AP Calculus AB/BC · Unit 8 · Topic 8.8
Volumes with Cross Sections: Triangles and Semicircles
Translate the base distance into the correct triangle or semicircle area formula.
1. Topic Focus
Apply definite integrals to average value, motion, net change, area, volume, and BC arc length or distance problems.
This topic: Translate the base distance into the correct triangle or semicircle area formula.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
If the region width is a semicircle diameter d, use πd²/8, not πd²/2.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
From a base segment to volume
As in every slicing problem, volume is the accumulation of cross-sectional area:
The planar base supplies a segment length
The wording of the problem determines what that segment represents in the triangle or semicircle. Converting \(s\) to the correct area formula is the central step.
Semicircular cross sections
If the base segment is the diameter \(d=s\), then the radius is \(s/2\):
If the base segment is explicitly the radius \(r=s\), then
These formulas differ by a factor of four. A semicircle whose diameter is \(s\) is not the same as one whose radius is \(s\).
Triangular cross sections
Begin with \(A=\tfrac12 bh\), then express both dimensions in terms of the base-region segment \(s\).
| Description | Area in terms of s |
|---|---|
| Base \(s\), height \(ks\) | \(\displaystyle A=\frac{k}{2}s^2\) |
| Base \(s\), fixed height \(h\) | \(\displaystyle A=\frac h2s\) |
| Equilateral triangle, side \(s\) | \(\displaystyle A=\frac{\sqrt3}{4}s^2\) |
| Right isosceles triangle, leg \(s\) | \(\displaystyle A=\frac12s^2\) |
| Right isosceles triangle, hypotenuse \(s\) | \(\displaystyle A=\frac14s^2\) |
For the hypotenuse case, each leg has length \(s/\sqrt2\), so \(A=\tfrac12(s/\sqrt2)^2=s^2/4\).
A reliable workflow
- Sketch the base and identify the slicing variable from the stated perpendicular direction.
- Find the base-region segment \(s\) using top minus bottom or right minus left.
- Decide what \(s\) represents: diameter, radius, triangle base, leg, hypotenuse, or side.
- Write the complete geometric area \(A\) in terms of \(s\).
- Substitute the segment function, integrate over the correct bounds, and include cubic units.
Orientation and changing boundaries
Sections perpendicular to the \(x\)-axis use \(A(x)dx\); sections perpendicular to the \(y\)-axis use \(A(y)dy\). If the boundary segment changes formula, split the volume integral before applying the geometric area formula.
Numerical data
When a table provides diameters, radii, or triangle dimensions, calculate each cross-sectional area first. Then approximate \(\int A\) with the requested numerical method. Averaging lengths first and inserting that average into a nonlinear area formula generally gives a different result.
Units and reasonableness
A segment length has linear units, a cross-sectional area has square units, and integrating along the slicing direction produces cubic units. Cross-sectional area must remain nonnegative. When \(s=0\) at a base endpoint, the corresponding triangular or semicircular section should also have zero area.
6. Detailed Worked Example and Error Check
Example 1: Semicircles with a given diameter
A solid has base under \(y=\sqrt{x}\) above the \(x\)-axis on \([0,4]\). Perpendicular cross sections are semicircles whose diameter is \(d(x)=\sqrt{x}\). Thus
and
Example 2: Semicircles between two curves
The base lies between \(y=x\) and \(y=x^2\) on \([0,1]\), and the vertical base segment is each semicircle's diameter. Since \(d(x)=x-x^2\),
Example 3: Equilateral triangle sections
The base lies between \(y=1\) and \(y=x^2\) for \(-1\le x\le1\). If the segment \(s(x)=1-x^2\) is the side of an equilateral triangle, then
Therefore
Example 4: Right isosceles triangles with a leg in the base
The base is between \(y=2x\) and \(y=x^2\) on \([0,2]\). If \(s(x)=2x-x^2\) is a leg,
Example 5: The same segment used as a hypotenuse
Using the same base region, suppose \(s(x)=2x-x^2\) is instead the hypotenuse of each right isosceles triangle. Then
The volume is half the leg-based result because the cross-sectional area coefficient is half as large.
