Washer method around other axes — AP Calculus BC
1. Core Concepts for Shifted Axes ★☆☆☆☆ ⏱ 3 min
The washer method around other axes is an extension of the basic washer method for calculating the volume of a solid of revolution, where the axis of rotation is a horizontal or vertical line that is not $x=0$ or $y=0$. It accounts for approximately 2-4% of the total AP Calculus BC exam score, appearing in both MCQ and FRQ sections.
The core logic matches the basic washer method: when you rotate a region between two curves around an axis, each cross-section perpendicular to the axis forms a washer (a disk with a concentric hole), with area $\pi(R^2 - r^2)$ where $R$ is the outer radius and $r$ the inner radius. The total volume is the integral of this cross-sectional area along the axis.
2. Rotation around Horizontal Axes $y = k, k \neq 0$ ★★☆☆☆ ⏱ 5 min
When rotating around a horizontal line $y = k$, the axis is parallel to the x-axis, so we integrate with respect to $x$ for the washer method. Radius is defined as the perpendicular distance between the curve and the axis of rotation, which is always non-negative.
For a region bounded above by $y = f(x)$ and below by $y = g(x)$ on $x \in [a, b]$: if the entire region lies above the axis $y = k$, the outer radius is $R = f(x) - k$, and the inner radius is $r = g(x) - k$. If the entire region lies below the axis, reverse the subtraction to get positive distance.
V = \pi \int_a^b \left[ R^2 - r^2 \right] dx
Exam tip: Always calculate radius as the absolute value of $(y_{curve} - k)$. Writing the subtraction in the order that gives a positive value eliminates 90% of common sign errors on this topic.
3. Rotation around Vertical Axes $x = k, k \neq 0$ ★★★☆☆ ⏱ 6 min
When rotating around a vertical line $x = k$, the axis is parallel to the y-axis, so the washer method requires integrating with respect to $y$. This is because washers are perpendicular to the axis of rotation: a vertical axis means perpendicular slices are horizontal, with thickness $dy$. You must rewrite all boundary curves as functions of $y$ ($x = f(y)$) first.
Radius is again the non-negative perpendicular distance from the curve to the axis $x = k$. If the entire region lies to the right of $x = k$, $R = f(y) - k$ and $r = g(y) - k$. If the entire region lies to the left of $x = k$, reverse the subtraction to get positive distance.
V = \pi \int_c^d \left[ R^2 - r^2 \right] dy
Exam tip: When rotating around a vertical axis with the washer method, always confirm you are integrating with respect to $y$, not $x$. If you end up with an integral in $dx$, you are either using the shell method by mistake or set up the problem incorrectly.
4. Axis Cuts Through the Region ★★★★☆ ⏱ 4 min
A common AP exam variation is when the axis of rotation passes through the interior of the bounded region, rather than lying entirely on one side. In this case, there is no hole in the cross-section, so a single washer formula with inner and outer radius does not apply.
To solve this, split the region into two sub-regions, one on each side of the axis. Each sub-region lies entirely on one side of the axis, so we calculate the volume of each separately as a solid disk (inner radius $r=0$ for each sub-region), then add the two volumes to get the total.
Exam tip: If you end up with a negative radius when you calculate $R - r$, that is a clear sign the axis cuts through the region and you need to split your integral.
5. Concept Check ★★☆☆☆ ⏱ 2 min
Common Pitfalls
Why: Students get used to rotation around $y=0$, where radius equals $f(x)$, and forget to adjust for a shifted axis.
Why: Students confuse washer method with shell method, or default to integrating with respect to $x$ since most functions are given as $y = f(x)$.
Why: Students assume 'left = inner, right = outer' even when the axis is to the right of the entire region.
Why: Students incorrectly assume there is a hole between the lower curve and the axis, but the entire region between the curves is rotated, so there is no hole.
Why: Students rush to integrate after setup and skip simplifying the integrand.