From a Base Slice to a Square Cross Section
A base region describes where a solid begins, but it does not by itself determine the solid’s volume. To specify the solid, we also need to know the shape and size of the cross sections. In this tutorial, every cross section is a square, and a slice through the base supplies one side of that square.
The key step is to find the length of the base region cut by a slice perpendicular to the chosen axis. That length becomes the side length of the square. Squaring it gives the cross-sectional area; integrating those areas across the region gives the volume.
The slice length is a distance, so it must be nonnegative. For vertical slices, it is usually the top boundary minus the bottom boundary. For horizontal slices, it is usually the right boundary minus the left boundary. The squared expression is the area of the square—not the length of its diagonal.
Why the Integral Gives Volume
Consider vertical slices from \(x=a\) to \(x=b\). At each \(x\), the square has side length \(s(x)\), so its area is \(s(x)^2\). A very thin piece of the solid near that \(x\)-value has approximate thickness \(\Delta x\) and approximate volume \(s(x)^2\Delta x\). Adding many such pieces and taking the limit of thinner slices gives the definite integral.
This argument explains why the integrand is a cross-sectional area. The square’s side length is squared first; that area is then accumulated over the length of the interval. The same reasoning applies to horizontal slices, with thickness \(\Delta y\) and integration with respect to \(y\).
Setting Up the Integral
Before integrating, make sure the slice direction, side length, and bounds match. The following sequence is useful when the base is bounded by curves:
Use vertical slices for cross sections perpendicular to the \(x\)-axis, or horizontal slices for cross sections perpendicular to the \(y\)-axis.
For vertical slices, subtract bottom \(y\)-value from top \(y\)-value. For horizontal slices, subtract left \(x\)-value from right \(x\)-value.
The cross-sectional area is the side length squared because each cross section is a square.
Integrate with \(dx\) across the \(x\)-interval for vertical slices, or with \(dy\) across the \(y\)-interval for horizontal slices.
This is a setup method, not a rule to use without checking the geometry. The boundaries must describe the same slice throughout the interval, and the slice length must be nonnegative there. If the boundary order changes, the region may need to be split into intervals or described with a different slice direction.
Worked Example: Vertical Slices Under a Line
Worked Example: Vertical Slices Under a Line
The base region is bounded by \(y=0\), \(y=2-x\), \(x=0\), and \(x=2\). Each cross section perpendicular to the \(x\)-axis is a square whose side is the vertical slice length. Set up and evaluate the volume.
Step 1: Identify the bounds. The vertical slices run from \(x=0\) to \(x=2\), so the volume integral uses \(dx\) on that interval.
Step 2: Find the side length. The top boundary is \(y=2-x\), and the bottom boundary is \(y=0\). Therefore,
For \(0\le x\le2\), this length is nonnegative.
Step 3: Square the side and integrate.
An antiderivative is \(4x-2x^2+\frac{x^3}{3}\). Evaluating at both bounds gives
Check: Differentiating the antiderivative gives \(4-4x+x^2=(2-x)^2\), the integrand. The side length is zero at \(x=2\) and positive before that endpoint, as expected. The volume is \(\frac{8}{3}\) cubic units.
Worked Example: Horizontal Slices Between Curves
Worked Example: Horizontal Slices Between a Parabola and a Line
The base region is bounded by \(x=y^2\) and \(x=4\). Each cross section perpendicular to the \(y\)-axis is a square whose side is the horizontal slice length. Set up and evaluate the volume.
Step 1: Find the \(y\)-bounds. Intersections satisfy \(y^2=4\), so \(y=-2\) or \(y=2\). The horizontal slices run from \(y=-2\) to \(y=2\).
Step 2: Find the side length. At each \(y\), the left boundary is \(x=y^2\) and the right boundary is \(x=4\). Thus,
On \([-2,2]\), \(y^2\le4\), so the slice length is nonnegative.
Step 3: Square and integrate.
An antiderivative is \(16y-\frac{8}{3}y^3+\frac{y^5}{5}\). Therefore,
Check: Differentiating the antiderivative gives \(16-8y^2+y^4=(4-y^2)^2\). The side length is zero at both \(y\)-endpoints and positive between them. The result, \(\frac{512}{15}\), is in cubic units.
Worked Example: Squaring a Difference of Curves
Worked Example: Squares Between Two Curves
The base region is bounded by \(y=\sqrt{x}\) and \(y=\frac{x}{2}\), for \(0\le x\le4\). Each cross section perpendicular to the \(x\)-axis is a square whose side is the vertical slice length. Set up and evaluate the volume.
Step 1: Confirm the boundary order. The curves meet when \(\sqrt{x}=\frac{x}{2}\). Since \(x\ge0\), squaring gives \(x=\frac{x^2}{4}\), so \(x(x-4)=0\). The intersections are \(x=0\) and \(x=4\). At \(x=1\), \(\sqrt{1}=1\) and \(\frac{1}{2}=0.5\), so \(\sqrt{x}\) is above \(\frac{x}{2}\) between the endpoints.
Step 2: Find the side length and set up the integral.
Expand the square carefully:
Step 3: Integrate and evaluate. An antiderivative is \(\frac{x^2}{2}-\frac{2}{5}x^{5/2}+\frac{x^3}{12}\). Hence,
Check: Differentiating the antiderivative gives \(\frac{x}{1}-x^{3/2}+\frac{x^2}{4}=x-x^{3/2}+\frac{x^2}{4}\), which is the expanded integrand. The side length is zero at both intersections and positive in between. The volume is \(\frac{8}{15}\) cubic units.
Common Mistake: Integrating the Side Length
A frequent error is to integrate the slice length itself, forgetting that the cross section is a square. In the first example, an incorrect setup would be
This integrates lengths across the interval; its units are square units, not cubic units. It does not add the areas of the square cross sections. The corrected setup squares the side length first:
Another tempting error is to square the two boundary functions separately and subtract, as in \(y_{\text{top}}^2-y_{\text{bottom}}^2\). That is not the area of a square with side \(y_{\text{top}}-y_{\text{bottom}}\). In general,
Find the slice length first, then square the entire difference.
What to Check Before You Finish
A reliable square-cross-section setup has three features: its bounds describe the full extent of the base in the chosen direction, its side length is a nonnegative distance, and its integrand is the square of that length. The differential must also match the slice direction: vertical slices use \(dx\), while horizontal slices use \(dy\).
If the side length is written as a difference, check which boundary is above or to the right. A negative distance over part of the interval signals that the order is wrong or that the interval must be split. Finally, a volume should have cubic units. That units check can reveal a missing square, even if the algebra otherwise looks reasonable.
Check Your Understanding
For each question, identify the side length before deciding what the volume integral should be.
- A base region has vertical slice length \(5-x\) for \(0\le x\le5\), and cross sections perpendicular to the \(x\)-axis are squares. Write the volume integral.
- A base region has horizontal slices from \(x=y^2\) to \(x=9\), for \(-3\le y\le3\). What is the square side length, and what integral gives the volume?
- Why must the slice length be squared before it is integrated for square cross sections?
- A student uses \(\int_a^b s(x)\,dx\) for square cross sections with side \(s(x)\). What quantity and units does this expression represent, and what should replace it for volume?
- If a vertical slice has top boundary \(y=3\) and bottom boundary \(y=x\), write the side length and square area for that slice.
