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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Finding volumes of solids with known square and rectangular cross sections perpendicular to the x-axis or y-axis, deriving cross-sectional area formulas, setting up and evaluating definite integrals for volume, and identifying bounds for the base region of the solid.

You should already know: How to find the area between two curves, how to evaluate definite integrals, how to rewrite functions in terms of the opposite 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 AB 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 part of Unit 8: Applications of Integration in the AP Calculus AB CED, accounting for roughly 4-6% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. The core idea is that we can find the volume of an irregular solid by slicing it into infinitely many thin parallel slices, each with a known cross-sectional shape (here, squares or rectangles), then summing the volume of all slices via integration. The base of the solid is always a bounded region in the xy-plane, and each cross section is perpendicular to either the x-axis or the y-axis, per AP exam convention. For any slice at a given x or y, the side length (or base/height for non-square rectangles) of the cross section is determined by the distance between the two curves that bound the base region. Unlike finding the area between two curves, where we sum the area of thin slices, here we calculate the volume of each thin slice as cross-sectional area multiplied by slice thickness (dx for vertical slices, dy for horizontal slices), then integrate to accumulate the total volume across the entire base region.

2. Square cross sections perpendicular to the x-axis

When cross sections are perpendicular to the x-axis, each slice is vertical, so we integrate with respect to x. First, we find the bounds of integration: the x-values that span the entire base region, calculated by finding the intersection points of the curves that bound the base, or using given boundaries like the x-axis or y-axis. Next, for any x between the bounds, the side length of the square cross section $s(x)$ equals the vertical distance between the upper bounding curve $y_{top}(x)$ and the lower bounding curve $y_{bottom}(x)$: $$s(x)=y_{top}(x)-y_{bottom}(x)$$ Because the cross section is a square, its area $A(x)$ is side length squared: $$A(x)=[s(x){]}^2={\left(y_{top}(x)-y_{bottom}(x)\right)}^2$$ The volume of each thin slice is approximately $A(x)\cdot dx$, so the total volume $V$ is the definite integral from the leftmost bound $x=a$ to the rightmost bound $x=b$: $$V={\int }_a^{b}A(x)dx={\int }_a^b{\left(y_{top}(x)-y_{bottom}(x)\right)}^{2}dx$$ Intuition: If you stack sheets of paper vertically to build a solid, each sheet is a thin cross section. Thicker sheets add more volume, which is why we multiply area by dx. Infinitely many infinitely thin sheets give the exact total volume.

Worked Example

Base of a solid is bounded by $y=x^2$, $y=4$, and the y-axis, for $x\ge 0$. Cross sections perpendicular to the x-axis are squares. Find the volume of the solid.

1. Find bounds of integration for x: Intersection of $y=x^2$ and $y=4$ is at $x^2=4$, so $x=2$ (since $x\ge 0$). Left bound is $x=0$, so $a=0$, $b=2$.
2. Calculate side length: Upper curve is $y=4$, lower curve is $y=x^2$, so $s(x)=4-x^2$.
3. Calculate cross-sectional area: $A(x)=(4-x^2{)}^2=16-8x^2+x^4$.
4. Set up and evaluate the integral: $$V={\int }_0^2(16-8x^2+x^4)dx={\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]}_0^2$$ $$=32-\frac{64}{3}+\frac{32}{5}=\frac{256}{15}\approx 17.07$$
5. The volume of the solid is $\frac{256}{15}$ cubic units.

Exam tip: Always expand the squared binomial for square cross sections before integrating—attempting to integrate the unexpanded binomial with incorrect substitution leads to nearly universal errors on this problem type.

3. Square cross sections perpendicular to the y-axis

When cross sections are perpendicular to the y-axis, slices are horizontal, so we integrate with respect to y. The core logic is identical to the perpendicular-to-x case, just swapped axes. First, find bounds of integration for y: the y-values that span the entire base region, from the bottommost bound $y=c$ to the topmost bound $y=d$. Next, the side length $s(y)$ is the horizontal distance between the right bounding curve $x_{right}(y)$ and the left bounding curve $x_{left}(y)$: $$s(y)=x_{right}(y)-x_{left}(y)$$ For square cross sections, area is still side length squared: $A(y)=[s(y){]}^2$, so total volume is: $$V={\int }_c^{d}A(y)dy={\int }_c^d{\left(x_{right}(y)-x_{left}(y)\right)}^{2}dy$$ The key difference from perpendicular-to-x cross sections is that all boundary curves must be rewritten as functions of y, rather than functions of x. This requires solving for x in terms of y if the original curve was given as $y=f(x)$, a step many students skip.

