Describe regions and reverse integration order

Try

Shade the triangle bounded by y=0, y=x and x=3. A vertical slice at a fixed x runs from y=0 to y=x, while x ranges from 0 to 3. A horizontal slice at a fixed y runs from x=y to x=3, while y ranges from 0 to 3. Both descriptions cover the same points. Reversing integration order means rebuilding these slice limits, not merely swapping the symbols dx and dy.

Predict

Predict the area’s value before integrating: it is a right triangle with legs 3, so its area is 9/2. The vertical-slice integral is ∫₀³∫₀ˣ1 dy dx, and the horizontal-slice version is ∫₀³∫ᵧ³1 dx dy. Each gives 9/2. This agreement is a useful check because the integrand 1 isolates the geometry. If two purported descriptions give different areas, they cannot represent the same region except in a special cancellation situation.

Experiment

For a nonconstant integrand y, the first setup gives ∫₀³x²/2 dx=9/2. The reversed setup gives ∫₀³y(3−y)dy=9/2. Track the inner variable explicitly: when integrating with respect to x, y is a constant multiplier.

Slice direction	Outer interval	Inner interval
Vertical	0≤x≤3	0≤y≤x
Horizontal	0≤y≤3	y≤x≤3

Draw one representative slice in each direction. A picture should answer which boundary is entered first and which is exited, rather than simply decorate the equations.

Understand

A vertically simple region has a≤x≤b and g₁(x)≤y≤g₂(x). Its integral is ∫ₐᵇ∫g₁(x)^g₂(x) f(x,y)dy dx under standard integrability hypotheses. A horizontally simple description uses corresponding left and right boundaries. Some regions admit both descriptions with one pair of formulas; others require a finite partition because the active boundary changes.

For the region under y=2−|x| and above y=0, vertical limits can be written with an absolute value or split at x=0. Horizontal slices instead have 0≤y≤2 and y−2≤x≤2−y. Neither direction is universally preferable. Choose the description that makes both bounds and integrand manageable, while preserving the actual region.

Apply

Repair “solve y=x² for x to get x=√y everywhere.” The parabola has two branches. A region that includes negative x may need x=−√y as well. State sign restrictions before choosing an inverse branch.

Order reversal can also remove a difficult antiderivative. Over 0≤y≤x≤2, integrating e^(x²) in y first gives ∫₀²x e^(x²)dx=(e⁴−1)/2. Integrating x first appears to require a nonelementary inner antiderivative. The region has not changed; only the computational path has. This example shows why a geometric setup can be more powerful than trying increasingly complicated one-variable integration techniques.

Continue

For optional depth, consider a region whose horizontal slices have two disjoint intervals. It cannot be represented by one inner interval without including unwanted points. Split the slice or split the region and sum the resulting integrals; do not silently fill the gap.

Save a reversed-order solution with an inequality description and a labeled sketch before either integral. Verify that boundary intersections give the endpoints of the outer interval. Numerical agreement for one integrand is supportive, but matching the point sets is the underlying justification. Continue to polar coordinates, where the same region-description discipline applies while the area element itself changes.