MTH2010 Chap.6 Double Integrals and Fubini's Theorem
Double Integrals and Fubini's Theorem
Week 6 defines the double integral and turns it into something computable. The integral of f over a region is the limit of Riemann sums built from small rectangles, and the slides note that it exists for continuous functions on rectangles and for continuous functions on closed bounded regions with reasonable boundaries.
When f is non-negative the integral is the volume between the graph and the plane, and integrating the constant 1 gives the area of the region. Fubini's theorem is the bridge to calculation. On a rectangle, the double integral of a continuous function equals either iterated integral, so the order of integration may be chosen freely.
For a general region the slides extend the function by zero to a surrounding rectangle and describe the region by inequalities. A type I region has constant limits in x and y between two curves; a type II region has constant limits in y and x between two curves.
A region of both types can be integrated in either order, which is how an awkward integral becomes possible: if the inner antiderivative is not elementary, sketch the region, describe it the other way, and integrate in the new order. Integrals are linear, preserve inequalities and add over regions with disjoint interiors, so a region that is neither type can be split into pieces that are.
The comparison property also gives upper and lower bounds without integrating, a task that appeared in the seminar polls. A homework example shows Fubini failing for a function that is not continuous.
What this chapter covers
- 01
Riemann sums: the definition of the double integral
- 02
Volume and area: the two geometric readings of the integral
- 03
Fubini on rectangles: either order for continuous integrands
- 04
Type I and type II regions: limits written as inequalities
- 05
Reversing the order: sketch, redescribe, re-integrate
- 06
Splitting regions: additivity over pieces with disjoint interiors
- 07
Bounds without integrating: minimum and maximum times area
- 08
Continuity hypothesis: the counterexample where the orders disagree
Worked example · free
Integrate over a parabolic region in type II form
- 1The curves meet where y² = 4, at y = -2 and y = 2, and the region is -2 ≤ y ≤ 2 with y² ≤ x ≤ 4.
- 1Inner integral: the integral of x from y² to 4 is (16 - y⁴)/2.
- 1Outer integral: (1/2)[16y - y⁵/5] from -2 to 2 = (1/2)(64 - 64/5).
- 1Simplify: (1/2)(256/5) = 128/5.
Key terms
- Riemann sum
- A sum of function values times small rectangle areas that approximates a double integral.
- Iterated integral
- Two single integrals performed in succession, the inner one with the outer variable held fixed.
- Type I region
- A plane region with constant x-limits and y between a lower and an upper curve.
Double Integrals and Fubini's Theorem FAQ
When can I change the order of integration?
When the integrand is continuous and the region can be described both ways, Fubini's theorem makes the two iterated integrals equal. Redraw the region from the given limits and rewrite them for the other order.
How do I know whether a region is type I or type II?
Ask whether every vertical line meets it in one interval between two curves, which makes it type I, or whether every horizontal line does, which makes it type II. Many regions are both.
Can a double integral be estimated without evaluating it?
Yes. If m ≤ f ≤ M on the region, the integral lies between m times the area and M times the area, a bound the seminar polls used repeatedly.
Why can the two iterated integrals disagree?
Fubini's theorem assumes continuity. The Week 6 homework gives a function that is unbounded at a corner of the square, and its two iterated integrals have opposite signs.
Exam move
Sketch every region before writing limits; most errors in this chapter are setup errors, not integration errors. Practise describing the same region both ways and integrating both, which gives a built-in check. Keep a short list of inner integrands that force a change of order, such as e to a power x² or y³ and sin(y²), and rehearse the swap on each.
Finally, attempt one bound question per session, choosing the minimum and maximum of the integrand on the region and multiplying by the area.
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