Do I need to memorize 8s2 and 34s2?

It helps to know them, but you can always build them. A semicircle is half of r2 with r = s2, and an equilateral triangle has height 32s, so its area is half of s times that height.

Unit 8 · Topic 8.8 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.8: Volumes with Cross Sections: Triangles and Semicircles

Q: How do I know whether to integrate with respect to x or y?

Read which axis the cross sections are perpendicular to. Perpendicular to the x-axis means each slice sits at one x, so you integrate dx. Perpendicular to the y-axis means each slice sits at one y, so you write its length in y and integrate dy.

Q: What if the base is between two curves?

The slice runs from the lower curve to the upper one, so its length is top minus bottom, as in the worked example. Find where the curves meet to get the limits.

Q: Is this the same as the disc method?

It is the same idea, integrating the area of one slice, but here the shape stands on a flat base. In 8.9 to 8.12 the slice is a disc or a washer, made by turning a region about a line.

Q: Can a cross section be a shape that is not on the list?

Yes. Any shape whose area you can write from the slice works, such as a full circle with its diameter across the base, a rectangle of given height, or an area function the problem hands you.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the volume of a solid whose cross sections are triangles, semicircles or other shapes by integrating the area of one cross section.

Key vocabulary

Term	What it means	The exam trap
Cross section	The flat shape where a plane cuts a solid; in this topic the plane is perpendicular to the $x$ -axis or the $y$ -axis, and the shape stands on a segment across the base.	The base region is not a cross section. Each cross section stands up out of the base.
Triangular cross section	A cross section shaped like a triangle with one side on the slice across the base, such as an equilateral triangle, an isosceles right triangle, or a triangle of stated height.	Name the side of the triangle that lies in the base before you write its area.
Semicircular cross section	A cross section shaped like half a disc, standing on the slice across the base with that slice as its diameter.	Write the radius from the slice before you write the area.
Volume by cross sections	The definite integral of the cross-section area, $V = \int_{a}^{b} A (x) d x$ , over the interval the base covers along the axis the slices are perpendicular to.	The limits are the ends of the base along that axis, not the heights of the curves.

The region between y = 4 - x squared over 4 and the x-axis for x from 0 to 4, crossed at x = 2 by a segment of length 3. Beside it, an equilateral triangle with side 3 and area 3.9 stands for the cross section there; the whole solid has volume about 14.78.

Part 1 Think first Same base, half-discs instead of squares

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A sculptor casts two bronze pieces on the same flat base. In the first, every slice cut straight across the base is a square standing on that slice, and the piece takes 48 liters of bronze. In the second, every slice is a half-disc standing on the same slice, with the slice as its straight edge. About how much bronze does the second piece take?

In short. The volume of a solid with known cross sections is the definite integral of the cross-section area, $V = \int_{a}^{b} A (x) d x$ , where $A (x)$ comes from the area formula of the shape that stands on the slice across the base.

The slice across the base at $x$ has length $s (x)$ , the top boundary minus the bottom one, and every area formula is written from it.
Read which part of the shape lies on the slice: a side of a triangle, a leg or a hypotenuse, or the diameter of a semicircle.
When the cross sections are perpendicular to the $y$ -axis, write the slice length in terms of $y$ and integrate with respect to $y$ .

Part 2 Learn it Three short ideas, each checked as you go

8a. Triangles: one slice, one area

Every solid in this topic is built the same way. Its base is a region in the $x y$ -plane. Cut the base with a segment perpendicular to the $x$ -axis at $x$ ; its length $s (x)$ is the top boundary minus the bottom one. A flat shape stands on that segment, straight up out of the plane, and its area is $A (x)$ . Adding the thin slabs from $x = a$ to $x = b$ gives

V = \int_{a}^{b} A (x) d x .

So the work is writing $A (x)$ from $s (x)$ . For a triangle, first read which side sits on the base:

an equilateral triangle with side $s$ has height $\frac{\sqrt{3}}{2} s$ , so $A = \frac{\sqrt{3}}{4} s^{2}$ ;
an isosceles right triangle with a leg on the base has both legs $s$ , so $A = \frac{1}{2} s^{2}$ ;
an isosceles right triangle with its hypotenuse on the base has legs $\frac{s}{\sqrt{2}}$ , so $A = \frac{1}{4} s^{2}$ ;
a triangle with base $s$ and a stated height $h$ has $A = \frac{1}{2} s h$ .

