From a Planar Base to a Solid
A solid can be described by specifying its planar base and the shape of each cross section. Here, every cross section is a square. The base tells us the length of a square’s side at each position; the square’s area then gives the area of the slice of the solid at that position.
The key is to match the slice direction to the way the base is described. A slice perpendicular to the \(x\)-axis runs across the base in the vertical direction, so its side length is a difference of \(y\)-coordinates. A slice perpendicular to the \(y\)-axis runs horizontally, so its side length is a difference of \(x\)-coordinates.
Once the side length has been found, the area is its square. Integrating these areas across the base gives the volume.
Side Length and Cross-Sectional Area
Suppose a base region lies between \(x=a\) and \(x=b\), and a slice perpendicular to the \(x\)-axis meets the base in a vertical segment. If the segment extends from \(y=g(x)\) to \(y=f(x)\), with \(f(x)\ge g(x)\) throughout the interval, its length is \(f(x)-g(x)\). When this segment is the side of a square, the square’s area is \([f(x)-g(x)]^2\).
For horizontal slices, suppose the base extends from \(y=c\) to \(y=d\), and a slice perpendicular to the \(y\)-axis runs from \(x=L(y)\) to \(x=R(y)\), with \(R(y)\ge L(y)\). Its length is \(R(y)-L(y)\), so the square’s area is \([R(y)-L(y)]^2\).
For a square of side length \(s\), the area is \(s^2\). Thus, if \(A(x)\) is the area of a slice perpendicular to the \(x\)-axis, the slice principle gives the volume by integrating \(A(x)\) across the interval of \(x\)-values occupied by the base. The corresponding statement holds for slices perpendicular to the \(y\)-axis.
The integrand is an area, while the differential represents a small distance along the slicing axis. Consequently, the integral has units of volume. If lengths are measured in metres, the result is measured in cubic metres.
Why Integrating Slice Areas Gives Volume
The square at one position has zero thickness, so its area alone is not a volume. Consider instead a thin slice of thickness \(\Delta x\). If its cross-sectional area is approximately \(A(x_i)\), its volume is approximately \(A(x_i)\Delta x\). Adding these estimates over a partition of the interval gives a Riemann sum. As the partition becomes finer, the sums approach the integral of the cross-sectional area.
At a fixed coordinate, identify the segment cut out by the base region.
If the segment length is \(s\), the cross-sectional area is \(s^2\).
Multiply each area by an infinitesimal thickness along the slicing axis and integrate across the full base.
This is the slice principle applied to square cross sections: the shape matters through its area formula, while the base and slice direction determine the side length and bounds.
Worked Examples
Worked Example: Squares Above a Square-Root Base Profile
The base region is bounded by \(y=\sqrt{x}\), the \(x\)-axis, and \(x=4\). Square cross sections are perpendicular to the \(x\)-axis. Find the volume.
At a fixed \(x\) between \(0\) and \(4\), the vertical segment in the base runs from \(y=0\) to \(y=\sqrt{x}\). Its length, and therefore the square’s side length, is
The cross-sectional area is
The base begins at \(x=0\), where the curve meets the \(x\)-axis, and ends at \(x=4\). Therefore,
To check the integration, differentiate the antiderivative: \(\frac{d}{dx}(x^2/2)=x\), which is the cross-sectional area function. The side length is nonnegative on the interval, and the resulting volume is positive. If the base lengths are in metres, the volume is \(8\text{ m}^3\).
Worked Example: Squares Between a Line and a Parabola
The base region is bounded above by \(y=4\) and below by \(y=x^2\), for \(-2\le x\le 2\). Square cross sections are perpendicular to the \(x\)-axis. Find the volume.
