Volume by Slicing

8 min read

Reference: Stewart §5.2

From Bread Loaves to Calculus

How do you find the volume of an irregularly shaped solid? You can’t just measure length times width times height when the shape keeps changing. But here’s an idea: slice it.

Think of a loaf of bread. Each slice has a certain area, and if you add up the volumes of all the thin slices (area × thickness), you get the total volume. The thinner the slices, the more accurate your answer. In the limit of infinitely thin slices, you get the exact volume, and that limit is a definite integral.

The key insight: Volume is to area what area was to length. Just as we built up area by integrating lengths, we build up volume by integrating areas.

Prerequisite Map

Before You Start

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

Question	If you struggle...
Evaluate $\int_0^3 x^2\,dx$	Review Definite Integrals
Explain why $\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x = \int_a^b f(x)\,dx$	Review Riemann Sums
Find the area of a circle with radius 5	Review geometric area formulas

Refresh: Definite Integral as a Limit

The definite integral is defined as: $$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$$

where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is a sample point in the $i$-th subinterval.

Key idea: We approximate the quantity (area, in the basic case) with a sum of simple pieces, then take the limit as the pieces become infinitely small.

Refresh: Basic Geometric Area Formulas

Shape	Area Formula
Rectangle	$A = \ell \times w$
Circle	$A = \pi r^2$
Triangle	$A = \frac{1}{2}bh$
Trapezoid	$A = \frac{1}{2}(b_1 + b_2)h$

These formulas are essential for computing cross-sectional areas.

Quick Reference

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

Key Concepts

The Slicing Principle

Consider a solid $S$ lying between $x = a$ and $x = b$. At each point $x$, slice the solid with a plane perpendicular to the $x$-axis. This produces a cross-section with area $A(x)$.

    y
    |     _____
    |    /     \      ← Solid S
    |   /   |   \
    |  /  A(x)   \    ← Cross-section at x has area A(x)
    | /     |     \
    +-------|------→ x
           x
           a         b

Definition of Volume

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

where:

$V$ = volume of the solid
$A(x)$ = cross-sectional area at position $x$
$[a, b]$ = interval over which the solid extends

Why This Works: The Riemann Sum Derivation

Divide the solid into $n$ thin slabs of equal width $\Delta x = \frac{b-a}{n}$.

Step 1: Approximate each slab as a cylinder with base area $A(x_i^*)$ and height $\Delta x$.

Step 2: The volume of the $i$-th slab is approximately $A(x_i^*)\Delta x$.

Step 3: Sum all slabs: $$V \approx \sum_{i=1}^n A(x_i^*)\Delta x$$

Step 4: Take the limit as $n \to \infty$: $$V = \lim_{n \to \infty} \sum_{i=1}^n A(x_i^*)\Delta x = \int_a^b A(x)\,dx$$

The approximation becomes exact because the integral captures the continuous variation in cross-sectional area.

Special Case: Cylinders

For a cylinder (where the cross-section doesn’t change), $A(x) = A$ for all $x$:

$$V = \int_a^b A\,dx = A(b - a) = A \cdot h$$

This confirms the familiar formula: Volume = Base Area × Height.

Choosing the Axis of Integration

You can slice perpendicular to any axis. If cross-sections are perpendicular to the $y$-axis:

$$V = \int_c^d A(y)\,dy$$

How do you choose? Pick the axis that makes $A$ easiest to express. If the solid’s shape varies more naturally along one direction, that’s usually your integration variable.

The Algorithm

Identify the solid and the axis perpendicular to the slices
Find the limits $a$ and $b$ (where the solid begins and ends)
Express $A(x)$ in terms of $x$ (the cross-sectional area)
Integrate $V = \int_a^b A(x)\,dx$

Common Pitfalls

Mistake	Why It’s Wrong	Correction
Wrong axis of integration	Cross-sections must be perpendicular to the axis of integration	If slices are perpendicular to $x$-axis, integrate with respect to $x$
Forgetting continuity requirement	The formula assumes $A(x)$ is continuous on $[a,b]$	Check for discontinuities; split the integral if needed
Wrong limits	Limits must be where the solid begins and ends, not the function	Identify the extent of the solid along the axis
Confusing $A(x)$ with $f(x)$	$A(x)$ is the cross-sectional area, not the function defining the boundary	Area often involves squaring: $A = \pi r^2$, $A = s^2$, etc.