In the picture at the top, the slice at $x = 2$ is 3 long, so the equilateral triangle on it has area $\frac{\sqrt{3}}{4} (3)^{2}$ , about 3.9.

Quick check · Match

No calculator. The slice across a solid's base at $x$ has length $s = 2 x$ . Tap each cross section, then its area $A (x)$ .

8b. Semicircles: the slice is the diameter

When the cross sections are semicircles, the problem says where their diameters lie. When the diameters are in the base, the slice of length $s$ is the diameter, the radius is $\frac{s}{2}$ , and

A = \frac{1}{2} π {(\frac{s}{2})}^{2} = \frac{π}{8} s^{2} .

Build it from the radius every time instead of trusting a memorized $\frac{π}{8}$ . The two slips students make are using $s$ as the radius, which makes every slice four times too big, and dropping the $\frac{1}{2}$ , which gives a whole disc.

Example. The base is the region between $y = x$ and the $x$ -axis for $0 \leq x \leq 4$ , and the cross sections perpendicular to the $x$ -axis are semicircles with diameters in the base. The slice is $s = x$ , so $A (x) = \frac{π}{8} x^{2}$ and

V = \int_{0}^{4} \frac{π}{8} x^{2} d x = \frac{π}{8} \cdot \frac{4^{3}}{3} = \frac{8 π}{3} .

Square $s (x)$ inside the integral. The integral of $s (x)$ alone is the area of the base, and squaring that number does not add up the areas of the slices.

Quick check · Calculate

No calculator. The base of a solid is the region in the first quadrant bounded by the line $y = 6 - 2 x$ and the coordinate axes. Cross sections perpendicular to the $x$ -axis are semicircles with diameters in the base. The volume of the solid is $k π$ . What is $k$ ?

Your answer

8c. Other shapes, and slices the other way

The same two steps work for any shape whose area you can write from the slice. A full circle with its diameter across the base has $A = π {(\frac{s}{2})}^{2} = \frac{π}{4} s^{2}$ , which is how you would find the volume of a funnel from its widths. Sometimes a problem simply gives the area of each cross section as a function, and then you integrate that function as it is.

The cross sections do not have to be perpendicular to the $x$ -axis. When they are perpendicular to the $y$ -axis, each slice is a horizontal segment at one height $y$ . Its length is the right boundary minus the left one, written in terms of $y$ , and the volume is $\int_{c}^{d} A (y) d y$ , from the lowest $y$ of the base to the highest.

Example. The base is the region between $x = y^{2}$ and $x = 9$ , and the cross sections perpendicular to the $y$ -axis are equilateral triangles. At height $y$ the slice runs from $x = y^{2}$ to $x = 9$ , so $s (y) = 9 - y^{2}$ , and the curves meet where $y^{2} = 9$ , at $y = - 3$ and $y = 3$ . The volume is

V = \int_{- 3}^{3} \frac{\sqrt{3}}{4} {(9 - y^{2})}^{2} d y .

Quick check · Multiple choice

No calculator. The base of a solid is the region bounded by $y = x^{2}$ and $y = 4$ . Cross sections perpendicular to the $y$ -axis are semicircles with diameters in the base. Which integral gives the volume of the solid?

Common mistakes

The mistake	Why it is wrong	Do this instead
Using the slice across the base as the radius of a semicircle.	The problem puts the diameter in the base, so the slice is the diameter and the radius is half of it. Using the slice as the radius makes every area four times too big.	Write $r = \frac{s}{2}$ first, then $A = \frac{1}{2} π r^{2}$ .
Treating a hypotenuse in the base as if it were a leg.	With hypotenuse $s$ , each leg of an isosceles right triangle is $\frac{s}{\sqrt{2}}$ , so the area is $\frac{1}{4} s^{2}$ , not $\frac{1}{2} s^{2}$ .	Name the side of the triangle that lies in the base before you write an area formula.
Integrating the length of the slice and squaring the result.	$\int_{a}^{b} s (x) d x$ is the area of the base. Squaring that number is not the sum of the areas of the slices.	Write $A (x)$ with $s (x)$ squared inside the integral, then integrate.

Part 3 Watch it done Triangles on a region between two curves

Watch each step, then do the last one yourself.

The base of a solid is the region bounded by $y = x + 2$ and $y = x^{2}$ . Cross sections perpendicular to the $x$ -axis are isosceles right triangles with one leg in the base. Find the volume of the solid.

