Unit 8Volumes with Known Cross-Sections

Concept 4.134Built for May 2027 AP Exam

Volumes with Known Cross-Sections

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A solid with known cross sections is fully described once you know its base region, the direction of the slices, and the shape of each slice. The volume is $V = \int A\,d(\text{slicing variable})$ , where $A$ is the slice’s area expressed through the base width $s$ . Three things can complicate a released problem, and all three are predictable: the base may need to be split because its upper boundary changes curve, the slice shape may specify a height rather than a side, and the slice may be described by its hypotenuse rather than its leg. Each complication is handled before integrating, and each leaves the integral itself unchanged in form.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A base region bounded by two or three curves, a named slice shape with an explicit orientation, and a volume request. Some stems specify a fixed height for rectangular slices, which turns the area into $s \cdot h$ .
What you’ll do: Establish the base width as a difference of boundaries, apply the shape’s area formula with attention to which segment the stem named, split the base if its boundary changes, and integrate.
Memory hook: Base, shape, axis. Name all three before writing the integral.
Use it on the exam: For MCQs: Compare choices by coefficient and by the power of $s$ . A rectangle contributes $s^{1}$ , most other shapes contribute $s^{2}$ . For FRQs: Write the area function and the limits as separate lines. Both are scored, and the integral is scored after them.

Concept Review

The Base Supplies Everything

Figure 1 — A base whose upper boundary changes curve, sliced by two different shapes.

The base region is the only source of length in the problem. At position $x$ , the available width is

$s(x) = \text{upper}(x) - \text{lower}(x)$

Every shape’s area is a fixed multiple of a power of $s(x)$ . The volume follows:

$V = \int_a^b k\,s(x)^{p}\,dx$

where $k$ depends on the shape and $p$ is $2$ for most shapes and $1$ for a rectangle of fixed height.

Triangular base region split at x equals 2, with a square slice drawn on the left piece and a semicircular slice drawn on the right piece

The Coefficient Table

Slice shape	Area in terms of $s$	Power of $s$
square, side $s$	$s^{2}$	$2$
rectangle, base $s$ , height $h$	$h\,s$	$1$
equilateral triangle, side $s$	$\frac{\sqrt{3}}{4}s^{2}$	$2$
isosceles right triangle, leg $s$	$\frac{1}{2}s^{2}$	$2$
isosceles right triangle, hypotenuse $s$	$\frac{1}{4}s^{2}$	$2$
semicircle, diameter $s$	$\frac{\pi}{8}s^{2}$	$2$
isosceles triangle, base $s$ , height $h$	$\frac{1}{2}h\,s$	$1$

The rows with power $1$ are the ones students miss. A rectangle of fixed height does not square the base width, and a fixed-height triangle does not either.

When the Base Needs a Split

A base bounded by three curves usually changes its upper boundary partway across. Topic 8.9 covers the mechanics; here the consequence is a volume written as two integrals:

$V = \int_a^c k\,s_1(x)^{2}dx + \int_c^b k\,s_2(x)^{2}dx$

The cut point is where the two upper boundary curves meet. Missing the cut does not produce a sign error, as it would in an area problem; it produces an interval where $s$ is measured to the wrong curve, so the volume is silently wrong with no visual signal.

Orientation Wording

The words that follow the shape name decide the formula:

“squares with sides on the base” means the base width is the side.
“isosceles right triangles with legs on the base” means the base width is a leg.
“isosceles right triangles with hypotenuse on the base” means the base width is the hypotenuse, and the area drops to $\frac{1}{4}s^{2}$ .
“semicircles with diameters on the base” means the base width is the diameter.
“rectangles of height $3$ ” means the area is $3s$ , with no squaring at all.

Comparing Volumes Without Integrating

Since the coefficient factors out of the integral, ratios of volumes are ratios of coefficients whenever the base and slicing direction are unchanged:

$\frac{V_{\text{semicircle}}}{V_{\text{square}}} = \frac{\pi}{8} \approx 0.393 \qquad\qquad \frac{V_{\text{equilateral}}}{V_{\text{square}}} = \frac{\sqrt{3}}{4} \approx 0.433$

That single observation answers a whole class of multiple-choice comparisons and provides a fast check on a computed answer.

Common Misconception Check

Myth 1: “Every cross-section problem squares the base width.” Truth: Rectangles and fixed-height triangles multiply it once.
Myth 2: “A base with three boundary curves can be integrated in one piece.” Truth: The upper boundary changes, so the base splits.
Myth 3: “Hypotenuse and leg give the same area.” Truth: They differ by a factor of two, so the volumes differ by two.

