Tags: volume by slicing, disk method, washer method, shell method, cylindrical shells, solid of revolution, cross-sectional area, rotating around axis, Texas A&M calculus
There are several ways to compute volumes using integration. The slicing method works for any solid with a known cross-sectional area. For solids of revolution, three specialised approaches exist: the disk method (no hole), the washer method (hole in the middle), and the shell method (cylindrical shells). Which one to pick depends on the axis of rotation and which variable makes the integral simplest.
Volume by slicing
A general method for finding the volume of a solid. Slice the solid perpendicular to an axis, find the cross-sectional area A(x) at each position, and integrate: V = ∫ A(x) dx.
Solid of revolution
A 3D solid formed by rotating a 2D region around an axis (typically the x-axis or y-axis).
Disk method
Used when rotating a region around an axis produces a solid with no hole. Each cross-section is a circular disk of radius f(x). V = ∫ π[f(x)]² dx.
Washer method
Used when rotating a region around an axis produces a solid with a hole (like a ring or washer). Each cross-section is an annulus. V = ∫ π([f(x)]² − [g(x)]²) dx, where f is the outer radius and g is the inner radius.
Shell method (cylindrical shells)
Instead of slicing perpendicular to the axis, you peel the solid into thin cylindrical shells parallel to the axis. V = ∫ 2π(radius)(height) dx. Especially useful when the other methods lead to difficult integrals.
Annulus
A ring-shaped region between two concentric circles. Its area is π(R² − r²), where R is the outer radius and r is the inner radius. This is the cross-section in the washer method.
The core idea: any solid whose cross-sectional area you can express as a function of position can have its volume found by integration.
Slice the solid into thin slabs of thickness dx (or dy). Each slab has volume ≈ A(x) · dx.
Summing all slabs and taking the limit gives:
V = ∫ from a to b of A(x) dx
This works for pyramids, wedges, and solids with non-circular cross-sections (triangles, squares, semicircles, etc.).
A pyramid with a known base shape and height can be sliced horizontally. At each height, the cross-section is a scaled version of the base.
The same principle applies whether the base is a square, triangle, circle, or ellipse; only the formula for A(x) changes.
A solid whose base is a disk of radius R with cross-sections that are isosceles right triangles (one leg on the base): find the leg length from the circle equation, compute the triangle area, then integrate.
The cross-sectional area formula changes depending on the shape (square, equilateral triangle, semicircle, etc.), but the integration framework is always V = ∫ A(x) dx.
When you rotate the graph of y = f(x) around the x-axis (with no gap between the curve and the axis), each cross-section is a full disk.
Radius of disk = f(x), so A(x) = π[f(x)]².
V = ∫ from a to b of π[f(x)]² dx
For rotation around the y-axis, rewrite as x = g(y) and integrate with respect to y:
V = ∫ from c to d of π[g(y)]² dy
Classic result: the volume of a sphere of radius R is derived by rotating y = √(R² − x²) around the x-axis, yielding (4/3)πR³.
When rotating a region between two curves around an axis, the solid has a hole and each cross-section is an annulus (washer).
Outer radius = f(x), inner radius = g(x).
V = ∫ from a to b of π([f(x)]² − [g(x)]²) dx
The washer method applies to both x-axis and y-axis rotations; just match the variable of integration to the slicing direction.
When rotating around a line other than an axis (e.g. y = 1), adjust the radii accordingly. Outer radius becomes |f(x) − 1| and inner radius becomes |g(x) − 1| (or similar, depending on the geometry).
Instead of slicing perpendicular to the axis, you "unwrap" the solid into thin cylindrical shells parallel to the axis.
A shell at position x has radius x, height h(x), and thickness dx. Its volume is approximately 2πx · h(x) · dx.
V = ∫ from a to b of 2πx · h(x) dx
This is the natural choice when rotating around the y-axis and the function is given as y = f(x), because you avoid having to invert the function.
For rotation around the x-axis using shells, integrate with respect to y:
V = ∫ from c to d of 2πy · h(y) dy
When rotating around x = c (a vertical line that is not the y-axis), the shell radius becomes |x − c| instead of x.
