Tutorial 14

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Stokes' theorem: $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}=\iint_{S}\left(\nabla\times\mathbf{F}\right)\cdot d\mathbf{S}$
• Choosing the easier side of the identity
• Orientation: how the normal direction fixes the direction of travel around the boundary
• Verifying both sides of the theorem
• Surface integrals via a parametrisation

Introduction

Stokes' theorem lets you replace a hard surface integral with an easy line integral along the boundary — or the other way round. The two sides are only equal when the orientations match: with the right-hand rule, the normal determines the positive direction of travel around $C$.

import sympy as sy
from sympy.abc import x, y, z, t, u, v, r, theta
from sympy import (sqrt, sin, cos, tan, pi, ln, exp, integrate, diff, simplify,
                   Rational, N, Matrix, Symbol)
sy.init_printing(use_latex=True)

def curl(F):
    """Curl of a 3D field given as a list/Matrix of components."""
    return Matrix([diff(F[2], y) - diff(F[1], z),
                   diff(F[0], z) - diff(F[2], x),
                   diff(F[1], x) - diff(F[0], y)])

def line_integral(F, rx, ry, rz, t, t0, t1):
    """Integral of F . dr along (x,y,z) = (rx, ry, rz) for t in [t0, t1]."""
    Fx = F[0].subs({x: rx, y: ry, z: rz})
    Fy = F[1].subs({x: rx, y: ry, z: rz})
    Fz = F[2].subs({x: rx, y: ry, z: rz})
    return integrate(Fx*diff(rx, t) + Fy*diff(ry, t) + Fz*diff(rz, t), (t, t0, t1))

Question 1

$\mathbf{F}=\langle-y,x,z\rangle$, $S$ is the part of $x^{2}+y^{2}+z^{2}=25$ below $z=4$, oriented with the outward normal. Compute $\iint_{S}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$.

The boundary is the circle $x^{2}+y^{2}=9$ in the plane $z=4$. The natural parametrisation $\mathbf{r}(t)=\langle3\cos t,3\sin t,4\rangle$ actually traverses $-C$, because with the outward normal the boundary must be walked clockwise.

F = Matrix([-y, x, z])
path = [3*cos(t), 3*sin(t), 4]

minus_C = line_integral(F, *path, t, 0, 2*pi)
display(minus_C)                                        # 18*pi  (this is the integral over -C)

display(-minus_C)                                       # -18*pi (over C)

$\iint_{S}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}=\int_{C}\mathbf{F}\cdot d\mathbf{r}=-18\pi$

Question 2

$\mathbf{F}=z^{2}\mathbf{i}-3xy\,\mathbf{j}+x^{3}y^{3}\mathbf{k}$, $S$ is the part of $z=5-x^{2}-y^{2}$ above $z=1$, oriented upwards.

The boundary is the circle $x^{2}+y^{2}=4$ in the plane $z=1$.

F = Matrix([z**2, -3*x*y, x**3*y**3])
I = line_integral(F, 2*cos(t), 2*sin(t), 1, t, 0, 2*pi)
display(simplify(I))                                    # 0

$\iint_{S}\operatorname{curl}\mathbf{F}\cdot d\mathbf{S}=0$

Question 3

$\mathbf{F}=z^{2}\mathbf{i}+y^{2}\mathbf{j}+x\mathbf{k}$ and $C$ is the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, counterclockwise.

F = Matrix([z**2, y**2, x])
display(simplify(curl(F)))                              # <0, 2*z - 1, 0>

# the plane is x + y + z = 1, so z = 1 - x - y and the normal is <1, 1, 1>
integrand = simplify((curl(F).subs({z: 1 - x - y})).dot(Matrix([1, 1, 1])))
display(integrand)                                      # 1 - 2*x - 2*y

I = integrate(integrand, (y, 0, 1 - x), (x, 0, 1))
display(I)                                              # -1/6

$\int_{C}\mathbf{F}\cdot d\mathbf{r}=-\frac{1}{6}$

Question 4

Verify Stokes' theorem for $\mathbf{F}=\left\langle x^{2},2x,z^{2}\right\rangle$ on the ellipse $S=\{4x^{2}+y^{2}\le4,\ z=0\}$.

