Topic 8.10 Volume with Disc Method: Revolving Around Other Axes: find the volume of a solid of revolution about a line other than a coordinate axis using the disc method.

What this topic is asking

The College Board (Topic 8.10) revolves a region about a line other than a coordinate axis, such as $y = -1$ or $x = 3$, using the disc method. The only change from Topic 8.9 is that the radius is the distance from the shifted axis to the curve.

Adjusting the radius for a shifted axis

A worked shifted-axis volume

Getting the radius sign and distance right

The single new skill is writing the distance from the axis of revolution to the curve, which depends on whether the curve is above/right of the axis or below/left. Revolving about a line below a region (like $y = -1$ under a positive curve) adds the shift: $R = f(x) + 1$. Revolving about a line above a region subtracts the curve from the line: $R = k - f(x)$. The radius must come out positive because it is a distance, so check the geometry: subtract the smaller coordinate from the larger. A sign error here is the dominant mistake, and because the radius is squared, it changes the answer.

When the shifted axis creates a hole

If the region does not touch the shifted axis, revolving leaves a gap, and the cross section is a washer rather than a disc; that is Topic 8.12. Before using a single disc about a shifted line, confirm the region actually reaches the axis of revolution so there is no inner empty radius. When it does touch, the disc method with the shifted radius applies directly. When it does not, you subtract an inner radius (also measured from the shifted axis) using the washer method. Recognizing which situation you are in, by sketching the region relative to the axis, is what determines disc versus washer.

A reliable way to write the shifted radius

A dependable habit is to write the radius as the larger coordinate minus the smaller coordinate along the direction perpendicular to the axis. For a horizontal axis $y = k$ with a curve at height $y = f(x)$, the radius is $|f(x) - k|$; resolve the absolute value by checking whether the curve is above or below the line. For $y = -1$ under a positive curve, the curve coordinate is larger, so $R = f(x) - (-1) = f(x) + 1$. For an axis $y = 5$ above a curve, the line is larger, so $R = 5 - f(x)$. Always sketching the axis and the curve, then subtracting the lower from the higher, produces a positive radius and removes the guesswork that causes sign errors.

Sketching makes the setup reliable

Because the radius now depends on a shifted reference line, a quick sketch is the single most valuable habit. Draw the region, draw the axis of revolution as a dashed line, and draw one representative disc perpendicular to the axis. The disc's radius is the segment from the axis to the curve, and its length, read off the sketch as a difference of coordinates, is what you square and integrate. The sketch also reveals immediately whether the region touches the axis (disc) or leaves a gap (washer). Spending a moment on the picture before writing the integral prevents the dominant errors of this topic: a mis-signed radius and an unnoticed hole.