Mass of Solids From Cross-Sectional Slices

From Cross Sections to Mass

A solid can have a shape that is difficult to describe all at once but simple cross sections. If the solid is divided into thin slices, each slice has approximately the area of its cross section multiplied by its thickness. Adding these small volumes and taking a limit leads to an integral for volume. For a solid with constant density, multiplying that volume by the density gives its mass.

This method is useful when the cross sections have familiar shapes, such as disks, squares, or rectangles, even if the whole solid is not a standard geometric object.

Definition: Suppose a solid extends along an axis from coordinate $a$ to coordinate $b$. Let $A(x)$ be the area of the cross section perpendicular to that axis at coordinate $x$, and let the solid have constant density $\rho$. If $A$ is integrable on $[a,b]$, the mass of the solid is $$ m=\rho\int_a^b A(x)\,dx. $$ Here $A(x)$ has units of area, $dx$ has units of length, and $\rho$ has units of mass per volume.

Why the Integral Gives Mass

First consider volume. Divide the interval $[a,b]$ into short subintervals. In a subinterval of width $\Delta x_i$, choose a coordinate $x_i^*$. When the subinterval is sufficiently narrow, the corresponding piece of the solid is approximated by a thin slab whose cross-sectional area is $A(x_i^*)$. Its approximate volume is $A(x_i^*)\Delta x_i$. Summing these approximate slab volumes gives a Riemann sum.

V\approx \sum_{i=1}^{n} A(x_i^*)\Delta x_i.

As the largest subinterval width tends to zero, the sum approaches the volume integral. This is precisely the definition of the integral of the cross-sectional area. For constant density, a piece with volume $\Delta V_i$ has mass $\rho\Delta V_i$. The same approximation therefore gives a mass sum, and its limit is $\rho$ times the volume integral.

V=\int_a^b A(x)\,dx, \qquad m=\rho V=\rho\int_a^b A(x)\,dx.

Derivation: For each thin slice, mass is approximately density times slice volume, and slice volume is approximately cross-sectional area times thickness. Thus $\Delta m_i\approx \rho A(x_i^*)\Delta x_i$. Summing and taking the limit gives $$ m=\lim_{\max \Delta x_i\to 0}\sum_{i=1}^n \rho A(x_i^*)\Delta x_i =\rho\int_a^b A(x)\,dx, $$ where the equality follows because $\rho$ is constant and can be factored out of the limit and integral.

The coordinate $x$ need not be a horizontal direction. It can measure distance along any chosen axis, provided the slices are perpendicular to that axis and the limits describe the solid’s full extent along it. The cross-sectional area may change with $x$; finding that area from the geometry is the essential setup step.

Worked Examples

Worked Example: A Cylinder With Circular Cross Sections

A cylinder is $5$ m long and has radius $2$ m. Its density is $3$ kg/m$^3$. Find its mass by slicing perpendicular to its length.

Let $x$ measure distance along the cylinder, with $0\le x\le 5$. Every perpendicular cross section is a circle of radius $2$ m, so its area is constant: $$ A(x)=\pi(2)^2=4\pi\ \text{m}^2. $$ The mass integral is $$ m=3\int_0^5 4\pi\,dx =12\pi[x]_0^5 =12\pi(5-0) =60\pi\ \text{kg}. $$ Thus the exact mass is $60\pi$ kg, approximately $188.50$ kg.

Check the integral by differentiating its antiderivative: $\frac{d}{dx}(4\pi x)=4\pi=A(x)$. Also, the volume is $4\pi\cdot5=20\pi$ m$^3$, so density times volume is $3(20\pi)=60\pi$ kg, as calculated.

Worked Example: A Solid Sphere

A sphere of radius $3$ m has constant density $2$ kg/m$^3$. Find its mass using circular slices perpendicular to the $x$-axis, with the sphere centered at $x=0$.

The sphere satisfies $x^2+y^2+z^2\le 9$. At a fixed $x$, the cross section is a disk in the $yz$-plane. Its radius $r$ satisfies $r^2=9-x^2$, so $$ A(x)=\pi r^2=\pi(9-x^2),\qquad -3\le x\le 3. $$ Therefore, $$ \begin{aligned} V&=\int_{-3}^{3}\pi(9-x^2)\,dx\\ &=\pi\left[9x-\frac{x^3}{3}\right]_{-3}^{3}\\ &=\pi\bigl((27-9)-(-27+9)\bigr)\\ &=36\pi\ \text{m}^3. \end{aligned} $$ Multiplying by the density gives $$ m=2(36\pi)=72\pi\ \text{kg}, $$ approximately $226.19$ kg.

For the integral check, the derivative of $9x-x^3/3$ is $9-x^2$, the area factor inside the integral. As a geometric check, the sphere’s volume is $\frac{4}{3}\pi(3)^3=36\pi$ m$^3$; multiplying this by $2$ kg/m$^3$ again gives $72\pi$ kg.

