Volumes with cross sections: squares and rectangles — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The general slicing method for volume, cross sections perpendicular to the x-axis and y-axis, volume calculation for cross sections that are squares (side or diagonal lying in the base region), and rectangles with proportional or constant heights.

You should already know: How to compute definite integrals of common functions. How to find the area between two curves. How to rewrite functions in terms of the opposite axis variable.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Volumes with cross sections: squares and rectangles?

This topic is a core application of definite integration, falling within Unit 8: Applications of Integration, which accounts for 6-12% of the total AP Calculus BC exam score per the official Course and Exam Description (CED). Unlike volumes of revolution (another common slicing application), cross-sectional volume problems involve a solid whose slices cut at any point along an axis are explicitly defined as squares or rectangles, rather than circles or annuli generated by rotation.

The core idea is that we approximate the total volume of the solid by slicing it into infinitely many thin slabs, each with volume equal to (area of cross section) times (thickness of the slab), then add (integrate) these volumes to get the exact total. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam: it commonly appears as a 3-5 point FRQ subpart or a 1-2 point MCQ question testing integral setup. Synonyms for this method include the "slicing method" and "method of cross sections."

2. Cross Sections Perpendicular to the x-axis

This is the most common case you will encounter on the AP exam. When cross sections are perpendicular to the x-axis, for any $x$ between the left bound $a$ and right bound $b$ of the base region (which lies in the $xy$-plane), we cut the solid with a vertical plane perpendicular to the x-axis, and the intersection of this plane with the solid is our square or rectangular cross section.

The base of the solid is always bounded between two curves $y=f(x)$ and $y=g(x)$, where $f(x)\ge g(x)$ for all $x$ in $[a,b]$. The length of the cross section edge lying in the base is the vertical distance between the two curves: $s(x)=f(x)-g(x)$. The thickness of each thin slice is $\Delta x$, which becomes $dx$ when we convert the Riemann sum to a definite integral. For any cross section shape, the total volume is given by: $$V={\int }_a^{b}A(x)dx$$ where $A(x)$ is the area of the cross section at position $x$.

Worked Example

The base of a solid is the region bounded by $y=x^2$ and $y=4$ on $[0,2]$. Cross sections perpendicular to the x-axis are squares with one side lying in the base region. Find the volume of the solid.

1. Identify bounds and side length function: Bounds are $a=0$, $b=2$, and the vertical side length at any $x$ is $s(x)=4-x^2$ (upper curve minus lower curve).
2. Calculate cross-sectional area: For squares, area is side length squared, so $A(x)=(4-x^2{)}^2=16-8x^2+x^4$.
3. Set up the volume integral: $V={\int }_0^2(16-8x^2+x^4)dx$.
4. Evaluate using the Fundamental Theorem of Calculus: $$V={\left[16x-\frac{8}{3}{x}^3+\frac{1}{5}{x}^5\right]}_0^2=32-\frac{64}{3}+\frac{32}{5}=\frac{256}{15}\approx 17.07$$

Exam tip: Always confirm the orientation of the cross section: if it is perpendicular to the x-axis, all functions and side lengths must be written in terms of x, and your bounds are x-values.

3. Cross Sections Perpendicular to the y-axis

When problems specify cross sections cut perpendicular to the y-axis, slices are horizontal rather than vertical, which changes how we calculate side length and set up the integral. For this case, we work with the base region bounded between $x=f(y)$ and $x=g(y)$, where $f(y)\ge g(y)$ for all $y$ between the lower bound $c$ and upper bound $d$.

The side length of the cross section lying in the base is now the horizontal distance between the two curves: $s(y)=f(y)-g(y)$. The thickness of each slice is $\Delta y$, which becomes $dy$ in the definite integral, so the total volume formula becomes: $$V={\int }_c^{d}A(y)dy$$ If your original curves are given as $y=f(x)$, you must invert them to get $x$ as a function of $y$ before calculating side length. This formula works for both squares and rectangles, just like the perpendicular-to-x case.

Worked Example

The base of a solid is the region bounded by $y=\sqrt{x}$, the x-axis, and $x=4$. Cross sections perpendicular to the y-axis are squares with one side lying in the base. Find the volume of the solid.

