Calculus · Grade 12 and undergraduate

Volumes of Revolution: Disks and Washers

Quick answer

Revolving a flat region around a line sweeps out a solid of revolution. Its slices perpendicular to the axis are disks, with area πR², or, where the region does not touch the axis, washers, with area π(R² − r²). So the volume is π∫ₐᵇ R² dx for disks and π∫ₐᵇ (R² − r²) dx for washers, where R and r are the outer and inner radii measured from the axis. Around a vertical axis, slice horizontally and integrate with respect to y.

What you'll learn

  • Find volumes of revolution with the disk method
  • Find volumes of revolution with the washer method
  • Measure radii from an axis that is not a coordinate axis
  • Revolve around a vertical axis by integrating in y

Disks

Revolve the region under y=f(x)y = f(x), for a≤x≤ba \le x \le b, around the xx-axis. It sweeps out a solid of revolution. Slice the solid perpendicular to the axis, and each slice is a disk whose radius is the height of the curve, f(x)f(x). Its area is A(x)=πf(x)2A(x) = \pi f(x)^2, so slicing gives the disk method:

V=π∫abf(x)2 dxV = \pi\int_a^b f(x)^2\,dx
The region under y = √x revolved around the x-axis A solid shaped like the bell of a horn lies along the x-axis from x = 0 to x = 4, widening from a point to a circular end of radius 2. The flat region under y = √x is shaded in its upper half. One thin disk is highlighted at x = 2.5; its radius r runs from the axis up to the curve. y r x
The region under y = √x revolved around the x-axis

Why the radius is the height of the curve

The slice at xx is swept out by one vertical segment of the region, the one from the axis up to the curve. As the segment turns around the axis, its top end traces a circle of radius f(x)f(x), and the segment fills the disk inside it. So the only new work in a volume of revolution is finding radii. Find the radius of one slice, square it, multiply by π\pi, and integrate.

Washers

If the region does not touch the axis, every slice has a hole. The segment at xx now runs from the nearer boundary to the farther one. It sweeps out a washer: a disk of radius R(x)R(x) with a disk of radius r(x)r(x) removed. Its area is πR2−πr2\pi R^2 - \pi r^2:

V=π∫ab(R(x)2−r(x)2) dxV = \pi\int_a^b \big(R(x)^2 - r(x)^2\big)\,dx

RR is the distance from the axis to the farther boundary, and rr the distance to the nearer one.

The region between y = x and y = x² revolved around the x-axis A cone along the x-axis from x = 0 to x = 1, widening to radius 1, with a hollow inside shaped like a narrower bowl, drawn with dashed lines. The thin region between the line y = x and the parabola y = x² is shaded. One washer is highlighted at x = 0.5, with outer radius R up to the line and inner radius r to the parabola. y R r x
The region between y = x and y = x² revolved around the x-axis

Other axes

A radius is a distance from the axis of revolution, which need not be the xx-axis. Around the line y=ky = k, a boundary at height yy is ∣y−k∣\lvert y - k \rvert away from the axis.

Around a vertical axis, the slices are horizontal. Write the boundaries as xx in terms of yy, measure the radii horizontally, and integrate with respect to yy.

Worked examples

Common mistakes

Practice problems

  1. Revolve the region under y=x2y = x^2, 0≤x≤20 \le x \le 2, around the xx-axis. Find the volume.

    Answer

    32π5\tfrac{32\pi}{5}

    Full solution

    V=π∫02(x2)2 dx=π∫02x4 dx=π⋅325V = \pi\int_0^2 (x^2)^2\,dx = \pi\int_0^2 x^4\,dx = \pi \cdot \tfrac{32}{5}.

  2. Revolve the region under y=1xy = \tfrac{1}{x}, 1≤x≤31 \le x \le 3, around the xx-axis. Find the volume.

    Answer

    2π3\tfrac{2\pi}{3}

    Full solution

    V=π∫13x−2 dx=π[−x−1]13=π(1−13)V = \pi\int_1^3 x^{-2}\,dx = \pi\big[-x^{-1}\big]_1^3 = \pi\left(1 - \tfrac{1}{3}\right).

  3. Revolve the region under y=sin⁡xy = \sin x, 0≤x≤π0 \le x \le \pi, around the xx-axis. Find the volume.

    Answer

    π22\tfrac{\pi^2}{2}

    Full solution

    V=π∫0πsin⁡2x dxV = \pi\int_0^{\pi} \sin^2 x\,dx. Use sin⁡2x=1−cos⁡2x2\sin^2 x = \tfrac{1 - \cos 2x}{2}: ∫0π1−cos⁡2x2 dx=[x2−sin⁡2x4]0π=π2\int_0^{\pi} \tfrac{1 - \cos 2x}{2}\,dx = \big[\tfrac{x}{2} - \tfrac{\sin 2x}{4}\big]_0^{\pi} = \tfrac{\pi}{2}. So V=π22V = \tfrac{\pi^2}{2}.

  4. Revolve the region under y=exy = e^x, 0≤x≤10 \le x \le 1, around the xx-axis. Find the volume.

    Answer

    π2(e2−1)≈10.0\tfrac{\pi}{2}(e^2 - 1) \approx 10.0

    Full solution

    V=π∫01e2x dx=π[12e2x]01=π2(e2−1)V = \pi\int_0^1 e^{2x}\,dx = \pi\big[\tfrac{1}{2}e^{2x}\big]_0^1 = \tfrac{\pi}{2}(e^2 - 1).

