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Volumes with Known Cross-Sections

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Reference: Stewart §5.2

Beyond Circles: Triangles, Squares, and More

Not every interesting solid is a solid of revolution. Some solids have cross-sections that are squares, triangles, semicircles, or other shapes. The slicing principle still works: we just need to express the cross-sectional area in terms of the position variable.

The key is that the base of each cross-section often comes from a curve in the $xy$-plane, and the shape of the cross-section is given in the problem.

The key insight: Identify how the cross-section’s dimensions relate to the base curve. Usually, one dimension of the cross-section equals the width of the region at that point.

Prerequisite Map

PrerequisitesBasic Geometric FormulasVolume by Slicing
This skillVolumes with Known Cross-Sections
Leads tono further branch yet

Before You Start

Self-check: Can you answer these questions? If not, review the linked prerequisites first.

Question If you struggle...
Explain why $V = \int_a^b A(x)\,dx$ gives volume Review Volume by Slicing
Find the area of an equilateral triangle with side 4 You need: $A = \frac{\sqrt{3}}{4}s^2 = 4\sqrt{3}$
If $(x-1)^2 + y^2 = 4$, express $y$ in terms of $x$ Rearranging equations to find cross-section dimensions is essential
Refresh: Area Formulas You'll Need
Shape Area Formula Example
Square (side $s$) $s^2$ Side 3 → Area = 9
Equilateral triangle (side $s$) $\frac{\sqrt{3}}{4}s^2$ Side 4 → Area = $4\sqrt{3}$
Isosceles right triangle (legs $\ell$) $\frac{1}{2}\ell^2$ Legs 2 → Area = 2
Isosceles right triangle (hypotenuse $h$) $\frac{h^2}{4}$ Hypotenuse 4 → Area = 4
Semicircle (diameter $d$) $\frac{\pi d^2}{8}$ Diameter 4 → Area = $2\pi$

Tip: Memorize these: they appear constantly in known cross-section problems.

Refresh: Finding the Cross-Section Dimension

The “width” of the base region at position $x$ usually becomes the key dimension of each cross-section.

For a region between two curves $y = f(x)$ and $y = g(x)$: $$\text{width} = f(x) - g(x) \quad \text{(assuming } f(x) \geq g(x)\text{)}$$

For a circular base $x^2 + y^2 = r^2$: $$\text{width} = 2y = 2\sqrt{r^2 - x^2}$$

This width becomes the side of a square, the base of a triangle, or the diameter of a semicircle.

Quick Reference

Property Value
Concept Applications of Integration
Course MATH162
Section Stewart 5.2
Difficulty Advanced
Time ~25 minutes

Key Concepts

The General Setup

A solid has a known base (usually a region in the $xy$-plane) and cross-sections of a specified shape perpendicular to one axis.

    y
    |     _____
    |    /     \      ← Base region in xy-plane
    |   /       \
    |  /    ▲    \    ← Cross-sections (triangles, squares, etc.)
    | /     |     \      stand perpendicular to the plane
    +---------------→ x

    3D view:
                    /\
                   /  \
                  /    \      ← Triangular cross-section
                 /______\
                /        \
               /__________\   ← Base

The Formula

$$\boxed{V = \int_a^b A(x)\,dx}$$

where $A(x)$ is the area of the cross-section at position $x$, expressed using the shape’s area formula.

Common Cross-Section Shapes

Shape Area Formula If base/diameter = $w$
Square $s^2$ $A = w^2$
Equilateral Triangle $\frac{\sqrt{3}}{4}s^2$ $A = \frac{\sqrt{3}}{4}w^2$
Isosceles Right Triangle $\frac{1}{2}(\text{leg})^2$ $A = \frac{1}{2}w^2$ (if legs = $w$)
Semicircle $\frac{1}{2}\pi r^2$ $A = \frac{\pi}{8}w^2$ (diameter = $w$)
Circle $\pi r^2$ $A = \frac{\pi}{4}w^2$ (diameter = $w$)

The Algorithm

  1. Identify the base region in the $xy$-plane
  2. Determine the cross-section shape and its orientation
  3. Express the cross-section’s dimension (base, side, diameter) in terms of $x$ or $y$
  4. Write $A(x)$ using the appropriate area formula
  5. Find limits and integrate

Finding the Cross-Section Dimension

Key question: What determines the size of each cross-section?

Usually, the “width” of the base region at position $x$ becomes a dimension of the cross-section:

$$\text{width at } x = f(x) - g(x) \text{ (top curve minus bottom curve)}$$

or

$$\text{width at } x = 2y = 2\sqrt{\text{something}}$$

Practice Problems

Level 1 Square Cross-Sections on a Triangle

The base of a solid is a triangular region with vertices at $(0, 0)$, $(1, 0)$, and $(0, 1)$. Cross-sections perpendicular to the $x$-axis are squares. Find the volume.

Thought Process

The base is bounded by $y = 0$, $x = 0$, and the line from $(1, 0)$ to $(0, 1)$, which has equation $y = 1 - x$.

