Green’s Theorem

Cover Page

Artifacts

Artifact #1	Line Integral by Green’s Theorem (HW19 #3)
Artifact #2	Area of a Parametric Region by Green’s Theorem (HW19 #4)

Sub-outcome coverage

Sub Outcome	Artifact #1	Artifact #2
Do a double integral instead: calculate a line integral by using Green’s Theorem to evaluate an equivalent double integral.	Yes
Do a line integral instead: calculate a double integral by using Green’s Theorem to evaluate an equivalent line integral.		Yes
Area of a region: use Green’s Theorem to calculate the area of a two-dimensional region bounded by lines or curves.		Yes

Artifact 1

Problem

Use Green’s Theorem to evaluate \[ \oint_{C} \vec{F} \cdot d\vec{r}, \qquad \vec{F} = \left\langle \sqrt{x} + 5y, \; 2x + \sqrt{y} \right\rangle, \] where \(C\) consists of the arc of the curve \(y = x - x^{2}\) from \((0, 0)\) to \((1, 0)\) and the line segment from \((1, 0)\) back to \((0, 0)\).

Solution:

The two square roots make a direct line integral unpleasant, so we trade the line integral for a double integral with Green’s Theorem.

Green’s Theorem

For a positively oriented (counterclockwise) simple closed curve \(C\) bounding a region \(D\), with \(\vec{F} = \langle P, Q \rangle\), \[ \oint_{C} P\, dx + Q\, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \]

First we check the orientation, since the hint warns us to. The arc \(y = x - x^{2}\) is the top boundary and the segment along the \(x\)-axis is the bottom boundary. Traveling along the arc from \((0, 0)\) to \((1, 0)\) and then back along the axis from \((1, 0)\) to \((0, 0)\) runs clockwise, which is the negative orientation. Green’s Theorem is stated for the counterclockwise direction, so we attach a minus sign: \[ \oint_{C} \vec{F} \cdot d\vec{r} = -\iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \]

Interactive

The shaded region is \(D\). Drag the slider to send a point around \(C\) in the order the problem gives, first along the arc and then along the segment, and watch that it travels clockwise.

viewof tpos = Inputs.range([0, 1], {step: 0.01, value: 0.25, label: "position along C"})

boundaryPt = {
  if (tpos <= 0.5) {
    const s = tpos / 0.5;          // arc (0,0) -> (1,0)
    return [s, s - s * s];
  } else {
    const s = (tpos - 0.5) / 0.5;  // segment (1,0) -> (0,0)
    return [1 - s, 0];
  }
}

arcPts = Array.from({length: 81}, (_, i) => { const x = i / 80; return [x, x - x * x]; })

Plot.plot({
  width: 460,
  height: 320,
  marginLeft: 45,
  style: {background: "transparent", color: "#334155"},
  x: {label: "x", grid: true, domain: [-0.1, 1.1]},
  y: {label: "y", grid: true, domain: [-0.1, 0.32]},
  marks: [
    Plot.areaY(arcPts, {x: d => d[0], y: d => d[1], fill: "#cfe3f7", fillOpacity: 0.7}),
    Plot.line(arcPts, {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2}),
    Plot.line([[1, 0], [0, 0]], {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2}),
    Plot.dot([boundaryPt], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 7})
  ]
})

Now we read off the partial derivatives. With \(P = \sqrt{x} + 5y\) and \(Q = 2x + \sqrt{y}\), \[ \frac{\partial Q}{\partial x} = 2, \qquad \frac{\partial P}{\partial y} = 5, \qquad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 5 = -3. \]

The square roots are gone, and the integrand is a constant. The region is \[ D = \left\{ (x, y) : 0 \le x \le 1, \; 0 \le y \le x - x^{2} \right\}, \] so \[ \begin{aligned} \oint_{C} \vec{F} \cdot d\vec{r} &= -\iint_{D} (-3)\, dA \\[4pt] &= 3 \iint_{D} dA \\[4pt] &= 3 \int_{0}^{1} \int_{0}^{\,x - x^{2}} 1 \, dy \, dx \\[4pt] &= 3 \int_{0}^{1} (x - x^{2}) \, dx \\[4pt] &= 3 \left[ \frac{x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{1} \\[4pt] &= 3 \left( \frac{1}{2} - \frac{1}{3} \right) \\[4pt] &= 3 \cdot \frac{1}{6} \\[4pt] &= \frac{1}{2}. \end{aligned} \]

Artifact 2

Problem

Find the area of the region enclosed by \[ x = t - t^{5}, \qquad y = t - t^{7}, \qquad 0 \le t \le 1, \] using Green’s Theorem.

