From Base Width to Triangular Slices
A solid can be difficult to describe using a familiar three-dimensional shape. A useful alternative is to examine its cross sections: slices made by cutting the solid with planes. If the cross sections are perpendicular to a chosen axis, each one has an area that can vary with position along that axis.
For a solid whose base lies in the \(xy\)-plane, imagine a slice at position \(x\). The part of the base cut by that slice has a width \(w(x)\). If the cross section above that base segment is a triangle, the width is one of the triangle's specified dimensions. The cross-sectional area then follows from the relevant triangle formula.
The key is not to treat every triangular cross section alike. For an equilateral triangle, the width is its side length. For a right triangle, the width might be one leg, the hypotenuse, or another specified dimension. The area formula depends on that geometric information.
Area Formulas for Triangular Cross Sections
An equilateral triangle has three equal sides. If each side has length \(s\), drawing an altitude divides it into two right triangles. Each half has hypotenuse \(s\) and horizontal leg \(s/2\), so the altitude \(h\) satisfies \(h^2+(s/2)^2=s^2\). Thus \(h=\sqrt{3}s/2\). Using area equal to one half base times height gives the equilateral-triangle area formula.
A right triangle has a right angle, so its perpendicular legs can serve as base and height. If their lengths are \(a\) and \(b\), its area is \(A=ab/2\). When one leg has length \(w\) and the other is a known multiple \(k w\), the area is \(k w^2/2\). If instead the width \(w\) is the hypotenuse of a right isosceles triangle, the equal legs each have length \(w/\sqrt{2}\), and the area is \(w^2/4\).
These formulas give area, not volume. The volume is found by accumulating the cross-sectional areas across the interval occupied by the solid. If the cross-sectional area \(A(x)\) is continuous on \([a,b]\), the slice principle gives \(V=\int_a^b A(x)\,dx\). The integral adds area times a small thickness, which has units of volume.
Why the Slice Integral Works
The slice principle gives a precise reason for integrating the area. Divide the interval \([a,b]\) into many short subintervals. In a subinterval of thickness \(\Delta x\), the solid is approximately a thin slab. If the slice area near a sample point \(x_i^*\) is \(A(x_i^*)\), that slab's approximate volume is \(A(x_i^*)\Delta x\).
At position \(x_i^*\), the triangular cross section has area \(A(x_i^*)\).
Over a small thickness \(\Delta x\), the volume is approximately \(A(x_i^*)\Delta x\).
The sum \(\sum_i A(x_i^*)\Delta x\) approximates the volume of the whole solid.
As the subintervals become arbitrarily thin, the sums approach \(\int_a^b A(x)\,dx\), provided \(A\) is continuous on the interval.
For triangular cross sections, the specific work is to express \(A(x)\) in terms of the base width and any additional dimensions the description supplies. Once that area function is correct, the integral follows from the slice principle.
Worked Examples
Worked Example: Equilateral Cross Sections Above a Parabolic Base
The base region lies between \(y=0\) and \(y=4-x^2\), for \(-2\le x\le2\). Perpendicular to the \(x\)-axis, the solid has equilateral-triangular cross sections whose sides are the vertical widths of the base. Find the volume.
First, the width is the upper curve minus the lower curve:
On \([-2,2]\), this width is nonnegative. Since it is the side of an equilateral triangle, the slice area is
Therefore the volume is
At the upper endpoint, the antiderivative has value
At the lower endpoint, it has value
Subtracting the endpoint values gives
Thus the volume is \(\frac{128\sqrt{3}}{15}\) cubic units. As a check, differentiating \(16x-\frac{8x^3}{3}+\frac{x^5}{5}\) gives \(16-8x^2+x^4=(4-x^2)^2\), the integrand before its constant factor. The width vanishes at both endpoints, as expected for this base.
