Calculus AB · CED Unit 8: Applications of Integration · 14 min read · Updated 2026-05-10
Disc method around other axes — AP Calculus AB
AP Calculus AB · CED Unit 8: Applications of Integration · 14 min read
1. Core Concept: Disc Method for Non-Coordinate Axes★★☆☆☆⏱ 3 min
The disc method is an integration technique to find the volume of a solid formed by rotating a planar region around a fixed axis, where each cross-section perpendicular to the axis is a solid disc. On the AP Calculus AB exam, you will frequently be asked to rotate around any horizontal or vertical line that is not a coordinate axis.
The core idea of the disc method remains the same regardless of the axis: volume is the integral of the area of cross-sectional discs. The only key difference from the basic method is how we calculate the radius $r$, which equals the distance between the bounding function and the axis of rotation.
V =
int \pi r^2 dx \quad (\text{integrate with respect to } x) \\ V = \int \pi r^2 dy \quad (\text{integrate with respect to } y)
2. Revolution Around a Horizontal Axis ($y = k$, $k \neq 0$)★★☆☆☆⏱ 4 min
When rotating around a horizontal line of the form $y = k$ where $k \neq 0$, we always integrate with respect to $x$. This is because cross-sections perpendicular to a horizontal axis are vertical slices, which are integrated along the $x$-axis.
The radius is the vertical distance between the bounding curve and the axis of rotation. Since radius must be positive, $r = |f(x) - k|$, and squaring removes the absolute value, so $r^2 = (f(x) - k)^2$ regardless of whether the curve is above or below the axis.
V = \pi \int_{a}^{b} \left[ f(x) - k \right]^2 dx
3. Revolution Around a Vertical Axis ($x = c$, $c \neq 0$)★★★☆☆⏱ 5 min
When rotating around a vertical line of the form $x = c$ where $c \neq 0$, the axis of rotation is vertical, so cross-sections perpendicular to the axis are horizontal slices, meaning we integrate with respect to $y$.
The radius here is the horizontal distance between the curve $x = g(y)$ (expressed as a function of $y$) and the axis $x=c$. Again, $r = |g(y) - c|$, so $r^2 = (g(y)-c)^2$ regardless of which side of the axis the curve sits on. If your original function is given as $y = f(x)$, you rearrange it to solve for $x$ as a function of $y$.
V = \pi \int_{d}^{e} \left[ g(y) - c \right]^2 dy
4. Disc Method for Regions Bounded by Two Intersecting Curves★★★☆☆⏱ 5 min
Many AP exam problems ask you to rotate a region bounded by two intersecting curves (not a curve and a given coordinate bound) around a non-coordinate axis. When one side of the region is the axis of rotation, the disc method still applies (there is no hole in the middle, so inner radius is zero, unlike the washer method).
The key extra step here is finding the bounds of integration by solving for the points of intersection of the two curves. Once you have bounds, you identify which curve forms the outer edge of the region, then calculate the radius as the distance between this outer curve and the axis of rotation.
Common Pitfalls
Why: Students memorize the basic disc formula for rotation around $y=0$, so they forget to adjust the radius for a non-zero axis.
Why: Students get used to always integrating with respect to $x$, and forget that slicing is always perpendicular to the axis of rotation.
Why: Students confuse disc method for two-curve regions with washer method, and misidentify what the radius measures.
Why: Students avoid inverse function rearrangement and incorrectly keep the original variable.
Why: Students don't think of radius as a positive distance, and just subtract in the order they read the problem.