---
title: "AP Calc 8.7 Volumes with Cross Sections"
description: "Review AP Calc 8.7 volumes with cross sections: square and rectangular slices, side-length expressions, bounds, dx vs. dy setup, and definite integrals."
canonical: "https://fiveable.me/ap-calc/unit-8/volumes-with-cross-sections-squares-rectangles/study-guide/djttfP0mZkJ7Nn8QrB7r"
type: "study-guide"
subject: "AP Calculus AB/BC"
unit: "Unit 8 – Applications of Integration"
lastUpdated: "2026-06-09"
---

# AP Calc 8.7 Volumes with Cross Sections

## Summary

Review AP Calc 8.7 volumes with cross sections: square and rectangular slices, side-length expressions, bounds, dx vs. dy setup, and definite integrals.

## Guide

To find the volume of a solid with square or rectangular cross sections, integrate the cross-section [area](/ap-calc/unit-6/applying-properties-definite-integrals/study-guide/lUbcVbDG5QVysAn9 "fv-autolink") along the axis the slices are perpendicular to: $V = \int_a^b A(x)\,dx$. For squares, $A = s^2$, and for rectangles, $A = w \cdot h$, where the side or width usually comes from the distance between two curves. For AP Calculus, write the side length or dimensions before setting up the integral.

## AP Calc 8.7: Volumes with Cross Sections

In [AP Calc](/ap-calc "fv-autolink") 8.7, you find volume by slicing a base region into known shapes and adding their areas with a definite integral. The setup is usually $V=\int_a^b A(x)\,dx$ or $V=\int_c^d A(y)\,dy$, where $A$ is the area of each square or rectangular [cross section](/ap-calc/key-terms/cross-section "fv-autolink").

The most important move is writing the side length or dimensions correctly. For slices perpendicular to the x-axis, use top minus bottom and integrate with $dx$; for slices perpendicular to the y-axis, use right minus left and integrate with $dy$.

## Why This Matters for the AP Calculus Exam

This topic shows up when you need to turn a flat region between curves into a 3D solid and find its volume. The core skill is setting up a definite integral from a geometric description: read the problem, figure out the cross-section shape, write its area as a [function](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink") of $x$ or $y$, and integrate over the correct bounds.

These problems test the same procedure-selection and setup skills that the AP Calculus exam rewards. You'll see them in both multiple-choice (often as a setup or quick evaluation) and free-response, where writing a correct integral expression with proper notation matters for clear, complete work. Getting comfortable with cross-section volumes also builds the foundation for the disc and washer methods in later topics.

## Key Takeaways

- Volume of a solid with known [cross sections](/ap-calc/key-terms/cross-sections "fv-autolink"): $V = \int_a^b A(x)\,dx$, where $A(x)$ is the cross-section area and $dx$ is the slice thickness.
- Square cross sections use $A = s^2$; rectangular cross sections use $A = w \cdot h$.
- The side length or width usually equals the distance between the two bounding curves (top minus bottom, or right minus left).
- If cross sections are perpendicular to the x-axis, integrate with respect to $x$. If perpendicular to the y-axis, integrate with respect to $y$ and write everything in terms of $y$.
- Find bounds where the curves intersect, or use a given line like $y = 0$ or $x = 2$.
- Sketching the base region and one slice helps you set up the correct side length and [limits](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/defining-convergent-divergent-infinite-series/study-guide/CIVFHStGQM90EJ4GtIDB "fv-autolink").

## Solids with Cross Sections

When a 3D object is hard to handle with basic geometry, slice it into infinitely thin pieces that are easier to work with. The volume of a solid with known cross sections is:

$$
V = \int_a^b A(x)\ dx
$$

Here $A(x)$ is the area of a cross section (a 2D shape) perpendicular to the x-axis on the [interval](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/finding-taylor-polynomial-approximations-functions/study-guide/LszguYzKz0M6GdqTRSr6 "fv-autolink") $[a,b]$, and $dx$ is the slice thickness. Each slice is basically a very thin prism, and integrating adds up all the slices.

### Square Cross Sections

For square cross sections, use $A(x) = s^2$. Substituting into the volume formula:

$$
V = \int_a^b s^2\ dx
$$

You're stacking up rectangular prisms whose thickness is infinitely thin, represented by $dx$.

### Rectangular Cross Sections

The area of a rectangle is $w \cdot h$, where $w$ is the width and $h$ is the height. So the volume of a solid with rectangular cross sections is:

$$
V = \int_a^b w \cdot h\ dx
$$

Again, $dx$ is your thickness.

---

## How to Use This on the AP Calculus Exam

### Problem Solving

Once you have the formulas, the real work is figuring out $s$, $w$, or $h$ and the bounds. Use these steps:

1. Identify the cross-section shape and pick the matching area formula.
2. Decide whether slices are perpendicular to the x-axis (integrate with respect to $x$) or the y-axis (integrate with respect to $y$).
3. Write the side length or dimensions using the bounding curves.
4. Find the [limits of integration](/ap-calc/key-terms/limits-of-integration "fv-autolink") from [intersection points](/ap-calc/key-terms/intersection-points "fv-autolink") or given boundary lines.
5. Set up the integral, then evaluate.