At position $x$, the width of the base (from $y = 0$ to $y = 1 - x$) is $1 - x$.

Each square cross-section has side length $s = 1 - x$, so area $A(x) = (1-x)^2$.

Show Answer

The base extends from $x = 0$ to $x = 1$.

At position $x$, the width of the region is $(1 - x) - 0 = 1 - x$.

Each square has side $s = 1 - x$, so: $$A(x) = s^2 = (1 - x)^2$$

$$V = \int_0^1 (1 - x)^2\,dx$$

Let $u = 1 - x$, $du = -dx$. When $x = 0$, $u = 1$; when $x = 1$, $u = 0$.

$$V = -\int_1^0 u^2\,du = \int_0^1 u^2\,du = \left[\frac{u^3}{3}\right]_0^1 = \boxed{\frac{1}{3}}$$

Level 2 Equilateral Triangle Cross-Sections

A solid has a circular base of radius 2 (the disk $x^2 + y^2 \leq 4$). Cross-sections perpendicular to the $x$-axis are equilateral triangles with one side in the base. Find the volume.

Thought Process

At position $x$, the chord of the circle has length $2y = 2\sqrt{4 - x^2}$ (from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$).

This chord is the base of the equilateral triangle. For an equilateral triangle with side $s$, the area is $\frac{\sqrt{3}}{4}s^2$.

Show Answer

The circle $x^2 + y^2 = 4$ gives $y = \pm\sqrt{4 - x^2}$.

At position $x$, the chord (base of the triangle) has length: $$s = 2\sqrt{4 - x^2}$$

Area of equilateral triangle with side $s$: $$A(x) = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(2\sqrt{4-x^2})^2 = \frac{\sqrt{3}}{4} \cdot 4(4 - x^2) = \sqrt{3}(4 - x^2)$$

The solid extends from $x = -2$ to $x = 2$: $$V = \int_{-2}^{2} \sqrt{3}(4 - x^2)\,dx$$

Since the integrand is even: $$V = 2\sqrt{3}\int_0^2 (4 - x^2)\,dx = 2\sqrt{3}\left[4x - \frac{x^3}{3}\right]_0^2 = 2\sqrt{3}\left(8 - \frac{8}{3}\right) = \boxed{\frac{32\sqrt{3}}{3}}$$

Level 3 Semicircular Cross-Sections

The base of a solid is the region bounded by $y = x^2$ and $y = 1$. Cross-sections perpendicular to the $y$-axis are semicircles with diameters in the base. Find the volume.

Thought Process

Since cross-sections are perpendicular to the $y$-axis, we integrate with respect to $y$.

At height $y$, the parabola gives $x = \pm\sqrt{y}$, so the width (diameter) is $2\sqrt{y}$.

For a semicircle with diameter $d$, the radius is $r = d/2$ and the area is $\frac{1}{2}\pi r^2 = \frac{\pi d^2}{8}$.

Show Answer

The region extends from $y = 0$ to $y = 1$.

At height $y$, from $y = x^2$ we get $x = \pm\sqrt{y}$.

The diameter of the semicircle is: $$d = 2\sqrt{y}$$

The radius is $r = \sqrt{y}$, and the area of the semicircle is: $$A(y) = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(\sqrt{y})^2 = \frac{\pi y}{2}$$

$$V = \int_0^1 \frac{\pi y}{2}\,dy = \frac{\pi}{2}\left[\frac{y^2}{2}\right]_0^1 = \frac{\pi}{2} \cdot \frac{1}{2} = \boxed{\frac{\pi}{4}}$$

Level 4 Isosceles Right Triangle Cross-Sections

The base of a solid is the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Cross-sections perpendicular to the $x$-axis are isosceles right triangles with the hypotenuse in the base. Find the volume.

Thought Process

From the ellipse equation, $y = \pm\frac{2}{3}\sqrt{9 - x^2}$, so the width at position $x$ is $2 \cdot \frac{2}{3}\sqrt{9-x^2} = \frac{4}{3}\sqrt{9-x^2}$.

This is the hypotenuse $h$ of the isosceles right triangle. For such a triangle, if the legs have length $\ell$, then $h = \ell\sqrt{2}$, so $\ell = \frac{h}{\sqrt{2}}$.

The area is $\frac{1}{2}\ell^2 = \frac{1}{2}\left(\frac{h}{\sqrt{2}}\right)^2 = \frac{h^2}{4}$.

