Tutorial 9

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Area between curves
• Work: pumping a tank, Hooke's law
• Motion problems: velocity, acceleration, distance travelled
• Arc length
• Centroid of a plane area
• Hydrostatic force

Introduction

integrate(f, (x, a, b)) computes a definite integral, and definite integrals can be nested just like the iterated integrals in the tutorial. Definite integrals are the easiest to check, because there is no arbitrary constant to worry about.

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

Question 1

Find the area bounded by the curves.

i. $y=x^{4}-x^{2}$ and $y=x^{2}$, to the right of the $y$-axis.

# intersection: x^4 - x^2 = x^2  ->  x^2 (x^2 - 2) = 0
display(sy.solve(x**4 - x**2 - x**2, x))         # [0, -sqrt(2), sqrt(2)]

# on (0, sqrt(2)) the line y = x^2 is above y = x^4 - x^2, so the integrand is
# (upper - lower) = x^2 - (x^4 - x^2) = 2x^2 - x^4
A = integrate(2*x**2 - x**4, (x, 0, sqrt(2)))
display(A, sy.N(A, 6))                           # 8*sqrt(2)/15 = 0.754247

$A=\int_{0}^{\sqrt{2}}\left(2x^{2}-x^{4}\right)dx=\frac{2^{7/2}}{15}\approx0.754$

note

The printed solution integrates $x^{4}-x^{2}-x^{2}$ (upper minus lower in the wrong order) and quotes $\frac{2^{7/2}}{15}-2$, which is negative. An area must be positive; the correct integrand is $2x^{2}-x^{4}$.

ii. $x=y^{3}$ and $x=y^{2}$.

display(sy.solve(y**3 - y**2, y))                # [0, 1]

# on (0, 1) the curve x = y^2 is to the right of x = y^3
A = integrate(y**2 - y**3, (y, 0, 1))
display(A)                                       # 1/12

$A=\int_{0}^{1}\left(y^{2}-y^{3}\right)dy=\frac{1}{12}$

iii. $y=\cos\left(\dfrac{\pi x}{2}\right)$ and $y=1-x^{2}$, in the first quadrant.

A = integrate(1 - x**2 - cos(pi*x/2), (x, 0, 1))
display(A, sy.N(A, 6))                           # 2/3 - 2/pi = 0.030047

$A=\int_{0}^{1}\left(1-x^{2}-\cos\frac{\pi x}{2}\right)dx=\frac{2}{3}-\frac{2}{\pi}\approx0.03$

Question 2

Area bounded by the parabola $x=8+2y-y^{2}$, the $y$-axis, $y=-1$ and $y=3$.

display(sy.factor(8 + 2*y - y**2))               # -(y - 4)*(y + 2)

A = integrate(8 + 2*y - y**2, (y, -1, 3))
display(A)                                       # 92/3

$A=\int_{-1}^{3}\left(8+2y-y^{2}\right)dy=\frac{92}{3}$

note

The printed solution says the parabola "cuts the $y$-axis at $y=-4$, $y=-2$". The factorisation is $(4-y)(2+y)$, so the roots are $y=4$ and $y=-2$.

Question 3

A cylindrical tank of radius $4\ \text{m}$ and height $10\ \text{m}$ is filled to a depth of $8\ \text{m}$. Find the work needed to pump the water out of the top.

Each horizontal layer at depth $x$ (measured downward from the top) has volume $16\pi\,\Delta x$ and must be lifted a distance $x$, so the total work is $\int_{2}^{10}9800\cdot16\pi\,x\,dx$.

W = 9800 * 16 * pi * integrate(x, (x, 2, 10))
display(W, sy.N(W, 8))                           # 7526400*pi = 23644883

$W=7\,526\,400\pi\approx2.36\times10^{7}\ \text{J}$

Question 4

An object moves with acceleration $a(t)=-\cos t$, $s(0)=1$ and $v(0)=0$. Find the maximum distance from zero and the maximum speed.

tau = Symbol('tau', positive=True)               # dummy variable of integration

v = integrate(-cos(tau), (tau, 0, t)) + 0        # apply v(0) = 0
display(v)                                       # -sin(t)

s = 1 + integrate(v.subs(t, tau), (tau, 0, t))   # apply s(0) = 1
display(simplify(s))                             # cos(t)

The position is $s(t)=\cos t$ and the speed is $|v(t)|=|\sin t|$. Both the maximum distance from the origin and the maximum speed are $1$.

note

The printed solution stops at $s=\cos t$ without answering the question. The object oscillates between $x=-1$ and $x=1$, so the maximum distance from zero is $1\ \text{m}$ and the maximum speed is $1\ \text{m/s}$.

Question 5

A particle's velocity is $v(t)=3.2t$ for $t\le5$, then $v(t)=16.0-1.5(t-5)$ for $5\le t\le11$, then constant at $7.0$ for $t>11$.

(a) The acceleration is the slope of each segment:

display(diff(3.2*t, t))                          # 3.2   for t <= 5
display(diff(16.0 - 1.5*(t - 5), t))             # -1.5  for 5 <= t <= 11
display(diff(7.0, t))                            # 0     for t > 11

So $a(t)=3.2\ \text{m/s}^{2}$ for $t\le5$, $a(t)=-1.5\ \text{m/s}^{2}$ for $5\le t\le11$, and $a(t)=0$ afterwards.

note

The printed solution gives $a(t)=+1.5\ \text{m/s}^{2}$ on the middle interval. The velocity is decreasing there ($v(11)=16-1.5(6)=7$), so the acceleration must be $-1.5\ \text{m/s}^{2}$.

