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Disc Method:
Revolving Around Other Axes

Master the technique for finding volumes of revolution when the axis of rotation is not the \(x\)-axis or \(y\)-axis.

AP Calculus AB

What Is the Disc Method?

When a region in the plane is revolved around a line, it sweeps out a solid of revolution. The disc method slices this solid into thin circular discs perpendicular to the axis of rotation and sums their volumes with an integral.

Quick Review: Standard Axes

Around the \(x\)-axis
\(V = \pi\!\int_a^b [f(x)]^2\,dx\)
Radius = \(f(x)\)
Around the \(y\)-axis
\(V = \pi\!\int_c^d [g(y)]^2\,dy\)
Radius = \(g(y)\)

Why Other Axes Matter

Real-world solids are not always symmetric about the coordinate axes. A bowl, a dome, or a pipe wall may be modeled by revolving a curve around a line like \(y = -1\), \(y = 5\), or \(x = 3\). The disc method still works: you just need to adjust the radius expression.

\(y = -1\) \(y = 0\) \(y = \sqrt{x}\) \(r\)

The General Disc Formula

The key: radius = distance from curve to axis, which is \(|f(x) - k|\) or \(|g(y) - k|\).

Visualizing the Radius

\(y = 4\) \(y = 0\) 0 1 2 3 4 \(y = x^2\) \(r = 4 - x^2\)

Axis Below vs. Axis Above

Axis Below Curve Axis Above Curve
Example axis \(y = -2\) \(y = 5\)
Radius \(f(x) - k\) \(k - f(x)\)
Integrand \([f(x) - k]^2\) \([k - f(x)]^2\)
After squaring Identical: \([f(x)-k]^2\)

Step-by-Step Approach

1
Sketch the region and identify the axis of rotation (e.g., \(y = k\) or \(x = k\)).
2
Write the radius: distance from the curve to the axis. This is \(|f(x) - k|\) or \(|g(y) - k|\).
3
Determine limits of integration from the region boundaries.
4
Set up the integral: \(V = \pi\!\int [\text{radius}]^2\,dx\) (or \(dy\)).
5
Evaluate the integral. Expand the square, integrate term by term, and multiply by \(\pi\).

Example 1: Revolving Around \(y = -1\)

Find the volume when the region bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\) is revolved about the line \(y = -1\).

\(y = -1\) 0 1 2 1 2 3 4 \(y = \sqrt{x}\) \(r = \sqrt{x}+1\)

Example 1: Solution

1
Radius: distance from curve to axis: \(r = \sqrt{x} - (-1) = \sqrt{x} + 1\)
2
Limits: \(x = 0\) to \(x = 4\)
3
Integral: \(V = \pi\!\int_0^4 (\sqrt{x}+1)^2\,dx = \pi\!\int_0^4 (x + 2\sqrt{x} + 1)\,dx\)
4
Evaluate: \(\pi\!\left[\frac{x^2}{2}+\frac{4x^{3/2}}{3}+x\right]_0^4 = \pi\!\left[8 + \frac{32}{3} + 4\right] = \frac{68\pi}{3}\)

Example 2: Revolving Around \(y = 4\)

Find the volume when the region bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) is revolved about \(y = 4\).

\(y = 4\) 0 1 2 3 4 1 2 \(y = x^2\) \(r = 4 - x^2\)

Example 2: Solution

1
Radius: axis is above the curve: \(r = 4 - x^2\)
2
Limits: \(x = 0\) to \(x = 2\)
3
Integral: \(V = \pi\!\int_0^2 (4 - x^2)^2\,dx = \pi\!\int_0^2 (16 - 8x^2 + x^4)\,dx\)
4
Evaluate: \(\pi\!\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = \pi\!\left[32 - \frac{64}{3} + \frac{32}{5}\right] = \frac{256\pi}{15}\)

Example 3: Revolving Around \(x = 3\)

Find the volume when the region bounded by \(x = y^2\) and \(x = 1\) (for \(y \geq 0\)) is revolved about \(x = 3\).

\(x = 3\) 0 1 2 3 0 1 \(x = y^2\) \(x = 1\) \(r = 3 - y^2\)

Example 3: Solution

Since there is a gap between \(x = 1\) and \(x = 3\), we use the washer method (disc with a hole). Outer radius: \(R = 3 - y^2\). Inner radius: \(r = 3 - 1 = 2\).

Common Mistakes

✕ Using \(f(x)\) as the radius
The radius is the distance to the axis, not the function value. Use \(f(x) - k\), not just \(f(x)\).
✕ Forgetting to square
The disc area is \(\pi r^2\). Forgetting to square the radius before integrating gives the wrong answer.
✕ Wrong variable of integration
Revolving around a vertical line usually requires integrating with respect to \(y\), not \(x\). Match the variable to the axis direction.

Summary Table

Axis of Rotation Radius Variable
\(y = 0\) (\(x\)-axis) \(f(x)\) \(dx\)
\(x = 0\) (\(y\)-axis) \(g(y)\) \(dy\)
\(y = k\) (below curve) \(f(x) - k\) \(dx\)
\(y = k\) (above curve) \(k - f(x)\) \(dx\)
\(x = k\) (right of curve) \(k - g(y)\) \(dy\)
\(x = k\) (left of curve) \(g(y) - k\) \(dy\)

Visual Explanation

The same region 0 ≤ y ≤ √x, 0 ≤ x ≤ 4 forms discs about y = 0 and washers about y = −1. About y = 0 the radius is √x. About y = −1 the outer radius is √x + 1 and the inner radius is 1 because the axis lies one unit below the region. The corresponding volumes are π∫₀⁴ x dx = 8π and π∫₀⁴[(√x + 1)² − 1] dx = 56π/3. The filled region is identical in both panels. Rotating it about y = 0 gives discs of radius √x. Rotating it about y = −1 gives washers with outer radius √x + 1 and inner radius 1; subtract the inner-circle area. Each radius is a non-negative distance from the axis to a boundary of the region. If the region reaches the axis, a perpendicular slice produces a disc. If a gap separates the region from the axis, the slice produces a washer: use the distance to the farther boundary as the outer radius and the distance to the nearer boundary as the inner radius.

Two solids made from the same region under y=√x. Rotation about y=0 gives discs; rotation about y=−1 gives washers with inner radius 1. Their volumes are 8π and 56π/3.
Moving the axis from y=0 to y=−1 leaves the region unchanged but opens an inner radius of 1, so discs become washers.
The same region 0≤y≤√x, 0≤x≤4, shown with axes of rotation y=0 and y=−1. For the shifted axis, outer radius is √x+1 and inner radius is 1.
The same region 0≤y≤√x, 0≤x≤4, shown with axes of rotation y=0 and y=−1. For the shifted axis, outer radius is √x+1 and inner radius is 1.

Question 1 of 5

The region bounded by \(y = x\), \(y = 0\), and \(x = 2\) is revolved about \(y = -1\). What is the radius of a typical disc?

Question 2 of 5

Which integral gives the volume when \(y = x^2\) (from \(x = 0\) to \(x = 1\)) is revolved about \(y = 2\)?

Question 3 of 5

The region bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 9\) is revolved about \(y = -2\). What is the volume?

Question 4 of 5

A student sets up the integral \(\pi\!\int_0^3 (x^2)^2\,dx\) for the region \(y = x^2\), \(y = 0\), \(x = 3\) revolved about \(y = -1\). What did they do wrong?

Question 5 of 5

The region bounded by \(x = y\) and \(x = 0\) for \(0 \leq y \leq 2\) is revolved about \(x = 4\). Which expression gives the outer radius?