From Cross-Sections to a Volume Integral
A solid can be difficult to describe by its outside surface, yet its cross-sections may be straightforward. If the area of each cross-section is known, volume can be built by adding the contributions of many thin slices.
The essential setup is to match three things: the area function, the direction of the slices, and the interval the slices cover. The integral then adds cross-sectional area over distance. Its result is volume.
Area, Slice Direction, and Bounds
For vertical slices, the coordinate \(x\) measures position from left to right. A vertical slice is perpendicular to the \(x\)-axis, so a thin slice has thickness \(dx\). If its cross-sectional area at position \(x\) is \(A(x)\), its approximate volume is \(A(x)\,dx\).
For horizontal slices, the coordinate \(y\) measures position from bottom to top. A horizontal slice is perpendicular to the \(y\)-axis, so its thickness is \(dy\). If its area at height \(y\) is \(A(y)\), its approximate volume is \(A(y)\,dy\).
Here \(A(x)\) or \(A(y)\) is the area of a whole cross-section, not the area of the base region and not the length of a side of the cross-section. Finding that area may require a geometric formula or information about the base; in this tutorial, the area function is supplied or already determined.
Why Integrating Area Gives Volume
The volume formula comes from approximating the solid with thin slices. Consider slices perpendicular to the \(x\)-axis between \(x=a\) and \(x=b\). Divide this interval into \(n\) smaller intervals, each with width \(\Delta x_i\). In the \(i\)th interval, choose a sample position \(x_i^*\). If the cross-sectional area does not change much across this thin interval, the slice's volume is approximately \(A(x_i^*)\Delta x_i\).
Adding these approximate slice volumes gives a Riemann sum. As the slices become thinner and more numerous, the approximation approaches the exact volume. When \(A\) is integrable on the interval, this limit is the definite integral.
The same reasoning applies to horizontal slices: use sample heights \(y_i^*\), thicknesses \(\Delta y_i\), and sum \(A(y_i^*)\Delta y_i\). In either direction, the integral adds the volumes of the slices across the full interval.
Worked Example: Vertical Slices with a Linear Area Function
Worked Example: Vertical Slices with a Linear Area Function
A solid has vertical cross-sections from \(x=0\) cm to \(x=4\) cm. At position \(x\), the cross-sectional area is \(A(x)=(2+x)\text{ cm}^2\). Set up and calculate the volume.
The slices are vertical, so they are stacked along the \(x\)-axis. The bounds are \(0\) and \(4\), and each slice has thickness \(dx\). Therefore,
An antiderivative of \(2+x\) is \(2x+\frac{x^2}{2}\). Evaluating at the bounds,
Check: differentiating the antiderivative gives \(\frac{d}{dx}(2x+\frac{x^2}{2})=2+x\), the stated area function. The area is \(2\text{ cm}^2\) at \(x=0\) and \(6\text{ cm}^2\) at \(x=4\); its average over this linear change is \(4\text{ cm}^2\). Multiplying that average by the \(4\text{ cm}\) length gives \(16\text{ cm}^3\), confirming the result.
When the Area Formula Changes
A solid may have different cross-sectional descriptions on different parts of its length. If the area function changes at a boundary, split the integral at that point. Each piece uses the area formula that applies there. The values at a single shared boundary do not affect the integral, provided the function remains integrable.
Worked Example: Horizontal Slices with Piecewise Areas
A solid has horizontal cross-sections from \(y=0\) cm to \(y=4\) cm. Their areas are \(A(y)=(y+1)\text{ cm}^2\) for \(0\le y\le2\), and \(A(y)=(5-y)\text{ cm}^2\) for \(2<y\le4\). Find the volume.
Horizontal slices are stacked along the \(y\)-axis, so the integral uses \(dy\). Because the formula changes at \(y=2\), split the volume into two integrals:
For the first part, an antiderivative is \(\frac{y^2}{2}+y\):
For the second part, an antiderivative is \(5y-\frac{y^2}{2}\):
Thus,
Check: the derivatives of the two antiderivatives are \(y+1\) and \(5-y\), respectively. The area formulas agree at the boundary: \(2+1=5-2=3\text{ cm}^2\). The two positive contributions add to the total volume, \(8\text{ cm}^3\).
Interpreting Units and Geometric Meaning
A cross-sectional area has units of length squared. The differential \(dx\) or \(dy\) represents a small distance, with units of length. Therefore, area multiplied by slice thickness has units of volume:
For example, if \(A(x)\) is measured in square metres and \(x\) in metres, then \(\int A(x)\,dx\) is measured in cubic metres. The differential is not decorative: it identifies the direction and units of the distance over which area is accumulated.
Worked Example: Interpreting a Volume from an Area Function
A solid has vertical cross-sections from \(x=0\) m to \(x=4\) m, with area \(A(x)=x(4-x)\text{ m}^2\). Find its volume and interpret the setup.
The slices are stacked along the \(x\)-axis, so the bounds are \(0\) and \(4\), and the setup is
An antiderivative is \(2x^2-\frac{x^3}{3}\). Therefore,
Check: \(\frac{d}{dx}(2x^2-\frac{x^3}{3})=4x-x^2=x(4-x)\), the given area function. It is nonnegative on \(0\le x\le4\), since both \(x\) and \(4-x\) are nonnegative there. Thus the integral adds nonnegative slice volumes and has the correct units, cubic metres.
A Common Setup Error
The bounds must describe the slice positions in the same direction as the chosen variable. A frequent mistake is to use bounds from the base region's other coordinate, rather than the interval along which the slices are stacked.
Even if the symbols \(x\) and \(y\) can be changed as dummy variables in an integral, their meaning in a geometric setup matters: they must refer to the coordinate along which the slices are stacked. Always identify the slice direction before choosing the differential and limits.
A Reliable Setup Procedure
Before evaluating, check the structure of the integral. These steps help keep the geometry, calculus, and units consistent.
Decide whether slices are stacked along \(x\) or along \(y\).
Use \(dx\) for slices stacked along the \(x\)-axis and \(dy\) for slices stacked along the \(y\)-axis.
Use the starting and ending coordinates of the interval along the slice direction.
Use the area of a full cross-section at the selected coordinate; split the integral if its formula changes.
Check the antiderivative, confirm the areas are nonnegative, and verify that the final units are cubic.
Check Your Understanding
For each question, focus first on matching the area function to the direction and interval of the slices.
- A solid has vertical slices from \(x=1\) to \(x=3\) with area \(A(x)=2x\). Write the volume integral and evaluate it.
- Horizontal slices run from \(y=0\) to \(y=2\) and have area \(A(y)=y^2+1\). Which differential belongs in the integral, and what are the bounds?
- A cross-sectional area is measured in \(\text{cm}^2\), while its position coordinate is measured in cm. What units does the volume integral have?
- If the area formula changes at \(x=2\), what should you do when setting up the volume integral?
- Explain why a sum of terms \(A(x_i^*)\Delta x_i\) approximates volume.
