AP® Calculus BC · Unit 8 · Topic 8.5

Finding the Area Between Curves Expressed as Functions of y

When a region is bounded by curves that are easy to solve for x but awkward to solve for y, slice it into horizontal strips and integrate right minus left with respect to y.

Some regions are a nightmare to slice vertically. If the boundary curves double back on themselves, a single x value can hit the region twice, so there's no clean top function and bottom function. But solve each curve for x in terms of y instead, and the trouble often disappears: for each height y, the region runs from one clear left boundary to one clear right boundary. That's your cue to slice horizontally.

A horizontal sliver has width dy and length equal to x_right(y) minus x_left(y). Stack these slivers from the lowest y value to the highest, and the integral of (x_right - x_left) dy adds them up into the total area — same logic as top minus bottom, just rotated ninety degrees. Find the limits by setting x_right(y) = x_left(y) and solving for the y values where the curves cross.

This is not just swapping letters in your old setup. You have to actually re-solve each curve as x = g(y), decide which one sits to the right, and reset your limits as y values instead of x values. Done right, though, it gives the exact same number as slicing the same region vertically in x — area doesn't care which way you cut it. Choosing the easier direction is really the whole skill: it's the same region, the same area, found by whichever slicing avoids splitting the integral into pieces.

Key terms

right minus left
The horizontal-slicing analog of top minus bottom, giving the length of each strip as x_right(y) - x_left(y).
x = g(y)
A curve written with x isolated as a function of y, needed before integrating with respect to y.
horizontal strip
A thin rectangle of width dy stretching from the left boundary to the right boundary at a given height.
limits of integration (in y)
The y values where the two curves intersect, found by setting their x-expressions equal.

Watch it explained

  1. How it worksHow does a horizontal sliver of width right minus left build the very same area?
  2. Worked exampleFind the area between x = y2 and x = y + 2 by integrating with respect to y.
  3. Common mistakeWhy isn't integrating with respect to y just the same problem with the letters swapped?
  4. Ties togetherHow does integrating in y connect to integrating that same region in x from the topic before?

Check yourself

Your turn1 of 2
Find the area between x = y2 and x = 6y by integrating with respect to y (for 6 > 0).

End of topic.