Tutorial 10

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Double integrals over rectangles and general regions
• Order of integration and why it matters for the difficulty of the calculation
• Triple integrals over boxes and solids
• Mass and centre of mass of a lamina and of a solid

Introduction

SymPy evaluates iterated integrals directly: write integrate(f, (y, y_low, y_high), (x, x_low, x_high)) and it works from the innermost variable outwards, exactly as you would by hand.

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

Question 1

Evaluate the iterated integrals.

a. $\displaystyle\int_{1}^{2}\int_{2}^{3}(2x-y)\,dy\,dx$

display(integrate(2*x - y, (y, 2, 3), (x, 1, 2)))        # 1/2

b. $\displaystyle\int_{1}^{5}\int_{-1}^{2}(x-5y)\,dx\,dy$

display(integrate(x - 5*y, (x, -1, 2), (y, 1, 5)))       # -174

c. $\displaystyle\int_{1}^{3}\int_{2}^{5}(2x-3y)\,dx\,dy$

display(integrate(2*x - 3*y, (x, 2, 5), (y, 1, 3)))      # 6

Question 2

Evaluate over the given region $D$.

a. $\displaystyle\iint_{D}\left(1-\frac{1}{2}x^{2}-\frac{1}{2}y^{2}\right)dA$, $D$ is the unit square $0\le x\le1$, $0\le y\le1$.

display(integrate(1 - Rational(1, 2)*x**2 - Rational(1, 2)*y**2,
                  (y, 0, 1), (x, 0, 1)))                 # 2/3

b. $\displaystyle\iint_{D}\left(4xy-y^{3}\right)dA$, where $D$ is bounded by $y=\sqrt{x}$ and $y=x^{3}$.

The curves meet at $x=0$ and $x=1$, and on that interval $x^{3}\le y\le\sqrt{x}$.

# solve over the reals only, so the complex roots are not returned
display(list(sy.solveset(sqrt(x) - x**3, x, domain=S.Reals)))     # [0, 1]
display(integrate(4*x*y - y**3, (y, x**3, sqrt(x)), (x, 0, 1)))   # 55/156

Question 3

$\displaystyle\iint_{D}42y^{2}-12x\,dA$ where $D=\{(x,y)\mid 0\le x\le4,\ (x-2)^{2}\le y\le6\}$.

display(integrate(42*y**2 - 12*x, (y, (x - 2)**2, 6), (x, 0, 4)))  # 11136

Question 4

$\displaystyle\iint_{R}12x-18y\,dA$ over $R=[-1,4]\times[2,3]$, both ways round.

display(integrate(12*x - 18*y, (x, -1, 4), (y, 2, 3)))   # -135
display(integrate(12*x - 18*y, (y, 2, 3), (x, -1, 4)))   # -135

Because the region is a rectangle and the integrand is a sum of separate functions of $x$ and $y$, both orders give the same answer.

Question 5

Compute the double integrals over the indicated rectangles.

a. $\displaystyle\iint_{R}6y\sqrt{x}-2y^{3}\,dA$, $R=[1,4]\times[0,3]$

display(integrate(6*y*sqrt(x) - 2*y**3, (x, 1, 4), (y, 0, 3)))    # 9/2

b. $\displaystyle\iint_{R}y\,e^{\,y^{2}-4x}\,dA$, $R=[0,2]\times[0,\sqrt{8}]$

R5 = integrate(y*exp(y**2 - 4*x), (y, 0, sqrt(8)), (x, 0, 2))
display(expand(R5), sy.N(R5, 8))                         # -1/4 + e^-8/8 + e^8/8 = 372.36979

$\iint_{R}y\,e^{\,y^{2}-4x}\,dA=\frac{1}{8}\left(e^{8}+e^{-8}-2\right)\approx372.37$

note

The printed solution for 5(b) is a copy-paste of 5(a)'s working. The value $372.3698$ is correct, but the integral shown above it is not the one being evaluated.

Question 6

Evaluate the triple integrals.

a. $\displaystyle\int_{-3}^{1}\int_{-1}^{2}\int_{0}^{1}6xyz\,dy\,dx\,dz$

display(integrate(6*x*y*z, (y, 0, 1), (x, -1, 2), (z, -3, 1)))   # -18

b. $\displaystyle\int_{0}^{4}\int_{-2}^{-1}\int_{1}^{2}xy\,dx\,dy\,dz$

display(integrate(x*y, (x, 1, 2), (y, -2, -1), (z, 0, 4)))       # -9

Question 7

Evaluate the triple integrals over the given boxes.

a. $\displaystyle\iiint_{B}xyz^{2}\,dV$, $0\le x\le1$, $-1\le y\le2$, $0\le z\le3$

display(integrate(x*y*z**2, (x, 0, 1), (y, -1, 2), (z, 0, 3)))   # 27/4

b. $\displaystyle\iiint_{B}\left(4x^{2}y-z^{3}\right)dz\,dy\,dx$, $2\le x\le3$, $-1\le y\le4$, $0\le z\le1$