Topic-Specific Study Method

Base, Shape, Axis, Then Integrate

Borrowed idea: These problems are decided before any integration happens. Name the base, name the shape, name the axis. What happens if the base changes its upper boundary halfway across? Then it splits, and each piece gets its own width expression. What happens if the stem names a height instead of a side? Then the area stops being quadratic in $s$ , and squaring anyway inflates the answer. Read the orientation words carefully, because "hypotenuse on the base" halves the coefficient compared with "leg on the base". Compute the base width first, apply the coefficient second, and integrate last.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“squares perpendicular to the $x$ -axis”	a base and a shape	$A = s^{2}$	integrate in $x$
“rectangles of height $2$ ”	a base and a fixed height	$A = 2s$	linear, not quadratic
“isosceles right triangles, hypotenuse on the base”	orientation wording	$A = \frac{1}{4}s^{2}$	halve the leg version
“the base is bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ ”	three boundaries	base splits at $x = 2$	two integrals
“semicircles on the base of the region”	a base and a diameter	$A = \frac{\pi}{8}s^{2}$	keep $\frac{\pi}{8}$ outside

① Write the base width as a difference of two named curves at a fixed position. If two different pairs are needed, note the cut immediately.

② Apply the coefficient from the shape table, then verify the power of $s$ against the shape’s dimensions.

③ Check the answer against the square-slice volume using the ratio table. A ratio far from the expected constant means the coefficient is wrong.

Training & Error Diagnosis

Core Archetype (Simplest Form): The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Standard Execution:

Base shape. The two lines meet at $x = 2$ , so the upper boundary is $y = x$ on $[0,2]$ and $y = 4 - x$ on $[2,4]$ .

First piece. $\int_0^2 x^{2}dx = \frac{8}{3}$ .

Second piece. $\int_2^4 (4 - x)^{2}dx = \left[-\frac{(4-x)^{3}}{3}\right]_2^4 = \frac{8}{3}$ .

Total. $V = \frac{16}{3}$ cubic units.

Exam Variation (Mixed Shapes and a Fixed Height / Context Shift): For the same base, compare three solids: square slices, semicircular slices with diameters on the base, and rectangular slices of height $2$ . Find each volume.

Solution: The square-slice volume is $\frac{16}{3}$ . Semicircles multiply it by $\frac{\pi}{8}$ , giving $\frac{2\pi}{3}$ . Rectangles of height $2$ need $\int_0^4 2s(x)\,dx$ , where $s$ is piecewise: $\int_0^2 2x\,dx = 4$ and $\int_2^4 2(4-x)\,dx = 4$ , so the volume is $8$ . Note that the rectangle volume is larger than the square volume here, because the fixed height $2$ exceeds the base width over most of the region, which is a reminder that the coefficient comparison only applies between shapes of the same power in $s$ .

Common Traps & Failure Signals:

Split missed (Signal: one integral on a base with three boundaries): find the meeting point of the upper curves.
Orientation ignored (Signal: $\frac{1}{2}s^{2}$ when the stem said hypotenuse): the coefficient depends on which segment lies on the base.
Fixed height squared (Signal: $h^{2}s^{2}$ in a rectangle problem): rectangles give $h\,s$ .

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Split Missed: You integrated across a change of boundary.
- Orientation Ignored: You used the leg coefficient when the stem had placed the hypotenuse on the base, which doubles the volume without changing anything else in the setup.
- Height Squared: You treated a fixed height as if it were a second side of the base region, so the area function gained a power of $s$ that the geometry never had.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Four of the nine released exams since 2017 ask for a solid whose cross sections are a named shape on a stated base region
10-Year Total Count: 4 of 9 Exams · 4 FRQ Parts; Each of the four appears as the final part of a question that already established the base region in an earlier part
FRQ Years Appeared: 2022 (Q2c), 2024 (Q6b), 2025 (Q2B), 2026 (Q2B)

Practice Examples

MCQ 1

A Base With a Split

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. The volume is

MCQ 2

Semicircles on the Same Base

With the same base, cross sections perpendicular to the $x$ -axis are semicircles with diameters on the base. The volume is

MCQ 3

Fixed Height

With the same base, cross sections perpendicular to the $x$ -axis are rectangles of height $2$ . The volume is

FRQ 4

Three Solids on One Base

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ .

(a) Explain why the base must be split and state the cut point.

(b) Find the volume of the solid whose cross sections perpendicular to the $x$ -axis are squares.

(c) Find the volume of the solid whose cross sections are semicircles with diameters on the base.

(d) Compare the two volumes using the coefficient ratio and explain why the ratio does not depend on the base region.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name the base region and its boundaries	the width has no source
2	check whether the upper boundary changes	the base is integrated as one piece when it is two
3	write $s$ as a difference of two curves	the width is a single function
4	match the shape and its orientation wording	the coefficient is wrong by a factor
5	confirm the power of $s$	rectangles become quadratic
6	integrate and compare with the coefficient ratio	a coefficient error passes

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Strengthen what you learned with focused AP-style practice.

Start Topic Practice

Unit 8Volumes with Known Cross-Sections

Concept 4.134Built for May 2027 AP Exam

Volumes with Known Cross-Sections

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A base region bounded by two or three curves, a named slice shape with an explicit orientation, and a volume request. Some stems specify a fixed height for rectangular slices, which turns the area into $s \cdot h$ .
What you’ll do: Establish the base width as a difference of boundaries, apply the shape’s area formula with attention to which segment the stem named, split the base if its boundary changes, and integrate.
Memory hook: Base, shape, axis. Name all three before writing the integral.
Use it on the exam: For MCQs: Compare choices by coefficient and by the power of $s$ . A rectangle contributes $s^{1}$ , most other shapes contribute $s^{2}$ . For FRQs: Write the area function and the limits as separate lines. Both are scored, and the integral is scored after them.

Concept Review

The Base Supplies Everything

Figure 1 — A base whose upper boundary changes curve, sliced by two different shapes.

The base region is the only source of length in the problem. At position $x$ , the available width is

$s(x) = \text{upper}(x) - \text{lower}(x)$

Every shape’s area is a fixed multiple of a power of $s(x)$ . The volume follows:

$V = \int_a^b k\,s(x)^{p}\,dx$

where $k$ depends on the shape and $p$ is $2$ for most shapes and $1$ for a rectangle of fixed height.

Triangular base region split at x equals 2, with a square slice drawn on the left piece and a semicircular slice drawn on the right piece

The Coefficient Table

Slice shape	Area in terms of $s$	Power of $s$
square, side $s$	$s^{2}$	$2$
rectangle, base $s$ , height $h$	$h\,s$	$1$
equilateral triangle, side $s$	$\frac{\sqrt{3}}{4}s^{2}$	$2$
isosceles right triangle, leg $s$	$\frac{1}{2}s^{2}$	$2$
isosceles right triangle, hypotenuse $s$	$\frac{1}{4}s^{2}$	$2$
semicircle, diameter $s$	$\frac{\pi}{8}s^{2}$	$2$
isosceles triangle, base $s$ , height $h$	$\frac{1}{2}h\,s$	$1$

The rows with power $1$ are the ones students miss. A rectangle of fixed height does not square the base width, and a fixed-height triangle does not either.

When the Base Needs a Split

A base bounded by three curves usually changes its upper boundary partway across. Topic 8.9 covers the mechanics; here the consequence is a volume written as two integrals:

$V = \int_a^c k\,s_1(x)^{2}dx + \int_c^b k\,s_2(x)^{2}dx$

Orientation Wording

The words that follow the shape name decide the formula:

“squares with sides on the base” means the base width is the side.
“isosceles right triangles with legs on the base” means the base width is a leg.
“isosceles right triangles with hypotenuse on the base” means the base width is the hypotenuse, and the area drops to $\frac{1}{4}s^{2}$ .
“semicircles with diameters on the base” means the base width is the diameter.
“rectangles of height $3$ ” means the area is $3s$ , with no squaring at all.

Comparing Volumes Without Integrating

Since the coefficient factors out of the integral, ratios of volumes are ratios of coefficients whenever the base and slicing direction are unchanged:

$\frac{V_{\text{semicircle}}}{V_{\text{square}}} = \frac{\pi}{8} \approx 0.393 \qquad\qquad \frac{V_{\text{equilateral}}}{V_{\text{square}}} = \frac{\sqrt{3}}{4} \approx 0.433$

That single observation answers a whole class of multiple-choice comparisons and provides a fast check on a computed answer.

