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Kernel: SageMath 10.3

Green's Theorem in the Plane

Green's Theorem allows us to convert the line integral (closed curve CC and nonconservative field F⃗\vec F) into a double integral over the region enclosed by CC.

We introduce two new ideas for Green's theorem: circulation density around an axis perpendicular to the plane and divergence (or flux density).

Spin around an axis: the k⃗\vec k-component of curl

Top and Bottom:

(M(x,y)−M(x,y+Δy))Δx≈−(∂M∂yΔy)Δx\begin{align*} &(M(x,y)-M(x, y+\Delta y))\Delta x\\ &\approx -({\partial M\over\partial y}\Delta y)\Delta x \end{align*}

Left and Right:

(N(x+Δx,y)−N(x,y))Δy≈(∂N∂xΔx)Δy\begin{align*} &(N(x+\Delta x,y)-N(x, y))\Delta y\\ &\approx ({\partial N\over\partial x}\Delta x)\Delta y \end{align*}

Definition The circulation density of a vector field F⃗=Mi⃗+Nj⃗\vec F = M \vec i + N \vec j at the point (x,y)(x, y) is the scalar expression ∂N∂x−∂M∂y.{\partial N\over \partial x}-{\partial M\over \partial y}. It is also called the k⃗\vec k-component of the curl (we will introduce curl later), denoted by (curl F⃗)⋅k⃗\vec F) \cdot \vec k.

Example: The vector fields represent the velocity of a gas flowing in the xyxy-plane. Find their circulation densities and interpret their physical meanings.

  • (a) Uniform expansion or compression: F⃗(x,y)=cxi⃗+cyj⃗\vec F(x, y) = cx \vec i + cy \vec j

  • (b) Uniform rotation: F⃗(x,y)=−cyi⃗+cxj⃗\vec F(x, y) = -cy\vec i + cx\vec j

  • (c) Shearing flow: F⃗(x,y)=yi⃗\vec F(x, y) = y\vec i

  • (d) Whirlpool effect: F⃗(x,y)=−yx2+y2i⃗+xx2+y2j⃗\vec F(x, y) = {-y\over x^2+y^2}\vec i + {x\over x^2+y^2}\vec j

Green's theorem (circulation-curl)

∮CF⃗(x,y)⋅dr⃗\oint_C\vec F(x,y)\cdot d\vec r

for the line integral when the simple closed curve CC is traversed counterclockwise, with its positive orientation. (the region is always to the left)

Theorem (Thm 4-Green's Theorem (Circulation-Curl /Tangential Form))

Let CC be a piecewise smooth, simple closed curve enclosing a region RR in the plane. Let F⃗=Mi⃗+Nj⃗\vec F = M \vec i + N \vec j be a vector field with MM and NN having continuous first partial derivatives in an open region containing RR. Then the counterclockwise circulation of F⃗\vec F around CC equals the double integral of (curl F⃗)⋅k⃗\vec F)\cdot \vec k over RR. ∮CF⃗⋅T⃗ds=∮CF⃗⋅dr⃗=∮CMdx+Ndy=∬R(∂N∂x−∂M∂y)dxdy\oint_C\vec F\cdot \vec T ds=\oint_C\vec F\cdot d\vec r=\oint_CMdx+Ndy=\iint_R\left({\partial N\over \partial x}-{\partial M\over\partial y}\right)dxdy

Divergence

Top and Bottom:

(N(x,y+Δy)−N(x,y))Δx≈(∂N∂yΔy)Δx\begin{align*} &(N(x,y+\Delta y)-N(x, y))\Delta x\\ &\approx ({\partial N\over\partial y}\Delta y)\Delta x \end{align*}

Left and Right:

(M(x+Δx,y)−M(x,y))Δy≈(∂M∂xΔx)Δy\begin{align*} &(M(x+\Delta x,y)-M(x, y))\Delta y\\ &\approx ({\partial M\over\partial x}\Delta x)\Delta y \end{align*}

Definition The divergence (flux density) of a vector field F⃗=Mi⃗+Nj⃗\vec F = M \vec i + N \vec j at the point (x,y)(x, y) is the scalar expression divF⃗=∂M∂x+∂N∂y.\mbox{div}\vec F={\partial M\over \partial x}+{\partial N\over \partial y}.

Example: Find the divergence and interpret what it means for each vector field representing the velocity of a gas flowing in the xyxy plane.

