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MTH2010 Chap.7 Polar Coordinates and Applications of Double Integrals

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Chapter 7 of 12 · MTH2010

Polar Coordinates and Applications of Double Integrals

Polar coordinates make discs, annuli and sectors as easy to integrate over as rectangles. With x = r cos(θ) and y = r sin(θ), a region bounded by circles about the origin and rays from it becomes a rectangle in the r-theta plane. The cost is the area element: a small polar box has sides of length dr and r dθ, so dA becomes r dr dθ, and leaving out that factor r is the most common error in the topic.

The Week 7 slides give two reasons to switch to polar form, either of which is worth quoting in a written answer: polar coordinates may simplify the region of integration, or they may simplify the integrand, as for any function of x² + y². The second half of the week applies double integrals.

A thin plate with density ρ has mass equal to the integral of ρ over its region, and a charge density integrates to total charge in the same way. The moments about the axes integrate y ρ and x rho, and dividing each moment by the mass gives the centre of mass, with symmetry often making one coordinate obvious.

Probability is the other application: a joint density is non-negative and integrates to 1 over the plane, the probability of an event is the integral of the density over the corresponding region, and expected values integrate x or y against the density. For independent variables the joint density is the product of the two separate densities.

In this chapter

What this chapter covers

  • 01

    Polar substitution: x = r cos(θ), y = r sin(θ)

  • 02

    Area element: dA = r dr dθ and why the r appears

  • 03

    Polar rectangles: exact area and separation into products

  • 04

    When to use polar: simpler region or simpler integrand

  • 05

    Converting boundaries: circles through the origin and rays

  • 06

    Mass and charge: integrating a density over the region

  • 07

    Moments and centre of mass: symmetry first

  • 08

    Joint densities: total probability one, events as regions, independence

Worked example · free

Gaussian integrand over a disc

Q [4 marks]. Evaluate the double integral of e^(-(x² + y²)) over the disc x² + y² ≤ 4. The mark allocation supports this practice solution; it is not an official university marking scheme.
A quarter annulus as a polar rectangle.A quarter annulus as a polar rectangle. 1 ≤ r ≤ 2 and 0 ≤ θ ≤ π/2; the shaded box has area close to r dr dθ.dA = r dr dθr = 1r = 2
A quarter annulus as a polar rectangle. 1 ≤ r ≤ 2 and 0 ≤ θ ≤ π/2; the shaded box has area close to r dr dθ.
  • 1In polar form the disc is 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2 π, and the integrand is e^(-r²) times r.
  • 1Separate: the integral equals 2 π times the integral of r e^(-r²) from 0 to 2.
  • 1With u = r² the radial integral is (1/2)(1 - e^(-4)).
  • 1Answer: π(1 - e^(-4)), about 3.084.
The integral is π(1 - e^(-4)), approximately 3.084. In Cartesian form the inner integral has no elementary antiderivative, so polar coordinates are the natural choice.
Sia tip — The factor r from dA is what makes the substitution u = r² work. If the radial integral looks impossible, check that the r has not been dropped.
Glossary

Key terms

Polar rectangle
A region described by constant bounds on the radius and on the angle.
Area element
The factor that converts a product of small coordinate changes into a small area, equal to r dr dθ in polar form.
Mass density
Mass per unit area of a thin plate, whose integral over the plate gives total mass.
Centre of mass
The point whose coordinates are the moments of a lamina divided by its mass.
Joint density
A non-negative function of two variables that integrates to one and gives probabilities as integrals over regions.
FAQ

Polar Coordinates and Applications of Double Integrals FAQ

Why does polar integration need the extra factor r?

A small polar box between radii r and r + dr and angles θ and θ + dθ has sides dr and r dθ, so its area is r dr dθ. Without the r, areas and every integral built on them come out wrong.

When should I switch to polar coordinates?

Switch when the region is bounded by circles centred at the origin and rays from it, or when the integrand depends on x² + y². Either reason alone usually makes the integral easier.

How do I find a centre of mass quickly?

Look for symmetry first, because a symmetric region with symmetric density has its centre on the axis of symmetry. Then compute the mass and only the moment you still need.

How do I check that a function is a joint probability density?

Confirm it is never negative and that its integral over the whole plane, or over its stated support, equals one. A seminar poll used a candidate that integrated to one half and so failed.

Study strategy

Exam move

Before integrating in polar form, write three conversions on separate lines: the region as bounds on r and theta, the integrand in r and theta, and dA = r dr dθ. When the bounds are constants and the integrand factorises, separate into a product of single integrals.

For applications, memorise the three formulas for mass and the two moments as one block, and practise a variable-density example where the centre moves towards the denser part. For probability, practise turning a word statement such as X < Y into a region before writing any integral.

Working through Polar Coordinates and Applications of Double Integrals in MTH2010? Sia is AskSia’s AI Mathematics tutor — ask any MTH2010 Polar Coordinates and Applications of Double Integrals question and get a clear, step-by-step explanation grounded in how MTH2010 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.

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