Aim for FiveCalculus BC

Unit 8 · Topic 8.5

8.5 Finding the Area Between Curves Expressed as Functions of y

Some regions are easier to slice horizontally. Then you write each curve as x in terms of y and integrate the right curve minus the left curve with respect to y, with y-values as the limits.

Key terms

  • horizontal slices (dy)
  • right minus left
  • x as a function of y
  • limits in y

When to switch to horizontal slices

Vertical slices (dx) work when every vertical line through the region enters at the same bottom curve and exits at the same top curve. If the bottom or top boundary changes partway across, you'd have to split the integral. Turning the problem sideways can avoid the split.

Horizontal slices (dy) work when every horizontal line through the region enters at the same left curve and exits at the same right curve. This is common when a boundary is given as x = (something with y), like x = y², or when a sideways parabola is involved.

The setup

Each horizontal strip has thickness dy and length (right curve) − (left curve). So

Area = ∫ from c to d of [(right curve) − (left curve)] dy,

where c and d are the lowest and highest y-values of the region. Every part of the integral must be in terms of y: the curves, the limits and the dy.

  • Solve each boundary for x: y = √x becomes x = y² (for y ≥ 0); y = x − 2 becomes x = y + 2.
  • Find intersection y-values by setting the x-expressions equal.
  • Decide which curve is on the right by testing a y-value between the limits.
  • Integrate right minus left, with respect to y.

Comparing dx and dy

SlicesStrip lengthLimitsIntegrate
vertical (dx)top − bottomx-valueswith respect to x
horizontal (dy)right − lefty-valueswith respect to y

Choosing well

Solving for x can produce two branches. The curve y = x² becomes x = √y on the right and x = −√y on the left. Use the branch that actually forms the boundary of your region, or both if the region spans both sides.

Both directions give the same area if you set them up correctly, so pick whichever needs fewer integrals and easier algebra. Sketch first: look at what the boundaries are as you sweep left to right, and then bottom to top. A region bounded by x = y² and x = y + 2 needs two dx integrals but only one dy integral.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    One dy integral instead of two dx integrals

    Find the area of the region bounded by x = y² and x = y + 2.

    Show the solution
    1. Step 1: Intersections: y² = y + 2 → y² − y − 2 = 0 → (y − 2)(y + 1) = 0, so y = −1 and y = 2.
    2. Step 2: Which is on the right? At y = 0: y + 2 = 2 and y² = 0, so the line x = y + 2 is on the right.
    3. Step 3: Area = ∫ from −1 to 2 of [(y + 2) − y²] dy.
    4. Step 4: Antiderivative: y²/2 + 2y − y³/3. At 2: 2 + 4 − 8/3 = 10/3. At −1: 1/2 − 2 + 1/3 = −7/6.
    5. Step 5: Area = 10/3 + 7/6 = 27/6 = 9/2.
    6. Step 6: With dx you'd need two integrals, because the bottom boundary changes at x = 1 (from y = −√x to y = x − 2). They also add to 9/2.

    Answer: 9/2

  2. Example 2

    Trap: limits must be y-values

    Region R is bounded by y = √x, the x-axis and the line x = 4. Write and evaluate an integral with respect to y for its area.

    Show the solution
    1. Step 1: Rewrite the curve: y = √x becomes x = y².
    2. Step 2: The region runs from y = 0 to y = √4 = 2. Using 0 to 4 as y-limits is the trap: 4 is an x-value.
    3. Step 3: For each y, the left boundary is x = y² and the right boundary is x = 4.
    4. Step 4: Area = ∫₀² (4 − y²) dy = 8 − 8/3 = 16/3.
    5. Step 5: Check with dx: ∫₀⁴ √x dx = (2/3)(8) = 16/3. ✓

    Answer: ∫₀² (4 − y²) dy = 16/3

Common mistakes

  • Using x-values as limits for a dy integral.
  • Subtracting top minus bottom in a dy integral. It's right minus left.
  • Leaving a curve written as y = f(x) inside a dy integral instead of solving for x.
  • Taking only the positive square root when the region extends below the x-axis (x = y² has both y = √x and y = −√x).

On the exam

  • Multiple-choice questions often ask which integral gives the area, with both dx and dy versions among the choices. Check the limits and the variable match.
  • If a free-response question says “write, but do not evaluate, an integral in terms of y,” the setup is the whole point. Don't switch to dx.

Connected topics

Videos

  • Calculus AB/BC – 8.5 Finding Area Between Curves Expressed as Functions of y

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  • Horizontal area between curves | Applications of definite integrals | AP Calculus AB | Khan Academy

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  • Area Between Curves With Respect to Y (KristaKingMath)

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  • Bounded Area: With Respect to y

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Check yourself

4 questions on 8.5 Finding the Area Between Curves Expressed as Functions of y. Pick an answer to see if you got it, and why.

Question 1 of 4

Let R be the region bounded by the graphs of x = y² − 4 and x = 2y − 1. Which of the following gives the area of R?

Question 2 of 4

What is the area of the region bounded by the graphs of x = y² and x = y + 2?

Question 3 of 4

Let R be the region bounded by the graph of y = √x, the x-axis, and the line x = 4. Which of the following integrals gives the area of R?

Question 4 of 4

Let R be the region bounded by the graph of y = ln x, the y-axis, the x-axis, and the line y = 2. What is the area of R?

0 of 4 answered