Worked Example

Base of a solid is bounded by $y=x^2$, $y=4$, and the y-axis, for $x\ge 0$. Cross sections perpendicular to the y-axis are squares. Find the volume of the solid.

1. Rewrite boundary curves as functions of y: $y=x^2\to x=\sqrt{y}$ (since $x\ge 0$). Left bound is $x=0$, right bound is $x=\sqrt{y}$.
2. Find y bounds: The base spans from $y=0$ (origin) to $y=4$, so $c=0$, $d=4$.
3. Calculate side length and area: $s(y)=\sqrt{y}-0=\sqrt{y}$, so $A(y)=(\sqrt{y}{)}^2=y$.
4. Integrate to find volume: $$V={\int }_0^{4}ydy={\left[\frac{1}{2}{y}^2\right]}_0^4=8$$ The volume of the solid is 8 cubic units.

Exam tip: Always solve for x explicitly before writing the side length for cross sections perpendicular to the y-axis. Skipping this step leaves you with a mix of x and y terms in the integral, which cannot be evaluated correctly.

4. Rectangular cross sections with given height

Squares are a special case of rectangular cross sections, where the height of the rectangle equals the base length (the side across the base region). AP exams frequently ask for general rectangular cross sections where the height is a constant, multiple of the base, or function of position. For any rectangular cross section, area equals base times height: $A=b\cdot h$. The base of the rectangle is always the distance across the base region, just like for squares: if perpendicular to the x-axis, $b(x)=y_{top}(x)-y_{bottom}(x)$, and if perpendicular to the y-axis, $b(y)=x_{right}(y)-x_{left}(y)$. The height is explicitly given in the problem statement, so you must read carefully to get the correct height expression. The general volume formulas are:

• Perpendicular to x-axis: $$V={\int }_a^{d}b(x)\cdot h(x)dx$$
• Perpendicular to y-axis: $$V={\int }_c^{d}b(y)\cdot h(y)dy$$ Squares follow this formula exactly, with $h=b$, so $A=b\cdot b=b^2$, which matches our earlier result.

Worked Example

Base of a solid is bounded by $y=\sin x$, $y=0$, $x=0$, and $x=\pi$. Cross sections perpendicular to the x-axis are rectangles where the height of each rectangle is 3 times the base length. Find the volume of the solid.

1. Bounds for x are given: $a=0$, $b=\pi$.
2. Calculate base length: Upper curve is $y=\sin x$, lower curve is $y=0$, so $b(x)=\sin x-0=\sin x$.
3. Calculate height and area: Height $h(x)=3\sin x$, so $A(x)=(\sin x)(3\sin x)=3{\sin }^{2}x$.
4. Evaluate the integral using the power-reduction identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$: $$V={\int }_0^{\pi }3\cdot \frac{1-\cos 2x}{2}dx=\frac{3}{2}{\left[x-\frac{1}{2}\sin 2x\right]}_0^{\pi }=\frac{3\pi }{2}\approx 4.71$$ The volume of the solid is $\frac{3\pi }{2}$ cubic units.

Exam tip: Never automatically assume the cross section is a square—always check if the problem says rectangle, and confirm the relationship between base and height before writing the area formula.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For cross sections perpendicular to the y-axis, you leave boundaries as $y=f(x)$ and use vertical distance for side length, same as for perpendicular to x-axis. Why: You don't stop to confirm the direction of the cross section, and rely on muscle memory instead of checking. Correct move: Always label the axis the cross section is perpendicular to, then confirm side length is vertical (perpendicular to x) or horizontal (perpendicular to y) before writing the area formula.
• Wrong move: You leave $(y_{top}-y_{bottom}{)}^2$ unexpanded and incorrectly integrate it as $\frac{(y_{top}-y_{bottom}{)}^3}{3(y_{top}^{\prime }-y_{bottom}^{\prime })}$. Why: You misapply the power rule for integration to composite functions without checking substitution requirements. Correct move: Always expand the squared binomial term by term before integrating.
• Wrong move: For rectangular cross sections, you use only the given height as the area, ignoring the base distance across the region. Why: You misread the problem and assume the given height is the full area. Correct move: For any rectangle, always calculate the base as the distance across the base region first, then multiply by the given height to get area.
• Wrong move: You automatically use the intersection points of two curves for bounds, ignoring a third boundary (like the y-axis or x-axis) when the base is bounded by three curves. Why: You don't sketch the base region, so you miss the boundary line. Correct move: Always sketch the base region and label all boundaries to confirm bounds before setting up the integral.
• Wrong move: For cross sections perpendicular to the y-axis, you integrate with respect to y but leave boundary curves as $y=f(x)$, leading to an integral with mixed variables. Why: You remember to integrate with respect to y, but forget to rewrite all curves as $x=f(y)$. Correct move: Always solve all boundaries for x in terms of y before writing the side length.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The base of a solid is the region bounded by $y=2x$ and $y=x^2$. Cross sections perpendicular to the x-axis are squares. What is the volume of the solid? A) $\frac{8}{15}$ B) $\frac{16}{15}$ C) $\frac{32}{15}$ D) $8$