  • (a) Uniform expansion or compression: F⃗(x,y)=cxi⃗+cyj⃗\vec F(x, y) = cx \vec i + cy \vec j

  • (b) Uniform rotation: F⃗(x,y)=−cyi⃗+cxj⃗\vec F(x, y) = -cy\vec i + cx\vec j

  • (c) Shearing flow: F⃗(x,y)=yi⃗\vec F(x, y) = y\vec i

  • (d) Whirlpool effect: F⃗(x,y)=−yx2+y2i⃗+xx2+y2j⃗\vec F(x, y) = {-y\over x^2+y^2}\vec i + {x\over x^2+y^2}\vec j

Green's theorem (flux-divergence)

Theorem (Green’s Theorem (Flux-Divergence or Normal Form)) Let CC be a piecewise smooth, simple closed curve enclosing a region RR in the plane. Let F⃗=Mi⃗+Nj⃗\vec F = M \vec i + N \vec j be a vector field with MM and NN having continuous first partial derivatives in an open region containing RR. Then, the outward flux of F⃗\vec F around CC equals the double integral of div F⃗\vec F over RR. ∮CF⃗⋅n⃗ds=∮CMdy−Ndx=∬R(∂M∂x+∂N∂y)dxdy.\oint_C\vec F\cdot \vec n ds=\oint_CMdy-Ndx=\iint_R\left({\partial M\over \partial x}+{\partial N\over\partial y}\right)dxdy.

Green's theorem: summary

circulation-curl ∮CF⃗⋅dr⃗=∮CMdx+Ndy=∬R(∂N∂x−∂M∂y)dxdy\oint_C\vec F\cdot d\vec r=\oint_CMdx+Ndy=\iint_R\left({\partial N\over \partial x}-{\partial M\over\partial y}\right)dxdyflux-divergence ∮CF⃗⋅n⃗ds=∮CMdy−Ndx=∬R(∂M∂x+∂N∂y)dxdy\oint_C\vec F\cdot \vec n ds=\oint_CMdy-Ndx=\iint_R\left({\partial M\over \partial x}+{\partial N\over\partial y}\right)dxdy

Example: Verify both forms of Green's theorem for the vector field F⃗(x,y)=(x−y)i⃗+xj⃗\vec F(x,y)=(x-y)\vec i+x\vec jand the region RR bounded by the unit circle C: r⃗(t)=(cos⁡t)i⃗+(sin⁡t)j⃗, 0≤t≤2π.C:~\vec r(t)=(\cos t)\vec i+(\sin t)\vec j,~0\leq t\leq 2\pi.

Solution:

∮CF⃗⋅dr⃗=∫02π(cos⁡t−sin⁡t)(−sin⁡t)+cos⁡tcos⁡tdt=2π∬R(∂N∂x−∂M∂y)dxdy=∬R(1+1)dxdy=2π∮CF⃗⋅n⃗ds=∫02π(cos⁡t−sin⁡t)(cos⁡t)+cos⁡tsin⁡tdt=π∬R(∂M∂x+∂N∂y)dxdy=∬R1dxdy=π\begin{align*} \oint_C\vec F\cdot d\vec r=&\int_0^{2\pi}(\cos t-\sin t)(-\sin t)+\cos t\cos t dt=2\pi\\ \iint_R\left({\partial N\over \partial x}-{\partial M\over\partial y}\right)dxdy=&\iint_R\left(1+1\right)dxdy=2\pi\\ \oint_C\vec F\cdot\vec n ds=&\int_0^{2\pi}(\cos t-\sin t)(\cos t)+\cos t\sin t dt=\pi\\ \iint_R\left({\partial M\over \partial x}+{\partial N\over\partial y}\right)dxdy=&\iint_R1dxdy=\pi \end{align*}
reset()

var('x, y, t')

r(t) = vector((cos(t), sin(t)))
show('r(t) =', r(t), 'for 0 <= t <= 2*pi')

F(x,y) = vector((x-y, x)) # Note that the index starts at 0.

