Questions: Double Integrals over Rectangular Regions

Question 1

A student is computing ∬_R xy dA over R = [1,3]×[0,2], set up as ∫_1^3 ∫_0^2 xy dy dx. After evaluating the inner integral ∫_0^2 xy dy, they obtain 2x. What does this intermediate result represent?

• A. The final answer to the double integral
• B. A function of x giving the cross-sectional 'slice area' of the solid at each fixed x-value
• C. An error — x should have been substituted with a constant before integrating
• D. The volume of the region below z = xy along the y-axis only

Answer: (B)

During the inner integration, x is treated as a fixed parameter — not a variable, not substituted away. The result 2x is a function of x that represents the area of the cross-section (a slice perpendicular to the x-axis) at each value of x. The outer integral ∫_1^3 2x dx then sums all these cross-sections to get the total volume. This two-step process — slicing, then summing slices — is the geometric heart of iterated integration.

Question 2

Why is it valid to switch the order of integration for a rectangular region, but not always for a non-rectangular region without changing the bounds?

• A. Rectangular regions have more symmetry, which guarantees that both orders give equal results by symmetry
• B. On rectangles all four bounds are constants, so neither bound depends on the other variable; on general regions, inner bounds depend on the outer variable and must be re-derived when switching
• C. Fubini's theorem only applies to rectangular regions and has no extension to general shapes
• D. Switching order on a non-rectangular region changes the numerical value of the integral

Answer: (B)

The key is constant bounds. On R = [a,b]×[c,d], the limits a, b, c, d are all fixed numbers — no bound depends on x or y. Whether you integrate y first (with x as the outer variable) or x first (with y as the outer variable), the same rectangle is covered. On a non-rectangular region, the inner limits are functions of the outer variable (e.g., ∫_0^1 ∫_0^x f dy dx). Switching order requires re-deriving the bounds to describe the same region correctly — the region itself hasn't changed, but the description has.

Question 3

When evaluating the inner integral ∫_c^d f(x,y) dy in an iterated double integral over a rectangle, the variable x is treated as a varying quantity that changes as you integrate.

• T. True
• F. False

Answer: False

During the inner integration over y, x is held fixed — treated as a constant parameter. You are computing the area of a single cross-sectional slice at one specific x-value. Only after the inner integral is fully evaluated (yielding a function of x) does x become the variable of integration in the outer integral. Treating x as varying during the inner step is the most common computational error in setting up iterated integrals.

Question 4

For a rectangular region R = [a,b]×[c,d], ∬_R f dA = ∫_a^b ∫_c^d f(x,y) dy dx = ∫_c^d ∫_a^b f(x,y) dx dy, regardless of the form of f.

• T. True
• F. False

Answer: True

This is exactly what Fubini's theorem guarantees for rectangles (assuming f is continuous, or more generally, integrable on R). Because the bounds are all constants, either order of integration covers the entire rectangle, and both iterated integrals compute the same double integral. In practice, one order may produce a simpler antiderivative than the other — the computational flexibility is a major advantage of working on rectangular regions.

Question 5

Explain what happens mathematically when you evaluate the inner integral in an iterated double integral over a rectangle. What does the result represent, and how does it lead to the final answer?

Model answer: The inner integral ∫_c^d f(x,y) dy treats x as a fixed constant and integrates f over the y-interval [c,d]. The result is a function A(x) representing the area of the cross-sectional slice of the solid z = f(x,y) at that fixed x-value. The outer integral ∫_a^b A(x) dx then sums all these cross-sectional areas across [a,b], accumulating them to produce the total volume (or signed volume if f takes negative values). This is the 'slice-then-sum' interpretation of iterated integration.

The two-step structure mirrors how single-variable integration works: the inner integral solves a single-variable problem (with x as a parameter), and the outer integral solves another single-variable problem using the result. The rectangle's constant bounds make this clean — no bounds need adjustment between steps. For non-rectangular regions, the bounds of the inner integral are functions of the outer variable, adding a layer of complexity that rectangular regions avoid.