Questions: Double Integrals over Rectangular Regions

Question 1

You want to evaluate ∬_R f(x,y) dA over the rectangle R = [0,1]×[0,1] where f(x,y) = sin(y²). Which iterated integral is computable in closed form?

• A. ∫₀¹ ∫₀¹ sin(y²) dy dx — integrate y first, then x
• B. ∫₀¹ ∫₀¹ sin(y²) dx dy — integrate x first, then y
• C. Both orders produce the same computable antiderivative
• D. Neither order produces a closed-form antiderivative, so the integral cannot be evaluated

Answer: (B)

sin(y²) has no elementary antiderivative with respect to y, so the dy-first order (option A) leaves an impossible inner integral. But sin(y²) is independent of x, so the dx-first inner integral ∫₀¹ sin(y²) dx = sin(y²)·[x]₀¹ = sin(y²). The outer integral then becomes ∫₀¹ sin(y²) dy, which still lacks a closed form on its own — but the key point is that the correct first step is to integrate with respect to x, yielding sin(y²), not to fight the impossible inner integral. Choosing the better order of integration is one of the most practical skills the theorem enables.

Question 2

A student sets up ∫₀² ∫₀³ f(x,y) dy dx and gets the answer 12. Her partner switches the order to ∫₀³ ∫₀² f(x,y) dx dy. If f is continuous, what answer should her partner get?

• A. −12, because reversing the order negates the integral
• B. 12, because Fubini's theorem guarantees both orders give the same value
• C. It depends on whether f is positive or negative over the rectangle
• D. Cannot be determined without knowing what f is

Answer: (B)

Fubini's theorem states that for a continuous function f on a rectangle, the double integral equals either iterated integral and both produce the same number. The double integral ∬_R f dA is a single well-defined value (the signed volume under the surface) — the order of integration is just a computational strategy to reach that value. Both students are computing the same geometric quantity by slicing it in perpendicular directions, so the answer must be 12.

Question 3

A double integral ∬_R f(x,y) dA where f can take negative values represents a signed volume — regions where f < 0 contribute negatively.

• T. True
• F. False

Answer: True

Just as a single-variable integral can be negative when the function dips below the x-axis, a double integral sums signed contributions: columns where f(x,y) > 0 contribute positive volume, columns where f(x,y) < 0 contribute negative volume. The total is the net signed volume. This is analogous to the 1D case — the double integral of f = 1 over R gives the area of R, while ∬_R (−1) dA gives the negative of that area.

Question 4

Switching the order of integration in a double integral over a rectangle changes the value of the integral.

• T. True
• F. False

Answer: False

For a continuous function on a closed rectangle, Fubini's theorem guarantees that both orders of integration produce the same value. The order of integration is a computational choice, not a mathematical one — both iterated integrals are evaluating the same double integral ∬_R f dA. The order can only matter in degenerate cases (discontinuous or non-integrable functions), not for the continuous functions encountered in standard multivariable calculus.

Question 5

Explain why computing a double integral over a rectangle can be reduced to two successive single-variable integrals, and what Fubini's theorem adds to this.

Model answer: The double integral sums infinitely many thin rectangular columns of height f(x,y) and base area dA = dx dy. By first integrating with respect to y (for a fixed x), you compute the area of a cross-sectional slice of the solid at that x. Then integrating those slice areas with respect to x sweeps through all cross-sections and accumulates the total volume. Fubini's theorem adds the guarantee that the two possible orders (integrate y first or x first) yield the same result for continuous f, so you can choose whichever order produces a simpler antiderivative.

The geometric interpretation is key: both orders slice the same solid — one cuts it with vertical planes parallel to the yz-plane, the other with planes parallel to the xz-plane. Since there is only one correct volume regardless of how you slice it, both orders must agree. Without Fubini, you would need to prove this agreement for each specific function; the theorem does it once and for all under mild conditions.