(dx,dy) = diff(r,t)
nds = vector((dy,-dx))    #Get the normal vector by swapping the order and changing the sign.

res = integral(F(r(t)[0],r(t)[1]).dot_product(nds), t, 0, 2*pi)
show('The flux over C is  ', res.full_simplify(), ' and the divergence is ', integral(F.div((x,y)), y, -sqrt(1-x^2), sqrt(1-x^2)).integral(x, -1, 1))

show(F.curl((x,y)))
res = integral(F(r(t)[0],r(t)[1]).dot_product(diff(r,t)), t, 0, 2*pi)
show('The circulation over C is  ', res.full_simplify(), ' and the divergence is ', integral(F.curl((x,y)), y, -sqrt(1-x^2), sqrt(1-x^2)).integral(x, -1, 1))

r(t) =(cos⁡(t), sin⁡(t))for 0 <= t <= 2*pi\displaystyle \verb|r(t)|\verb| |\verb|=| \left(\cos\left(t\right),\,\sin\left(t\right)\right) \verb|for|\verb| |\verb|0|\verb| |\verb|<=|\verb| |\verb|t|\verb| |\verb|<=|\verb| |\verb|2*pi|

The flux over C isπ and the divergence isπ\displaystyle \verb|The|\verb| |\verb|flux|\verb| |\verb|over|\verb| |\verb|C|\verb| |\verb|is| \pi \verb| |\verb|and|\verb| |\verb|the|\verb| |\verb|divergence|\verb| |\verb|is| \pi

(x,y) ↦ 2\displaystyle \left( x, y \right) \ {\mapsto} \ 2

The circulation over C is2 π and the divergence is2 π\displaystyle \verb|The|\verb| |\verb|circulation|\verb| |\verb|over|\verb| |\verb|C|\verb| |\verb|is| 2 \, \pi \verb| |\verb|and|\verb| |\verb|the|\verb| |\verb|divergence|\verb| |\verb|is| 2 \, \pi

Using Green's theorem to evaluate line integrals

Example: Evaluate the line integral ∮Cxydy−y2dx\oint_Cxydy-y^2dx where CC is the square cut from the first quadrant by the lines x=1x=1 and y=1y=1.

Solution:

Let M=xyM=xy, N=y2N=y^2 ∮CMdy−Ndx=∮CF⃗⋅n⃗ds=∫01∫01(y+2y)dxdy=32\begin{align*} \oint_CMdy-Ndx=\oint_C\vec F\cdot\vec n ds=\int_0^1\int_0^1 (y+2y)dxdy={3\over2} \end{align*} Let M=−y2M=-y^2, N=xyN=xy ∮CNdy+Mdx=∮CF⃗⋅dr⃗=∫01∫01(y+2y)dxdy=32\begin{align*} \oint_CNdy+Mdx=\oint_C\vec F\cdot d\vec r=\int_0^1\int_0^1 (y+2y)dxdy={3\over2} \end{align*}

Example: Calculate the outward flux of the vector field F⃗(x,y)=2exyi⃗+y3j⃗\vec F(x, y) = 2e^{xy} \vec i + y^3 \vec j across the square bounded by the lines x=±1x = \pm1 and y=±1y = \pm1.

Solution:

∮CF⃗⋅n⃗ds=∫−11∫−11(2exyy+3y2)dxdy=∫−11(2exy+3xy2)∣−11dy=∫−11(2ey−2e−y+6y2)dy=(2ey+2e−y+2y3)∣−11=4\begin{align*} \oint_C\vec F\cdot\vec n ds=&\int_{-1}^1\int_{-1}^1 (2e^{xy}y+3y^2) dxdy=\int_{-1}^1 (2e^{xy}+3xy^2)\Big|_{-1}^1 dy\\ =&\int_{-1}^1 (2e^{y}-2e^{-y}+6y^2) dy=(2e^y+2e^{-y}+2y^3)\Big|_{-1}^1=4 \end{align*}

Proof of Green's theorem on special regions

We prove the components with MM only, and the part with NN can be proved in a similar way.

∫ab∫f1(x)f2(x)∂M∂ydydx=∫abM(x,f2(x))−M(x,f1(x))dx=−∫baM(x,f2(x))dx−∫abM(x,f1(x))dx=−∫C2Mdx−∫C1Mdx\begin{align*} &\int_a^b \int_{f_1(x)}^{f_2(x)}{\partial M\over \partial y}dydx\\ =&\int_a^b M(x,f_2(x))-M(x,f_1(x))dx\\ =&-\int_b^a M(x,f_2(x))dx-\int_a^bM(x,f_1(x))dx\\ =&-\int_{C_2} Mdx-\int_{C_1}Mdx \end{align*}