Common Misconception Check

Myth 1: “Every cross-section problem squares the base width.” Truth: Rectangles and fixed-height triangles multiply it once.
Myth 2: “A base with three boundary curves can be integrated in one piece.” Truth: The upper boundary changes, so the base splits.
Myth 3: “Hypotenuse and leg give the same area.” Truth: They differ by a factor of two, so the volumes differ by two.

Topic-Specific Study Method

Base, Shape, Axis, Then Integrate

Exam disguise	What you see on the page	What your brain must translate	Required execution
“squares perpendicular to the $x$ -axis”	a base and a shape	$A = s^{2}$	integrate in $x$
“rectangles of height $2$ ”	a base and a fixed height	$A = 2s$	linear, not quadratic
“isosceles right triangles, hypotenuse on the base”	orientation wording	$A = \frac{1}{4}s^{2}$	halve the leg version
“the base is bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ ”	three boundaries	base splits at $x = 2$	two integrals
“semicircles on the base of the region”	a base and a diameter	$A = \frac{\pi}{8}s^{2}$	keep $\frac{\pi}{8}$ outside

① Write the base width as a difference of two named curves at a fixed position. If two different pairs are needed, note the cut immediately.

② Apply the coefficient from the shape table, then verify the power of $s$ against the shape’s dimensions.

③ Check the answer against the square-slice volume using the ratio table. A ratio far from the expected constant means the coefficient is wrong.

Training & Error Diagnosis

Core Archetype (Simplest Form): The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Standard Execution:

Base shape. The two lines meet at $x = 2$ , so the upper boundary is $y = x$ on $[0,2]$ and $y = 4 - x$ on $[2,4]$ .

First piece. $\int_0^2 x^{2}dx = \frac{8}{3}$ .

Second piece. $\int_2^4 (4 - x)^{2}dx = \left[-\frac{(4-x)^{3}}{3}\right]_2^4 = \frac{8}{3}$ .

Total. $V = \frac{16}{3}$ cubic units.

Solution: The square-slice volume is $\frac{16}{3}$ . Semicircles multiply it by $\frac{\pi}{8}$ , giving $\frac{2\pi}{3}$ . Rectangles of height $2$ need $\int_0^4 2s(x)\,dx$ , where $s$ is piecewise: $\int_0^2 2x\,dx = 4$ and $\int_2^4 2(4-x)\,dx = 4$ , so the volume is $8$ . Note that the rectangle volume is larger than the square volume here, because the fixed height $2$ exceeds the base width over most of the region, which is a reminder that the coefficient comparison only applies between shapes of the same power in $s$ .

Common Traps & Failure Signals:

Split missed (Signal: one integral on a base with three boundaries): find the meeting point of the upper curves.
Orientation ignored (Signal: $\frac{1}{2}s^{2}$ when the stem said hypotenuse): the coefficient depends on which segment lies on the base.
Fixed height squared (Signal: $h^{2}s^{2}$ in a rectangle problem): rectangles give $h\,s$ .

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Split Missed: You integrated across a change of boundary.
- Orientation Ignored: You used the leg coefficient when the stem had placed the hypotenuse on the base, which doubles the volume without changing anything else in the setup.
- Height Squared: You treated a fixed height as if it were a second side of the base region, so the area function gained a power of $s$ that the geometry never had.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Four of the nine released exams since 2017 ask for a solid whose cross sections are a named shape on a stated base region
10-Year Total Count: 4 of 9 Exams · 4 FRQ Parts; Each of the four appears as the final part of a question that already established the base region in an earlier part
FRQ Years Appeared: 2022 (Q2c), 2024 (Q6b), 2025 (Q2B), 2026 (Q2B)

Practice Examples

MCQ 1

A Base With a Split

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. The volume is

MCQ 2

Semicircles on the Same Base

With the same base, cross sections perpendicular to the $x$ -axis are semicircles with diameters on the base. The volume is

MCQ 3

Fixed Height

With the same base, cross sections perpendicular to the $x$ -axis are rectangles of height $2$ . The volume is

FRQ 4

Three Solids on One Base

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ .

(a) Explain why the base must be split and state the cut point.

(b) Find the volume of the solid whose cross sections perpendicular to the $x$ -axis are squares.

(c) Find the volume of the solid whose cross sections are semicircles with diameters on the base.