Worked Example: Right Isosceles Cross Sections with Width as a Leg
The base region is bounded by \(y=0\) and \(y=2-x\) for \(0\le x\le2\). Each cross section perpendicular to the \(x\)-axis is a right isosceles triangle, and the base-region width is the length of each equal leg. Find the volume.
The width is \(w(x)=(2-x)-0=2-x\). Because both perpendicular legs have length \(w(x)\), the cross-sectional area is
Integrate this area from \(x=0\) to \(x=2\):
The volume is \(\frac{4}{3}\) cubic units. Differentiating the antiderivative inside the brackets gives \(4-4x+x^2=(2-x)^2\), so the integral has the required area integrand after multiplication by \(1/2\). At \(x=2\), the width and slice area are zero; at \(x=0\), the width is \(2\) and the area is \(2\) square units.
Worked Example: Right Isosceles Cross Sections with Width as the Hypotenuse
The base is a disk of radius \(2\), described by \(x^2+y^2\le4\). Cross sections perpendicular to the \(x\)-axis are right isosceles triangles whose hypotenuse is the full width of the base at \(x\). Find the volume.
At each \(x\) from \(-2\) to \(2\), the upper and lower boundaries of the disk are \(y=\sqrt{4-x^2}\) and \(y=-\sqrt{4-x^2}\). Their difference gives the width:
Here \(w(x)\) is the hypotenuse, not a leg. In a right isosceles triangle with hypotenuse \(w(x)\), each leg has length \(w(x)/\sqrt{2}\). Thus
Integrating gives
Correction: the endpoint simplification in the third line must be evaluated carefully. \(8-\frac{8}{3}=\frac{16}{3}\), while \(-8+\frac{8}{3}=-\frac{16}{3}\), so their difference is \(\frac{32}{3}\). Therefore the volume is \(\frac{32}{3}\) cubic units. Differentiating \(4x-x^3/3\) gives \(4-x^2=A(x)\). The area is zero at \(x=\pm2\), consistent with the disk narrowing to a point there.
Common Mistake: Confusing a Hypotenuse with a Leg
In the disk example, a tempting error is to use the right-isosceles area formula \(A=w^2/2\) with \(w\) equal to the hypotenuse. That formula applies when \(w\) is each leg, not when it is the hypotenuse. Since each leg is \(w/\sqrt{2}\), the correct area is \(w^2/4\).
Using the incorrect formula would give \(A(x)=w(x)^2/2=2(4-x^2)\), and hence the incorrect volume
The correct volume is \(\frac{32}{3}\), half that incorrect value. The difference comes directly from confusing which triangle dimension the base width represents. Before integrating, label the width as a side, leg, or hypotenuse and derive the area from that relationship.
Conditions and a Reliable Setup
A cross-sectional volume integral requires a valid area function across the full interval. In these examples, the width is continuous, the geometric relationship determines a nonnegative area, and the interval covers the solid from one end to the other. If the width or cross-sectional specification changes, the area formula must reflect that change on each part of the interval.
A practical setup is to proceed in this order:
- Identify the axis along which the slices are stacked and state the interval of positions.
- Find the base-region width at a typical position by subtracting the appropriate boundary values.
- Specify how that width relates to the triangle's dimensions.
- Write the cross-sectional area as a function of position.
- Integrate that area over the full interval and check the units and endpoint behavior.
Each slice area has units of square units. The integration variable contributes one more unit of length, so the result has cubic units. That dimensional check can reveal an error, but it cannot replace checking whether the width has been used as the correct side, leg, or hypotenuse.
Check Your Understanding
Use the width-to-area relationship before setting up each integral.
- A base width is \(w(x)=3-x\) on \(0\le x\le3\). What is the area function for equilateral cross sections?
- For a right isosceles cross section, the base width is the length of each leg. What area formula should be used?
- For a right isosceles cross section whose hypotenuse is \(w(x)\), what is the area in terms of \(w(x)\)?
- Why does integrating cross-sectional area with respect to position produce volume?
- What geometric information must be checked before using a right-triangle area formula with a base-region width?
