← Return Home Calculus Starships

Volume of Rotation: Custom Axis Simulator

To revolve a region about a line other than the x-axis, such as y = k, build each radius as top boundary minus bottom boundary measured from that line, then apply the washer method with those adjusted radii.

When you rotate a solid around an axis other than the x-axis, you can treat this as another application of the Washer Method.

However, setting up the radii is a little trickier because the distances depend on the rotation axis y = K.

To make this simple and foolproof, always use the "Top Curve minus Bottom Curve" rule to find the radii:

Scenario 1: Rotation Axis is BELOW the X-Axis

Here, the region lies entirely above the axis y = K.

• For the Inner Radius: The top boundary is the x-axis (y = 0) and the bottom is the rotation axis (y = K). Subtracting them gives:

r(x)=0−K=−Kr(x) = 0 - K = -K

• For the Outer Radius: The top boundary is the function (y = f(x)) and the bottom is the rotation axis (y = K). Subtracting them gives:

R(x)=f(x)−KR(x) = f(x) - K

(Logically, we are adding the extra distance below the x-axis to the function's height, which mathematically works out via -K since K is negative.)

Mathematically, thinking in terms of Top minus Bottom makes the signs fit the logic naturally!

How do you find volume revolving about a line like y = k?

Rotation axis y=K is below the x-axis:V=π∫ab([f(x)−K]2−[−K]2)dx(where K=rotation axis)\begin{gathered} \text{Rotation axis } y = K \text{ is below the x-axis:} \\ V = \pi \int_{a}^{b} \left( [f(x) - K]^2 - [-K]^2 \right) dx \\ \\ (\text{where } K = \text{rotation axis}) \end{gathered}

Custom-axis worked example: revolving about y = negative 1

Find the volume of the solid generated by revolving the region between y = 2x and y = x² about the line y = -1.

1. The boundary curves intersect at x = 0 and x = 2.

2. The axis of rotation is y = -1. The outer curve is y = 2x and the inner is y = x².

3. Since y = -1 is below the curves, add 1 to each radius:

R(x)=2x−(−1)=2x+1andr(x)=x2−(−1)=x2+1R(x) = 2x - (-1) = 2x + 1 \quad \text{and} \quad r(x) = x^2 - (-1) = x^2 + 1

4. Set up the volume integral:

V=π∫02((2x+1)2−(x2+1)2)dx=π∫02(4x2+4x+1−x4−2x2−1)dxV = \pi \int_{0}^{2} \left( (2x + 1)^2 - (x^2 + 1)^2 \right) dx = \pi \int_{0}^{2} (4x^2 + 4x + 1 - x^4 - 2x^2 - 1) dx
V=π∫02(2x2+4x−x4)dx=π[2x33+2x2−x55]02=136π15V = \pi \int_{0}^{2} (2x^2 + 4x - x^4) dx = \pi \left[ \frac{2x^3}{3} + 2x^2 - \frac{x^5}{5} \right]_{0}^{2} = \frac{136\pi}{15}

Another worked example

Revolve the region under y = x, from x = 0 to x = 2, about the line y = negative 1. Step 1: the outer radius is the curve minus the axis, x minus negative 1, which is x plus 1; the inner radius is 0 minus negative 1, which is 1. Step 2: integrate pi times ((x plus 1) squared minus 1 squared) from 0 to 2. Step 3: that expands to pi times (x squared plus 2x), giving pi times (8 over 3 plus 4), which equals 20 pi over 3.

Aligned with the College Board AP Calculus CED — Unit 8 (Applications of Integration), Topics 8.11 and 8.12 (revolving around other axes). College Board ↗

Interactive Preview

Interactive simulator is loading...

Engaged with the simulation?

Practice actual exam problems and master calculus guaranteed. Get a 5 on the AP Calculus exam or get an A in the calculus class, or we'll pay you $100.

Accept the Mission