Show Answer

Step 1: From $\frac{x^2}{9} + \frac{y^2}{4} = 1$, solve for $y$: $$y = \pm\frac{2}{3}\sqrt{9 - x^2}$$

Step 2: The width (hypotenuse) at position $x$ is: $$h = 2 \cdot \frac{2}{3}\sqrt{9 - x^2} = \frac{4}{3}\sqrt{9 - x^2}$$

Step 3: For an isosceles right triangle with hypotenuse $h$:

  • Legs: $\ell = \frac{h}{\sqrt{2}}$
  • Area: $A = \frac{1}{2}\ell^2 = \frac{1}{2} \cdot \frac{h^2}{2} = \frac{h^2}{4}$

So: $$A(x) = \frac{1}{4}\left(\frac{4}{3}\sqrt{9-x^2}\right)^2 = \frac{1}{4} \cdot \frac{16}{9}(9 - x^2) = \frac{4}{9}(9 - x^2)$$

Step 4: The ellipse extends from $x = -3$ to $x = 3$: $$V = \int_{-3}^{3} \frac{4}{9}(9 - x^2)\,dx = \frac{8}{9}\int_0^3 (9 - x^2)\,dx$$

$$= \frac{8}{9}\left[9x - \frac{x^3}{3}\right]_0^3 = \frac{8}{9}\left(27 - 9\right) = \frac{8}{9} \cdot 18 = \boxed{16}$$

Level 5 Designing a Solid with a Given Volume

A decorative pillar has a square base with side length 2 meters. The pillar tapers upward with square cross-sections. If the cross-section at height $h$ (measured in meters from the base) has side length $s(h)$, and the pillar has total height $H$:

  1. If $s(h) = 2 - \frac{2h}{H}$ (linear taper), find the volume in terms of $H$.
  2. What height $H$ gives a volume of exactly $4$ cubic meters?
  3. Compare to a rectangular box with the same base and height. What fraction of the box's volume does the pillar occupy?
Thought Process

The cross-section at height $h$ is a square with side $s(h) = 2(1 - h/H)$. The area is $s(h)^2$. Integrate from $h = 0$ to $h = H$.

This is similar to a pyramid that tapers to a point, which has volume $\frac{1}{3}(\text{base area})(\text{height})$.

Show Answer

(a) Finding the volume:

The side length is $s(h) = 2 - \frac{2h}{H} = 2\left(1 - \frac{h}{H}\right)$.

The cross-sectional area is: $$A(h) = s(h)^2 = 4\left(1 - \frac{h}{H}\right)^2$$

The volume is: $$V = \int_0^H 4\left(1 - \frac{h}{H}\right)^2\,dh$$

Let $u = 1 - \frac{h}{H}$, so $du = -\frac{1}{H}dh$, meaning $dh = -H\,du$.

When $h = 0$, $u = 1$; when $h = H$, $u = 0$.

$$V = 4\int_1^0 u^2 \cdot (-H)\,du = 4H\int_0^1 u^2\,du = 4H \cdot \frac{1}{3} = \boxed{\frac{4H}{3}}$$

(b) Finding $H$ for $V = 4$ m³:

$$\frac{4H}{3} = 4 \implies H = 3 \text{ meters}$$

(c) Comparison to a box:

A rectangular box with base $2 \times 2$ and height $H$ has volume: $$V_{\text{box}} = 2 \times 2 \times H = 4H$$

The pillar’s volume is $\frac{4H}{3}$.

$$\text{Fraction} = \frac{V_{\text{pillar}}}{V_{\text{box}}} = \frac{4H/3}{4H} = \boxed{\frac{1}{3}}$$

Insight: The pillar (which tapers linearly to a point) has exactly $\frac{1}{3}$ the volume of the enclosing box, the same ratio as a pyramid to its enclosing prism. This $\frac{1}{3}$ factor appears whenever a solid tapers linearly from a base to a point.

Common Misconceptions

Common misconception

when cross-sections are squares with side equal to the chord length, the cross-sectional area is the chord length itself.

This is the height-vs-slope error. The chord length at position $x$ gives the side $s$ of the square, but the cross-sectional area is $s^2$, not $s$. For a solid with base $x^2 + y^2 \leq 4$ and square cross-sections perpendicular to the $x$-axis, the chord at position $x$ has length $2\sqrt{4 - x^2}$, so the area is $(2\sqrt{4 - x^2})^2 = 4(4 - x^2)$, not $2\sqrt{4 - x^2}$. Omitting the squaring step yields an integral of a square root rather than a polynomial, which also gives the wrong numerical answer.

Mastery Checklist

Mental Model

The Variable Cookie Cutter: Imagine a cookie cutter that changes shape (or size) as you move along the solid. At each position, it cuts a cross-section of a specific shape. The volume is built up by stacking infinitely many of these cookies, each with area $A(x)$ and thickness $dx$.

Reference: Area Formulas

Shape Area Notes
Square (side $s$) $s^2$
Equilateral triangle (side $s$) $\frac{\sqrt{3}}{4}s^2$ Height = $\frac{\sqrt{3}}{2}s$
Isosceles right (legs $\ell$) $\frac{1}{2}\ell^2$ Hypotenuse = $\ell\sqrt{2}$
Isosceles right (hypotenuse $h$) $\frac{h^2}{4}$ Legs = $\frac{h}{\sqrt{2}}$
Semicircle (diameter $d$) $\frac{\pi d^2}{8}$ Radius = $\frac{d}{2}$
Circle (diameter $d$) $\frac{\pi d^2}{4}$ Radius = $\frac{d}{2}$

Connections

Looking back:

Looking ahead:

Real-world applications:

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Last updated: 2026-01-23