Integrating the constant function 1 over R sums all infinitesimal area elements dA across R, giving the total area of R. This is the 2D analogue of ∫_a^b 1 dx = b − a (the length of [a,b]). It also equals the volume under the surface z = 1 above R — a prism of height 1 — which gives base area × height = Area × 1 = Area, confirming the result.
Question 2 Multiple Choice
A student calculates the area of a unit disk (radius 1) in polar coordinates as ∫₀²π ∫₀¹ 1 dr dθ = 2π, but the actual area is π. What went wrong?
AThe outer limits should be 0 to π, not 0 to 2π
BThe student forgot the factor of r in the polar area element; the correct integrand is r, not 1
CThe inner limits for r should be -1 to 1
DThe constant function 1 cannot be integrated in polar coordinates — f(r,θ) must depend on r
In polar coordinates the area element is dA = r dr dθ, not dr dθ. A small polar 'rectangle' at radius r has width dr and arc length r dθ, so its area is r dr dθ — the factor r appears because arcs get longer as you move away from the origin. The correct integral is ∫₀²π ∫₀¹ r dr dθ = ∫₀²π [r²/2]₀¹ dθ = ∫₀²π ½ dθ = π. Forgetting the factor of r is the single most common error in polar integration.
Question 3 True / False
The formula V = ∬_R f(x,y) dA for volume under z = f(x,y) above R requires that f(x,y) > 0 everywhere on R.
TTrue
FFalse
Answer: False
The formula applies regardless of the sign of f. When f(x,y) < 0, those regions contribute negatively to the integral, giving a 'signed volume.' The integral ∬_R f(x,y) dA is always defined; it just doesn't equal the total geometric volume between the surface and the xy-plane when f changes sign. For unsigned geometric volume where f < 0 somewhere, you would integrate |f(x,y)| or split the region.
Question 4 True / False
For a continuous function f over a rectangular region [a,b] × [c,d], the double integral can be evaluated as ∫_a^b(∫_c^d f(x,y) dy) dx or as ∫_c^d(∫_a^b f(x,y) dx) dy, and both give the same result.
TTrue
FFalse
Answer: True
Fubini's theorem guarantees that for a continuous function over a rectangle, the order of integration is interchangeable — both iterated integrals equal the double integral. This is a practical tool: you can choose whichever order makes the computation simpler. The theorem also extends to non-rectangular regions under appropriate conditions, but the rectangular case is the cleanest statement.
Question 5 Short Answer
When setting up a double integral over a non-rectangular region R, what is the key step before writing down the limits of integration?
Think about your answer, then reveal below.
Model answer: Draw the region R and identify its boundary curves. Then decide whether to treat R as x-simple (for each fixed x in [a,b], y runs from a lower boundary g₁(x) to an upper boundary g₂(x)) or y-simple (for each fixed y in [c,d], x runs between two boundary functions h₁(y) and h₂(y)). The inner limits are functions, the outer limits are constants — and the sketch reveals which description is simpler.
Setting up the limits correctly is the core skill of this topic. Students who skip the sketch often write incorrect limits or choose an order that leads to an impossible inner integral. The sketch reveals which boundaries are functions of which variable. Sometimes one order produces elementary integrals while the other is intractable — and you can only see this by understanding the region geometrically first.