### Example 1: Solids with Square Cross Sections

Suppose a region bounded by $y = x^2$ and $y = \sqrt{x}$ forms the base of a solid, and each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

*Question courtesy of [Flipped Math](https://calculus.flippedmath.com/uploads/1/1/3/0/11305589/calc_8.7_packet.pdf)*

Since the cross sections are squares, use $V = \int_a^b s^2\ dx$.

Start by visualizing the region. A picture may or may not be given, but sketching the graphs helps you see the region and choose correct bounds.

![Untitled](https://storage.googleapis.com/static.prod.fiveable.me/images/Untitled.png-1706235836854-53795)
###### Graph of region bounded by $x^2$ and $\sqrt{x}$. Image courtesy of Flipped Math

Here $y = \sqrt{x}$ is the upper curve and $y = x^2$ is the lower curve. Let $h(x) = \sqrt{x}$ and $g(x) = x^2$. The grey area is the base, and the purple line is one cross section coming toward you along a z-axis not drawn here.

The side length of the square at a given $x$ is the difference between the upper and lower curves, so $s = \sqrt{x} - x^2$.

Next, find the bounds. The curves intersect at $x = 0$ and $x = 1$, so the interval is $[0, 1]$. You can confirm this algebraically:

$$
h(x)=g(x)\\
\sqrt{x}=x^2\\
x=x^4\\
\therefore \ x=0\text{ or }x=1
$$

Now plug in:

$$
V = \int_0^1 (\sqrt{x}-x^2)^2\ dx
$$

Evaluating gives $\frac{9}{70}$, about **0.1285**.

**The steps:**

First, expand and integrate.

$$
\int_0^1(\sqrt{x}-x^2)^2\ dx=\int_0^1(x^{1/2}-x^2)^2\ dx=\int_0^1x-2x^{5/2}+x^4\ dx=\\ \int_0^1x\ dx-2\int_0^1x^{5/2}\ dx+\int_0^1x^4\ dx=\frac{x^5}{5}-\frac{4x^{7/2}}{7}+\frac{x^2}{2}
$$

Then evaluate over the bounds.

$$
\Bigg(\frac{1^5}{5}-\frac{4(1)^{7/2}}{7}+\frac{1^2}{2}\Bigg)-\Bigg(\frac{0^5}{5}-\frac{4(0)^{7/2}}{7}+\frac{0^2}{2}\Bigg)=\frac{1}{5}-\frac{4}{7}+\frac{1}{2}=\frac{14-40+35}{70}=\boxed{\frac{9}{70}}
$$

### Example 2: Solids with Rectangular Cross Sections

The base of a solid is bounded by $y = x^3$, $y = 0$, and $x = 2$. Find the volume if the cross sections, taken perpendicular to the y-axis, form a rectangle whose height is 6.

*Question courtesy of [Flipped Math](https://calculus.flippedmath.com/uploads/1/1/3/0/11305589/calc_8.7_packet.pdf)*

Two important [differences](/ap-calc/unit-1/determining-limits-using-algebraic-properties-limits/study-guide/HjStgVKViPGZj1CxYwEB "fv-autolink") from Example 1: the cross sections are perpendicular to the y-axis, and three boundaries are given. Since the slices are perpendicular to the y-axis, integrate with respect to $y$ using $V = \int_a^b w \cdot h\ dy$.

![Untitled](https://storage.googleapis.com/static.prod.fiveable.me/images/Untitled_2.png-1706235836867-51013)

###### Graph of given problem for visualization purposes. Image courtesy of Flipped Math.

Here $y = x^3$ is the left curve and $x = 2$ is the right curve. The purple line is a cross section perpendicular to the y-axis. To find the width at a given $y$, rewrite $y = x^3$ as $x = \sqrt[3]{y}$. The width is the right boundary minus the left boundary: $w = 2 - \sqrt[3]{y}$. The problem gives the height as $h = 6$.

For the bounds, the curves meet at $y = 8$, and the boundary $y = 0$ gives the lower limit, so the interval is $[0, 8]$. Plugging in:

$$
V = \int_0^8 (2-\sqrt[3]y) \cdot 6\ dy
$$

This evaluates to 24.

$$
\int_0^86(2-\sqrt[3]y)\ dy
$$

$$
\int_0^812-6\sqrt[3]y\ dy
$$

$$
\int_0^812\ dy-6\int_0^8\sqrt[3]y\ dy
$$

$$
[12y]_0^8-6\int_0^8y^{1/3}\ dy
$$

$$
\Bigg[12y-\frac{18y^{4/3}}{4}\Bigg]_0^8
$$

$$
\Bigg[12y-\frac{9y^{4/3}}{2}\Bigg]_0^8
$$

$$
\Bigg[12(8)-\frac{9(8)^{4/3}}{2}\Bigg]-\Bigg[12(0)-\frac{9(0)^{4/3}}{2}\Bigg]=[96-72]-[0-0]=\boxed{24}
$$

### Common Trap

Mixing up which variable to integrate with respect to. If slices are perpendicular to the x-axis, your integral uses $dx$ and everything must be in terms of $x$. If slices are perpendicular to the y-axis, switch to $dy$ and rewrite functions in terms of $y$.