1. Rewrite curves as functions of y: $y=\sqrt{x}$ becomes $x=y^2$, and bounds for $y$ are $c=0$ (at $x=0$) to $d=2$ (at $x=4$).
2. Calculate side length: The horizontal distance between the right curve $x=4$ and left curve $x=y^2$ is $s(y)=4-y^2$.
3. Calculate cross-sectional area: $A(y)=(4-y^2{)}^2=16-8y^2+y^4$.
4. Set up and evaluate the integral: $$V={\int }_0^2(16-8y^2+y^4)dy={\left[16y-\frac{8}{3}{y}^3+\frac{1}{5}{y}^5\right]}_0^2=\frac{256}{15}\approx 17.07$$

Exam tip: If cross sections are perpendicular to the y-axis, always invert your functions to get x in terms of y before calculating side length — never leave the side length in terms of x when integrating with respect to y.

4. Special Cases: Squares with Diagonal in Base and Rectangular Cross Sections

Not all problems use the default case of squares with their side lying in the base. Two common tested variations are squares with their diagonal lying in the base, and rectangles with height proportional to the base side length. For both cases, you just need to correctly relate the given base length to the cross-sectional area:

• For a square with diagonal $d$ (equal to the distance between the two base curves), we use the geometric relationship $d=s\sqrt{2}$, so $s=d/\sqrt{2}$, and area $A=s^2=d^2/2$.
• For a rectangle with base side $s$ (equal to the distance between the two base curves) and height $h=ks$ (where $k$ is a constant given in the problem), area $A=s\times h=k{s}^2$. If height is a constant $h$, then $A=sh$.

Worked Example

The base of a solid is the region bounded by $y=x$ and $y=x^2$ on $[0,1]$. Cross sections perpendicular to the x-axis are squares whose diagonal lies in the base region. Find the volume of the solid.

1. Find diagonal length at any x: $d(x)=x-x^2$, since $y=x$ is above $y=x^2$ on $[0,1]$.
2. Relate diagonal to cross-sectional area: $A(x)=\frac{d(x{)}^2}{2}=\frac{(x-x^2{)}^2}{2}=\frac{x^2-2x^3+x^4}{2}$.
3. Set up and evaluate the integral: $$V={\int }_0^1\frac{x^2-2x^3+x^4}{2}dx=\frac{1}{2}{\left[\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right]}_0^1=\frac{1}{2}\left(\frac{10-15+6}{30}\right)=\frac{1}{60}$$

Exam tip: Always read whether the side, diagonal, or diameter of the square lies in the base — these give different area formulas, and the AP exam will always test whether you check this detail.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When the diagonal of the square is in the base, you calculate area as $d^2$ instead of $d^2/2$. Why: You confuse side length with diagonal length, defaulting to the common case where the side is in the base without reading the problem. Correct move: Underline whether the given segment in the base is the side or diagonal of the square before starting your calculation.
• Wrong move: For cross sections perpendicular to the y-axis, you leave the side length in terms of x and integrate with respect to y. Why: You forget that the variable of integration must match the axis the cross sections are perpendicular to, and skip the step of inverting functions. Correct move: After reading the cross section orientation, write all functions and side lengths in terms of the axis variable (x for perpendicular to x, y for perpendicular to y) before setting up the integral.
• Wrong move: You subtract the curves in the wrong order, getting a negative side length, then forget to take the absolute value. Why: You just subtract top from bottom or right from left by habit, but don't check which curve is larger in the interval. Correct move: After finding the side length, confirm it is positive for all points in the interval of integration; swap the order of subtraction if needed.
• Wrong move: For a rectangle with height proportional to the base side, you use area $=ks$ instead of $k{s}^2$. Why: You forget that area of a rectangle is base times height, not just base times the constant of proportionality. Correct move: Write down "base = s, height = k s, area = s * k s" explicitly before moving on, to avoid skipping the multiplication step.
• Wrong move: When integrating the square of a binomial like $(a-b{x}^n{)}^2$, you expand it to $a^2-b^2{x}^{2n}$ instead of $a^2-2ab{x}^n+b^2{x}^{2n}$. Why: You rush the algebra step, skipping the middle term of the square of a difference. Correct move: Always expand the square of a binomial term-by-term before integrating to avoid missing the cross term.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The base of a solid is the region bounded by $y=e^x$, $y=0$, $x=0$, and $x=1$. Cross sections perpendicular to the x-axis are rectangles where the height of each rectangle is equal to 3 times the length of its base in the xy-plane. What is the volume of the solid? A) $\frac{3}{2}(e^2-1)$ B) $3(e^2-1)$ C) $\frac{3}{2}(e-1)$ D) $3(e-1)$

Worked Solution: First, the base side length at any $x$ is $s(x)=e^x-0=e^x$. The height of the rectangle is given as 3 times the base side length, so $h=3e^x$. The cross-sectional area is $A(x)=s\cdot h=e^x\cdot 3e^x=3e^{2x}$. The volume is the integral of $A(x)$ from 0 to 1: $V={\int }_0^{1}3e^{2x}dx={\left[\frac{3}{2}{e}^{2x}\right]}_0^1=\frac{3}{2}{e}^2-\frac{3}{2}{e}^0=\frac{3}{2}(e^2-1)$. The correct answer is A.

