Difficulty: Intermediate | Prerequisites: Disk and washer methods (see companion notes), area between curves.
Not every volume problem involves spinning a region around an axis. Some solids have cross-sections of a known geometric shape (squares, equilateral triangles, semicircles) perpendicular to an axis, and you find their volume the same way: integrate the cross-sectional area. This set of notes also covers the common exam twist of rotating about a line that is not one of the coordinate axes, which changes how you compute your radii. Both topics build directly on the general slicing principle from the disk/washer notes and round out the full toolkit for Calc II volume problems.
When a solid has cross-sections of a known shape (squares, triangles, etc.) perpendicular to an axis, find the side length from the bounding curves, compute the area formula for that shape, and integrate. When the axis of rotation is shifted away from the coordinate axes, adjust each radius by measuring the distance from the curve to that line before plugging into the disk or washer formula.
Known cross-section
A solid whose slices perpendicular to one axis are all the same type of geometric figure (square, equilateral triangle, semicircle, etc.), though the size changes with position. Think of it as a loaf of bread where every slice is a square, but the squares get bigger and then smaller.
Side length function, S(x)
The expression that gives the side length (or diameter, depending on the shape) of each cross-section as a function of position along the axis. You read this off the distance between the bounding curves.
Shifted axis of rotation
An axis of rotation that is not the x-axis or y-axis, such as y = 4 or x = -1. In simple terms, the solid is being spun around a line that sits away from the origin, so every radius must be recalculated as a distance to that line.
The general formula is always the same:
V = \int_a^b A(x)\,dxwhat changes is how you build A(x) from the shape of the cross-section.
Identify the base region in the xy-plane (the curves that bound it).
Determine the shape of each cross-section (square, equilateral triangle, semicircle, etc.).
Write the side length S(x) as the vertical or horizontal distance between the bounding curves.
Plug S(x) into the area formula for that shape.
Integrate from one end of the base region to the other.
Square cross-sections: A(x) = [S(x)]^2
Equilateral triangle cross-sections: A(x) = (sqrt(3)/4) [S(x)]^2
Semicircle cross-sections (diameter = S(x)): A(x) = (pi/8) [S(x)]^2
Isosceles right triangle (hypotenuse = S(x)): A(x) = (1/4) [S(x)]^2
Find the volume of a solid whose base is the region bounded by y = x and y = -x (from x = 0 to x = 1), and whose cross-sections perpendicular to the x-axis are squares.
The side length of each square is the distance between the two curves: S(x) = x - (-x) = 2x.
The cross-sectional area is A(x) = (2x)^2 = 4x^2.
V = \int_0^1 4x^2\,dx = \frac{4x^3}{3}\bigg|_0^1 = \frac{4}{3}Find the volume of a solid whose base is bounded by y = 3x^2 and y = 1 - x^2, with square cross-sections perpendicular to the x-axis.
First find where the curves intersect: 3x^2 = 1 - x^2, so 4x^2 = 1, giving x = +/- 1/2.
The side length is S(x) = (1 - x^2) - 3x^2 = 1 - 4x^2.
The cross-sectional area is A(x) = (1 - 4x^2)^2.
V = \int_{-1/2}^{1/2} (1 - 4x^2)^2\,dx = \int_{-1/2}^{1/2} (1 - 8x^2 + 16x^4)\,dx= \left[x - \frac{8x^3}{3} + \frac{16x^5}{5}\right]_{-1/2}^{1/2} = \frac{8}{15}The function is even, so you can also compute 2 times the integral from 0 to 1/2.
When the axis of rotation is not the x-axis or y-axis, the key adjustment is simple: every radius is a distance from the curve to that axis, not the function value itself.
If rotating about y = k, the radius from a curve y = f(x) to the axis is |f(x) - k|.
If rotating about x = k, the radius from a curve x = g(y) to the axis is |g(y) - k|.
Decide which curve is farther from the axis (outer radius R) and which is closer (inner radius r) by sketching or testing a point.
Determine the volume obtained by rotating the region bounded by y = x^2 - 2x, y = x, about the line y = 4.
First find intersection points: x^2 - 2x = x gives x^2 - 3x = 0, so x = 0 and x = 3.
On this interval, y = x is the upper curve and y = x^2 - 2x is the lower curve. Since we are rotating about y = 4 (above both curves), the curve closer to y = 4 is y = x (smaller distance), and the curve farther from y = 4 is y = x^2 - 2x (larger distance).
Inner radius: r = 4 - x (distance from y = x to y = 4)
Outer radius: R = 4 - (x^2 - 2x) = 4 - x^2 + 2x (distance from y = x^2 - 2x to y = 4)
V = \pi \int_0^3 \left[(4 - x^2 + 2x)^2 - (4 - x)^2\right] dxExpanding, simplifying, and evaluating gives V = 153pi/5.
The critical step is correctly identifying which curve gives the outer and inner radii relative to the axis of rotation, not relative to the x-axis.