B7 = integrate(4*x**2*y - z**3, (z, 0, 1), (y, -1, 4), (x, 2, 3))
display(B7)                                                      # 755/4
display(sy.N(B7, 8))                                             # 188.75

$\iiint_{B}\left(4x^{2}y-z^{3}\right)dV=\frac{755}{4}=188.75$

note

The printed solution flips the sign after the $z$-integration, writing $\frac{1}{4}-4x^{2}y$ instead of $4x^{2}y-\frac{1}{4}$, and so obtains $-755/4$. Substituting the $z$-limits gives $\left[4x^{2}yz-\frac{z^{4}}{4}\right]_{0}^{1}=4x^{2}y-\frac{1}{4}$. Note this integrand is not everywhere positive: it is negative for $y<\frac{1}{16x^{2}}$ (for instance $-65/4$ at $x=2,y=-1$), so the printed sign flip cannot be argued away by a positivity claim. The correct answer is $+755/4$.

Question 8

Integrate $f(x,y,z)=xy$ over the volume enclosed by $z=x+y$, $z=0$, $y=x^{2}$ and $x=y^{2}$.

The curves $y=x^{2}$ and $x=y^{2}$ meet at $(0,0)$ and $(1,1)$; integrating $z$ first gives the limits $0\le z\le x+y$.

V8 = integrate(x*y, (z, 0, x + y), (y, x**2, sqrt(x)), (x, 0, 1))
display(V8)                                                      # 3/28

Question 9

Mass of the unit-square lamina with density $\rho(x,y)=x+y+2\ \text{g/cm}^{2}$.

m = integrate(x + y + 2, (x, 0, 1), (y, 0, 1))
display(m)                                                       # 3 g

Question 10

Mass and centre of mass of the tetrahedron $Q$ bounded by $x+2y+3z=6$ and the coordinate planes, with density $\rho(x,y,z)=x^{2}yz$.

The tetrahedron meets the axes at $(6,0,0)$, $(0,3,0)$ and $(0,0,2)$; projecting onto the $xy$-plane gives $0\le x\le6$, $0\le y\le\frac{1}{2}(6-x)$ and $0\le z\le\frac{1}{3}(6-x-2y)$.

ztop = Rational(1, 3)*(6 - x - 2*y)
ytop = Rational(1, 2)*(6 - x)

m    = integrate(x**2*y*z,     (z, 0, ztop), (y, 0, ytop), (x, 0, 6))
Mxy  = integrate(x**2*y*z**2,  (z, 0, ztop), (y, 0, ytop), (x, 0, 6))   # about xy-plane
Mxz  = integrate(x**2*y**2*z,  (z, 0, ztop), (y, 0, ytop), (x, 0, 6))   # about xz-plane
Myz  = integrate(x**3*y*z,     (z, 0, ztop), (y, 0, ytop), (x, 0, 6))   # about yz-plane

display(m, Mxy, Mxz, Myz)          # 108/35, 54/35, 81/35, 243/35
display(simplify(Myz/m), simplify(Mxz/m), simplify(Mxy/m))     # 9/4, 3/4, 1/2

$$m=\frac{108}{35},\qquad (\bar{x},\bar{y},\bar{z})=\left(\frac{9}{4},\frac{3}{4},\frac{1}{2}\right)=(2.25,\ 0.75,\ 0.5)$$

note

The printed solution labels the last coordinate $\bar{y}=\dfrac{M_{xy}}{m}$. By definition $\bar{z}=M_{xy}/m$, $\bar{y}=M_{xz}/m$ and $\bar{x}=M_{yz}/m$; the numbers are right, only the label is wrong.

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Extra Learning Resources

Key Concepts to Master

1. Fubini's theorem: for a rectangle the order of integration can be swapped freely
2. General regions: find the outer limits from the projection, then the inner limits from the bounding curves
3. Order matters for effort: some integrands have no elementary antiderivative in one order but are easy in the other
4. Triple integrals: integrate the innermost variable first, then work outwards
5. Centre of mass: $\bar{x}=M_{yz}/m$, $\bar{y}=M_{xz}/m$, $\bar{z}=M_{xy}/m$

SymPy Resources

• SymPy integrate — nested definite integrals
• SymPy Rational — exact fractions for limits such as $\frac{1}{3}(6-x-2y)$
• SymPy solve — intersection points of the bounding curves

Common Pitfalls

• Use Rational(1, 2), not 1/2, inside an integrand or SymPy will switch to floats
• The inner limits may depend on the outer variables — get the nesting order right
• Watch the sign when the integrand changes sign over the region (as in 1(b), where the answer is $-174$)
• Double-check which moment belongs to which coordinate: $\bar{z}$ pairs with $M_{xy}$