What is the shell method volume?
Type the curves that bound a region, the interval [a, b], and a vertical axis x = k. The page turns the region about the axis and gives the volume by cylindrical shells, exactly when it can.
- Volume
- 8.37758041
The shell method volume for y = 2x - x^2 from x = 0 to x = 2 about the line x = 0 is 8.37758041.
- Exact volume
- 8π/3
Volume: 8.37758041. The shell method volume for y = 2x - x^2 from x = 0 to x = 2 about the line x = 0 is 8.37758041.
How to calculate
Finds the volume of the solid made by turning the region between y = f(x) and y = g(x), a ≤ x ≤ b, about a vertical line x = k.
Example with the default inputs (Top curve y = f(x) 2x - x^2, From x = a 0, To x = b 2, Axis x = k 0): The shell method volume for y = 2x - x^2 from x = 0 to x = 2 about the line x = 0 is 8.37758041.
Method: V = 2π ∫ₐᵇ |x − k| |f(x) − g(x)| dx: shells of radius |x − k| and height |f − g|.
- The axis x = k is outside the open interval (a, b).
- Angles in radians; an answer that fails its check is not shown.
Worked examples
Each example is checked against the calculator on every build.
- Top curve y = f(x) 2x - x^2, From x = a 0, To x = b 2, Axis x = k 0 gives Volume 8.37758, Exact volume 8π/3.Source: OpenStax, Calculus Volume 1, section 6.3 Volumes of Revolution: Cylindrical Shells (https://openstax.org/books/calculus-volume-1/pages/6-3-volumes-of-revolution-cylindrical-shells), Example 6.13 (about the y-axis)
- Top curve y = f(x) x, From x = a 1, To x = b 2, Axis x = k -1 gives Volume 24.085544, Exact volume 23π/3.Source: OpenStax, Calculus Volume 1, section 6.3 Volumes of Revolution: Cylindrical Shells (https://openstax.org/books/calculus-volume-1/pages/6-3-volumes-of-revolution-cylindrical-shells), Example 6.15 (about the line x = −1)
- Top curve y = f(x) sqrt(x), Bottom curve y = g(x) (empty is 0) 1/x, From x = a 1, To x = b 4, Axis x = k 0 gives Volume 59.061942, Exact volume 94π/5.Source: OpenStax, Calculus Volume 1, section 6.3 Volumes of Revolution: Cylindrical Shells (https://openstax.org/books/calculus-volume-1/pages/6-3-volumes-of-revolution-cylindrical-shells), Example 6.16 (between √x and 1/x, about the y-axis)
- Top curve y = f(x) 1/x, From x = a 1, To x = b 3, Axis x = k 0 gives Volume 12.566371, Exact volume 4π.Source: OpenStax, Calculus Volume 1, section 6.3 Volumes of Revolution: Cylindrical Shells (https://openstax.org/books/calculus-volume-1/pages/6-3-volumes-of-revolution-cylindrical-shells), Example 6.12
How it works
The region lies between y = f(x) and y = g(x) for a ≤ x ≤ b (an empty g is the x-axis, g = 0). It turns about the vertical line x = k. a, b and k are numbers or constants such as pi, with a < b and k not strictly between a and b (k ≤ a or k ≥ b).
V = 2π ∫ₐᵇ |x − k| |f(x) − g(x)| dx.
- Because x − k keeps one sign on [a, b], |x − k| |f − g| = |(x − k)(f(x) − g(x))|. The page finds where (x − k)(f − g) changes sign inside (a, b) (where the curves cross), each crossing exact where the algebra gives it, by the same zero search as the area between curves calculator (a grid of 8,001 points and bisection).
- Between each two crossings, a computer algebra system (nerdamer, open source) finds ∫ (x − k)(f(x) − g(x)) dx, checked against a numeric integral. The absolute values are added and multiplied by 2π.
- The exact volume is 2π times the sum of the exact pieces, each with the sign that makes it positive, simplified by the algebra. It shows when it holds no rounded number and equals the decimal volume to 10⁻⁹; its value is then the one shown.
Angles are in radians. There is no answer when f or g has a pole or is not real on [a, b], an end point included. Typed decimals go to the algebra as exact fractions (1e-9 as 1/10^9).
Worked examples by hand
f(x) = 2x − x² on [0, 2], about the y-axis (OpenStax Calculus Volume 1, Example 6.13). V = 2π ∫₀² x(2x − x²) dx = 2π [2x³/3 − x⁴/4]₀² = 2π (16/3 − 4) = 8π/3 ≈ 8.37758041.
f(x) = x on [1, 2], about x = −1 (Example 6.15). The radius is x + 1. V = 2π ∫₁² (x + 1)x dx = 2π [x³/3 + x²/2]₁² = 2π (14/3 − 5/6) = 2π × 23/6 = 23π/3 ≈ 24.08554368.
Between √x and 1/x on [1, 4], about the y-axis (Example 6.16). V = 2π ∫₁⁴ x(√x − 1/x) dx = 2π [2x^(5/2)/5 − x]₁⁴ = 2π (64/5 − 4 − 2/5 + 1) = 2π × 47/5 = 94π/5.
f(x) = 1/x on [1, 3], about the y-axis (Example 6.12). V = 2π ∫₁³ x × (1/x) dx = 2π × 2 = 4π.
Other questions people ask
What is the shell method?
A way to find the volume of a solid of revolution. Cut the region into thin vertical strips; turning a strip at x about the axis makes a thin cylindrical shell of radius |x − k|, height |f(x) − g(x)| and thickness dx, with volume about 2π × radius × height × dx. Adding the shells gives V = 2π ∫ₐᵇ |x − k| |f(x) − g(x)| dx.
When should I use the shell method instead of washers?
Use shells when the region is easy to describe with functions of x and the axis is vertical (the y-axis or x = k). Washers about a vertical axis would need the curves written as functions of y. For y = 2x − x² about the y-axis, shells need no inverse function.
How do I revolve about the y-axis?
Use k = 0. The radius of each shell is then |x|. For f(x) = 2x − x² on [0, 2] that gives V = 2π ∫₀² x(2x − x²) dx = 8π/3.
How do I revolve about another vertical line?
Type its x value as k. About x = −1, the radius is x + 1: the region under y = x on [1, 2] gives V = 2π ∫₁² (x + 1) x dx = 23π/3.
Why must the axis not cut the region?
If the axis x = k lies strictly between a and b, the two sides of the region sweep through the same space, and adding their shells counts that space twice. Split the region at the axis and use the larger side, or choose a k outside (a, b).
How do I revolve about a horizontal axis?
With shells, a horizontal axis needs the region written as functions of y. Use the washer method calculator instead: it turns a region in x about a line y = k.
How is the answer checked?
Each definite integral comes from a computer algebra system and is checked against a numeric integral. The exact volume shows only when it equals the decimal volume to 10⁻⁹.