Difficulty: Intermediate to Advanced | Prerequisites: Definite integrals, area between curves, basic understanding of 3D solids
Once you can find the area of a region between curves, the next step in AP Calculus AB is to use that region as the base of a three-dimensional solid. The idea: if you know the shape of every cross section (squares, semicircles, equilateral triangles, etc.), you can integrate those cross-sectional areas along an axis to get the total volume. This sits in the same unit as disc/washer and shell methods, but it is conceptually simpler because you never revolve anything.
This set of notes also covers how to write (but not solve) an integral equation whose solution gives a particular value, such as a line that divides a region into equal areas. That type of question tests whether you understand what an integral represents well enough to set it up as a condition, not just as a computation.
You should be comfortable with area between curves and evaluating definite integrals before tackling this material.
To find the volume of a solid with known cross sections, integrate the area of a single cross-sectional slice along the axis perpendicular to it. For "write but do not solve" integral equation questions, set up the integral expression that equals the required condition (e.g., half the total area) and leave it unsolved. Both question types reward the setup more than the arithmetic.
Cross section
A slice of a solid taken perpendicular to a given axis. The shape of that slice (square, semicircle, triangle, etc.) is stated in the problem.
Think of it as cutting through a loaf of bread: each slice has a specific shape.
Known cross sections (volumes with known cross-sectional area)
A class of volume problems where you are told the shape of every cross section and must integrate the area of that shape along the axis to find the total volume.
Side length of the cross section
The distance across the base region at a given x (or y) value. For cross sections perpendicular to the x-axis, this is typically the vertical distance between the top and bottom curves at that x.
"Write, but do not evaluate" / "Write, but do not solve"
An AP exam instruction meaning you must produce the correct integral expression (or equation) with the right integrand and limits, but you do not need to compute the numerical answer. Full marks are awarded for the setup alone.
Equal-area condition
A problem type where a horizontal or vertical line divides a region into two parts of equal area, and you must write an integral equation whose solution gives the position of that line.
The setup is always the same:
You have a region in the xy-plane (your base).
Cross sections perpendicular to one axis have a stated geometric shape.
You integrate the area of that shape along the axis.
Finding the side length. The side (or diameter, or base) of each cross section equals the distance across the region at that point. If cross sections are perpendicular to the x-axis, the side length at a given x is the vertical distance between the bounding curves: top function minus bottom function.
Computing cross-sectional area. Once you have the side length s(x):
Square cross section: A(x) = [s(x)]²
Semicircular cross section: A(x) = (π/2) × [s(x)/2]² = πs(x)²/8
Equilateral triangle cross section: A(x) = (√3/4) × [s(x)]²
Isosceles right triangle (leg = s): A(x) = s(x)²/2
Integrating for volume:
V = \int_a^b A(x)\,dxRegion R is bounded by the x-axis, y = ln x, and y = 5 − x in the first quadrant. Cross sections perpendicular to the x-axis are squares.
The intersection of y = ln x and y = 5 − x occurs at x ≈ 3.69344 (call it A). Also, ln x = 0 at x = 1 and 5 − x = 0 at x = 5.
From x = 1 to x = A: the top curve is y = ln x, the bottom is y = 0 (the x-axis). The side length of each square is ln x.
From x = A to x = 5: the top curve is y = 5 − x, the bottom is y = 0. The side length of each square is 5 − x.
Since each cross section is a square, A(x) = (side length)².
\text{Volume} = \int_1^A (\ln x)^2\,dx + \int_A^5 (5-x)^2\,dxThis is a "write, but do not evaluate" problem, so you stop here. No need to compute the number.
The College Board awards 2 points for getting both integrands correct and 1 point for combining them into a single expression for total volume.
Some FRQ parts ask you to write an equation involving integrals whose solution gives a specific value. The 2012 Question 2(c) is a classic example.
The problem: The horizontal line y = k divides region R into two regions of equal area. Write (but do not solve) an equation whose solution gives k.
The approach:
You already know the total area of R (from part a): approximately 2.986.
The line y = k cuts R horizontally. The area below y = k must equal half the total area.
Using the dy setup from part (a), the area from y = 0 to y = k is:
\int_0^k \bigl[(5-y) - e^y\bigr]\,dySet this equal to half the total:
\int_0^k \bigl[(5-y) - e^y\bigr]\,dy = \frac{1}{2} \cdot 2.986That is the equation. You do not solve for k.
The College Board awards 1 point each for the correct integrand, the correct limits (0 to k), and a valid equation (setting the integral equal to half the total area).
