Questions: Iterated Integrals and Fubini's Theorem

Question 1

A student sets up ∫₀¹ ∫₀ˣ f(x,y) dy dx over a triangular region. To switch the order of integration, what are the correct new bounds?

• A. ∫₀¹ ∫₀^y f(x,y) dx dy
• B. ∫₀¹ ∫_y^1 f(x,y) dx dy
• C. ∫₀¹ ∫₀¹ f(x,y) dx dy
• D. ∫₀¹ ∫₀ˣ f(x,y) dx dy

Answer: (B)

In the original, x goes 0 to 1 and y goes 0 to x — covering the triangle below the diagonal y = x. To reverse the order, describe the same triangle from y's perspective: y ranges from 0 to 1, and for each fixed y, x ranges from y to 1. This gives ∫₀¹ ∫_y^1 f(x,y) dx dy. Option A covers a different triangle (above the diagonal). Option C integrates over the full unit square — too large a region.

Question 2

Fubini's theorem guarantees that for a continuous function over a rectangle [a,b] × [c,d]:

• A. The iterated integral with dy dx always gives a larger value than the one with dx dy
• B. The order of integration can be reversed without changing the result
• C. The double integral equals the product of two single integrals
• D. The outer integral must be evaluated before the inner integral

Answer: (B)

Fubini's theorem states that ∫ₐᵇ ∫_c^d f(x,y) dy dx = ∫_c^d ∫ₐᵇ f(x,y) dx dy for continuous f on a rectangle. The order can be reversed without changing the value. This is useful because one order may be algebraically much simpler. Option C is wrong — that would require f(x,y) to factor as g(x)h(y).

Question 3

When switching the order of integration for a non-rectangular region, the numerical bounds of integration stay the same — primarily the variable labels change.

• T. True
• F. False

Answer: False

The bounds change substantively, not just in label. For example, ∫₀¹ ∫₀ˣ f dy dx becomes ∫₀¹ ∫_y^1 f dx dy — the inner bound shifts from 'x' (a function of the outer variable) to 'y' and '1'. You must re-describe the same 2D region from the new outer variable's perspective, which requires sketching the region and reading the new bounds directly from the geometry.

Question 4

The key strategy behind iterated integrals is to compute a double integral as two sequential single-variable integrations, treating one variable as constant while integrating over the other.

• T. True
• F. False

Answer: True

This is exactly the method: fix x, integrate f(x,y) over y to get a cross-sectional area function A(x), then integrate A(x) over x to accumulate total volume. Each step is a standard single-variable integral. The two-pass strategy reduces a 2D problem to sequential 1D problems that you already know how to handle.

Question 5

Why does switching the order of integration for an iterated integral over a non-rectangular region require changing the integration bounds? What is the reliable technique for finding the new bounds?

Model answer: The bounds encode a description of the integration region from the perspective of the outer variable. Switching which variable is outer requires re-describing the same region from the new perspective. For a triangle like 0 ≤ y ≤ x ≤ 1, x as outer variable means y runs 0 to x; y as outer variable means x runs y to 1. The reliable technique is to sketch the region, label corners and boundary curves, and read the new bounds directly from the sketch — asking 'for each fixed value of the outer variable, what range does the inner variable span?'

Algebraic manipulation of the bounds alone is error-prone. The sketch-first approach directly reads the geometric constraints of the region and translates them into integration limits, which is far more reliable especially when dealing with curved boundaries.