---

## Common Misconceptions

- **Forgetting to square the side length.** For square cross sections, $A = s^2$, not just $s$. If $s = \sqrt{x} - x^2$, you must square the entire difference before integrating.
- **Squaring incorrectly.** $(\sqrt{x} - x^2)^2$ is not $x - x^4$. Expand it fully: $x - 2x^{5/2} + x^4$.
- **Using the wrong variable.** Cross sections perpendicular to the y-axis require integrating with respect to $y$, so rewrite your curves as functions of $y$ first.
- **Picking bounds that don't match the slicing direction.** When integrating with respect to $y$, your limits are $y$-values, not $x$-values.
- **Confusing cross-section volume with [solids of revolution](/ap-calc/key-terms/solids-of-revolution "fv-autolink").** This topic uses a flat base region with shapes built on top of it. It is not the disc or [washer method](/ap-calc/unit-8/volume-with-washer-method-revolving-around-other-axes/study-guide/LlG9jCFLxe4kpDTDNtHX "fv-autolink"), which spins a region around an axis.
- **Treating the side length as one curve.** The side or width is usually the distance between two curves (top minus bottom or right minus left), not just one function's value.

## Summary

To find the volume of a solid with known cross sections:

$$
V = \int_a^b A(x)\ dx
$$

where $A(x)$ is the area of a cross section perpendicular to the x-axis on $[a, b]$.

- Square cross sections: $A(x) = s^2$, where $s$ is the side length.
- Rectangular cross sections: $A(x) = w \cdot h$, where $w$ is width and $h$ is height.

Use the methods for finding areas between curves to write $s$, or $w$ and $h$, then find the bounds from intersection points or given boundary lines. Plug in $A(x)$, integrate, and you have the volume.

## Related AP Calculus Guides

- [Unit 8 Overview: Applications of Integration](/ap-calc/unit-8/review/study-guide/95uuVjdtA80roOMvV8IK)
- [8.1 Finding the Average Value of a Function on an Interval](/ap-calc/unit-8/finding-average-value-function-on-an-interval/study-guide/HjiYTRAnQdY0eCQpqtpg)
- [8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals](/ap-calc/unit-8/connecting-position-velocity-acceleration-functions-using-integrals/study-guide/k9tY28YXs7YDVu1uqFuw)
- [8.4 Finding the Area Between Curves Expressed as Functions of x](/ap-calc/unit-8/finding-area-between-curves-expressed-as-functions-x/study-guide/Zyj7XJuPfoWBuAJ96ZAG)
- [8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts](/ap-calc/unit-8/using-accumulation-functions-definite-integrals-applied-contexts/study-guide/nUlJKvXqRcsfLnVMd5fG)
- [8.6 Finding the Area Between Curves That Intersect at More Than Two Points](/ap-calc/unit-8/finding-area-between-curves-that-intersect-at-more-than-two-points/study-guide/QVBQ9TQDM4ZObJl6o0ad)

## Vocabulary

- **cross section**: Two-dimensional slices of a three-dimensional solid, perpendicular to an axis, used to build up the volume through integration.
- **definite integral**: The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.
- **rectangular cross sections**: Two-dimensional rectangular slices of a solid whose areas can be integrated to find the total volume.
- **square cross sections**: Two-dimensional square slices of a solid whose areas can be integrated to find the total volume.
- **volumes of solids**: The measure of three-dimensional space occupied by a solid object, calculated using integration techniques.

## FAQs

### What is AP Calc 8.7 about?

AP Calc 8.7 covers finding volumes of solids with known square or rectangular cross sections using definite integrals.

### How do I set up volume with cross sections?

Write the volume as an integral of cross-section area: V = integral A(x) dx or V = integral A(y) dy. The area formula depends on the cross-section shape.

### What formula do I use for square cross sections?

For square cross sections, the area is A = s^2, where s is the side length. The side length often comes from the distance between two curves.

### What formula do I use for rectangular cross sections?

For rectangular cross sections, use A = width times height. One dimension may come from the distance between curves, while the other may be given as a constant or ratio.

### How do I know whether to integrate with dx or dy?

If slices are perpendicular to the x-axis, integrate with dx and write lengths in terms of x. If slices are perpendicular to the y-axis, integrate with dy and write lengths in terms of y.

### How are cross-section volumes different from washers?

Cross-section problems build known shapes on a base region. Washer and disk methods involve rotating a region around an axis, so the area formulas and setup are different.

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