From a Base-Region Chord to a Square
A solid can be described by its base region and the shape of its cross-sections. In this tutorial, every cross-section is a square. The base region determines one side of each square: take a slice through the base, measure the chord cut out by that slice, and use that chord length as the square’s side.
The orientation of the slices determines which chord to measure. A slice perpendicular to the \(x\)-axis is vertical in the base plane, so its chord runs in the \(y\)-direction. A slice perpendicular to the \(y\)-axis is horizontal in the base plane, so its chord runs in the \(x\)-direction.
It is essential to distinguish the chord from the square’s area. The chord gives a length, which is the square’s side. Squaring that length gives the area, which is what gets integrated to find volume.
Setting Up the Side Length
Suppose a vertical slice at position \(x\) passes through a base region bounded below by \(y=g(x)\) and above by \(y=f(x)\). When \(f(x)\ge g(x)\), the chord length is the upper \(y\)-value minus the lower \(y\)-value:
For square cross-sections perpendicular to the \(x\)-axis, this gives \(A(x)=[f(x)-g(x)]^2\). If instead the region is described by left and right boundaries \(x=L(y)\) and \(x=R(y)\), a horizontal slice at height \(y\) has chord length \(R(y)-L(y)\). Squares perpendicular to the \(y\)-axis then have area \(A(y)=[R(y)-L(y)]^2\).
In either orientation, the bounds must cover the positions where the slices pass through the base. When each slice intersects the base in one segment and the resulting area function is continuous on the interval, the volume is the integral of that area over the interval.
Why the Volume Is an Integral of Squared Chords
Consider first slices perpendicular to the \(x\)-axis. Divide the interval \([a,b]\) into many thin subintervals, each of width \(\Delta x\). In a thin slab near \(x_i\), the cross-section is approximately a square with side \(s(x_i)\), so its area is approximately \(s(x_i)^2\). Its volume is therefore approximately \(s(x_i)^2\Delta x\).
Adding the slab volumes gives an approximation to the whole volume. As the subinterval widths shrink, the sum approaches the definite integral of the cross-sectional area. This explains both why the chord must be squared and why the integration variable matches the direction along which the slices are stacked.
For horizontal slices, the same reasoning uses thin slab thickness \(\Delta y\) and sums areas \(s(y_i)^2\). The limiting integral is then taken with respect to \(y\).
Worked Example: Vertical Chords from Two Curves
Worked Example: Vertical Chords from Two Curves
The base region is bounded by \(y=x\) and \(y=x^2\) for \(0\le x\le1\). Square cross-sections are perpendicular to the \(x\)-axis. Find the volume.
Step 1: Identify the chord. On \(0\le x\le1\), \(x\ge x^2\), since \(x-x^2=x(1-x)\ge0\). Thus \(y=x\) is the upper curve and \(y=x^2\) is the lower curve. The vertical chord has length $$ s(x)=x-x^2. $$
Step 2: Find the square’s area. $$ A(x)=[s(x)]^2=(x-x^2)^2=x^2-2x^3+x^4. $$
Step 3: Integrate over the given interval. $$ V=\int_0^1(x^2-2x^3+x^4)\,dx =\left[\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right]_0^1 =\frac13-\frac12+\frac15 =\frac{10-15+6}{30} =\frac1{30}. $$
Check: Differentiating the antiderivative gives \(x^2-2x^3+x^4\), the area integrand. The volume is \(\frac1{30}\) cubic units.
Worked Example: Horizontal Chords
Worked Example: Horizontal Chords
The base region is bounded by \(x=y^2\) and \(x=2\), with \(0\le y\le\sqrt{2}\). Square cross-sections are perpendicular to the \(y\)-axis. Find the volume.
Step 1: Read the horizontal chord. At each fixed \(y\) in the given interval, the left endpoint is \(x=y^2\) and the right endpoint is \(x=2\). The chord length is $$ s(y)=2-y^2. $$ It is nonnegative because \(y^2\le2\) on this interval.