When rotating around y = c, the shell radius becomes |y − c|.
Rotating around the x-axis: disk/washer integrates with respect to x; shells integrate with respect to y.
Rotating around the y-axis: disk/washer integrates with respect to y; shells integrate with respect to x.
The rule of thumb: if the axis of rotation is parallel to the slicing direction, use shells. If perpendicular, use disk/washer.
Pick whichever method avoids splitting the integral or inverting a difficult function.
General slicing:
V = ∫ from a to b of A(x) dx, where A(x) = cross-sectional area
Disk (around x-axis):
V = ∫ from a to b of π[f(x)]² dx
Disk (around y-axis):
V = ∫ from c to d of π[g(y)]² dy
Washer (around x-axis):
V = ∫ from a to b of π([f(x)]² − [g(x)]²) dx
Shell (around y-axis):
V = ∫ from a to b of 2πx · h(x) dx
Shell (around x-axis):
V = ∫ from c to d of 2πy · h(y) dy
Volume of a sphere:
V = (4/3)πR³
⚠️ The most common mistake in the washer method is subtracting the radii before squaring: π(f − g)² is wrong. It must be π(f² − g²). Outer radius squared minus inner radius squared, not (outer minus inner) squared.
⚠️ When the axis of rotation is not the x-axis or y-axis (e.g. y = 2 or x = 1), you must adjust every radius by the offset. Draw the cross-section and measure from the axis of rotation to the curve.
⚠️ Shell method questions often appear when the disk/washer setup would be painful (e.g. rotating y = sin(x²) around the y-axis). Recognise when shells save you from inverting a function.
⚠️ For cross-section problems (non-revolution), read the shape carefully: "isosceles right triangle with a side on the base" is different from "equilateral triangle with a side on the base." The area formula changes.
⚠️ Always identify your limits of integration from the geometry of the region, not from memory.
Q: A region bounded by y = x, y = x², rotated around the x-axis. Do you use disk or washer?
A: Washer, because between x = 0 and x = 1 there is a gap between the two curves. The outer radius is x (the line) and the inner radius is x² (the parabola). V = ∫₀¹ π(x² − x⁴) dx.
Q: How do you find the volume when y = x³, y = 8, x = 0 is rotated around the y-axis?
A: Since you are rotating around the y-axis, express x as a function of y: x = y^(1/3). Use the disk method with respect to y. V = ∫₀⁸ π[y^(1/3)]² dy = ∫₀⁸ πy^(2/3) dy.
Q: Why might you choose the shell method for rotating y = sin(x²), y = 0, x = 0, x = √π around the y-axis?
A: The washer method would require expressing x in terms of y, which means inverting sin(x²), a very difficult inversion. Shells integrate with respect to x directly: V = ∫₀^√π 2πx · sin(x²) dx, which is solvable with a simple u-substitution (u = x²).
Q: What is the shell formula for rotating y = x², y = 0, x = 1, x = 2 around x = 1?
A: The shell radius is (x − 1), and the height is x². V = ∫₁² 2π(x − 1)(x²) dx.
Q: A solid has a circular base of radius R and cross-sections perpendicular to the base that are isosceles right triangles with one leg on the base. What is the volume?
A: At position x, the chord length (leg) is 2√(R² − x²). The area of an isosceles right triangle with leg s is s²/2, so A(x) = [2√(R² − x²)]²/2 = 2(R² − x²). Then V = ∫ from −R to R of 2(R² − x²) dx = (8/3)R³.
Q: State the key difference between disk/washer and shell methods in terms of the direction of integration relative to the axis of rotation.
A: Disk/washer slices are perpendicular to the axis of rotation (so the variable of integration matches the axis direction). Shell slices are parallel to the axis of rotation (so the variable of integration is the other variable).
volume of revolution, disk method, washer method, shell method, cylindrical shells, slicing method, cross-sectional area, annulus, solid of revolution, rotating around x-axis, rotating around y-axis, rotating around a line, volume integral, sphere volume derivation, pyramid volume, MATH 151, MATH 152, Texas A&M