(d) Compare the two volumes using the coefficient ratio and explain why the ratio does not depend on the base region.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name the base region and its boundaries	the width has no source
2	check whether the upper boundary changes	the base is integrated as one piece when it is two
3	write $s$ as a difference of two curves	the width is a single function
4	match the shape and its orientation wording	the coefficient is wrong by a factor
5	confirm the power of $s$	rectangles become quadratic
6	integrate and compare with the coefficient ratio	a coefficient error passes

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen what you learned with focused AP-style practice.

Start Topic Practice

Unit 8Volumes with Known Cross-Sections

Concept 4.134Built for May 2027 AP Exam

Volumes with Known Cross-Sections

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A base region bounded by two or three curves, a named slice shape with an explicit orientation, and a volume request. Some stems specify a fixed height for rectangular slices, which turns the area into $s \cdot h$ .
What you’ll do: Establish the base width as a difference of boundaries, apply the shape’s area formula with attention to which segment the stem named, split the base if its boundary changes, and integrate.
Memory hook: Base, shape, axis. Name all three before writing the integral.
Use it on the exam: For MCQs: Compare choices by coefficient and by the power of $s$ . A rectangle contributes $s^{1}$ , most other shapes contribute $s^{2}$ . For FRQs: Write the area function and the limits as separate lines. Both are scored, and the integral is scored after them.

Concept Review

The Base Supplies Everything

Figure 1 — A base whose upper boundary changes curve, sliced by two different shapes.

The base region is the only source of length in the problem. At position $x$ , the available width is

$s(x) = \text{upper}(x) - \text{lower}(x)$

Every shape’s area is a fixed multiple of a power of $s(x)$ . The volume follows:

$V = \int_a^b k\,s(x)^{p}\,dx$

where $k$ depends on the shape and $p$ is $2$ for most shapes and $1$ for a rectangle of fixed height.

Triangular base region split at x equals 2, with a square slice drawn on the left piece and a semicircular slice drawn on the right piece

The Coefficient Table

Slice shape	Area in terms of $s$	Power of $s$
square, side $s$	$s^{2}$	$2$
rectangle, base $s$ , height $h$	$h\,s$	$1$
equilateral triangle, side $s$	$\frac{\sqrt{3}}{4}s^{2}$	$2$
isosceles right triangle, leg $s$	$\frac{1}{2}s^{2}$	$2$
isosceles right triangle, hypotenuse $s$	$\frac{1}{4}s^{2}$	$2$
semicircle, diameter $s$	$\frac{\pi}{8}s^{2}$	$2$
isosceles triangle, base $s$ , height $h$	$\frac{1}{2}h\,s$	$1$

The rows with power $1$ are the ones students miss. A rectangle of fixed height does not square the base width, and a fixed-height triangle does not either.

When the Base Needs a Split

A base bounded by three curves usually changes its upper boundary partway across. Topic 8.9 covers the mechanics; here the consequence is a volume written as two integrals:

$V = \int_a^c k\,s_1(x)^{2}dx + \int_c^b k\,s_2(x)^{2}dx$

Orientation Wording

The words that follow the shape name decide the formula:

“squares with sides on the base” means the base width is the side.
“isosceles right triangles with legs on the base” means the base width is a leg.
“isosceles right triangles with hypotenuse on the base” means the base width is the hypotenuse, and the area drops to $\frac{1}{4}s^{2}$ .
“semicircles with diameters on the base” means the base width is the diameter.
“rectangles of height $3$ ” means the area is $3s$ , with no squaring at all.

Comparing Volumes Without Integrating

Since the coefficient factors out of the integral, ratios of volumes are ratios of coefficients whenever the base and slicing direction are unchanged:

$\frac{V_{\text{semicircle}}}{V_{\text{square}}} = \frac{\pi}{8} \approx 0.393 \qquad\qquad \frac{V_{\text{equilateral}}}{V_{\text{square}}} = \frac{\sqrt{3}}{4} \approx 0.433$

That single observation answers a whole class of multiple-choice comparisons and provides a fast check on a computed answer.

Common Misconception Check

Myth 1: “Every cross-section problem squares the base width.” Truth: Rectangles and fixed-height triangles multiply it once.
Myth 2: “A base with three boundary curves can be integrated in one piece.” Truth: The upper boundary changes, so the base splits.
Myth 3: “Hypotenuse and leg give the same area.” Truth: They differ by a factor of two, so the volumes differ by two.

