256\pi}{3}. \]

Step 6: Combine.

\[ \bar{z} = \frac{M_{z}}{M} = \frac{\dfrac{256\pi}{3}}{16\pi} = \frac{256}{48} = \frac{16}{3}. \]

So the center of mass is \[ \boxed{\left( 0, \, 0, \, \tfrac{16}{3} \right)}. \]

Interactive

Drag to rotate the solid. The blue bowl is the paraboloid \(z = 2r^{2}\), the gray disk is the upper bound \(z = 8\), and the orange dot is the center of mass at \(\left( 0, 0, \tfrac{16}{3} \right)\). It sits on the axis of symmetry, which is the symmetry argument made visible.

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

comView = {
  const R = 2;
  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 rs = lin(0, R, 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 bowl = grid((r, t) => r * Math.cos(t), (r, t) => r * Math.sin(t), (r) => 2 * r * r, rs, thetas);
  const lid = grid((r, t) => r * Math.cos(t), (r, t) => r * Math.sin(t), () => 8, rs, thetas);

  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 com = {
    type: "scatter3d", mode: "markers",
    x: [0], y: [0], z: [16 / 3],
    marker: {size: 6, color: "#d97706"}, hoverinfo: "skip"
  };

  const div = html`<div></div>`;
  Plotly.newPlot(div, [
    surf(lid, "#94a3b8", 0.95),
    surf(bowl, "#2563eb", 0.85),
    com
  ], {
    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", range: [0, 8], color: "#475569", gridcolor: "#d8e6f3", backgroundcolor: "rgba(0,0,0,0)"},
      camera: {eye: {x: 1.6, y: 1.6, z: 1.2}}
    }
  }, {displayModeBar: false, responsive: true});
  return div;
}

Essay

I evaluate, analyze, and synthesize information and ideas in order to construct informed, meaningful, and justifiable conclusions.

The information in this problem was almost entirely geometric: two surfaces, a paraboloid and a horizontal plane, bound a solid, and the question asks for one point, the center of mass. Before setting up any integral I read the shape. The paraboloid depends on \(x\) and \(y\) only as \(x^{2} + y^{2}\), so the solid is round about the \(z\)-axis, and the density is constant. Evaluating those two facts first, rather than rushing into three integrals, is the justified conclusion the definition asks for, because together they force the center of mass onto the axis of symmetry, giving \(\bar{x} = \bar{y} = 0\) before any computation. My strength is this opening analysis: I read a problem for symmetry and structure before I commit to a method.

I recognize when multiple models or methods are needed to solve a problem and am able to synthesize them.

A center of mass problem looks like it needs three integrals, one for each coordinate. By recognizing symmetry I replaced two of them with a one line argument, that every point \((x, y, z)\) is matched by \((-x, -y, z)\) of equal mass, then switched models to a direct cylindrical integral for the one coordinate symmetry cannot give, the height \(\bar{z}\). Choosing cylindrical coordinates was its own method decision, driven by the circular boundaries that turn the paraboloid into \(z = 2r^{2}\) and make \(dV = r \, dz \, dr \, d\theta\) clean. Combining a symmetry argument with a triple integral, rather than grinding out all three integrals in rectangular coordinates, is the synthesis the definition names. The area I want to keep improving is staying as careful in the mechanical computation as in the clever setup, since the \(z\) moment still needs an accurate antiderivative and an accurate evaluation at both limits, and the artifact shows both that synthesis and where I have to slow down.