Solids With a Prescribed Cross-Sectional Area

From Cross-Sections to Volume

A solid can be described by the areas of its slices, even when its three-dimensional shape is difficult to visualize. If every slice perpendicular to an axis has a known area, those areas can be accumulated along the solid’s extent to obtain its volume.

The key is to integrate the area of each cross-section with respect to distance along the solid. The resulting volume depends both on the area function and on the interval over which the solid exists. An antiderivative alone does not determine the correct answer: the bounds must match the solid’s full extent.

Definition: Let $A(x)$ be the area of the cross-section perpendicular to the $x$-axis at position $x$. If the solid extends from $x=a$ to $x=b$, its volume is $$ V=\int_a^b A(x)\,dx, $$ provided $A(x)$ is integrable on $[a,b]$. Since cross-sectional area cannot be negative, $A(x)\geq 0$ throughout the interval.

If $A(x)$ is measured in square centimetres and $x$ in centimetres, then $A(x)\,dx$ has units of cubic centimetres. In general, area units multiplied by length units give volume units.

Why the Area Integral Gives Volume

Here is a brief justification of the formula. Divide the interval from $a$ to $b$ into small subintervals. In a subinterval of width $\Delta x_i$, choose a sample position $x_i^*$. The slice there has area $A(x_i^*)$, so a thin slab of that width has approximate volume $A(x_i^*)\Delta x_i$. Adding these approximate slab volumes gives a Riemann sum.

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

As the widths of the subintervals approach zero, the approximation accounts for the changing area with increasing precision. When $A$ is integrable, these sums approach the definite integral, giving the volume formula. Continuity is a sufficient condition for integrability; a bounded area function that is piecewise continuous on the interval is also integrable.

V=\lim_{\max \Delta x_i\to 0}\sum_{i=1}^{n} A(x_i^*)\Delta x_i =\int_a^b A(x)\,dx.

This reasoning explains why a cross-sectional area is integrated over a length. It also clarifies what the bounds mean: $a$ and $b$ are the starting and ending positions of the solid along the slicing direction, not values chosen merely because they simplify the antiderivative.

Choosing the Bounds and Setting Up the Integral

Before evaluating an integral, identify the coordinate that measures distance along the solid and determine where the solid begins and ends in that coordinate. If the solid occupies $a\leq x\leq b$, those are the bounds for slices perpendicular to the $x$-axis. Use the complete interval of the solid, including endpoints where the cross-sectional area may be zero.

Then substitute the given area function directly into the volume formula. Do not integrate a side length or a perimeter unless it is first used to obtain the cross-sectional area. Keep units visible: $A(x)$ has area units, while $dx$ contributes one length unit.

Identify the slice coordinate.
Use the variable that measures position along the direction in which the cross-sections are stacked.

Read the solid’s extent.
Record its starting and ending coordinates as the integration bounds.

Integrate the area.
Evaluate the definite integral of $A(x)$ over those bounds.

Check the result.
Confirm the antiderivative, the endpoint substitution, and the volume units.

Worked Example: A Polynomial Area Function

A solid extends from $x=0$ cm to $x=2$ cm. Its cross-sectional area is $A(x)=3x^2+2$ cm$^2$. Find its volume.

The bounds are the endpoints of the solid’s extent, $0$ cm and $2$ cm. Integrate the area function:

V=\int_0^2(3x^2+2)\,dx.

An antiderivative of $3x^2+2$ is $x^3+2x$, since its derivative is $3x^2+2$. Evaluate at both bounds:

\begin{aligned} V &=\left[x^3+2x\right]_0^2\\ &=(2^3+2\cdot 2)-(0^3+2\cdot 0)\\ &=(8+4)-0\\ &=12\text{ cm}^3. \end{aligned}

Check: Differentiating $x^3+2x$ gives $3x^2+2$, the stated area function. The area is in cm$^2$ and the coordinate is in cm, so the volume is correctly in cm$^3$.

Worked Example: A Trigonometric Area Function

A solid extends from $x=0$ m to $x=4$ m. Its cross-sectional area is $A(x)=2+\sin(\pi x/4)$ m$^2$. Find its volume. The argument $\pi x/4$ is dimensionless when $x$ is expressed in metres.

Use the full extent as the bounds and integrate each term. An antiderivative of $\sin(\pi x/4)$ is $-\frac{4}{\pi}\cos(\pi x/4)$, by the chain rule:

\begin{aligned} V &=\int_0^4 \left(2+\sin\left(\frac{\pi x}{4}\right)\right)\,dx\\ &=\left[2x-\frac{4}{\pi}\cos\left(\frac{\pi x}{4}\right)\right]_0^4\\ &=\left(8-\frac{4}{\pi}\cos\pi\right) -\left(0-\frac{4}{\pi}\cos 0\right)\\ &=\left(8+\frac{4}{\pi}\right)-\left(-\frac{4}{\pi}\right)\\ &=8+\frac{8}{\pi}\text{ m}^3. \end{aligned}

Check: Differentiating $2x-\frac{4}{\pi}\cos(\pi x/4)$ gives $2+\sin(\pi x/4)$. Also, the area stays between $1$ and $3$ m$^2$ over a length of $4$ m, so a volume between $4$ and $12$ m$^3$ is reasonable. Since $8+8/\pi$ is between those values, the result passes this estimate.