Step 2: Square the side length. $$ A(y)=(2-y^2)^2=4-4y^2+y^4. $$
Step 3: Integrate with respect to \(y\). $$ V=\int_0^{\sqrt{2}}(4-4y^2+y^4)\,dy =\left[4y-\frac{4y^3}{3}+\frac{y^5}{5}\right]_0^{\sqrt{2}}. $$
Since \((\sqrt{2})^3=2\sqrt{2}\) and \((\sqrt{2})^5=4\sqrt{2}\), $$ V=4\sqrt{2}-\frac{8\sqrt{2}}{3}+\frac{4\sqrt{2}}{5} =\frac{60\sqrt{2}-40\sqrt{2}+12\sqrt{2}}{15} =\frac{32\sqrt{2}}{15}. $$
Check: The derivative of \(4y-\frac{4y^3}{3}+\frac{y^5}{5}\) is \(4-4y^2+y^4\), the squared chord. The volume is \(\frac{32\sqrt{2}}{15}\) cubic units.
Worked Example: A Linear Chord on a Triangle
Worked Example: A Linear Chord on a Triangle
The base is the triangle bounded by \(y=0\), \(x=0\), and \(x+y=2\). Square cross-sections are perpendicular to the \(x\)-axis. Find the volume.
Step 1: Describe the vertical chord. Rearranging the sloping boundary gives \(y=2-x\). The triangle extends from \(x=0\) to \(x=2\), and each vertical chord runs from \(y=0\) to \(y=2-x\). Therefore $$ s(x)=(2-x)-0=2-x. $$
Step 2: Find the square’s area. $$ A(x)=(2-x)^2=4-4x+x^2. $$
Step 3: Integrate from the left endpoint to the right endpoint. $$ V=\int_0^2(4-4x+x^2)\,dx =\left[4x-2x^2+\frac{x^3}{3}\right]_0^2 =8-8+\frac83 =\frac83. $$
Check: Differentiating \(4x-2x^2+\frac{x^3}{3}\) gives \(4-4x+x^2=(2-x)^2\). Thus the volume is \(\frac83\) cubic units.
A Common Mistake: Using Only One Boundary
A chord is the full distance between the two sides of the base region, not the distance from a curve to an axis unless that axis is actually the other boundary. In the first example, taking \(s(x)=x\) would use the upper curve alone and ignore the lower curve \(y=x^2\).
That incorrect choice would give $$ V_{\text{wrong}}=\int_0^1x^2\,dx=\frac13. $$ But the actual chord is \(x-x^2\), so the correct volume is $$ V_{\text{correct}}=\int_0^1(x-x^2)^2\,dx=\frac1{30}. $$
The two answers differ because the wrong calculation builds squares with side \(x\), rather than with the actual chord length \(x-x^2\). Before squaring, identify both endpoints of the chord and subtract the smaller coordinate from the larger one.
Choosing the Integral and Checking It
For a square cross-section problem, keep three quantities separate:
- Chord length: a length read from the base region.
- Cross-sectional area: the square of that length.
- Volume: the integral of the area with respect to the coordinate along which slices are stacked.
A useful setup check is to ask whether the dimensions and variable match the slice orientation. Vertical chords give a side length described using \(x\), and perpendicular-to-\(x\)-axis slices lead to an integral in \(dx\). Horizontal chords give a side length described using \(y\), and perpendicular-to-\(y\)-axis slices lead to an integral in \(dy\). If the side is measured in units of length, its square has units of area; integrating with respect to another length gives units of volume.
Check Your Understanding
For each question, identify the chord carefully before forming the volume integral.
- A base region lies between \(y=3\) and \(y=x\) for \(0\le x\le3\). What is the side length for square cross-sections perpendicular to the \(x\)-axis?
- A base region lies between \(x=1\) and \(x=5-y\) for \(0\le y\le4\). Write the volume integral for squares perpendicular to the \(y\)-axis.
- Why is the cross-sectional area \(s(x)^2\), rather than \(s(x)\), when the chord length is \(s(x)\)?
- For squares perpendicular to the \(x\)-axis, which coordinate measures the thickness of a thin slab?
- What information is lost if you use the value of only one boundary curve instead of subtracting the two boundary coordinates?
