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MZB127 Chap.3 Multiple Integrals and Region Setup

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Chapter 3 of 11 · MZB127

Multiple Integrals and Region Setup

Double integrals accumulate height, density or another quantity across a planar region. Their limits are produced by a family of vertical or horizontal slices, so geometry must be settled before antiderivatives begin. This chapter builds regions from inequalities, compares both integration orders, explains when a split is necessary, and checks area, volume and mass results through support, units and bounding values.

An iterated integral is therefore a geometric description as well as a calculation. The outer variable chooses a slice, while the inner limits describe where that slice enters and leaves the region. Reversing integration order requires rebuilding the inequalities rather than swapping symbols. A single set of vertical slices may cover a region cleanly even when horizontal slices require two pieces, or the opposite may occur.

The integrand determines what is accumulated: one gives area, a height gives volume, and a surface density gives mass. Units combine the integrand units with the two differential lengths. Positive integrands over finite regions also support strong magnitude checks because the result must lie between minimum times area and maximum times area when those bounds apply.

These connections make the sketch, limits, integration and final interpretation mutually checking parts of one solution.

In this chapter

What this chapter covers

  • 01

    Planar regions as inequalities

  • 02

    Vertical and horizontal slice descriptions

  • 03

    Iterated integrals and differential order

  • 04

    Changing integration order

  • 05

    Area, volume and density applications

Worked example · free

Integrate over a triangular region

Q [4 marks]. The mark allocation shown here is not an official university assessment scheme. Evaluate the integral of x+y over zero to two in x and x to two in y.
The shaded triangle realises zero to two for x and x to two for y, so the inner limits follow a vertical slice.The shaded triangle realises zero to two for x and x to two for y, so the inner limits follow a vertical slice.x≤y≤20≤x≤2y=xxy
The shaded triangle realises zero to two for x and x to two for y, so the inner limits follow a vertical slice.
  • 1Identify the vertical slice and its boundaries.
  • 1Integrate in y with x fixed.
  • 1Integrate the resulting expression in x.
  • 1Check the positive volume against base area and height.
The iterated integral equals four cubic units. Reversing the order to y from zero to two and x from zero to y produces the same value.
Sia tip — Test one interior point against both descriptions after reversing the order; matching points are stronger evidence than similar-looking bounds.
Glossary

Key terms

Iterated integral
A multiple integral evaluated as nested single integrals in a stated order.
Inner integral
The first integration, whose limits describe one slice while the outer variable is fixed.
Integration region
The planar set over which a double integral accumulates its integrand.
Order reversal
A geometric rewrite of the same region using the other slice orientation.
FAQ

Multiple Integrals and Region Setup FAQ

How do inner limits relate to a sketch?

They give the entry and exit points of one slice. If y is integrated first, those limits are the bottom and top boundaries as functions of the fixed x coordinate.

When must a region be split?

Split when one pair of inner boundary functions cannot describe every slice across the full outer range. The change usually occurs at an intersection where the active left, right, lower or upper boundary changes.

Why can a positive volume integral be checked by bounds?

If the surface height stays between known minimum and maximum values, volume must lie between each height multiplied by the base area. A result outside that range signals a region or algebra error.

How can reversed bounds be detected before integration?

Test slices at the two outer endpoints and one interior value. Each inner lower bound should lie below its upper bound, or each left boundary should lie left of its right boundary. If that ordering changes, the region probably needs a split or the boundary functions have been assigned in the wrong order.

Study strategy

Exam move

For every problem, sketch and label the boundary equations, choose a slice and write the region as inequalities before writing an integral. At extreme outer values, check that the inner interval matches the visible edge. Integrate inside first and keep brackets around bound substitutions. For density or height problems, carry units and compare the answer with minimum-times-area and maximum-times-area bounds.

Train region setup separately from integration. Take a collection of shaded regions and write both vertical-slice and horizontal-slice inequalities without evaluating anything. Mark every boundary intersection and use those coordinates to determine the outer interval.

At the left, middle and right of a proposed vertical description, ask which curve is lower and which is upper; for horizontal slices, ask which is left and which is right. If the identity changes, locate the change point and split the integral. Next convert inequalities back into a sketch to test whether the description reproduces the intended region.

During evaluation, keep the inner antiderivative and its bound substitution on separate lines, then simplify before the outer integration. For area checks, temporarily replace the integrand by one. For volume or mass, write the compound units before calculating and estimate a plausible range from the region area and integrand bounds.

Practise reversing order on the same problem and confirm that both forms describe the same set even if only one is evaluated. In timed work, never let familiar antiderivative rules lure you into integrating before the geometry is fixed. A clean labelled sketch, explicit slice inequalities and one magnitude check usually expose reversed bounds, missing pieces or an implausible sign before those errors reach the final line.

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