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Disc Method:
Revolving Around Other Axes

Master the technique of computing volumes of revolution when the axis of rotation is not the \(x\)-axis or \(y\)-axis: horizontal lines \(y = k\) and vertical lines \(x = k\).

AP Calculus BC

What Is the Disc Method?

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

You already know the standard formulas for revolving around the \(x\)-axis and \(y\)-axis. Now we extend the method to any horizontal or vertical line.

Why Other Axes Matter

AP Exam Frequency
Free-response questions regularly specify axes like \(y = -1\) or \(x = 3\). You must adapt the radius formula.
Real-World Shapes
Most physical objects (vases, turbines, bells) are not symmetric about the coordinate axes: their axis of symmetry is offset.
Deeper Understanding
Handling a shifted axis forces you to think about what the radius truly represents: the distance from curve to axis.

The Radius Is a Distance

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

A transparent lathe reveals the hole inside a washer

A hollow wooden bowl on a stationary lathe accompanies a calibrated generating region and a separate annular cross-section.For 0 ≤ x ≤ 2, rotate the region from y = −1 to y = x² about y = −1. Each full disk has radius x² + 1, so V = π∫₀²(x² + 1)² dx = 206π/15.

A disk reaches the shifted axis

For 0 ≤ x ≤ 2, rotate the region from y = −1 to y = x² about y = −1. Each full disk has radius x² + 1, so V = π∫₀²(x² + 1)² dx = 206π/15.For 0 ≤ x ≤ 2, rotate the region from y = −1 to y = x² about y = −1. Each full disk has radius x² + 1, so V = π∫₀²(x² + 1)² dx = 206π/15.

Core Formula: Horizontal Axis \(y = k\)

When the axis is below the region (\(k < f(x)\)), the radius is \(f(x) - k\). When the axis is above (\(k > f(x)\)), the radius is \(k - f(x)\). Either way, squaring removes sign issues.

Core Formula: Vertical Axis \(x = k\)

When the axis of rotation is vertical, slices are horizontal discs. The radius of each disc is the horizontal distance from the curve to the line \(x = k\).

\(x = 5\) \(y\)-axis \(x = y^2\) \(r\) \(r = 5 - g(y) = 5 - y^2\)

Standard vs. Shifted Axis

Standard Axis Shifted Axis
Axis \(y = 0\) or \(x = 0\) \(y = k\) or \(x = k\)
Radius (horiz.) \(r = f(x)\) \(r = f(x) - k\)
Radius (vert.) \(r = g(y)\) \(r = g(y) - k\)
Formula \(\pi \int [f(x)]^2 \, dx\) \(\pi \int [f(x) - k]^2 \, dx\)
Common error Forgetting to square Using \(f(x)\) instead of \(f(x) - k\)

Step-by-Step Approach

1
Sketch the region and the axis. Draw the curve(s), shade the region, and clearly mark the axis of rotation \(y = k\) or \(x = k\).
2
Determine integration variable. If the axis is horizontal, integrate with respect to \(x\). If vertical, integrate with respect to \(y\).
3
Write the radius. Radius = (outer boundary) \(-\) (axis value): \(r = f(x) - k\) or \(r = g(y) - k\).
4
Find the limits. Determine where the region starts and ends along the integration variable.
5
Set up and evaluate. Write \(V = \pi \int_a^b [r]^2 \, dx\) (or \(dy\)), expand, and integrate.

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 \(y = -1\).

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

Example 1: Solution

1
Expand: \((\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1\)
2
Integrate term by term:
\(\displaystyle\int_0^4 (x + 2\sqrt{x} + 1)\,dx = \left[\frac{x^2}{2} + \frac{4x^{3/2}}{3} + x\right]_0^4\)
3
Evaluate: \(\displaystyle\frac{16}{2} + \frac{4 \cdot 8}{3} + 4 = 8 + \frac{32}{3} + 4 = \frac{68}{3}\)
4
Multiply by \(\pi\): \(\displaystyle V = \frac{68\pi}{3}\)

Example 2: Revolving Around \(y = 3\)

Find the volume when the region bounded by \(y = x^2\) and \(y = 0\) on \([0, 1]\) is revolved about \(y = 3\).

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

Example 2: Solution

1
Identify Radii: Outer radius \(R = 3 - 0 = 3\). Inner radius \(r = 3 - x^2\).
2
Set up Integral: \(V = \pi \displaystyle\int_0^1 \bigl[R^2 - r^2\bigr]\,dx = \pi \displaystyle\int_0^1 \bigl[3^2 - (3-x^2)^2\bigr]\,dx\)
3
Expand: \(9 - (9 - 6x^2 + x^4) = 6x^2 - x^4\)
4
Integrate: \(\displaystyle\int_0^1 (6x^2 - x^4)\,dx = \left[2x^3 - \frac{x^5}{5}\right]_0^1\)
5
Evaluate & Finalize: \(\left(2 - \frac{1}{5}\right) - 0 = \frac{9}{5}\). So, \(V = \frac{9\pi}{5}\).

Example 3: Revolving Around \(x = 2\)

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

\(x = 2\) \(x = 1\) \(x = y^2\) \(R = 2 - y^2\) \(r = 2 - 1 = 1\)

Example 3: Solution

1
Expand \((2 - y^2)^2\): \(4 - 4y^2 + y^4\)
2
Subtract inner: \((4 - 4y^2 + y^4) - 1 = 3 - 4y^2 + y^4\)
3
Integrate: \(\displaystyle\int_0^1 (3 - 4y^2 + y^4)\,dy = \left[3y - \frac{4y^3}{3} + \frac{y^5}{5}\right]_0^1\)
4
Evaluate: \(3 - \frac{4}{3} + \frac{1}{5} = \frac{45 - 20 + 3}{15} = \frac{28}{15}\)
5
Final: \(\displaystyle V = \frac{28\pi}{15}\)

Common Mistakes to Avoid

✕ Forgetting the Shift
Using \(r = f(x)\) instead of \(r = f(x) - k\). If the axis is \(y = -1\), the radius is not just the function value.
✕ Wrong Sign Convention
Writing \(r = k - f(x)\) when \(f(x) > k\), giving a negative radius. Always take the absolute distance or rely on squaring.
✕ Disc vs. Washer Confusion
Using the disc formula when the axis does not touch the region. If there is a gap, you need the washer formula with outer and inner radii.

Summary of Disc Formulas

Scenario Formula Radius
About \(x\)-axis \(\pi \int [f(x)]^2\,dx\) \(r = f(x)\)
About \(y = k\) (below) \(\pi \int [f(x) - k]^2\,dx\) \(r = f(x) - k\)
About \(y = k\) (above) \(\pi \int [k - f(x)]^2\,dx\) \(r = k - f(x)\)
About \(y\)-axis \(\pi \int [g(y)]^2\,dy\) \(r = g(y)\)
About \(x = k\) \(\pi \int [g(y) - k]^2\,dy\) \(r = g(y) - k\)

Question 1 of 5

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

Question 2 of 5

Which integral gives the volume when \(y = x^2\) on \([0, 2]\) is revolved about \(y = 5\)?

Question 3 of 5

The region under \(y = \sqrt{x}\) from \(x = 0\) to \(x = 4\) is revolved about \(y = -1\). What is \(V\)?

Question 4 of 5

When revolving a region about a vertical line \(x = k\), the discs are horizontal. What must you do with the curve equation?

Question 5 of 5

The region bounded by \(y = x^2\) and \(y = 0\) on \([0, 1]\) is revolved about \(y = 3\). What is the volume?