Worked Example: A Cone With Circular Slices

A right circular cone has height $6$ m, base radius $2$ m, and constant density $5$ kg/m$^3$. Set $x=0$ at the tip and $x=6$ at the base. Find its mass.

At distance $x$ from the tip, the cross-sectional radius grows in proportion to $x$. The ratio of radius to distance is $2/6=1/3$, so $r(x)=x/3$ m. The cross section is a disk: $$ A(x)=\pi\left(\frac{x}{3}\right)^2=\frac{\pi x^2}{9},\qquad 0\le x\le 6. $$ Then $$ \begin{aligned} V&=\int_0^6\frac{\pi x^2}{9}\,dx\\ &=\frac{\pi}{9}\left[\frac{x^3}{3}\right]_0^6\\ &=\frac{\pi}{27}(216-0)\\ &=8\pi\ \text{m}^3. \end{aligned} $$ The mass is $$ m=5(8\pi)=40\pi\ \text{kg}, $$ approximately $125.66$ kg.

Differentiate the antiderivative $x^3/3$ to get $x^2$, confirming the integral. The cone volume formula gives $\frac{1}{3}\pi(2)^2(6)=8\pi$ m$^3$; its product with the stated density is $40\pi$ kg, which verifies the result.

Worked Example: A Solid With Square Cross Sections

A solid extends from $x=0$ to $x=2$ m. Each perpendicular cross section is a square of side length $2-x$ m. The solid has constant density $4$ kg/m$^3$. Find its mass.

A square’s area is side length squared, so $$ A(x)=(2-x)^2\ \text{m}^2. $$ Thus $$ \begin{aligned} V&=\int_0^2(2-x)^2\,dx\\ &=\int_0^2(4-4x+x^2)\,dx\\ &=\left[4x-2x^2+\frac{x^3}{3}\right]_0^2\\ &=8-8+\frac{8}{3}\\ &=\frac{8}{3}\ \text{m}^3. \end{aligned} $$ The mass is $m=4(8/3)=32/3$ kg.

The derivative of $4x-2x^2+x^3/3$ is $4-4x+x^2=(2-x)^2$, confirming the antiderivative. The volume is positive, and multiplying $8/3$ m$^3$ by $4$ kg/m$^3$ gives $32/3$ kg.

Choosing and Checking the Setup

A reliable setup follows a short sequence. First choose the axis along which the solid extends and identify the endpoint coordinates $a$ and $b$. Next identify the shape of a typical perpendicular cross section and express its dimensions in terms of the slicing coordinate. Use the appropriate area formula to obtain $A(x)$. Finally integrate over the entire interval and multiply by the constant density.

Set the limits.
Use the coordinates of the two ends of the solid along the slicing axis.

Find a typical section.
Describe its dimensions at coordinate $x$, then calculate its area $A(x)$.

Integrate for volume.
Evaluate $V=\int_a^b A(x)\,dx$.

Convert volume to mass.
For constant density $\rho$, calculate $m=\rho V$.

Units provide a useful check. If lengths are measured in metres, then $A(x)\,dx$ has units $\text{m}^2\cdot\text{m}=\text{m}^3$. Multiplying by density in kg/m$^3$ gives kilograms. A result with units of area or volume alone is not yet a mass.

Common Mistake: Using Perimeter Instead of Area

For a cylinder of radius $2$ m, a mistaken setup might use the circumference $2\pi r=4\pi$ as the cross-sectional quantity and write $m=3\int_0^5 4\pi\,dx$, claiming the $4\pi$ came from circumference. This calculation happens to produce the cylinder’s correct numerical expression only because $4\pi$ is also the area of a radius-$2$ disk in this particular example. But circumference has units of metres, not square metres; integrating it over length would produce square metres, and multiplying by density would not give kilograms.

The correction is to use the area of each perpendicular section: $A=\pi r^2=4\pi$ m$^2$. Then $\int_0^5 A\,dx=20\pi$ m$^3$, and multiplying by $3$ kg/m$^3$ gives $60\pi$ kg. The area formula and its units matter even when a coincidental numerical value obscures the mistake.

Check Your Understanding

For each question, identify the cross-sectional area before evaluating or interpreting the mass integral.

A cylinder has length $L$, radius $r$, and constant density $\rho$. What integral gives its mass when sliced perpendicular to its length?
A sphere of radius $R$ is centered at the origin. What is the area of its disk cross section at coordinate $x$, and what are the integration limits?
A solid has cross-sectional area $A(x)=3x$ on $0\le x\le 2$, with constant density $4$. Set up and evaluate its mass.
What are the units of $\int_a^b A(x)\,dx$ when $A$ is in square metres and $x$ is measured in metres?
Why is a cross-sectional perimeter not a substitute for cross-sectional area in the mass integral?