For each \(x\) in this interval, the vertical base segment runs from \(x^2\) to \(4\). The inequality \(x^2\le4\) holds throughout \([-2,2]\), so the side length is
Its square area is
Integrate over the full interval of \(x\)-values:
For a check, differentiate the antiderivative: $$ \frac{d}{dx}\left(16x-\frac{8x^3}{3}+\frac{x^5}{5}\right)=16-8x^2+x^4=(4-x^2)^2. $$ The derivative recovers the cross-sectional area. The side length is zero at \(x=-2\) and \(x=2\), and positive between them, consistent with a positive volume of \(\frac{512}{15}\) cubic units.
Worked Example: Horizontal Squares in a Parabolic Base
The base region is bounded by \(x=y^2\) and \(x=1\). Square cross sections are perpendicular to the \(y\)-axis. Find the volume.
The curves meet when \(y^2=1\), giving \(y=-1\) and \(y=1\). For a fixed \(y\) in this interval, a horizontal segment in the base runs from \(x=y^2\) to \(x=1\). The side length is
The square area is
Because the slices are perpendicular to the \(y\)-axis, integrate with respect to \(y\):
Differentiating the antiderivative gives \(1-2y^2+y^4=(1-y^2)^2\), as required. The side length is nonnegative on \([-1,1]\) and vanishes at both endpoints, so the bounds and positive volume are consistent.
Choosing the Side Length and Bounds
The slice direction determines which coordinate changes across a base segment. For slices perpendicular to the \(x\)-axis, hold \(x\) fixed and measure vertically: top curve minus bottom curve. For slices perpendicular to the \(y\)-axis, hold \(y\) fixed and measure horizontally: right boundary minus left boundary.
| Slice orientation | Segment in the base | Square area | Integration variable |
|---|---|---|---|
| Perpendicular to the \(x\)-axis | Vertical length \(f(x)-g(x)\) | \([f(x)-g(x)]^2\) | \(x\) |
| Perpendicular to the \(y\)-axis | Horizontal length \(R(y)-L(y)\) | \([R(y)-L(y)]^2\) | \(y\) |
The bounds cover the complete range of the coordinate along the slicing axis. They are not automatically the left and right endpoints of the region: for horizontal slices, the relevant bounds are the lowest and highest \(y\)-values. Likewise, the side length must describe the segment in the base, not a distance measured along the slicing axis.
A Common Mistake: Integrating Side Lengths
A frequent error is to integrate \(s(x)\) rather than the square area \(s(x)^2\). For the first example, that incorrect calculation would be
This is not the volume: it integrates lengths over a length, rather than areas over a length. The square cross section has area \((\sqrt{x})^2=x\), so the correct calculation is
A useful check is dimensional: if \(s\) is measured in metres, then \(s\) has units of metres, while \(s^2\) has units of square metres. Integrating \(s^2\) with respect to a length gives cubic metres, the correct units for volume.
Why the Setup Matters
In these problems, the integration is often straightforward once the cross-sectional area is correct. The modelling work is deciding which segment of the base becomes the square’s side, writing its length in the correct order, and using bounds that cover the entire solid. A reversed difference can produce a negative side length even though a geometric length cannot be negative. Use top minus bottom for vertical segments and right minus left for horizontal segments, after checking which boundary is actually above or to the right.
Finally, the square area formula applies because the specified cross sections are squares. The planar base does not itself have to be square. Its role is to provide the side length at each slice; the square cross section then determines the area used in the volume integral.
Check Your Understanding
For each question, identify the base segment, the square area, and the appropriate integral before calculating.
- A base lies between \(y=3\) and \(y=x\) for \(0\le x\le3\). What is the side length for square slices perpendicular to the \(x\)-axis?
- A base lies between \(y=0\) and \(y=2-x^2\) for \(-\sqrt{2}\le x\le\sqrt{2}\). Write the volume integral for square slices perpendicular to the \(x\)-axis.
- A base lies between \(x=y^2\) and \(x=4\) for \(-2\le y\le2\). What is the side length for square slices perpendicular to the \(y\)-axis?
- Why does integrating the square’s side length alone fail to calculate volume?
- What units result when side lengths are measured in centimetres and the slices are integrated across a length measured in centimetres?