Step 1 · Find the limits
The curves meet where $x + 2 = x^{2}$ , so $x^{2} - x - 2 = (x - 2) (x + 1) = 0$ , at $x = - 1$ and $x = 2$ . Between them the line is on top: at $x = 0$ it is at 2 and the parabola is at 0.
Step 2 · Write the slice and the area
$s (x) = (x + 2) - x^{2}$ . A leg lies in the base, so both legs are $s$ and $A (x) = \frac{1}{2} {(x + 2 - x^{2})}^{2}$ .
Why this step?
Why is the leg $(x + 2) - x^{2}$ and not just $x + 2$ ?
The triangle stands on the segment across the base, which runs from the lower curve $y = x^{2}$ up to the upper one, $y = x + 2$ . Its length is the difference. On its own, $x + 2$ is the height of the line above the $x$ -axis, and the $x$ -axis is not the bottom of this base.
Step 3 · Set up and expand
$V = \int_{- 1}^{2} \frac{1}{2} {(x + 2 - x^{2})}^{2} d x$ , and ${(x + 2 - x^{2})}^{2} = x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4$ .
Step 4 · Integrate the square
$\int_{- 1}^{2} (x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4) d x = {[\frac{x^{5}}{5} - \frac{x^{4}}{2} - x^{3} + 2 x^{2} + 4 x]}_{- 1}^{2} = 6.4 - (- 1.7) = 8.1$ .
Your step · Calculate

No calculator. Finish it. For this solid, $\int_{- 1}^{2} {(x + 2 - x^{2})}^{2} d x = 8.1$ , and each cross section is an isosceles right triangle with a leg in the base. What is the volume of the solid?

Your answer

Part 4 Try it A new base, and a comparison

Try it · Calculate

No calculator. The base of a solid is the region between $y = \sqrt{\cos x}$ and $y = - \sqrt{\cos x}$ for $- \frac{π}{2} \leq x \leq \frac{π}{2}$ . Cross sections perpendicular to the $x$ -axis are equilateral triangles. The volume of the solid is $k \sqrt{3}$ . What is $k$ ?

Your answer

Try it · Decide and justify

No calculator. Two solids have the same base. In one, the cross sections perpendicular to the $x$ -axis are semicircles with diameters in the base. In the other, they are equilateral triangles with a side in the base. A classmate says the semicircle solid has the greater volume. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. A garden designer is pouring a concrete sculpture. Its base is the region $R$ bounded by the graph of $y = 2 \sqrt{x}$ , the $x$ -axis and the line $x = 4$ , where $x$ and $y$ are measured in feet.

(a) In the first design, the cross sections of the sculpture perpendicular to the $x$ -axis are semicircles with diameters in $R$ . Find the volume of the sculpture. Indicate units of measure.

(b) In a second design with the same base, the cross sections perpendicular to the $x$ -axis are equilateral triangles with one side in $R$ . Find the volume of this sculpture. Indicate units of measure.

(c) For the first design, the vertical plane $x = k$ cuts the sculpture into two pieces of equal volume. Find the value of $k$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(a) The slice at $x$ is $2 \sqrt{x}$ , the diameter, so the radius is $\sqrt{x}$ and $A (x) = \frac{1}{2} π {(\sqrt{x})}^{2} = \frac{π x}{2}$ . Then $V = \int_{0}^{4} \frac{π x}{2} d x = {[\frac{π x^{2}}{4}]}_{0}^{4} = 4 π$ cubic feet.

(b) $A (x) = \frac{\sqrt{3}}{4} {(2 \sqrt{x})}^{2} = \sqrt{3} x$ , so $V = \int_{0}^{4} \sqrt{3} x d x = {[\frac{\sqrt{3} x^{2}}{2}]}_{0}^{4} = 8 \sqrt{3}$ cubic feet.

(c) The piece from $x = 0$ to $x = k$ holds half of $4 π$ . Since $\int_{0}^{k} \frac{π x}{2} d x = \frac{π k^{2}}{4}$ , set $\frac{π k^{2}}{4} = 2 π$ . Then $k^{2} = 8$ , and because $0 < k < 4$ , $k = 2 \sqrt{2}$ feet.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes the cross-section area $\frac{1}{2} π {(\sqrt{x})}^{2}$ , with the radius half of $2 \sqrt{x}$ .
Writes $\int_{0}^{4} \frac{π x}{2} d x$ and gets $4 π$ cubic feet.
Writes $\int_{0}^{4} \frac{\sqrt{3}}{4} {(2 \sqrt{x})}^{2} d x$ and gets $8 \sqrt{3}$ cubic feet.
Sets $\int_{0}^{k} \frac{π x}{2} d x$ equal to $2 π$ , half the volume from part (a).
Solves to get $k = 2 \sqrt{2}$ , keeping the root between 0 and 4.

