In AP Calculus, the outer radius R(x) is the distance from the axis of revolution to the curve farther from that axis. In the washer method, volume equals π∫(R² − r²) dx, where the outer radius is squared first and the inner radius squared is subtracted from it (CHA-5.C.3, Topic 8.11).
When you revolve a region around an axis and the solid has a hole in it, every cross section is a ring (a washer). The outer radius is the distance from the axis of revolution to the curve that sits farther away. It traces the outside edge of each ring. The inner radius traces the hole.
The washer method formula is V = π∫[R(x)² − r(x)²] dx, where R is the outer radius and r is the inner radius. Here's the part that trips people up. The outer radius is NOT just "the top function." It's measured from the axis of revolution, so if you revolve around y = -1 instead of the x-axis, the outer radius becomes (top function) − (−1), which adds 1 to everything. Think of it as the long arm of the washer, swinging from the axis out to the far curve. If you're revolving around the y-axis, the radii are horizontal distances and everything gets written in terms of y instead.
The outer radius lives in Topic 8.11 (Volume with Washer Method) in Unit 8: Applications of Integration. It directly supports learning objective 8.11.A, calculating volumes of solids of revolution with definite integrals, and essential knowledge CHA-5.C.3, which says ring-shaped cross sections are handled with the washer method. Unit 8 is one of the heaviest free-response units on the AP exam, and a solids-of-revolution part shows up constantly. Setting up R and r correctly is usually worth a point on its own, even before you integrate anything. Most washer-method mistakes on the exam are radius mistakes, not integration mistakes.
Keep studying AP Calculus Unit 8
Visual cheatsheet
view galleryInner radius (Unit 8)
The outer radius is half of the washer setup. The inner radius r(x) measures from the axis to the closer curve, and the washer's area is π(R² − r²). If the inner radius is zero, the hole disappears and you're back to the plain disc method.
Cross Sections (Unit 8)
The washer method is really just the cross-sections idea (Topics 8.7-8.8) with a specific shape. Each slice is a ring, its area is πR² − πr², and integrating that area gives volume. The outer radius is what defines the ring's outside edge.
Definite integrals as accumulation (Unit 6)
The volume integral π∫(R² − r²) dx is accumulation in action. You're adding up infinitely many thin washer volumes, area times thickness dx. The outer radius determines how big each contribution is.
On multiple choice, you're usually shown a region and an axis of revolution and asked which integral represents the volume. The wrong answers are built from radius mistakes, like forgetting to shift the radii when revolving around y = -1, or writing π∫(R − r)² instead of π∫(R² − r²). For example, revolving the region under y = √x around y = -1 makes the outer radius √x + 1 and the inner radius 1, not √x and 0. Revolving around the y-axis flips everything to y. For x = y² and x = 4 revolved around the y-axis, the volume is ∫π(4² − y⁴) dy from -2 to 2, where 4 is the outer radius and y² is the inner one. On FRQs, the setup is where the points live. You typically earn credit for correct radii and bounds even if the arithmetic afterward goes sideways, so write R and r explicitly before you build the integral.
Both are measured from the axis of revolution, but the outer radius reaches the farther curve while the inner radius reaches the closer one. The classic mix-up happens when the axis isn't the x-axis. Around y = -1, the function closer to the line y = -1 gives the inner radius even if it's the "bottom" function on your graph. Always ask which curve is farther from the axis, not which curve is on top.
The outer radius R is the distance from the axis of revolution to the curve farther from that axis, and it forms the outside edge of each washer.
The washer method volume formula is V = π∫(R² − r²) dx, and you must square each radius separately. π∫(R − r)² dx is wrong.
When the axis of revolution is shifted, like y = -1, adjust both radii. The outer radius becomes the function minus the axis value, such as √x − (−1) = √x + 1.
When revolving around the y-axis, write the outer radius as a function of y and integrate with respect to y, like the outer radius of 4 in ∫π(4² − y⁴) dy.
If the inner radius is zero, the washer method collapses to the disc method, so the disc method is just a washer with no hole.
It's the distance from the axis of revolution to the curve farther from that axis. In the formula V = π∫(R² − r²) dx, the outer radius is R, and it defines the outside edge of each ring-shaped cross section (CHA-5.C.3).
No. The outer radius is whichever curve is farther from the axis of revolution. If you revolve around a line below the region, like y = -1, the top function is the outer radius, but if you revolve around a line above the region, the bottom function becomes the outer one. Always measure from the axis.
The outer radius goes from the axis to the farther curve and the inner radius goes from the axis to the closer curve. The inner radius creates the hole in the washer, and the volume comes from subtracting πr² from πR² for each slice.
Subtract the axis value from the function. For the region under y = √x revolved around y = -1, the outer radius is √x − (−1) = √x + 1, and the inner radius is 0 − (−1) = 1. Forgetting that +1 shift is the most common washer-method error.
No, those are not equal. (R − r)² expands to R² − 2Rr + r², which is a different value. Square the outer radius and inner radius separately, then subtract. AP multiple-choice distractors are built on exactly this mistake.
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