Question 2 (Free Response)

Let $R$ be the region bounded by $y=4-x^2$ and the x-axis ($y=0$). (a) The base of a solid $S$ is $R$. Cross sections perpendicular to the x-axis are squares with one side lying in $R$. Write, but do not evaluate, an integral expression that gives the volume of $S$. (b) A different solid $T$ has base $R$, with cross sections perpendicular to the y-axis that are rectangles with height equal to $y$. Find the volume of $T$. (c) How would the integral for volume change if cross sections for solid $S$ were perpendicular to the x-axis and squares whose diagonal lies in $R$? Justify your answer.

Worked Solution: (a) First, find bounds of $R$: set $4-x^2=0$ to get $x=\pm 2$, so bounds are $x=-2$ to $x=2$. The side length at any $x$ is $s(x)=4-x^2$, so area $A(x)=(4-x^2{)}^2$. The volume integral is: $$V_S={\int }_{-2}^2(4-x^2{)}^{2}dx$$

(b) Rewrite $R$ as functions of $y$: $y=4-x^2⟹x=\pm \sqrt{4-y}$, with bounds $y=0$ to $y=4$. The base side length is $s(y)=\sqrt{4-y}-(-\sqrt{4-y})=2\sqrt{4-y}$. Area is $A(y)=s(y)\cdot y=2y\sqrt{4-y}$. Use substitution $u=4-y$, $y=4-u$, $dy=-du$: $$V_T={\int }_0^{4}2y\sqrt{4-y}dy=2{\int }_0^4(4u^{1/2}-u^{3/2})du=2{\left[\frac{8}{3}{u}^{3/2}-\frac{2}{5}{u}^{5/2}\right]}_0^4=\frac{128}{3}\approx 42.67$$

(c) If the diagonal lies in the base instead of the side, the area formula becomes $A(x)=\frac{(4-x^2{)}^2}{2}$, so the new integral is $V_S={\int }_{-2}^2\frac{(4-x^2{)}^2}{2}dx$, which is half the value of the original integral.

Question 3 (Application / Real-World Style)

A structural engineer is designing a tapered concrete support pillar for a pedestrian bridge. The pillar has a height of 4 meters along the vertical z-axis (from $z=0$ at the base to $z=4$ at the top). Horizontal cross sections cut perpendicular to the z-axis are rectangles. At height $z$, the base of the rectangle is $(4-z)$ meters, and the width of the rectangle is half the base. What is the total volume of concrete needed to cast the pillar, in cubic meters?

Worked Solution: Cross sections are perpendicular to the z-axis, so we integrate with respect to $z$ from 0 to 4. At any height $z$, the base of the rectangle is $s(z)=4-z$, and the width is $h(z)=\frac{1}{2}(4-z)$. The cross-sectional area is $A(z)=s(z)\cdot h(z)=(4-z)\cdot \frac{1}{2}(4-z)=\frac{1}{2}(4-z{)}^2$. The total volume is: $$V={\int }_0^4\frac{1}{2}(4-z{)}^{2}dz=\frac{1}{2}{\left[-\frac{(4-z{)}^3}{3}\right]}_0^4=\frac{1}{2}\left(0+\frac{64}{3}\right)=\frac{32}{3}\approx 10.67$$ Interpretation: The pillar requires approximately 10.67 cubic meters of concrete, which weighs roughly 25,600 kilograms for structural load calculations.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational introduction to the slicing method for volume, which you will immediately extend to other cross-sectional shapes and volumes of revolution next. Volumes of revolution (disks/washers and cylindrical shells) are just special cases of the slicing method: disks are circular cross sections, so you replace the square/rectangle area formula with the area of a circle. Without mastering the setup of cross-sectional volume integrals, it is very easy to mix up variables and formulas for volumes of revolution, which are a larger tested portion of the AP Calculus BC exam. This topic also reinforces your core skill of finding the distance between two curves, which is used in nearly all applications of integration. The follow-on topics build directly on the skills you learned here: Volumes of revolution: disks and washers Volumes of revolution: cylindrical shells Arc length of planar curves Area between parametric curves