F = Matrix([x**2, 2*x, z**2])
path = [cos(t), 2*sin(t), 0]               # counterclockwise ellipse, normal upwards

lhs = line_integral(F, *path, t, 0, 2*pi)
display(simplify(lhs))                                  # 4*pi

display(simplify(curl(F)))                              # <0, 0, 2>
rhs = 2 * (pi * 1 * 2)                                  # 2 x area of the ellipse (pi*a*b)
display(rhs)                                            # 4*pi

display(simplify(lhs - rhs))                            # 0 -> verified

Both sides equal $4\pi$, so Stokes' theorem is verified.

Question 5

Verify Stokes' theorem for $\mathbf{F}=\left\langle y^{2},-x,5z\right\rangle$ on the paraboloid $z=x^{2}+y^{2}$ with boundary $x^{2}+y^{2}=1$.

F = Matrix([y**2, -x, 5*z])
display(simplify(curl(F)))                              # <0, 0, -1 - 2*y>

# surface side:  normal < -2x, -2y, 1 >,  integrand (-1 - 2y),  over the unit disc
rhs = integrate((-1 - 2*r*sin(theta))*r, (r, 0, 1), (theta, 0, 2*pi))
display(simplify(rhs))                                  # -pi

# line side:  x = cos t, y = sin t, z = 1
lhs = line_integral(F, cos(t), sin(t), 1, t, 0, 2*pi)
display(simplify(lhs))                                  # -pi

display(simplify(lhs - rhs))                            # 0 -> verified

Both sides equal $-\pi$.

Question 6

$\mathbf{F}=\left\langle xz^{2},x^{3},\cos(xz)\right\rangle$, $S$ is the part of the ellipsoid $x^{2}+y^{2}+3z^{2}=1$ below the $xy$-plane, and $\hat{n}$ is the lower normal.

The boundary is the circle $x^{2}+y^{2}=1$ in the plane $z=0$. A lower normal induces the clockwise orientation on the boundary.

F = Matrix([x*z**2, x**3, sy.cos(x*z)])

ccw = line_integral(F, cos(t), sin(t), 0, t, 0, 2*pi)   # counterclockwise
display(ccw)                                            # 3*pi/4

display(-ccw)                                           # -3*pi/4 (clockwise)

$\iint_{S}\left(\nabla\times\mathbf{F}\right)\cdot\hat{n}\,dS=-\frac{3\pi}{4}$

note

The counterclockwise integral is $+\frac{3\pi}{4}$, but the lower normal induces the clockwise orientation, so the answer is $-\frac{3\pi}{4}$. The downloadable solution PDF still reports $+\frac{3\pi}{4}$, having missed the orientation flip.

Question 7

$\mathbf{F}=\left(4e^{x^{2}}-y\right)\mathbf{i}+\left(16\sin\left(y^{2}\right)+3x\right)\mathbf{j}+\left(4y-2x-e^{z}\right)\mathbf{k}$, and $C$ is the intersection of the cylinder $x^{2}+y^{2}=16$ with the plane $z=2x+4y$, counterclockwise when viewed from above.

