AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/8/8-8)
Unit 8 · Topic 8.8
8.8 Volumes with Cross Sections: Triangles and Semicircles
This is the same slicing idea as 8.7 with different shapes: semicircles, equilateral triangles and right triangles. Find the side length from the base region, plug it into the shape's area formula, and integrate.
Key terms
- semicircle
- equilateral triangle
- isosceles right triangle
- area formula
- cross section
Area formulas in terms of the side s in the base
In every case, s is the length of the slice's edge that lies in the base: top − bottom (or right − left).
| Cross section | s is the… | Area |
|---|---|---|
| square | side | s² |
| semicircle | diameter | (π/8)s² |
| equilateral triangle | side | (√3/4)s² |
| isosceles right triangle | leg | s²/2 |
| isosceles right triangle | hypotenuse | s²/4 |
| rectangle with height h·s | base | h·s² |
Where those formulas come from
- Semicircle: the radius is s/2, so the area is ½π(s/2)² = πs²/8. The most common error is using s as the radius.
- Equilateral triangle: the height is (√3/2)s, so the area is ½ · s · (√3/2)s = (√3/4)s².
- Isosceles right triangle with a leg in the base: both legs are s, so the area is ½s².
- Isosceles right triangle with the hypotenuse in the base: each leg is s/√2, so the area is ½(s/√2)² = s²/4.
Steps (same as 8.7)
- Sketch the base and one slice.
- Find s in terms of x (or y).
- Write A in terms of s using the table.
- Integrate A over the base's extent.
Common base widths
- Region between y = f(x) and the x-axis: s = f(x).
- Region between two curves: s = top − bottom.
- Disk x² + y² ≤ r², sliced perpendicular to the x-axis: s = 2√(r² − x²).
- Region between a curve and the y-axis, sliced perpendicular to the y-axis: s = x expressed in terms of y.
A check you can trust
Here's a case where you know the answer from geometry. Let the base be the disk x² + y² ≤ r², with semicircle cross sections perpendicular to the x-axis. Each slice has diameter s = 2√(r² − x²), so the volume is (π/8) ∫ from −r to r of 4(r² − x²) dx = (π/2)(4r³/3) = (2/3)πr³. That's exactly the volume of a hemisphere, which is what those stacked half-discs form. Checks like this build trust in the formulas.
Read the wording carefully
“Semicircles whose diameters lie in the base” means s is the diameter. “Isosceles right triangles with one leg in the base” means s is a leg. “With the hypotenuse in the base” means s is the hypotenuse. These change the constant in front of s², and therefore the final answer, so read them closely. Pull the constant out of the integral and integrate s² once.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Semicircle cross sections
The base of a solid is the region bounded by y = √x, y = 0 and x = 4. Cross sections perpendicular to the x-axis are semicircles with diameters in the base. Find the volume.
Show the solutionHide the solution
- Step 1: Diameter s = √x.
- Step 2: Area = (π/8)s² = (π/8)x.
- Step 3: Volume = (π/8) ∫₀⁴ x dx = (π/8)(8) = π.
Answer: π cubic units
- Example 2
Equilateral triangle cross sections
The base of a solid is the region between y = x and y = x² for 0 ≤ x ≤ 1. Cross sections perpendicular to the x-axis are equilateral triangles. Find the volume.
Show the solutionHide the solution
- Step 1: On (0, 1), x > x², so s = x − x².
- Step 2: Area = (√3/4)(x − x²)².
- Step 3: Expand: (x − x²)² = x² − 2x³ + x⁴. ∫₀¹ (x² − 2x³ + x⁴) dx = 1/3 − 1/2 + 1/5 = 1/30.
- Step 4: Volume = (√3/4)(1/30) = √3/120 ≈ 0.0144.
Answer: √3/120 cubic units
- Example 3
Trap: hypotenuse in the base
The base of a solid is the disk x² + y² ≤ 9. Cross sections perpendicular to the x-axis are isosceles right triangles with the hypotenuse in the base. Find the volume.
Show the solutionHide the solution
- Step 1: At x, the disk runs from y = −√(9 − x²) to y = √(9 − x²), so s = 2√(9 − x²).
- Step 2: The hypotenuse is s, so the area is s²/4 = 4(9 − x²)/4 = 9 − x².
- Step 3: Volume = ∫ from −3 to 3 of (9 − x²) dx = 2(27 − 9) = 36.
- Step 4: If you used the leg formula s²/2 by mistake, you'd get 72.
Answer: 36 cubic units
Common mistakes
- Using the diameter as the radius for semicircles: π s²/2 instead of π s²/8.
- Using the leg formula when the hypotenuse is in the base, or the reverse.
- Forgetting that a full circle region has width 2√(r² − x²), not √(r² − x²).
- Mixing up the order: compute s, then square it, then multiply by the constant.
On the exam
- Expect these as the last part of a region question. You're often asked to write the integral without evaluating it, so the constant and s² must be exactly right.
- If allowed a calculator, still write the integral first, like (π/8) ∫₀⁴ (f(x) − g(x))² dx.
Connected topics
Videos
Check yourself
4 questions on 8.8 Volumes with Cross Sections: Triangles and Semicircles. Pick an answer to see if you got it, and why.
The base of a solid is the region bounded by the graph of y = 4 − x² and the x-axis. Each cross section of the solid perpendicular to the x-axis is a semicircle whose diameter lies in the base. What is the volume of the solid?
The base of a solid is the triangular region bounded by the line y = 2 − x and the coordinate axes. Each cross section of the solid perpendicular to the x-axis is an equilateral triangle. What is the volume of the solid?
The base of a solid is the region bounded by the graphs of y = x and y = x². Each cross section of the solid perpendicular to the x-axis is a semicircle whose diameter lies in the base. What is the volume of the solid?
The base of a solid is the disk bounded by the circle x² + y² = 4. Each cross section of the solid perpendicular to the x-axis is an isosceles right triangle with one leg in the base. What is the volume of the solid?
0 of 4 answered