Topic-Specific Study Method

Base, Shape, Axis, Then Integrate

Exam disguise	What you see on the page	What your brain must translate	Required execution
“squares perpendicular to the $x$ -axis”	a base and a shape	$A = s^{2}$	integrate in $x$
“rectangles of height $2$ ”	a base and a fixed height	$A = 2s$	linear, not quadratic
“isosceles right triangles, hypotenuse on the base”	orientation wording	$A = \frac{1}{4}s^{2}$	halve the leg version
“the base is bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ ”	three boundaries	base splits at $x = 2$	two integrals
“semicircles on the base of the region”	a base and a diameter	$A = \frac{\pi}{8}s^{2}$	keep $\frac{\pi}{8}$ outside

① Write the base width as a difference of two named curves at a fixed position. If two different pairs are needed, note the cut immediately.

② Apply the coefficient from the shape table, then verify the power of $s$ against the shape’s dimensions.

③ Check the answer against the square-slice volume using the ratio table. A ratio far from the expected constant means the coefficient is wrong.

Training & Error Diagnosis

Core Archetype (Simplest Form): The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Standard Execution:

Base shape. The two lines meet at $x = 2$ , so the upper boundary is $y = x$ on $[0,2]$ and $y = 4 - x$ on $[2,4]$ .

First piece. $\int_0^2 x^{2}dx = \frac{8}{3}$ .

Second piece. $\int_2^4 (4 - x)^{2}dx = \left[-\frac{(4-x)^{3}}{3}\right]_2^4 = \frac{8}{3}$ .

Total. $V = \frac{16}{3}$ cubic units.

Solution: The square-slice volume is $\frac{16}{3}$ . Semicircles multiply it by $\frac{\pi}{8}$ , giving $\frac{2\pi}{3}$ . Rectangles of height $2$ need $\int_0^4 2s(x)\,dx$ , where $s$ is piecewise: $\int_0^2 2x\,dx = 4$ and $\int_2^4 2(4-x)\,dx = 4$ , so the volume is $8$ . Note that the rectangle volume is larger than the square volume here, because the fixed height $2$ exceeds the base width over most of the region, which is a reminder that the coefficient comparison only applies between shapes of the same power in $s$ .

Common Traps & Failure Signals:

Split missed (Signal: one integral on a base with three boundaries): find the meeting point of the upper curves.
Orientation ignored (Signal: $\frac{1}{2}s^{2}$ when the stem said hypotenuse): the coefficient depends on which segment lies on the base.
Fixed height squared (Signal: $h^{2}s^{2}$ in a rectangle problem): rectangles give $h\,s$ .

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Split Missed: You integrated across a change of boundary.
- Orientation Ignored: You used the leg coefficient when the stem had placed the hypotenuse on the base, which doubles the volume without changing anything else in the setup.
- Height Squared: You treated a fixed height as if it were a second side of the base region, so the area function gained a power of $s$ that the geometry never had.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Four of the nine released exams since 2017 ask for a solid whose cross sections are a named shape on a stated base region
10-Year Total Count: 4 of 9 Exams · 4 FRQ Parts; Each of the four appears as the final part of a question that already established the base region in an earlier part
FRQ Years Appeared: 2022 (Q2c), 2024 (Q6b), 2025 (Q2B), 2026 (Q2B)

Practice Examples

MCQ 1

A Base With a Split

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ . Cross sections perpendicular to the $x$ -axis are squares. The volume is

MCQ 2

Semicircles on the Same Base

With the same base, cross sections perpendicular to the $x$ -axis are semicircles with diameters on the base. The volume is

MCQ 3

Fixed Height

With the same base, cross sections perpendicular to the $x$ -axis are rectangles of height $2$ . The volume is

FRQ 4

Three Solids on One Base

The base of a solid is the region bounded by $y = x$ , $y = 4 - x$ , and $y = 0$ .

(a) Explain why the base must be split and state the cut point.

(b) Find the volume of the solid whose cross sections perpendicular to the $x$ -axis are squares.

(c) Find the volume of the solid whose cross sections are semicircles with diameters on the base.

(d) Compare the two volumes using the coefficient ratio and explain why the ratio does not depend on the base region.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name the base region and its boundaries	the width has no source
2	check whether the upper boundary changes	the base is integrated as one piece when it is two
3	write $s$ as a difference of two curves	the width is a single function
4	match the shape and its orientation wording	the coefficient is wrong by a factor
5	confirm the power of $s$	rectangles become quadratic
6	integrate and compare with the coefficient ratio	a coefficient error passes

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