General volume with known cross sections (perpendicular to the x-axis):
V = \int_a^b A(x)\,dxwhere A(x) is the area of the cross section at position x.
Common cross-sectional areas (where s is the side length at a given x):
Shape | Area formula |
|---|---|
Square | s² |
Semicircle (diameter = s) | πs²/8 |
Equilateral triangle (side = s) | (√3/4)s² |
Isosceles right triangle (leg = s) | s²/2 |
The side length s(x) is the distance across the base region at that x, which is the vertical distance between the bounding curves: top function minus bottom function.
Volumes with known cross sections model real physical objects. A dam with a trapezoidal cross section, an airplane fuselage with a roughly circular cross section, or a bridge pier with a varying rectangular profile can all be computed this way. Integral equations show up in engineering whenever you need to find a threshold, such as the water level at which a tank is half full.
Students often forget to square the side length for square cross sections. The side length is (top − bottom), but the cross-sectional area is (top − bottom)². Missing the square loses the integrand point.
Confusing the side length with the area. For a semicircular cross section where the diameter equals the distance across the base, students frequently plug in the full distance as the radius. The radius is half the distance.
Splitting the volume integral in the wrong place (or not at all). If the base region has two different expressions for the height, the volume integral must split at the same x-value.
For equal-area problems, students sometimes set the integral equal to the total area instead of half the total area. Read the condition carefully.
Writing an integral from 0 to k using dx when the dividing line is horizontal (y = k). A horizontal cut is naturally handled with a dy integral, and the upper limit is k as a y-value.
⚠️ Volumes with known cross sections appear regularly as part (b) of an AP Calculus AB FRQ. The problem nearly always says "write, but do not evaluate," so full credit depends entirely on the setup.
⚠️ The most common cross-section shape on the exam is the square. Semicircles and equilateral triangles have also appeared. Know all the area formulas.
⚠️ "Write but do not solve" integral equations (like the equal-area problem) are worth 3 points and test conceptual understanding. You must show the correct integrand, correct limits, and a correct equation. No calculator work is needed.
⚠️ When the problem says "one or more integrals," it is signalling that you may need to split. Always check whether the bounding functions change across the interval.
True or False: For a solid with square cross sections perpendicular to the x-axis, you integrate [f(x) − g(x)] dx to find the volume. Answer: False. You integrate [f(x) − g(x)]² dx, because the area of each square is the side length squared.
Fill in the blank: If the cross sections are semicircles with diameter equal to the distance between the curves, the area of each cross section is ______. Answer: πs²/8 (where s is the diameter, i.e. the distance between the curves)
True or False: In a "write, but do not solve" problem, you lose points if you do not compute the numerical answer. Answer: False. You earn full marks for a correct setup. Evaluating is not required (and attempting it can introduce errors).
Fill in the blank: To write an equation for y = k dividing a region into equal areas, set the integral from 0 to k equal to ______ of the total area. Answer: one half
Q: Region R is bounded by y = √x and y = x/3 from x = 0 to x = 9. Cross sections perpendicular to the x-axis are squares. Write an expression for the volume of the solid.
A: The side length at each x is √x − x/3 (since √x ≥ x/3 on [0, 9]). Volume = ∫ from 0 to 9 of (√x − x/3)² dx.
Q: Using the same region R, suppose cross sections perpendicular to the x-axis are equilateral triangles. Write the volume expression.
A: The side length is still √x − x/3. The area of an equilateral triangle with side s is (√3/4)s². Volume = ∫ from 0 to 9 of (√3/4)(√x − x/3)² dx.
Q: The region S is bounded by y = x² and y = 4. The vertical line x = c divides S into two regions of equal area. Write an equation whose solution gives c. (Assume c > 0.)
A: Total area = ∫ from 0 to 2 of (4 − x²) dx (using symmetry or the right half). Setting the area from 0 to c equal to half the total: ∫ from 0 to c of (4 − x²) dx = (1/2) × ∫ from 0 to 2 of (4 − x²) dx.
Q: In the 2012 FRQ, part (c), why is a dy integral more natural for the equal-area condition y = k than a dx integral?
A: The dividing line y = k is horizontal, and using dy means the upper limit of integration changes directly with k. With dx, the dividing line would create an awkward piecewise boundary that depends on which curve is on top, making the integral harder to express cleanly.
This connects to area between curves, which provides the base region and the side-length function for every cross-section problem. It also connects to volumes of revolution (disc/washer and shell methods), which are a special case where the cross section is always a circle or a washer. The "write, but do not solve" skill reappears in differential equations (setting up but not solving a separable DE) and in accumulation function problems.
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