Tutorial 12

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Line integrals $\int_{C}\mathbf{A}\cdot d\mathbf{r}$ along parametric curves
• Path dependence: the same field gives different answers along different paths
• Green's theorem: $\oint_{C}M\,dx+N\,dy=\iint_{R}\left(N_{x}-M_{y}\right)dA$
• Work done on a particle in a force field

Introduction

The line integral of a field $\mathbf{F}=(F_{1},F_{2},F_{3})$ along a curve $\mathbf{r}(t)$ is

$$\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{t_{0}}^{t_{1}} \left(F_{1}\frac{dx}{dt}+F_{2}\frac{dy}{dt}+F_{3}\frac{dz}{dt}\right)dt$$

so we can build a reusable helper that takes the field and the parametrisation:

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

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

$\mathbf{A}=\left(3x^{2}+6y\right)\hat{i}-14yz\,\hat{j}+20xz^{2}\,\hat{k}$ from $(0,0,0)$ to $(1,1,1)$.

a. Along $x=t$, $y=t^{2}$, $z=t^{3}$.

A = Matrix([3*x**2 + 6*y, -14*y*z, 20*x*z**2])
display(line_integral(A, t, t**2, t**3, t, 0, 1))        # 5

b. Along the straight segments $(0,0,0)\to(1,0,0)\to(1,1,0)\to(1,1,1)$.

seg1 = line_integral(A, t, 0, 0, t, 0, 1)                # (0,0,0) -> (1,0,0)
seg2 = line_integral(A, 1, t, 0, t, 0, 1)                # (1,0,0) -> (1,1,0)
seg3 = line_integral(A, 1, 1, t, t, 0, 1)                # (1,1,0) -> (1,1,1)
display(seg1, seg2, seg3, seg1 + seg2 + seg3)            # 1, 0, 20/3, 23/3

c. Along the straight line joining $(0,0,0)$ and $(1,1,1)$.

display(line_integral(A, t, t, t, t, 0, 1))              # 13/3

The three answers $5$, $\frac{23}{3}$ and $\frac{13}{3}$ all differ, which shows the field is path dependent (not conservative).

Question 2

Work done by $\mathbf{F}=3xy\,\hat{i}-5z\,\hat{j}+10x\,\hat{k}$ along $x=t^{2}+1$, $y=2t^{2}$, $z=t^{3}$ from $t=1$ to $t=2$.

F = Matrix([3*x*y, -5*z, 10*x])
W = line_integral(F, t**2 + 1, 2*t**2, t**3, t, 1, 2)
display(W)                                               # 303

Question 3

Work done counterclockwise once around the circle of radius $3$ centred at the origin for $\mathbf{F}=(2x-y+z)\hat{i}+\left(x+y-z^{2}\right)\hat{j}+(3x-2y+4z)\hat{k}$.

On the circle $z=0$, and the counterclockwise parametrisation is $x=3\cos t$, $y=3\sin t$, $0\le t\le2\pi$.

F = Matrix([2*x - y + z, x + y - z**2, 3*x - 2*y + 4*z])
display(line_integral(F, 3*cos(t), 3*sin(t), 0, t, 0, 2*pi))     # 18*pi

The positive (counterclockwise) direction gives $18\pi$; traversing the circle clockwise would give $-18\pi$.

Question 4

$\displaystyle\oint_{C}\left(xy+y^{2}\right)dx+x^{2}\,dy$ around the region bounded by $y=x$ and $y=x^{2}$, traversed positively.

M, N = x*y + y**2, x**2
display(simplify(diff(N, x) - diff(M, y)))               # x - 2y

I = integrate(x - 2*y, (y, x**2, x), (x, 0, 1))
display(I)                                               # -1/20

By Green's theorem the line integral equals $\iint_{R}(x-2y)\,dA=-\dfrac{1}{20}$, which matches the direct computation in the tutorial ($\frac{19}{20}-1$).

Question 5

$\displaystyle\oint_{C}(y-\sin x)\,dx+\cos x\,dy$ around the triangle with vertices $(0,0)$, $\left(\frac{\pi}{2},0\right)$ and $\left(\frac{\pi}{2},1\right)$.

The hypotenuse joins $(0,0)$ to $\left(\frac{\pi}{2},1\right)$, so $y=\frac{2x}{\pi}$.

M, N = y - sin(x), cos(x)
display(simplify(diff(N, x) - diff(M, y)))               # -sin(x) - 1

I = integrate(-sin(x) - 1, (y, 0, 2*x/pi), (x, 0, pi/2))
display(simplify(I))                                     # -pi/4 - 2/pi

Question 6

$\displaystyle\oint_{C}-x^{2}y\,dx+xy^{2}\,dy$ around the circle of radius $2$ centred on the origin.