Worked Solution: First find intersection points to get bounds: set $2x=x^2$, so $x(x-2)=0$, giving bounds $x=0$ to $x=2$. The side length of each square is upper curve minus lower curve: $2x-x^2$. Cross-sectional area is $(2x-x^2{)}^2=4x^2-4x^3+x^4$. Integrate from 0 to 2: $V={\left[\frac{4x^3}{3}-x^4+\frac{x^5}{5}\right]}_0^2=\frac{32}{3}-16+\frac{32}{5}=\frac{16}{15}$. The correct answer is B.

Question 2 (Free Response)

The base of a solid is bounded by $y=e^x$, $y=1$, $x=0$, and $x=1$. (a) Cross sections perpendicular to the x-axis are squares. Set up, but do not evaluate, the definite integral that gives the volume of the solid. (b) Cross sections perpendicular to the x-axis are rectangles with height equal to $x^2$ times the base length of the cross section. Set up and evaluate the integral for the volume of the solid. (c) The volume of the solid with square cross sections is equal to $V=k$, where $k$ is a constant. A second solid has cross sections perpendicular to the x-axis that are squares with all side lengths doubled. What is the volume of the second solid in terms of $k$?

Worked Solution: (a) Bounds for $x$ are $0$ to $1$, side length is $e^x-1$, so the volume integral is: $$V={\int }_0^1(e^x-1{)}^{2}dx$$ (b) Base length is $e^x-1$, height is $x^2(e^x-1)$, so area is $x^2(e^x-1{)}^2=x^2(e^{2x}-2e^x+1)$. Using integration by parts to evaluate: $$V={\int }_0^1(x^2{e}^{2x}-2x^2{e}^x+x^2)dx=\frac{e^2}{4}-2e+\frac{47}{12}\approx 1.05$$ (c) If all side lengths are doubled, the area of each cross section becomes $(2s{)}^2=4s^2=4A(x)$, so the volume scales by a factor of 4: $V_{new}=4k$.

Question 3 (Application / Real-World Style)

A civil engineer is designing a water channel for a new development. A cross section of the channel perpendicular to its length is bounded on the top by $y=0$ (ground level) and below by $y=-0.25x^2$ for $x$ between $-4$ and $4$ (units in meters). The channel extends 100 meters along its length, and cross sections perpendicular to the length are rectangles: the depth at any $x$ is the distance from ground to the channel bottom, and the channel length is the height of the rectangle. What is the total volume of excavated material needed to build the channel?

Worked Solution: For any $x$ between $-4$ and $4$, the base (depth) of the rectangle is $0-(-0.25x^2)=0.25x^2$. The height of the rectangle is the constant 100 meters (channel length), so cross-sectional area $A(x)=(0.25x^2)(100)=25x^2$. Volume is the integral of area from $x=-4$ to $x=4$: $$V={\int }_{-4}^{4}25x^{2}dx=25{\left[\frac{x^3}{3}\right]}_{-4}^4=25\left(\frac{64}{3}-\frac{-64}{3}\right)=\frac{3200}{3}\approx 1066.67$$ Interpretation: The total volume of material that must be excavated to build the channel is approximately 1067 cubic meters.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of cross-section volumes is a critical prerequisite for the next core topic in Unit 8: volumes of revolution. Volumes of revolution can be thought of as a special case of cross-section volumes, where cross sections are circles (or washers) perpendicular to an axis. The skills you learned here—finding integration bounds, identifying the distance between bounding curves, setting up an integral to accumulate cross-sectional area to get volume—transfer directly to the disk/washer method for volumes of revolution. If you can set up a cross-section volume for squares, you can almost immediately set up a volume of revolution with washers, since the integral structure is identical. This topic also reinforces the core idea of integration as an accumulation tool, which is central to all applications of integration in AP Calculus AB. Without mastering the cross-section setup here, you will struggle to correctly set up volume integrals for revolution problems.

• Volumes of revolution: disk and washer method
• Area between two curves
• Integration by parts for definite integrals