From a Base Slice to a Triangle
A solid can be described by a two-dimensional base region and a rule for the cross sections perpendicular to a chosen axis. At each position along that axis, a segment in the base determines one side of a cross-sectional triangle. The crucial question is which side: is that segment a leg of the triangle, or is it the hypotenuse?
For an isosceles right triangle, confusing those roles changes the area by a factor of two. The base-region slice gives a length, but it does not by itself tell you which area formula to use. The problem’s cross-section description must specify the role of that length.
Area Formulas for the Two Cases
The area of a right triangle is half the product of its perpendicular legs. If the base slice is a leg, the isosceles condition makes the other leg equal in length. If the base slice is the hypotenuse, the Pythagorean theorem determines the length of each leg. These two interpretations lead to different formulas.
- If \(L\) is a leg, the cross-sectional area is \(A=\frac{1}{2}L^2\).
- If \(L\) is the hypotenuse, the cross-sectional area is \(A=\frac{1}{4}L^2\).
To justify the leg formula, both perpendicular legs have length \(L\), so the area is \(\frac{1}{2}L\cdot L=\frac{1}{2}L^2\). For the hypotenuse formula, call each equal leg \(a\). The Pythagorean theorem gives \(L^2=a^2+a^2=2a^2\), so \(a^2=L^2/2\). The area is therefore \(\frac{1}{2}a^2=\frac{1}{2}(L^2/2)=\frac{1}{4}L^2\). The formulas require that the cross sections are isosceles right triangles and that \(L\) is assigned the stated role.
Once the area \(A\) of the cross section at each position is expressed in terms of the position, the volume is found by integrating those areas along the axis of slicing. In a setup with vertical slices, the position variable is \(x\), and the volume is \(\int_a^b A(x)\,dx\). With horizontal slices, it is \(y\), and the volume is \(\int_c^d A(y)\,dy\). The limits must cover the base region in the chosen direction, and the slice length must be nonnegative throughout the interval.
Worked Examples
Worked Example: Vertical Slices Are a Leg
The base region is bounded by \(y=x\), the \(x\)-axis, \(x=0\), and \(x=2\). Perpendicular to the \(x\)-axis, the cross sections are isosceles right triangles whose leg lies in the base region. Find the volume.
At a fixed \(x\) from \(0\) to \(2\), a vertical slice runs from the lower curve \(y=0\) to the upper curve \(y=x\). Its length is
This slice is a leg, so the cross-sectional area is
Integrate across the full interval of \(x\)-values:
Check the antiderivative: \(\frac{d}{dx}\left(\frac{x^3}{6}\right)=\frac{x^2}{2}\), which is the integrand. The slice length is zero at \(x=0\) and \(2\) at \(x=2\); the calculated volume is positive, as expected.
Worked Example: Vertical Slices Are a Hypotenuse
The base region lies above the \(x\)-axis and below \(y=\sqrt{x}\), for \(0\le x\le4\). Perpendicular to the \(x\)-axis, each cross section is an isosceles right triangle whose hypotenuse lies in the base region. Find the volume.
At a fixed \(x\), the vertical slice extends from \(y=0\) to \(y=\sqrt{x}\), so its length is
Here the slice is the hypotenuse, not a leg. Therefore, the area at \(x\) is
The volume integral and its evaluation are
Check: \(\frac{d}{dx}\left(\frac{x^2}{8}\right)=\frac{x}{4}\), the cross-sectional area. At \(x=4\), the hypotenuse length is \(2\), giving area \(2^2/4=1\), which agrees with \(A(4)=4/4=1\). The area is zero at \(x=0\), consistent with the vanishing slice there.
Worked Example: Horizontal Slices Are a Leg
The base region is bounded by \(x=y^2\), \(x=4\), \(y=0\), and \(y=2\). Cross sections perpendicular to the \(y\)-axis are isosceles right triangles, with the horizontal slice serving as a leg. Find the volume.
For each \(y\) between \(0\) and \(2\), the horizontal slice starts at \(x=y^2\) and ends at \(x=4\). Its length is
Because this length is a leg, the cross-sectional area is
The volume is integrated with respect to \(y\), since the slices are horizontal:
Check the antiderivative by differentiating: \(\frac{d}{dy}\left(8y-\frac{4}{3}y^3+\frac{1}{10}y^5\right)=8-4y^2+\frac{1}{2}y^4=\frac{1}{2}(4-y^2)^2\). Also, \(L(0)=4\) and \(L(2)=0\), so the area decreases from \(8\) to \(0\) over the interval, consistent with the geometry.
Choosing the Correct Formula
The direction of slicing and the role of the slice are separate decisions. “Perpendicular to the \(x\)-axis” tells you to use vertical slices and integrate with respect to \(x\). It does not tell you whether the slice is a leg or a hypotenuse. That must come from the cross-section description.
For cross sections perpendicular to the \(x\)-axis, use vertical slices. For cross sections perpendicular to the \(y\)-axis, use horizontal slices.
Subtract the lower \(y\)-value from the upper \(y\)-value for vertical slices, or the left \(x\)-value from the right \(x\)-value for horizontal slices.
Use \(A=L^2/2\) if the base slice is a leg, and \(A=L^2/4\) if it is the hypotenuse.
Use bounds that cover the base region in the chosen direction and integrate the resulting cross-sectional area.
The slice length must be the actual distance across the base region, so boundary order matters. For a vertical slice, for example, use upper curve minus lower curve. If the boundaries or the stated side of the triangle change within the interval, the area formula must be reconsidered on the affected intervals rather than applied without checking.
A Common Factor-of-Two Mistake
Suppose a base slice has length \(6\) and is specified as the hypotenuse. A frequent error is to use the leg formula without checking the description:
That calculation treats \(6\) as a leg. But if \(6\) is the hypotenuse, each leg has length \(6/\sqrt{2}\), so the correct area is
Equivalently, the hypotenuse formula gives \(A=6^2/4=9\). The incorrect answer is twice the correct one. Always label the slice as a leg or hypotenuse before squaring its length.
Why the Side Identification Matters
Cross-sectional area determines how much volume accumulates along the base region. Since both formulas square the slice length, an error in interpreting the side affects every cross section, not just one position. A reliable setup therefore records three things before integrating: the direction of slicing, the length of the base-region slice, and the role of that length in the triangle.
The final integral should have a nonnegative area throughout its bounds. For a leg of length \(L\), the area is \(L^2/2\); for a hypotenuse of length \(L\), it is \(L^2/4\). In either case the area is zero when the slice collapses and positive when the slice has positive length. These checks can reveal a reversed boundary subtraction or a mismatched side formula.
Check Your Understanding
For each question, identify the correct slice length and triangle-area formula before calculating or setting up an integral.
- A base slice has length \(5\) and is a leg of an isosceles right triangle. What is the cross-sectional area?
- A base slice has length \(5\) and is the hypotenuse. What is the cross-sectional area?
- A base region has vertical slices from \(y=0\) to \(y=x^2\), for \(0\le x\le2\). The slice is a leg. Write the volume integral.
- A base region has horizontal slices of length \(3-y\), for \(0\le y\le3\). The slice is the hypotenuse. Write the volume integral.
- Why does “perpendicular to the \(x\)-axis” determine the integration variable but not whether the base slice is a leg or a hypotenuse?