Solution:

We start with the definition of area as a double integral over the enclosed region \(D\): \[ A = \iint_{D} 1 \, dA. \]

Setting this up directly is the hard part. The boundary is given parametrically, and to write \(D\) as \(0 \le x \le x_{\max}\), \(g(x) \le y \le h(x)\) we would need \(y\) as a function of \(x\). Eliminating \(t\) from \(x = t - t^{5}\) is not possible in any elementary closed form, so there is no clean pair of \(y\)-limits to integrate between. This is exactly the situation Green’s Theorem is built for: it turns the awkward double integral into a line integral around the boundary, which we can evaluate straight from the parametrization.

Area from Green’s Theorem

Taking \(P = -\tfrac{1}{2} y\) and \(Q = \tfrac{1}{2} x\) gives \(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = \tfrac{1}{2} + \tfrac{1}{2} = 1\), so \[ A = \iint_{D} 1 \, dA = \frac{1}{2} \oint_{C} \left( x \, dy - y \, dx \right). \]

Before applying it we confirm \(C\) is a closed curve. At \(t = 0\) we get \((x, y) = (0, 0)\), and at \(t = 1\) we get \((1 - 1, \, 1 - 1) = (0, 0)\), so the curve starts and ends at the same point.

Interactive

Drag the slider to trace the curve as \(t\) runs from \(0\) to \(1\). The orange dot is the current point, the blue curve is the path traced so far, and the shaded region is the area we are computing.

viewof tparam = Inputs.range([0, 1], {step: 0.01, value: 0.5, label: "t"})

fullCurve = Array.from({length: 121}, (_, i) => { const t = i / 120; return [t - t ** 5, t - t ** 7]; })

tracedCurve = fullCurve.filter((_, i) => i / 120 <= tparam + 1e-9)

curPt = [tparam - tparam ** 5, tparam - tparam ** 7]

Plot.plot({
  width: 460,
  height: 340,
  marginLeft: 45,
  style: {background: "transparent", color: "#334155"},
  x: {label: "x", grid: true},
  y: {label: "y", grid: true},
  marks: [
    Plot.areaY(fullCurve, {x: d => d[0], y: d => d[1], fill: "#cfe3f7", fillOpacity: 0.6}),
    Plot.line(tracedCurve, {x: d => d[0], y: d => d[1], stroke: "#2563eb", strokeWidth: 2.5}),
    Plot.dot([curPt], {x: d => d[0], y: d => d[1], fill: "#d97706", r: 7})
  ]
})

Now we set up the line integral from the parametrization. The differentials are \[ dx = \left( 1 - 5t^{4} \right) dt, \qquad dy = \left( 1 - 7t^{6} \right) dt. \]

Form the integrand \(x \, dy - y \, dx\): \[ \begin{aligned} x \, dy - y \, dx &= \left( t - t^{5} \right)\!\left( 1 - 7t^{6} \right) dt - \left( t - t^{7} \right)\!\left( 1 - 5t^{4} \right) dt \\[4pt] &= \left( t - 7t^{7} - t^{5} + 7t^{11} \right) dt - \left( t - 5t^{5} - t^{7} + 5t^{11} \right) dt \\[4pt] &= \left( 4t^{5} - 6t^{7} + 2t^{11} \right) dt. \end{aligned} \]

So the area is \[ \begin{aligned} A = \frac{1}{2} \oint_{C} \left( x \, dy - y \, dx \right) &= \frac{1}{2} \int_{0}^{1} \left( 4t^{5} - 6t^{7} + 2t^{11} \right) dt \\[4pt] &= \frac{1}{2} \left[ \frac{4t^{6}}{6} - \frac{6t^{8}}{8} + \frac{2t^{12}}{12} \right]_{0}^{1} \\[4pt] &= \frac{1}{2} \left( \frac{2}{3} - \frac{3}{4} + \frac{1}{6} \right) \\[4pt] &= \frac{1}{2} \cdot \frac{8 - 9 + 2}{12} \\[4pt] &= \frac{1}{2} \cdot \frac{1}{12} \\[4pt] &= \frac{1}{24}. \end{aligned} \]

The value is positive, which confirms the parametrization runs counterclockwise, so the formula returns the area directly with no sign correction. The enclosed area is \[ \boxed{\dfrac{1}{24}}. \]

Essay

My approach to a Green’s Theorem problem is to decide which direction across the theorem saves the most work, because the equality lets me trade a line integral for a double integral or the other way around. In the first artifact the field carried two square roots, which are awkward to integrate along a curve, but the combination \(\partial Q / \partial x - \partial P / \partial y\) collapsed to the constant \(-3\), so the double integral was just a constant times an area. In the second artifact the trade ran the other way: the area is a double integral by definition, but the region was pinned down only by a parametric curve I could not solve for \(y\), so converting to the line integral \(\tfrac{1}{2}\oint (x\, dy - y\, dx)\) turned an impossible setup into a routine polynomial integral. My strength is reading the problem for that decision and checking orientation before I commit, since the first curve was clockwise and the sign would have flipped the answer. The area I want to keep sharpening is the bookkeeping once the setup is done, especially tracking signs through the orientation check and keeping the parametric algebra in the second artifact straight, since those are the steps where a small slip would change the result.