Short on time? This page is the recap. For explanations, the diagram and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: ∫ (c to d) [R(y) − L(y)] dy means the definite integral of R(y) − L(y) from y = c to y = d.
Recap
- Slice the region into thin horizontal strips. Each strip has length right − left and thickness Δy.
- The Riemann sum Σ [R(yᵢ) − L(yᵢ)] Δy becomes ∫ (c to d) [R(y) − L(y)] dy.
- Write every boundary as x = (expression in y). For example, y = ln x becomes x = eʸ, and y = x³ becomes x = ∛y.
- Limits are the bottom and top edges: given horizontal lines, or the y-coordinates where the curves meet.
- Right curve: test one y-value strictly between the limits; the larger x is on the right.
- dx and dy give the same area. Choose the set-up that needs one integral and no ± square roots.
Key relationships
| Situation | Area integral |
|---|---|
| R(y) ≥ L(y) for c ≤ y ≤ d | ∫ (c to d) [R(y) − L(y)] dy |
| Region between x = g(y) and the y-axis, g(y) ≥ 0 | ∫ (c to d) [g(y) − 0] dy |
| Region left of the y-axis, g(y) ≤ 0 | ∫ (c to d) [0 − g(y)] dy |
| Curves meet at y = c and y = d | Solve R(y) = L(y) for c and d |
| Same region with vertical strips | ∫ (a to b) [top − bottom] dx (Topic 8.4) |
When dy is the better choice
- A boundary is a sideways parabola, such as x = 5 − y², or any curve given as x = g(y).
- Vertical strips would meet a different curve partway across, forcing a split.
- Rewriting y = f(x) as x = g(y) turns a hard antiderivative into an easy one (for example ln x into eʸ).
Mistakes to avoid
- Left minus right, giving a negative “area”.
- x-values as limits in a dy integral.
- An integrand still in x, such as ∫ ln x dy.
- Asking which curve is on top in a dy set-up. The question is which is on the right.
- Missing brackets or dy, so the integral is ambiguous.
- Rounding intersection values before using them on a calculator.
Quick self-check
- Find the area enclosed by x = y² and x = 4. (Limits y = −2 and 2; ∫ (−2 to 2) (4 − y²) dy = 32/3)
- The curves x = 6 − y² and x = 2 meet where? (6 − y² = 2, so y = ±2. The area is again 32/3: it is the same shape, reflected and moved.)
- Your dy integral gives −9. What went wrong? (You subtracted left − right. The area is 9.)
Next: practice questions.