F = Matrix([4*exp(x**2) - y, 16*sin(y**2) + 3*x, 4*y - 2*x - exp(z)])
cF = curl(F)
display(simplify(cF))                                   # <4, 2, 4>

# the plane z = 2x + 4y gives N dS = < -2, -4, 1 > dA
N = Matrix([-2, -4, 1])
display(simplify(cF.dot(N)))                            # -12

I = -12 * (pi * 4**2)                                   # area of the disc of radius 4
display(I)                                              # -192*pi

$\int_{C}\mathbf{F}\cdot d\mathbf{r}=\iint_{R}(-12)\,dA=-12\times16\pi=-192\pi$

note

Differentiating the field literally gives $\operatorname{curl}\mathbf{F}=\langle4,2,4\rangle$, so $\operatorname{curl}\mathbf{F}\cdot\langle-2,-4,1\rangle=-8-8+4=-12$ and the answer is $-192\pi$. The downloadable solution PDF instead reports the curl as $\langle-4,-2,4\rangle$, giving $+20$ and $320\pi$.

Question 8

$\mathbf{F}(x,y,z)=\langle xy,2z,3y\rangle$ along $C$, the intersection of the cylinder $x^{2}+y^{2}=9$ with the plane $x+z=5$.

Parametrise the enclosed surface by $\mathbf{r}(u,v)=(u\cos v,\ u\sin v,\ 5-u\cos v)$ with $0\le u\le3$, $0\le v\le2\pi$.

rv = Matrix([u*cos(v), u*sin(v), 5 - u*cos(v)])
cross = rv.diff(u).cross(rv.diff(v))
display(simplify(cross))                                # <u, 0, u>

F = Matrix([x*y, 2*z, 3*y])
display(simplify(curl(F)))                              # <1, 0, -x>

integrand = simplify(curl(F).subs({x: u*cos(v), y: u*sin(v), z: 5 - u*cos(v)}).dot(cross))
display(simplify(integrand))                            # u - u**2*cos(v)

I = integrate(integrand, (v, 0, 2*pi), (u, 0, 3))
display(I)                                              # 9*pi

$\int_{C}\mathbf{F}\cdot d\mathbf{r}=9\pi$

note

Differentiating the field literally gives $\operatorname{curl}\mathbf{F}=\langle1,0,-x\rangle$. The downloadable solution PDF writes $\langle1,0,x\rangle$, dropping the minus sign. The $\cos v$ term integrates to zero over a full period, so the answer $9\pi$ is unaffected by that sign error.

Question 9

$\iint\left(\nabla\times\mathbf{F}\right)\cdot\mathbf{n}\,dS$ where $\mathbf{F}(x,y,z)=\langle yz,xz,xy\rangle$ and $S$ is the part of $x^{2}+y^{2}+z^{2}=4$ inside $x^{2}+y^{2}=1$ and above the $xy$-plane.

The boundary is the circle of radius $1$ at height $z=\sqrt{3}$.

F = Matrix([y*z, x*z, x*y])
display(simplify(curl(F)))                              # <0, 0, 0> -> conservative field

I = line_integral(F, cos(t), sin(t), sqrt(3), t, 0, 2*pi)
display(simplify(I))                                    # 0

$\iint_{S}\left(\nabla\times\mathbf{F}\right)\cdot\mathbf{n}\,dS=0$

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Extra Learning Resources

Key Concepts to Master

1. Stokes' theorem: the flux of the curl through $S$ equals the circulation around $\partial S$
2. The surface is your choice: any surface with the same boundary gives the same flux
3. Orientation and the right-hand rule: with the normal up, the boundary is traversed counterclockwise
4. Lower normal, clockwise boundary: reversing the normal reverses the direction of travel and flips the sign
5. A zero curl means a conservative field, so the circulation around any closed loop is zero

SymPy Resources

• SymPy cross — for $\mathbf{r}_{u}\times\mathbf{r}_{v}$
• SymPy subs — substituting the surface or the parametrisation
• SymPy integrate — the resulting single or double integral

Common Pitfalls

• Sign of the curl: differentiate each component literally; a dropped minus sign changes the answer (Q7)
• Orientation: always check whether the given normal induces a clockwise or counterclockwise boundary
• A surface integral over a disc is often just a constant times the area — recognise it before integrating
• If $\operatorname{curl}\mathbf{F}=\mathbf{0}$ the answer is $0$ without any integration