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Week 8 Discussion Notes

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Green's Theorem

Theorem

Let D\mathscr{D} be a domain whose boundary, denoted ∂D\partial\mathscr{D}, is a simple closed curve (i.e., a loop that doesn't intersect itself) oriented counter-clockwise. If F1F_1 and F2F_2 are continuously differentiable in an open set containing D\mathscr{D}, then

∮∂DF1 dx+F2 dy=∬D∂F2∂x−∂F1∂y dA\oint_{\partial\mathscr{D}} F_1 \,\diff{x} + F_2 \,\diff{y} = \iint_{\mathscr{D}} \pderiv{F_2}{x} - \pderiv{F_1}{y} \,\diff{A}

or equivalently,

∮∂DF⃗⋅dr⃗=∬Dcurl⁡zF⃗ dA.\oint_{\partial\mathscr{D}} \vec{F} \cdot \diff\vec{r} = \iint_{\mathscr{D}} \curl_z{\vec{F}} \,\diff{A}.

Green's Theorem is a fundamental theorem of calculus. Compare it with the fundamental theorem of calculus in 1D:

∫abf′(x) dx=f(b)−f(a)\int_a^b f'\p{x} \,\diff{x} = f\p{b} - f\p{a}

This relates an integral of a derivative on the whole domain (f′f' on the interval [a,b]\br{a, b}) to an "integral" on the boundary of the function (ff at the points aa and bb).

Green's Theorem is a generalization of this: it relates an integral of a "derivative" on the whole domain (curl⁡zF⃗\curl_z{\vec{F}} on D\mathscr{D}) to an integral of the function on the boundary (F⃗\vec{F} on ∂D\partial\mathscr{D}). Every integral theorem we'll see in this class (i.e., Stokes' Theorem and the divergence theorem) will be in this form.

Example 1.

Using Green's Theorem, calculate ∮Cx2y dx\displaystyle\oint_{\mathscr{C}} x^2y \,\diff{x} where C\mathscr{C} is the unit circle centered at the origin oriented counterclockwise.
Solution.

In this problem, C=∂D\mathscr{C} = \partial\mathscr{D}, where D\mathscr{D} is the unit ball:

Also, F1=x2yF_1 = x^2y and F2=0F_2 = 0, so Green's Theorem tells us

∮Cx2y dx=∬D−∂∂y(x2y) dA=∬D−x2 dA=∫02π∫01−r2cos⁡2θ⋅r dr dθ=−π4.\begin{aligned} \oint_{\mathscr{C}} x^2y \,\diff{x} &= \iint_{\mathscr{D}} -\pderiv{}{y} \p{x^2y} \,\diff{A} \\ &= \iint_{\mathscr{D}} -x^2 \,\diff{A} \\ &= \int_0^{2\pi} \int_0^1 -r^2\cos^2\theta \cdot r \,\diff{r} \,\diff\theta \\ &= \boxed{-\frac{\pi}{4}}. \end{aligned}

(As usual, the hardest part of the problem was figuring out which integral you need to calculate, i.e., figuring out how to use Green's Theorem.)
Example 2.

Let D\mathscr{D} be the domain and C1,C2\mathscr{C}_1, \mathscr{C}_2 be the curves in the following figure:

Assume that

∮C2F⃗⋅dr⃗=12and∂F2∂x−∂F1∂y=−3 on D.\oint_{\mathscr{C}_2} \vec{F} \cdot \diff\vec{r} = 12 \quad\text{and}\quad \pderiv{F_2}{x} - \pderiv{F_1}{y} = -3 \text{ on } \mathscr{D}.

Using Green's Theorem, calculate ∮C1F⃗⋅dr⃗\oint_{\mathscr{C}_1} \vec{F} \cdot \diff\vec{r}.
Solution.

In our statement of Green's Theorem, we need ∂D\partial\mathscr{D} to be a single simple closed curve, but in this problem, ∂D\partial\mathscr{D} comprises of two such curves: ∂D=C1−C2\partial\mathscr{D} = \mathscr{C}_1 - \mathscr{C}_2. So, we can't apply Green's Theorem right away, but we can use the following trick: divide D\mathscr{D} like so:

We can apply Green's Theorem on both halves D1\mathscr{D}_1 and D2\mathscr{D}_2 of the domain to get

∬D1∂F2∂x−∂F1∂y dA=∮∂D1F⃗⋅dr⃗∬D2∂F2∂x−∂F1∂y dA=∮∂D2F⃗⋅dr⃗.\begin{aligned} \iint_{\mathscr{D}_1} \pderiv{F_2}{x} - \pderiv{F_1}{y} \,\diff{A} &= \oint_{\partial\mathscr{D}_1} \vec{F} \cdot \diff\vec{r} \\ \iint_{\mathscr{D}_2} \pderiv{F_2}{x} - \pderiv{F_1}{y} \,\diff{A} &= \oint_{\partial\mathscr{D}_2} \vec{F} \cdot \diff\vec{r}. \end{aligned}

Notice also that ∂D1+∂D2=∂D=C1−C2\partial\mathscr{D}_1 + \partial\mathscr{D}_2 = \partial\mathscr{D} = \mathscr{C}_1 - \mathscr{C}_2, so if we add everything up,

∬D∂F2∂x−∂F1∂y dA=∮C1F⃗⋅dr⃗−∮C2F⃗⋅dr⃗.\iint_{\mathscr{D}} \pderiv{F_2}{x} - \pderiv{F_1}{y} \,\diff{A} = \oint_{\mathscr{C}_1} \vec{F} \cdot \diff\vec{r} - \oint_{\mathscr{C}_2} \vec{F} \cdot \diff\vec{r}.

Plugging in the numbers we know, we get

∮∂C1F⃗⋅dr⃗=∬D∂F2∂x−∂F1∂y dA+∮∂C2F⃗⋅dr⃗=∬D−3 dA+12=−3⋅area⁡(D)+12=−3(60−4π)+12.\begin{aligned} \oint_{\partial\mathscr{C}_1} \vec{F} \cdot \diff\vec{r} &= \iint_{\mathscr{D}} \pderiv{F_2}{x} - \pderiv{F_1}{y} \,\diff{A} + \oint_{\partial\mathscr{C}_2} \vec{F} \cdot \diff\vec{r} \\ &= \iint_{\mathscr{D}} -3 \,\diff{A} + 12 \\ &= -3 \cdot \operatorname{area}\p{\mathscr{D}} + 12 \\ &= \boxed{-3\p{60 - 4\pi} + 12}. \end{aligned}