(b) The position is obtained by integrating each segment, matching the value at the junction.

x1 = integrate(3.2*tau, (tau, 0, t))                     # x(0) = 0
display(x1.subs(t, 2))                                   # 6.4 m

x2 = 40 + integrate(16.0 - 1.5*(tau - 5), (tau, 5, t))   # x(5) = 1.6*(5^2) = 40
display(x2.subs(t, 7))                                   # 69.0 m

x3 = 109 + integrate(7.0, (tau, 11, t))                  # x(11) = 109 m
display(x3.subs(t, 12))                                  # 116.0 m

$x(2)=6.4\ \text{m},\qquad x(7)=69\ \text{m},\qquad x(12)=116\ \text{m}$

Question 6

A force of $50\ \text{N}$ stretches a spring from $2\ \text{m}$ to $2.5\ \text{m}$.

i. Spring constant.

k = 50 / Rational(1, 2)                          # F = k x with x = 0.5 m
display(k)                                       # 100 N/m

ii. Work needed to stretch it by $0.5\ \text{m}$.

W = integrate(100*x, (x, 0, Rational(1, 2)))
display(W, sy.N(W, 4))                           # 25/2 = 12.5 J

$W=\int_{0}^{0.5}100x\,dx=\frac{100(0.5)^{2}}{2}=12.5\ \text{J}$

note

The printed solution gives $W=100\cdot0.5^{2}/2=0.25\ \text{J}$, which is an arithmetic slip: $100\times0.25/2=12.5\ \text{J}$.

Question 7

Arc length of $y=x^{3/2}$ from $x=0$ to $x=5$.

y7 = x**Rational(3, 2)
dy = diff(y7, x)
display(dy)                                      # 3*sqrt(x)/2

L = integrate(sqrt(1 + dy**2), (x, 0, 5))
display(L, sy.N(L, 6))                           # 335/27 = 12.4074

$L=\int_{0}^{5}\sqrt{1+\frac{9}{4}x}\;dx=\frac{335}{27}$

Question 8

Arc length of the catenary $y=\dfrac{a}{2}\left(e^{x/a}+e^{-x/a}\right)$ from $x=0$ to $x=a$.

a = Symbol('a', positive=True)                   # declare a positive so the square root collapses
y8 = a/2*(exp(x/a) + exp(-x/a))
dy8 = simplify(diff(y8, x))
display(dy8)                                     # sinh(x/a)

# 1 + sinh^2(x/a) = cosh^2(x/a), so the integrand reduces to cosh(x/a)
integrand = simplify(sqrt(1 + dy8**2))
display(integrand)                               # sqrt(cosh(x/a)**2)  =  cosh(x/a)

L = integrate(integrand, (x, 0, a))
display(simplify(L))                             # a*sinh(1)

# a*sinh(1) and a/2*(e - e^-1) are the same number, but `simplify` alone leaves
# sinh(1) unexpanded, so rewrite it into exponentials before subtracting.
display(simplify((L - a/2*(exp(1) - exp(-1))).rewrite(exp)))    # 0

$L=\frac{a}{2}\left(e-e^{-1}\right)=a\sinh 1$

Question 9

Centroid of the area bounded by $y=2x-x^{2}$ and $y=3x^{2}-6x$.

f, g = 2*x - x**2, 3*x**2 - 6*x
display(sy.solve(f - g, x))                      # [0, 2]

A  = integrate(integrate(1, (y, g, f)), (x, 0, 2))
My = integrate(integrate(x, (y, g, f)), (x, 0, 2))
Mx = integrate(integrate(y, (y, g, f)), (x, 0, 2))

display(A, My, Mx)                               # 16/3, 16/3, -64/15
display(simplify(My/A), simplify(Mx/A))          # 1, -4/5

The centroid is $\left(1,-\dfrac{4}{5}\right)$.

Question 10

Hydrostatic force on a circular plate of radius $2$ whose top is $6\ \text{m}$ below the surface. Taking $y$ from the centre of the plate, the depth is $8-y$ and the strip width is $2\sqrt{4-y^{2}}$.

F = 9810 * integrate((8 - y) * 2 * sqrt(4 - y**2), (y, -2, 2))
display(simplify(F), sy.N(F, 8))                 # 313920*pi = 986208.77

$F=19620\int_{-2}^{2}(8-y)\sqrt{4-y^{2}}\,dy=313\,920\pi\approx9.86\times10^{5}\ \text{N}$

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Extra Learning Resources

Key Concepts to Master

1. Area between curves: $\int(\text{upper}-\text{lower})\,dx$, with the limits at the intersection points
2. Work: $\int F\,dx$, or $\int(\text{weight density})(\text{area})(\text{distance})\,dx$ for pumping problems
3. Hooke's law: $F=kx$, so $W=\int kx\,dx$
4. Arc length: $L=\int\sqrt{1+(y')^{2}}\,dx$
5. Centroid: $\bar{x}=M_{y}/A$, $\bar{y}=M_{x}/A$
6. Hydrostatic force: $F=\int \rho g\,(\text{depth})(\text{width})\,dy$

SymPy Resources

• SymPy integrate — definite and nested integrals
• SymPy solve — finding intersection points
• SymPy Rational — keep fractions exact instead of using floats

Common Pitfalls

• Always check which curve is on top before writing the integrand; a negative area is a sign that the order is reversed
• Areas, masses and lengths are positive — treat a negative result as a warning, not an answer
• For piecewise motion, match the constants of integration at every junction
• Use Rational(3, 2) rather than 3/2 when the exponent must stay exact