Example 6: Triangle height proportional to its base
The base lies between \(y=4-x^2\) and the \(x\)-axis for \(-2\le x\le2\). Each triangle has base \(s=4-x^2\) and height \(2s\), so
Hence
Example 7: Triangle with fixed height
The base lies between \(y=1-x\) and \(y=0\) on \([0,1]\). Every triangular cross section has base \(s(x)=1-x\) and fixed height \(3\). Therefore
Because the height is fixed, the area is linear rather than quadratic in \(s\).
Example 8: Horizontal semicircular sections
The base lies between \(x=y^2\) and \(x=4\). Sections perpendicular to the \(y\)-axis are semicircles whose diameter is \(d(y)=4-y^2\), for \(-2\le y\le2\). Thus
Common errors
- Using a diameter as a radius, making a semicircular area four times too large.
- Using \(\pi s^2\) for a semicircle instead of half the corresponding circle area.
- Assuming every triangle has area proportional to \(s^2\), even when its height is fixed.
- Confusing a right-isosceles leg with its hypotenuse.
- Using the triangle's perimeter or a semicircle's arc length instead of area.
- Applying a numerical rule to lengths before converting them to areas.
- Reporting square rather than cubic units.
7. AP Reasoning Routine
Sketch and label the region, decide whether slices are vertical or horizontal, write a nonnegative geometric quantity, and split bounds when the geometry changes.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Find each volume using known cross sections. State what the base segment represents.
(a) Semicircular sections perpendicular to the \(x\)-axis have diameter \(2-x\) for \(0\le x\le2\).
(b) The base is between \(y=2x\) and \(y=x^2\) on \([0,2]\). Cross sections are semicircles whose diameter lies in the base.
(c) Use the base from part (b), but let the sections be equilateral triangles whose side lies in the base.
(d) Use the base from part (b), but let the sections be right isosceles triangles whose leg lies in the base.
(e) Use the base from part (b), but let the segment in the base be the hypotenuse of each right isosceles triangle.
(f) The base lies between \(y=1\) and \(y=x^2\) for \(0\le x\le1\). Each triangular height is three times its base.
(g) Use the base from part (f), but let every triangular height be fixed at \(2\).
(h) The base lies between \(x=y^2\) and \(x=4\). Horizontal sections are semicircles whose diameter lies in the base.
(i) Semicircular sections have diameters \(d(0)=2\), \(d(1)=4\), and \(d(3)=1\). Use the trapezoidal rule to estimate the volume.
(j) Explain how the area formulas differ when a base segment \(s\) is a semicircle's diameter versus its radius.
Check the solution
(a) \(A(x)=\pi(2-x)^2/8\), so \(V=(\pi/8)\int_0^2(2-x)^2dx=\pi/3\).
(b) \(d(x)=2x-x^2\), so \(V=(\pi/8)\int_0^2(2x-x^2)^2dx=2\pi/15\).
(c) \(A(x)=(\sqrt3/4)(2x-x^2)^2\), so \(V=4\sqrt3/15\).
(d) The segment is a leg, so \(A(x)=\tfrac12(2x-x^2)^2\) and \(V=8/15\).
(e) The segment is the hypotenuse, so \(A(x)=\tfrac14(2x-x^2)^2\) and \(V=4/15\).
(f) \(s(x)=1-x^2\) and \(A(x)=\tfrac32s^2\). Thus \(V=(3/2)(8/15)=4/5\).
(g) \(A(x)=\tfrac12(1-x^2)(2)=1-x^2\), so \(V=\int_0^1(1-x^2)dx=2/3\).
(h) \(d(y)=4-y^2\), so \(V=(\pi/8)\int_{-2}^{2}(4-y^2)^2dy=64\pi/15\).
(i) The areas are \(\pi/2,2\pi,\pi/8\). Thus \(V\approx1(\pi/2+2\pi)/2+2(2\pi+\pi/8)/2=27\pi/8\).
(j) If \(s=d\), then \(A=\pi s^2/8\). If \(s=r\), then \(A=\pi s^2/2\), which is four times as large.