Method | Formula | When to use |
|---|---|---|
General slicing (x) | V = integral of A(x) dx | Any solid with known cross-sectional area |
General slicing (y) | V = integral of A(y) dy | Cross-sections perpendicular to y-axis |
Disk (x-axis) | V = pi integral of [f(x)]^2 dx | Region touches the axis of rotation |
Disk (y-axis) | V = pi integral of [g(y)]^2 dy | Same, but integrating in y |
Washer | V = pi integral of [R^2 - r^2] dx (or dy) | Region does not touch the axis |
Square cross-section | A(x) = [S(x)]^2 | Cross-sections are squares |
Equilateral triangle | A(x) = (sqrt 3 / 4) [S(x)]^2 | Cross-sections are equilateral triangles |
Semicircle | A(x) = (pi / 8) [S(x)]^2 | Cross-sections are semicircles, diameter = S(x) |
Students often use the function value directly as the side length for cross-section problems. The side length is the distance between the two bounding curves, not the value of a single curve.
When cross-sections are perpendicular to the x-axis, students sometimes integrate with respect to y. Match the variable of integration to the axis the cross-sections are perpendicular to.
For shifted-axis problems, students frequently subtract in the wrong direction. Always compute radius as |curve value - axis value|. Sketch the setup to confirm which curve is closer and which is farther from the axis.
Students sometimes expand (R^2 - r^2) incorrectly. This is a difference of squares, not (R - r)^2. Keep the terms separate: compute R^2 and r^2 individually, then subtract.
Cross-section problems test whether you understand the general slicing principle beyond the disk/washer special case. They appear frequently as a standalone problem or as a bonus.
Shifted-axis problems are among the most common ways professors increase difficulty on a volume question. If you can handle these confidently, you can handle the hardest version of any disk/washer problem on the exam.
The formula table above is worth memorising. Having the area formulas for common shapes (square, equilateral triangle, semicircle) at your fingertips saves time under exam pressure.
True or false: For a solid with square cross-sections, the area of each slice is S(x), not [S(x)]^2.
False. The area of a square is side length squared, so A(x) = [S(x)]^2.
Fill in the blank: If rotating about y = 5, the radius from the curve y = f(x) to the axis is ______.
|f(x) - 5|, or equivalently 5 - f(x) when f(x) < 5.
True or false: When cross-sections are perpendicular to the x-axis, you should integrate with respect to y.
False. You integrate with respect to x.
Fill in the blank: The area of an equilateral triangle cross-section with side length S is ______.
(sqrt(3)/4) S^2.
Q: The base of a solid is the region bounded by y = 3x^2 and y = 1 - x^2. Cross-sections perpendicular to the x-axis are squares. Set up and evaluate the volume integral.
A: The curves meet at x = +/- 1/2. Side length S(x) = (1 - x^2) - 3x^2 = 1 - 4x^2. Area A(x) = (1 - 4x^2)^2. V = integral from -1/2 to 1/2 of (1 - 8x^2 + 16x^4) dx = 8/15.
Q: The region bounded by y = x^2 - 2x, y = x is rotated about y = 4. Set up the washer integral.
A: Intersections at x = 0 and x = 3. Inner radius r = 4 - x. Outer radius R = 4 - (x^2 - 2x) = 4 - x^2 + 2x. V = pi integral from 0 to 3 of [(4 - x^2 + 2x)^2 - (4 - x)^2] dx = 153pi/5.
Q: Explain in one sentence why rotating about y = 4 instead of y = 0 changes which curve gives the outer radius.
A: The outer radius is always the curve farther from the axis of rotation, so moving the axis above the region can swap which curve is farther and which is closer.
Q: A solid has a base bounded by y = sqrt(x) and y = -sqrt(x), from x = 0 to x = 4. Cross-sections perpendicular to the x-axis are equilateral triangles. Find the volume.
A: Side length S(x) = 2 sqrt(x). Area A(x) = (sqrt(3)/4)(2 sqrt(x))^2 = sqrt(3) x. V = sqrt(3) integral from 0 to 4 of x dx = sqrt(3) [x^2/2] from 0 to 4 = 8 sqrt(3).
The general slicing framework here is the same one used in the disk and washer methods (companion notes). Mastering it also prepares you for the method of cylindrical shells, which slices the solid differently and integrates in the other variable.
Known cross-section problems connect to later Calc II and Calc III material on computing volumes of more complex solids, including those described in polar or parametric form.
The idea of computing a geometric quantity by integrating a formula that depends on a cross-section also appears in arc length (integrating a length element) and surface area of revolution (integrating a circumference element).
Volumes by cross-sections, known cross-section method, square cross-sections, equilateral triangle cross-sections, semicircle cross-sections, shifted axis of rotation, rotation about y = k, rotation about x = k, non-standard axis, Calc II volume problems, M408D, integral calculus volumes, UT Austin Calc 2, cross-sectional area, side length function, washer method shifted axis