# Volumes with cross sections: squares and rectangles

> AP Calculus AB · Unit 8: Applications of Integration
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u8-volumes-with-cross-sections-squares/

This module teaches you to find volumes of irregular solids with known square and rectangular cross sections perpendicular to the x or y axis. You will learn to derive area formulas, set up and evaluate integrals, and identify correct bounds for base regions.

**Prerequisites:** [Finding the area between two curves](https://www.owlsprep.com/study/ap-calculus-ab-u8-area-between-curves/); Evaluating definite integrals; Rewriting functions in terms of the opposite variable

## Learning objectives

- Identify correct bounds of integration for cross-section volume problems
- Calculate cross-sectional area for squares and rectangles perpendicular to the x-axis or y-axis
- Set up and evaluate definite integrals to find total volume
- Avoid common exam errors in cross-section volume problems

## Core Concept: Cross-Sectional Volume Slicing

This topic accounts for roughly 4-6% of your total AP Calculus AB exam score, and appears in both multiple-choice and free-response sections.

The core idea is to split an irregular solid into infinitely many thin parallel slices, each with a known cross-sectional shape. The base of the solid is always a bounded region in the xy-plane, and cross sections are perpendicular to either the x-axis or y-axis per AP exam convention.

**Cross-Sectional Volume** — Total volume is found by accumulating the volume of thin slices. Each slice's volume equals cross-sectional area multiplied by slice thickness (dx for vertical slices, dy for horizontal slices).

*Notation:* V = \int_a^b A(x) \, dx

*Example:* For a vertical slice at position $x$, volume ≈ $A(x)dx$, so total volume is the integral of $A(x)$ across all bounds.

## Square Cross Sections Perpendicular to the X-Axis

When cross sections are perpendicular to the x-axis, slices are vertical, so we integrate with respect to x. First find bounds of integration, which are the x-values that span the entire base region.

$$s(x) = y_{top}(x) - y_{bottom}(x)$$

$$A(x) = [s(x)]^2 = \left(y_{top}(x) - y_{bottom}(x)\right)^2$$

$$V = \int_a^b \left(y_{top}(x) - y_{bottom}(x)\right)^2 dx$$

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

1. Find bounds of integration for x

   $$x^2 = 4 \implies x = 2 \ (x \geq 0), \text{ so } a=0, b=2$$
2. Calculate side length of the square cross section

   $$s(x) = 4 - x^2$$
3. Calculate cross-sectional area by expanding the squared side length

   $$A(x) = (4 - x^2)^2 = 16 - 8x^2 + x^4$$
4. Set up and evaluate the definite integral

   $$V = \int_0^2 (16 - 8x^2 + x^4) dx = \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2$$
5. Simplify to get the final volume

   $$V = 32 - \frac{64}{3} + \frac{32}{5} = \frac{256}{15} \approx 17.07$$

> **Exam tip:** Always expand the squared binomial for square cross sections before integrating—unexpanded binomials almost always lead to incorrect integration.

## 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 x-axis case, just with swapped axes. All boundary curves must be rewritten as functions of y, a step many students forget.

$$s(y) = x_{right}(y) - x_{left}(y)$$

$$V = \int_c^d \left(x_{right}(y) - x_{left}(y)\right)^2 dy$$

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

1. Rewrite all boundary curves as functions of y

   $$y = x^2 \implies x = \sqrt{y} \ (x \geq 0), \text{ left bound } x=0$$
2. Base spans from $y=0$ (origin) to $y=4$, so $c=0$, $d=4$
3. Calculate side length and cross-sectional area

   $$s(y) = \sqrt{y} - 0 = \sqrt{y}, \quad A(y) = (\sqrt{y})^2 = y$$
4. Integrate to find total volume

   $$V = \int_0^4 y dy = \left[\frac{1}{2}y^2\right]_0^4 = 8$$

> **Exam tip:** Always solve for x explicitly before calculating side length for cross sections perpendicular to the y-axis to avoid mixed variables in your integral.

## General Rectangular Cross Sections

Squares are a special case of rectangular cross sections, where the height of the rectangle equals the base length (the distance across the base region). AP exams frequently ask for general rectangles where height is a constant, multiple of the base, or function of position. For any rectangle, area equals base × height:

- Perpendicular to x-axis: Base $b(x) = y_{top}(x) - y_{bottom}(x)$, Volume $V = \int_a^b b(x) \cdot h(x) dx$
- Perpendicular to y-axis: Base $b(y) = x_{right}(y) - x_{left}(y)$, Volume $V = \int_c^d b(y) \cdot h(y) dy$

**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. $a=0$, $b=\pi$
2. Calculate base length of the rectangle

   $$b(x) = \sin x - 0 = \sin x$$
3. Calculate height and cross-sectional area

   $$h(x) = 3 \sin x, \quad A(x) = (\sin x)(3 \sin x) = 3 \sin^2 x$$
4. Use the power-reduction identity and evaluate

   $$\sin^2 x = \frac{1 - \cos 2x}{2} \\ V = \int_0^\pi 3 \cdot \frac{1 - \cos 2x}{2} dx = \frac{3}{2} \left[ x - \frac{1}{2} \sin 2x \right]_0^\pi = \frac{3\pi}{2} \approx 4.71$$

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. 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?

   - $\frac{8}{15}$
   - $\frac{16}{15}$
   - $\frac{32}{15}$
   - $8$

   *Answer:* $\frac{16}{15}$

   *Why:* Correct! The integral expands and evaluates to $\frac{16}{15}$. If you got a different answer, check that you correctly expanded $(2x - x^2)^2$ and evaluated the antiderivative at the bounds.

> **Exam tip:** Never automatically assume the cross section is a square—always check the problem statement and confirm the relationship between base and height before writing the area formula.

## Common pitfalls

- **Wrong:** For cross sections perpendicular to the y-axis, leave boundaries as $y=f(x)$ and use vertical distance for side length.
  - Why it fails: You rely on muscle memory instead of confirming the direction of the cross section.
  - Correct: 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:** Leave $(y_{top} - y_{bottom})^2$ unexpanded and incorrectly integrate it as $\frac{(y_{top} - y_{bottom})^3}{3(y'_{top} - y'_{bottom})}$.
  - Why it fails: You misapply the power rule for integration to composite functions without checking substitution requirements.
  - Correct: Always expand the squared binomial term by term before integrating.
- **Wrong:** For rectangular cross sections, use only the given height as the area, ignoring the base distance across the region.
  - Why it fails: You misread the problem and assume the given height is the full area.
  - Correct: 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:** Automatically use 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 it fails: You don't sketch the base region, so you miss the extra boundary line.
  - Correct: Always sketch the base region and label all boundaries to confirm bounds before setting up the integral.
- **Wrong:** Integrate with respect to y but leave boundary curves as $y=f(x)$, leading to an integral with mixed variables.
  - Why it fails: You remember to integrate with respect to y, but forget to rewrite all curves as $x=f(y)$.
  - Correct: Always solve all boundaries for x in terms of y before writing the side length expression.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Square cross sections (perpendicular to x-axis) | $V = \int_a^b (y_{top}(x) - y_{bottom}(x))^2 dx$, |  |

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