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Choosing dx or dy, and Splitting When You Must

You’ll be able to

The slice decides everything

Integrating with respect to x means slicing the region vertically, so each slice runs from a lower curve to an upper curve and its height is top minus bottom, both written as functions of x. Integrating with respect to y means slicing horizontally, so each slice runs from a left curve to a right curve with width right minus left, both written as functions of y. Choose whichever slicing lets a single integral describe the whole region. The test is mechanical: if a vertical line drawn anywhere across the region always enters through the same curve and exits through the same curve, dx works in one piece. If it does not — because the bottom boundary changes partway across — either split the region or switch to dy.

When to switch and when to split

A region bounded on the left by a curve and below by the x-axis is the standard case where dy is one integral and dx is two. Consider the region bounded by y = √x, the line x = 4, and the x-axis. Sliced vertically, every slice runs from y = 0 up to y = √x, so dx handles it in one piece. But the region bounded by y = √x, y = 0, and x = 4 to the left of the curve — that is, between the y-axis and the curve — requires two dx integrals or one dy integral. The general rule worth carrying: switch variables before splitting, because two integrals is twice the chance of an algebra error, and setting up in the other variable usually costs only the work of solving for the inverse. When you do have to split, the split point is where a boundary curve changes.

Limits come from intersections, and they must be found first

The limits of integration are the coordinates where the boundary curves meet, so solve for the intersections before writing any integral. Set the two functions equal and solve; if you are integrating with respect to y, the limits are y-coordinates, which means you may need to substitute back to find them. Two traps. First, a pair of curves can intersect more than twice, in which case the "upper" curve switches and the region genuinely requires splitting at the crossing — the exam includes this. Second, when a calculator gives an intersection like x = 1.8637, keep at least three decimal places through the computation and round only at the end, since rounding early moves the answer outside the accepted tolerance.

Area with horizontal slices
A = ∫_c^d [x_right(y) − x_left(y)] dy
Right minus left, both as functions of y, with limits that are y-values. The mirror of top minus bottom with dx.
Worked example

Find the area of the region bounded by y = x² and y = x + 2.

  1. 1.Find intersections: x² = x + 2, so x² − x − 2 = 0, which factors as (x − 2)(x + 1) = 0. So x = −1 and x = 2.
  2. 2.Decide which curve is on top between them. Test x = 0: the line gives 0 + 2 = 2, the parabola gives 0. The line is above.
  3. 3.Vertical slices work in one piece, since across the whole region each slice runs from the parabola up to the line.
  4. 4.Set up: A = ∫_{−1}^{2} [(x + 2) − x²] dx.
  5. 5.Antidifferentiate: x²/2 + 2x − x³/3.
  6. 6.At x = 2: 2 + 4 − 8/3 = 6 − 8/3 = 10/3. At x = −1: 1/2 − 2 + 1/3 = −7/6.
  7. 7.Subtract: 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2.
Answer: The area is 9/2 = 4.5. Vertical slices suffice because the line is above the parabola across the entire interval [−1, 2].
On the exam

Always determine which curve is on top by testing a point inside the interval, not by looking at the equations. Reversing top and bottom produces the negative of the right answer, and an area cannot be negative — so if your result comes out negative, that is the error.

Checkpoint

For the region bounded by x = y² and x = 4, integrating with respect to y is preferable because

Checkpoint

Two curves intersect at x = 0, x = 2, and x = 5, with f above g on (0, 2) and g above f on (2, 5). The total area between them on [0, 5] is

Answer the 2 checkpoints as you read.

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