Practice Problems

Level 1 Cylinder Verification

A cylinder has circular cross-sections of radius 3 and extends from $x = 0$ to $x = 5$. Use the slicing formula to find its volume.

Thought Process

The cross-sectional area is constant: $A(x) = \pi r^2 = \pi(3)^2 = 9\pi$ for all $x$. Apply the volume formula directly.

Show Answer

The cross-sectional area is $A(x) = \pi(3)^2 = 9\pi$.

$$V = \int_0^5 9\pi\,dx = 9\pi[x]_0^5 = 9\pi(5 - 0) = 45\pi$$

This matches $V = \pi r^2 h = \pi(9)(5) = 45\pi$.

Level 2 Volume of a Hemisphere

Find the volume of a hemisphere of radius 5 by integrating circular cross-sections. (The hemisphere is the upper half of a sphere.)

Thought Process

Place the flat face of the hemisphere on the $xy$-plane with center at the origin. The hemisphere extends from $z = 0$ to $z = 5$. At height $z$, the cross-section is a circle whose radius comes from $x^2 + y^2 + z^2 = 25$, giving radius $\sqrt{25 - z^2}$.

Show Answer

Step 1: Position the hemisphere with flat face on the $xy$-plane. It extends from $z = 0$ to $z = 5$.

Step 2: At height $z$, the cross-section is a circle with radius $\sqrt{25 - z^2}$ (from the sphere equation $x^2 + y^2 + z^2 = 25$).

Step 3: The cross-sectional area is: $$A(z) = \pi(\sqrt{25 - z^2})^2 = \pi(25 - z^2)$$

Step 4: Integrate: $$V = \int_0^5 \pi(25 - z^2)\,dz = \pi\left[25z - \frac{z^3}{3}\right]_0^5$$

$$= \pi\left(125 - \frac{125}{3}\right) = \pi \cdot \frac{250}{3} = \boxed{\frac{250\pi}{3}}$$

Verification: A full sphere of radius 5 has volume $\frac{4}{3}\pi(125) = \frac{500\pi}{3}$. Half of this is $\frac{250\pi}{3}$. ✓

Level 3 Pyramid with Rectangular Base

Find the volume of a pyramid with height 6 and a rectangular base measuring 4 by 8. Place the apex at the origin with the axis along the positive $z$-axis.

Thought Process

At distance $z$ from the apex, the cross-section is a rectangle similar to the base. Use similar triangles to find how both dimensions scale with $z$: each dimension is proportional to $z/h = z/6$.

Show Answer

Step 1: The pyramid extends from $z = 0$ (apex) to $z = 6$ (base).

Step 2: At height $z$, the cross-section is a rectangle. By similar triangles, the dimensions scale linearly: $$\text{width} = \frac{4z}{6} = \frac{2z}{3}, \qquad \text{length} = \frac{8z}{6} = \frac{4z}{3}$$

Step 3: The cross-sectional area is: $$A(z) = \frac{2z}{3} \cdot \frac{4z}{3} = \frac{8z^2}{9}$$

Step 4: Integrate: $$V = \int_0^6 \frac{8z^2}{9}\,dz = \frac{8}{9}\left[\frac{z^3}{3}\right]_0^6 = \frac{8}{9} \cdot \frac{216}{3} = \frac{8 \cdot 72}{9} = \boxed{64}$$

Verification: The base area is $4 \times 8 = 32$, and the formula $V = \frac{1}{3}Bh$ gives $\frac{1}{3}(32)(6) = 64$. ✓

Level 4 Wedge from a Cylinder

A wedge is cut from a circular cylinder of radius 3 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of $45°$ along a diameter of the cylinder. Find the volume of the wedge.

Thought Process

Place the $x$-axis along the diameter where the planes meet. The base is a semicircle $y = \sqrt{9 - x^2}$. At position $x$, the cross-section perpendicular to the $x$-axis is a right triangle with base $y$ and height $y\tan(45°) = y$.

Show Answer

Step 1: The wedge extends from $x = -3$ to $x = 3$.

Step 2: At position $x$, the base of the triangular cross-section is $y = \sqrt{9 - x^2}$ (from the semicircle), and the height is: $$\text{height} = y\tan(45°) = \sqrt{9 - x^2} \cdot 1 = \sqrt{9 - x^2}$$

Step 3: The cross-sectional area is: $$A(x) = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}\sqrt{9-x^2} \cdot \sqrt{9-x^2} = \frac{9 - x^2}{2}$$

Step 4: Integrate: $$V = \int_{-3}^3 \frac{9 - x^2}{2}\,dx$$

Since the integrand is even: $$V = 2 \cdot \frac{1}{2}\int_0^3 (9 - x^2)\,dx = \left[9x - \frac{x^3}{3}\right]_0^3$$

$$= 27 - 9 = \boxed{18}$$

Sanity check: Since $\tan(45°) = 1$, this wedge is “half” of a half-cylinder in some sense. The simpler angle gives a cleaner answer.