M, N = -x**2*y, x*y**2
display(simplify(diff(N, x) - diff(M, y)))               # x**2 + y**2

I = integrate((r**2)*r, (r, 0, 2), (theta, 0, 2*pi))
display(I)                                               # 8*pi

The integrand $x^{2}+y^{2}=r^{2}$ is exactly the reason polar coordinates are the natural choice here.

Question 7

Line integral of $\mathbf{F}(x,y)=\left\langle x^{3},4x\right\rangle$ along the closed path $C$.

Green's theorem gives $\iint_{D}\left(Q_{x}-P_{y}\right)dA=\iint_{D}4\,dA=4\times(\text{area of }D)$. The region enclosed by $C$ plus the closing segment $L$ has an area of $4$ unit squares, and $\int_{L}\mathbf{F}\cdot d\mathbf{r}=-4$.

P, Q = x**3, 4*x
display(simplify(diff(Q, x) - diff(P, y)))               # 4

area = 4
integral_D = 4*area
integral_L = -4
display(integral_D, integral_D - integral_L)             # 16, 20

So $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}=16-(-4)=20$.

Question 8

A particle starts at $(-2,0)$, moves along the $x$-axis to $(2,0)$, then back along the upper half of $x^{2}+y^{2}=4$. Find the work done by $\mathbf{F}(x,y)=\left\langle x,\ x^{3}+3xy^{2}\right\rangle$.

The path is closed and traversed counterclockwise, so Green's theorem applies with $P=x$ and $Q=x^{3}+3xy^{2}$.

P, Q = x, x**3 + 3*x*y**2
display(simplify(diff(Q, x) - diff(P, y)))               # 3*x**2 + 3*y**2

# over the upper half-disc of radius 2, using x^2 + y^2 = r^2
W = integrate(3*r**2 * r, (r, 0, 2), (theta, 0, pi))
display(W, sy.N(W, 6))                                   # 12*pi = 37.6991

$$W=\iint 3\left(x^{2}+y^{2}\right)dA =\int_{0}^{\pi}\int_{0}^{2}3r^{3}\,dr\,d\theta=12\pi\approx37.70$$

note

The downloadable solution PDF still works from $P=\sin(x^{3})$ and $Q=2y e^{x^{2}}$, which is not the field in the question and leads to a nonsensical $e^{4}-5$. The field is $\mathbf F=\langle x,\;x^{3}+3xy^{2}\rangle$, so $Q_x=3x^{2}+3y^{2}$ and $P_y=0$, giving $12\pi\approx37.70$.

Question 9

$\mathbf{F}(x,y)=\langle\sin x,\ \cos y\rangle$ along the top half of $x^{2}+y^{2}=1$ traversed counterclockwise from $(1,0)$ to $(-1,0)$, followed by the line segment from $(-1,0)$ to $(-2,3)$.

F = Matrix([sin(x), cos(y), 0])

# C1: the upper semicircle,  x = cos t,  y = sin t,  0 <= t <= pi
C1 = line_integral(F, cos(t), sin(t), 0, t, 0, pi)
display(simplify(C1))                                    # 0

# C2: the segment,  x = -1 - t,  y = 3t,  0 <= t <= 1
C2 = line_integral(F, -1 - t, 3*t, 0, t, 0, 1)
display(simplify(C2))                                    # -cos(2) + cos(1) + sin(3)

display(simplify(C1 + C2))                               # cos(1) - cos(2) + sin(3)

$\int_{C}\sin x\,dx+\cos y\,dy=\cos 1-\cos 2+\sin 3$

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Extra Learning Resources

Key Concepts to Master

1. Parametrise, then substitute: replace $x$, $y$, $z$ and $dx$, $dy$, $dz$ in terms of $t$
2. Path dependence: if the answers differ for different paths, the field is not conservative
3. Green's theorem: $\oint M\,dx+N\,dy=\iint\left(N_{x}-M_{y}\right)dA$, for a positively oriented simple closed curve
4. Positive orientation: counterclockwise for the outer boundary
5. Closing a path: to use Green's theorem on an open curve, add a straight segment and subtract its integral

SymPy Resources

• SymPy integrate — definite integrals in $t$
• SymPy subs — substituting a parametrisation into the field
• SymPy diff — partial derivatives for the Green's theorem integrand

Common Pitfalls

• A line integral is a scalar, not a vector — check you have a dot product, not a cross product
• In a subs dictionary, substitute $x$ and $y$ and $z$ in one call, otherwise SymPy substitutes sequentially and you get the wrong field
• Watch the orientation: reversing the direction of travel flips the sign of the integral
• For Green's theorem, the curve must be closed and positively oriented — add a closing segment if it is not