  5. Revolve the region between y=2xy = 2x and y=x2y = x^2 around the xx-axis. Find the volume.

    Answer

    64π15\tfrac{64\pi}{15}

    Full solution

    The curves meet at x=0x = 0 and x=2x = 2, with the line farther from the axis: R=2xR = 2x and r=x2r = x^2. V=π∫02(4x2−x4) dx=π(323−325)=64π15V = \pi\int_0^2 (4x^2 - x^4)\,dx = \pi\left(\tfrac{32}{3} - \tfrac{32}{5}\right) = \tfrac{64\pi}{15}.

  6. Revolve the region under y=xy = \sqrt{x}, 0≤x≤40 \le x \le 4, around the line y=2y = 2. Find the volume.

    Answer

    40π3\tfrac{40\pi}{3}

    Full solution

    The axis is above the region. The xx-axis is farther from it, R=2R = 2, and the curve is nearer, r=2−xr = 2 - \sqrt{x}. V=π∫04(4−(2−x)2)dx=π∫04(4x−x)dx=π(643−8)=40π3V = \pi\int_0^4 \left(4 - (2 - \sqrt{x})^2\right)dx = \pi\int_0^4 \left(4\sqrt{x} - x\right)dx = \pi\left(\tfrac{64}{3} - 8\right) = \tfrac{40\pi}{3}.

  7. Revolve the region between y=x3y = x^3 and y=8y = 8, for x≥0x \ge 0, around the yy-axis. Find the volume.

    Answer

    96π5\tfrac{96\pi}{5}

    Full solution

    At height yy the radius is x=y1/3x = y^{1/3}, for 0≤y≤80 \le y \le 8. V=π∫08y2/3 dy=π⋅35⋅85/3=π⋅35⋅32=96π5V = \pi\int_0^8 y^{2/3}\,dy = \pi \cdot \tfrac{3}{5} \cdot 8^{5/3} = \pi \cdot \tfrac{3}{5} \cdot 32 = \tfrac{96\pi}{5}.

  8. Revolve the region between y=2xy = 2x and y=x2y = x^2 around the yy-axis, using washers. Find the volume.

    Answer

    8π3\tfrac{8\pi}{3}

    Full solution

    In terms of yy, the parabola is x=yx = \sqrt{y} and the line is x=y2x = \tfrac{y}{2}, for 0≤y≤40 \le y \le 4. The parabola is farther from the axis: R=yR = \sqrt{y}, r=y2r = \tfrac{y}{2}. V=π∫04(y−y24)dy=π(8−163)=8π3V = \pi\int_0^4 \left(y - \tfrac{y^2}{4}\right)dy = \pi\left(8 - \tfrac{16}{3}\right) = \tfrac{8\pi}{3}.

  9. Show that a ball of radius aa has volume 43πa3\tfrac{4}{3}\pi a^3 by revolving the region under y=a2−x2y = \sqrt{a^2 - x^2} around the xx-axis.

    Answer

    π∫−aa(a2−x2) dx=43πa3\pi\int_{-a}^{a} (a^2 - x^2)\,dx = \tfrac{4}{3}\pi a^3

    Full solution

    The radius is a2−x2\sqrt{a^2 - x^2}, so V=π∫−aa(a2−x2) dx=π[a2x−x33]−aa=π(2a3−2a33)=43πa3V = \pi\int_{-a}^{a} (a^2 - x^2)\,dx = \pi\big[a^2x - \tfrac{x^3}{3}\big]_{-a}^{a} = \pi\left(2a^3 - \tfrac{2a^3}{3}\right) = \tfrac{4}{3}\pi a^3.

  10. A student revolves the region between y=xy = \sqrt{x} and y=xy = x around the xx-axis and writes V=π∫01(x−x)2 dx=π30V = \pi\int_0^1 (\sqrt{x} - x)^2\,dx = \tfrac{\pi}{30}. What went wrong?

    Hint

    What is the area of a washer with radii RR and rr?

    Answer

    The student squared the difference of the radii. The volume is π6\tfrac{\pi}{6}.

    Full solution

    On [0,1][0, 1], x≥x\sqrt{x} \ge x, so R=xR = \sqrt{x} and r=xr = x. A washer’s area is π(R2−r2)=π(x−x2)\pi(R^2 - r^2) = \pi(x - x^2).

    V=π∫01(x−x2) dx=π(12−13)=π6V = \pi\int_0^1 (x - x^2)\,dx = \pi\left(\tfrac{1}{2} - \tfrac{1}{3}\right) = \tfrac{\pi}{6}.

Frequently asked questions

What is the disk method?

Revolving the region under y = f(x) around the x-axis makes slices that are disks of radius f(x), so the volume is π times the integral of f(x)².

When do I use washers instead of disks?

When the region does not touch the axis. Each slice then has a hole, and its area is π(R² − r²): the outer disk minus the inner one.

Is the area of a washer π(R − r)²?

No. It is πR² − πr² = π(R² − r²). Squaring the difference gives a smaller, wrong answer.

How do I revolve around a line like y = −1?

Measure each radius from that line. A point at height y is y + 1 above the line y = −1.

How do I revolve around the y-axis with disks or washers?

Slice perpendicular to the y-axis. Write the boundaries as x in terms of y, and integrate with respect to y between y-limits.

What to learn next

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