Part 6 Check yourself A clay sculpture from a table

The three case questions use the same table. Answer them from it.

An artist builds a clay sculpture on a flat base. The base is the region between the graph of a function $w$ and the $x$ -axis for $0 \leq x \leq 6$ , where $x$ and $w (x)$ are in centimeters, so the slice across the base at $x$ is $w (x)$ centimeters long. Cross sections of the sculpture perpendicular to the $x$ -axis are semicircles with diameters in the base. Some values of $w$ are in the table.

Length of the slice across the base
$x$ (centimeters)	$w (x)$ (centimeters)
0	2
2	4
4	8
6	6

Case question 1 · Calculate

No calculator. The area of the sculpture's cross section at $x = 4$ is $k π$ square centimeters. What is $k$ ?

Your answer

Case question 2 · Multiple choice

No calculator. A second sculpture has the same base, but its cross sections perpendicular to the $x$ -axis are isosceles right triangles with the hypotenuse in the base. Which expression gives its volume in cubic centimeters?

Case question 3 · Calculate

No calculator. Use a right Riemann sum with the three subintervals the table gives to approximate the volume of the semicircle sculpture. The estimate is $k π$ cubic centimeters. What is $k$ ?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. The base of a solid is the region between $y = 2 x$ and the $x$ -axis for $0 \leq x \leq 3$ . Cross sections perpendicular to the $x$ -axis are isosceles right triangles with the hypotenuse in the base. A student finds the volume. Tap the first line with a mistake.

Check yourself · Sort it

No calculator. Sort each solid by the variable its volume integral uses. Tap a card, then a bin.

Integrate with respect to x

Integrate with respect to y

Part 7 Keep it Bring back earlier ideas

Every volume here starts from a base region and ends with a definite integral. Pulling those two skills back up keeps them ready.

Bring it back · Multiple choice

No calculator. What is the area of the region bounded by the graphs of $y = 6 x - x^{2}$ and $y = 2 x$ ?

From 6.9 Integrating Using Substitution · Calculate

No calculator. Evaluate $\int_{1}^{e^{2}} \frac{\ln x}{x} d x$ .

Your answer

In three lines

The volume of a solid with known cross sections is the integral of the area of one cross section, $\int_{a}^{b} A (x) d x$ .
For a triangle, name the side on the base: equilateral $\frac{\sqrt{3}}{4} s^{2}$ , a leg $\frac{1}{2} s^{2}$ , a hypotenuse $\frac{1}{4} s^{2}$ .
For a semicircle the slice is the diameter, so $A = \frac{π}{8} s^{2}$ ; slices perpendicular to the $y$ -axis integrate in $y$ .

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

Before you integrate, write the area of one cross section as a function of $x$ (or $y$ ), with the slice as top minus bottom and the radius or leg named. Then write the integral with its limits and $d x$ . On a part without a calculator, show the antiderivative before the number, and give the volume in cubic units.

Questions students ask

Do I need to memorize

\frac{π}{8} s^{2}

and

\frac{\sqrt{3}}{4} s^{2}

It helps to know them, but you can always build them. A semicircle is half of $π r^{2}$ with $r = \frac{s}{2}$ , and an equilateral triangle has height $\frac{\sqrt{3}}{2} s$ , so its area is half of $s$ times that height.

How do I know whether to integrate with respect to

x

y

Read which axis the cross sections are perpendicular to. Perpendicular to the $x$ -axis means each slice sits at one $x$ , so you integrate $d x$ . Perpendicular to the $y$ -axis means each slice sits at one $y$ , so you write its length in $y$ and integrate $d y$ .

What if the base is between two curves?

The slice runs from the lower curve to the upper one, so its length is top minus bottom, as in the worked example. Find where the curves meet to get the limits.

Is this the same as the disc method?

It is the same idea, integrating the area of one slice, but here the shape stands on a flat base. In 8.9 to 8.12 the slice is a disc or a washer, made by turning a region about a line.

Can a cross section be a shape that is not on the list?

Yes. Any shape whose area you can write from the slice works, such as a full circle with its diameter across the base, a rectangle of given height, or an area function the problem hands you.

Course alignment, for teachers

AP Calculus AB topic 8.8, Unit 8: Applications of Integration.

Skill 3.D: Apply an appropriate mathematical definition, theorem, or test.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.