Level 5 Cavalieri's Principle Proof

Cavalieri’s Principle: If two solids have equal cross-sectional areas at every height, they have equal volumes.

Prove Cavalieri's Principle using the slicing formula.
Use this principle to show that an oblique cylinder (whose axis is tilted) has the same volume as a right cylinder with the same base area and height.

Thought Process

For part (a): If $A_1(x) = A_2(x)$ for all $x \in [a,b]$, then the integrals $\int_a^b A_1(x)\,dx$ and $\int_a^b A_2(x)\,dx$ must be equal.

For part (b): An oblique cylinder has the same circular cross-section at each height as a right cylinder: the cross-section is just shifted horizontally, but its area is unchanged.

Show Answer

(a) Proof of Cavalieri’s Principle:

Let $S_1$ and $S_2$ be two solids, both lying between $x = a$ and $x = b$. Let $A_1(x)$ and $A_2(x)$ be their cross-sectional areas at position $x$.

Given: $A_1(x) = A_2(x)$ for all $x \in [a, b]$.

To prove: $V_1 = V_2$.

By the slicing formula: $$V_1 = \int_a^b A_1(x)\,dx \quad \text{and} \quad V_2 = \int_a^b A_2(x)\,dx$$

Since $A_1(x) = A_2(x)$ for all $x$: $$V_1 = \int_a^b A_1(x)\,dx = \int_a^b A_2(x)\,dx = V_2$$

Therefore $V_1 = V_2$. $\square$

(b) Application to Oblique Cylinders:

Consider a right cylinder with base area $A$ and height $h$, and an oblique cylinder with the same base area $A$ and vertical height $h$ (measured perpendicular to the base).

At any height $x$ between $0$ and $h$:

The right cylinder has cross-section = base, so $A_{\text{right}}(x) = A$
The oblique cylinder’s cross-section is the base shifted horizontally, but slicing perpendicular to the vertical axis still gives area $A_{\text{oblique}}(x) = A$

Since $A_{\text{right}}(x) = A_{\text{oblique}}(x) = A$ for all $x \in [0, h]$, Cavalieri’s Principle gives: $$V_{\text{oblique}} = V_{\text{right}} = Ah$$

Common Misconceptions

Common misconception

the cross-sectional area $A(x)$ is just the function value $f(x)$ at that position.

This is the height-vs-slope error applied to three dimensions. The function $f(x)$ gives a linear measurement such as a radius or half-width, while $A(x)$ is a two-dimensional area computed from that measurement using a geometric formula. For a hemisphere of radius 5, the function describing the boundary is $\sqrt{25 - z^2}$, but the cross-sectional area at height $z$ is $\pi(25 - z^2)$, which requires squaring and multiplying by $\pi$. Writing $A(z) = \sqrt{25 - z^2}$ instead produces an integral whose units are length, not area, and the result cannot be a volume.

Mastery Checklist

I can explain why volume is the integral of cross-sectional area
I can identify the limits of integration from the solid’s extent
I can express cross-sectional area $A(x)$ as a function of position
I can set up and evaluate $V = \int_a^b A(x)\,dx$
I can derive volume formulas for standard solids (sphere, pyramid, cone)
I understand when to integrate with respect to $x$ versus $y$

Mental Model

The Bread Slicer: Imagine an infinitely sharp bread slicer that can cut arbitrarily thin slices. Each slice is a flat disk of area $A(x)$ and thickness $dx$. The volume of each infinitesimal slice is $A(x)\,dx$, and the total volume is the “sum” (integral) of all these slices from one end of the loaf to the other.

Connections

Looking back:

The Definite Integral provides the framework for “summing” continuously varying quantities
Riemann Sums show how the volume formula arises as a limit

Looking ahead:

The Disk Method applies slicing to solids of revolution with circular cross-sections
The Washer Method handles hollow solids of revolution
Volumes with Known Cross-Sections uses non-circular cross-sections

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Properties of Integrals	Skills Index	Disk Method

Last updated: 2026-01-23