When the Area Formula Changes

Sometimes the cross-sectional area is specified by different formulas on different parts of the solid. The volume is still the integral of area over the entire extent. To evaluate it, divide the integral at each point where the formula changes, then add the resulting volumes. A value assigned at a single boundary does not change the integral, but the formulas on either side must cover the full interval.

Worked Example: A Piecewise Area Function

A solid extends from $x=0$ cm to $x=3$ cm. Its cross-sectional area is $A(x)=x+1$ cm$^2$ for $0\leq x\leq 1$, and $A(x)=3-x$ cm$^2$ for $1<x\leq 3$. Find its volume.

The formula changes at $x=1$, so split the integral there. The two pieces cover the entire extent:

\begin{aligned} V &=\int_0^1(x+1)\,dx+\int_1^3(3-x)\,dx\\ &=\left[\frac{x^2}{2}+x\right]_0^1 +\left[3x-\frac{x^2}{2}\right]_1^3\\ &=\left(\frac{1}{2}+1-0\right) +\left(\left(9-\frac{9}{2}\right)-\left(3-\frac{1}{2}\right)\right)\\ &=\frac{3}{2}+\left(\frac{9}{2}-\frac{5}{2}\right)\\ &=\frac{3}{2}+2\\ &=\frac{7}{2}\text{ cm}^3. \end{aligned}

Check: The derivative of $\frac{x^2}{2}+x$ is $x+1$, and the derivative of $3x-\frac{x^2}{2}$ is $3-x$. The second piece’s value falls to zero at $x=3$, while both formulas give area $2$ cm$^2$ at the change point $x=1$. The total is positive and has the correct volume units.

A Common Mistake: Using One Slice for the Whole Solid

A frequent error is to multiply the area at one endpoint by the full length, as though every cross-section had that same area. For the polynomial example, the area at $x=2$ is $3(2)^2+2=14$ cm$^2$. Multiplying this by the length $2$ cm gives $28$ cm$^3$, which is not the volume: it treats every slice as if its area were $14$ cm$^2$.

The correction is to account for the changing area by integration:

\int_0^2(3x^2+2)\,dx=12\text{ cm}^3, \qquad\text{not}\qquad 14\cdot 2=28\text{ cm}^3.

Multiplying one area by the length is valid only when the cross-sectional area is constant throughout the solid. In the general case, it is the definite integral that adds the varying areas across the whole extent.

What the Integral Tells You

A prescribed area function contains the information needed for volume, even if no familiar cross-sectional shape is named. The function may be polynomial, trigonometric, or given by different formulas on different intervals. The essential steps remain the same: use the solid’s extent for the bounds, integrate the area on that extent, and check that the result has volume units.

Keep in mind that the definite integral accumulates area with respect to distance. Reversing the bounds would reverse the sign of the result, so bounds should be recorded in increasing order from the solid’s start to its end. A physically meaningful cross-sectional area is nonnegative; a negative computed volume is a signal to check the setup or the order of the bounds.

Check Your Understanding

For each question, identify the bounds and the area function before evaluating any integral.

A solid extends from $x=1$ to $x=3$ cm, and $A(x)=2x$ cm$^2$. Write the volume integral and find the volume.
Why does $A(x)\,dx$ have volume units when $A(x)$ is an area?
If the area formula changes at $x=c$, how should the volume integral be written?
For a constant cross-sectional area, when is multiplying area by length equivalent to integrating?
What is one check that can reveal an incorrect antiderivative in a volume calculation?

Solids With a Prescribed Cross-Sectional Area

What You'll Learn

From Cross-Sections to Volume

Why the Area Integral Gives Volume

Choosing the Bounds and Setting Up the Integral

Worked Example: A Polynomial Area Function

Worked Example: A Polynomial Area Function

Worked Example: A Trigonometric Area Function

Worked Example: A Trigonometric Area Function

When the Area Formula Changes

Worked Example: A Piecewise Area Function

Worked Example: A Piecewise Area Function

A Common Mistake: Using One Slice for the Whole Solid

What the Integral Tells You

Check Your Understanding

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Vol. 2: Solutions Manual

Solids With a Prescribed Cross-Sectional Area

What You'll Learn

From Cross-Sections to Volume

Why the Area Integral Gives Volume

Choosing the Bounds and Setting Up the Integral

Worked Example: A Polynomial Area Function

Worked Example: A Polynomial Area Function

Worked Example: A Trigonometric Area Function

Worked Example: A Trigonometric Area Function

When the Area Formula Changes

Worked Example: A Piecewise Area Function

Worked Example: A Piecewise Area Function

A Common Mistake: Using One Slice for the Whole Solid

What the Integral Tells You

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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