# Volumes with cross sections: squares and rectangles

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

This guide covers the general slicing method for volumes of solids with square and rectangular cross sections, perpendicular to both axes, including common problem variations tested on the AP Calculus BC exam.

**Prerequisites:** Compute definite integrals of common functions; Find the area between two curves; Rewrite functions in terms of the opposite axis variable

## Learning objectives

- Set up volume integrals for solids with square/rectangular cross sections
- Solve for volume when cross sections are perpendicular to the x-axis or y-axis
- Handle special cases: squares with diagonal in base, rectangles with proportional height
- Avoid common exam traps in cross-section volume problems

## Cross Sections Perpendicular to the x-axis

This is the most common case on the AP exam. The base of the solid lies in the $xy$-plane, bounded between two curves $y=f(x)$ and $y=g(x)$ (with $f(x) \geq g(x)$ on $[a,b]$). We slice the solid with vertical planes perpendicular to the x-axis, each slice has thickness $dx$ and cross-sectional area $A(x)$, so total volume is the integral of $A(x)$ over the bounds of the base.

**General Volume Formula (x-perpendicular cross sections)** — Total volume equals the integral of cross-sectional area from the left bound $a$ to right bound $b$ of the base region.

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

**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 (vertical distance between curves):
2. $$a=0, \quad b=2, \quad s(x) = 4 - x^2$$
3. Calculate cross-sectional area for a square with side $s(x)$:
4. $$A(x) = (4 - x^2)^2 = 16 - 8x^2 + x^4$$
5. Set up and evaluate the integral using the Fundamental Theorem of Calculus:
6. $$V = \int_0^2 (16 - 8x^2 + x^4) dx = \left[16x - \frac{8}{3}x^3 + \frac{1}{5}x^5\right]_0^2 = \frac{256}{15} \approx 17.07$$

> **tip**
>
> If cross sections are perpendicular to the x-axis, all functions and side lengths must be written in terms of $x$, and your bounds will be x-values.

## Cross Sections Perpendicular to the y-axis

When cross sections are perpendicular to the y-axis, slices are horizontal instead of vertical, so we integrate with respect to $y$. The base is bounded between $x=f(y)$ and $x=g(y)$ (with $f(y) \geq g(y)$ on $[c,d]$), and side length is the horizontal distance between the two curves. If your original curves are given as $y=f(x)$, you must invert them to get $x$ as a function of $y$ first.

**General Volume Formula (y-perpendicular cross sections)** — Total volume equals the integral of cross-sectional area from the lower bound $c$ to upper bound $d$ of the base region along the y-axis.

*Notation:* V = \int_c^d A(y) dy

**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 $y$, find bounds:
2. $$y = \sqrt{x} \implies x = y^2, \quad c=0, \quad d=2$$
3. Calculate side length (horizontal distance between curves):
4. $$s(y) = 4 - y^2$$
5. Calculate cross-sectional area and integrate:
6. $$A(y) = (4 - y^2)^2, \quad V = \int_0^2 (16 - 8y^2 + y^4) dy = \frac{256}{15} \approx 17.07$$

> **tip**
>
> Never leave side lengths in terms of $x$ when integrating with respect to $y$ — always invert functions first to avoid variable mismatch.

## Special Case Variations

Two common tested variations change how you calculate cross-sectional area from the base side length:

1. **Squares with diagonal in the base**: If the distance between the base curves equals the diagonal $d$ of the square, use the relationship $d = s\sqrt{2}$, so area $A = \frac{d^2}{2}$.
2. **Rectangular cross sections**: If the base side is $s$, and height is proportional to $s$ ($h = ks$ for constant $k$), area $A = s \times ks = ks^2$. If height is a constant $h$, area $A = sh$.

**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 $x$:
2. $$d(x) = x - x^2$$
3. Calculate cross-sectional area for a square with diagonal $d(x)$:
4. $$A(x) = \frac{d(x)^2}{2} = \frac{(x - x^2)^2}{2}$$
5. Set up and evaluate the integral:
6. $$V = \int_0^1 \frac{x^2 - 2x^3 + x^4}{2} dx = \frac{1}{60}$$

> **tip**
>
> Always check whether the problem states the side or diagonal of the square lies in the base — this is a very common exam point that changes your area calculation.

## AP-Style Worked Practice

**Worked example:** 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?

1. Base side length at any $x$: $s(x) = e^x$, height $h = 3e^x$
2. Cross-sectional area:
3. $$A(x) = e^x \times 3e^x = 3e^{2x}$$
4. Integrate to find volume:
5. $$V = \int_0^1 3e^{2x} dx = \frac{3}{2}(e^2 - 1)$$

## Common pitfalls

- **Wrong:** When the diagonal of the square is in the base, calculate area as $d^2$ instead of $d^2/2$
  - Why it fails: You default to the common side-in-base case without reading the problem carefully
  - Correct: Underline whether the segment in the base is the side or diagonal before starting your calculation
- **Wrong:** For cross sections perpendicular to the y-axis, leave side length in terms of x and integrate with respect to y
  - Why it fails: You forget the variable of integration must match the cross section orientation, and skip inverting functions
  - Correct: Write all functions and side lengths in terms of the axis variable (x for x-perpendicular, y for y-perpendicular) before setting up the integral
- **Wrong:** Subtract curves in the wrong order, get a negative side length, and forget to take absolute value
  - Why it fails: You subtract top-from-bottom by habit without checking which curve is larger on the interval
  - Correct: Confirm your side length is positive for all points in the integration interval, swap subtraction order if needed
- **Wrong:** For a rectangle with height proportional to base side, use area $= ks$ instead of $ks^2$
  - Why it fails: You forget area of a rectangle is base × height, skipping the multiplication step
  - Correct: Explicitly write out $\text{base} = s, \text{height} = ks, \text{Area} = s \times ks$ before proceeding
- **Wrong:** When squaring $(a - bx^n)$, expand to $a^2 - b^2x^{2n}$ skipping the middle term
  - Why it fails: You rush the algebra step before integrating
  - Correct: Always expand the square of a binomial term-by-term to avoid missing the cross term $-2abx^n$

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Volume (perp to x-axis) | $V = \int_a^b A(x) dx$ | $a,b$ = x-bounds of base, $A(x)$ = cross-section area |
| Volume (perp to y-axis) | $V = \int_c^d A(y) dy$ | $c,d$ = y-bounds of base, $A(y)$ = cross-section area |
| Square (side in base) | $A = s^2$ | $s$ = distance between base boundary curves |
| Square (diagonal in base) | $A = \frac{s^2}{2}$ | $s$ = diagonal length = distance between boundaries |
| Rectangle (base $s$, height $h$) | $A = s \cdot h$ | If $h=ks$, $A=ks^2$; if $h$ constant, $A=hs$ |
| Vertical side length (x-perp) | $s(x) = f(x) - g(x), f \geq g$ | Between two y-boundary curves |
| Horizontal side length (y-perp) | $s(y) = f(y) - g(y), f \geq g$ | Between two x-boundary curves |

## 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.

- [Volumes with cross sections: triangles and semicircles](https://www.owlsprep.com/study/ap-calculus-bc-u8-volumes-with-cross-sections-triangles/)
- [Disc method around the x- or y-axis](https://www.owlsprep.com/study/ap-calculus-bc-u8-disc-method-around-the-x/)
- [Disc method around other axes](https://www.owlsprep.com/study/ap-calculus-bc-u8-disc-method-around-other-axes/)

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