Unit 8 · Topic 8.7 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 8.7: Volumes with Cross Sections: Squares and Rectangles

Q: Do I need to draw the solid?

No. Draw the base region in the xy-plane and one slice across it. The slice gives you the side, and the shape's area formula does the rest.

Q: When do I integrate with respect to y?

When the cross sections are perpendicular to the y-axis. Then each slice is horizontal, its length is right minus left written in terms of y, and the limits are the lowest and highest y in the base.

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 squares or rectangles with a definite integral of the cross-sectional area.

Key vocabulary

Term	What it means	The exam trap
Cross section	The flat shape you get when you slice a solid with a plane, here a plane perpendicular to the $x$ -axis or to the $y$ -axis.	Read which axis the slices are perpendicular to before you measure anything; it decides the variable you integrate in.
Solid with known cross sections	A solid whose cross sections all have one shape, such as squares, sized by a base region, so the area $A (x)$ of each can be written down.	The base region is where the solid stands, not a cross section of it.
Square cross section	A cross section that is a square; with side $s (x)$ , its area is $A (x) = (s (x))^{2}$ .	Find the side as one length across the base before you square anything.
Rectangular cross section	A cross section that is a rectangle; its area is its base times its height, $A (x) = s (x) \cdot h (x)$ .	The height comes from the problem's wording, so reread that sentence before you write the area.
Volume by cross sections	$V = \int_{a}^{b} A (x) d x$ , the definite integral of the cross-sectional area over the interval the solid covers.	The limits are the ends of the base along the axis the slices are perpendicular to.

Graph of a base region between the curve y = 4 - x squared over 4, on top, and the line y = x/2 - 2, below it, for x from 0 to 4. A slice across the region at x = 2 runs from y = -1 up to y = 3, so it is 4 long. Beside the graph, the square standing on that slice is drawn to the same scale: 4 by 4, with area 16. The caption gives the volume of the whole solid as 66.13.

Part 1 Think first How much glass is in the paperweight?

Make a guess first. Being wrong now still helps the right answer stick.

Predict first

A glass paperweight is a pyramid with a square base 6 centimeters on a side and a height of 9 centimeters. Every cross section parallel to the base is a square. How much glass is in it?

In short. A solid with known cross sections has volume $V = \int_{a}^{b} A (x) d x$ , where $A (x)$ is the area of the cross section at $x$ , found from the area formula for its shape.

The side of a square, or the base of a rectangle, is the slice across the base region: top minus bottom, or right minus left when the slices are perpendicular to the $y$ -axis.
A square of side $s (x)$ has area $(s (x))^{2}$ , and a rectangle has area base times height, with the height read from the problem.
Integrate the area, not the side, over the interval the base covers, in the variable of the axis the slices are perpendicular to.

Part 2 Learn it Slice, measure, integrate

7a. Volume is the integral of cross-sectional area

Think of a loaf of bread cut into thin slices. Each slice holds about the area of its face times its thickness, and the slices together are the loaf.

A solid with known cross sections works the same way. Lay it along the $x$ -axis from $x = a$ to $x = b$ , and let $A (x)$ be the area of its cross section perpendicular to the $x$ -axis at $x$ . A slab of thickness $Δ x$ there holds about $A (x) Δ x$ , so the volume is about $\sum A (x_{i}) Δ x$ . That is a Riemann sum, and as the slabs get thinner its limit is a definite integral:

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

When each cross section is a square with side $s (x)$ , its area is $A (x) = (s (x))^{2}$ .

Back to the paperweight. Lay it on its side with the tip at $x = 0$ and the base at $x = 9$ . The side grows steadily from 0 to 6, so $s (x) = \frac{2}{3} x$ ; at $x = 3$ the side is 2 and the area is 4. Then

V = \int_{0}^{9} {(\frac{2}{3} x)}^{2} d x = \int_{0}^{9} \frac{4}{9} x^{2} d x = {[\frac{4}{27} x^{3}]}_{0}^{9} = 108

cubic centimeters. Square the side first, then integrate. Integrating the side alone adds up lengths, not areas.

Quick check · Calculate

No calculator. A solid lies along the $x$ -axis from $x = 0$ to $x = 3$ . Its cross section perpendicular to the $x$ -axis at each $x$ is a square with side length $x + 2$ . What is the volume of the solid?

Your answer

7b. The side is the slice across the base

Most problems don't hand you the side. They give a base region in the $x y$ -plane and say the cross sections perpendicular to the $x$ -axis are squares. The solid stands on that base, and each square stands on one slice across it.

The slice at $x$ runs from the bottom boundary up to the top one, so its length is $f (x) - g (x)$ , top minus bottom. That length is the side of the square, and

V = \int_{a}^{b} (f (x) - g (x))^{2} d x,

where $a$ and $b$ are the left and right ends of the base, often where the curves meet.

In the picture at the top, the top is $f (x) = 4 - \frac{x^{2}}{4}$ and the bottom is $g (x) = \frac{x}{2} - 2$ . At $x = 2$ , $f (2) = 3$ and $g (2) = - 1$ , so the side is $3 - (- 1) = 4$ and the square's area is 16. A bottom boundary below the $x$ -axis makes the slice longer, not shorter.

Square the whole difference. $(f (x) - g (x))^{2}$ is not $(f (x))^{2} - (g (x))^{2}$ .

If the cross sections are perpendicular to the $y$ -axis instead, each slice is horizontal. Its length is the right boundary minus the left one, written in terms of $y$ , and you integrate with respect to $y$ from the lowest $y$ in the base to the highest.

Quick check · Multiple choice

No calculator. The base of a solid is the region bounded by $y = x$ and $y = x^{2}$ for $0 \leq x \leq 1$ . Cross sections perpendicular to the $x$ -axis are squares. Which integral gives the volume of the solid?

7c. Rectangles: base times height

A rectangle needs two lengths. Its base lies in the base region: the slice across it, $f (x) - g (x)$ , just as for a square. The problem tells you the height, and the area is base times height:

A (x) = (f (x) - g (x)) \cdot h (x) .

The height can be given in a few ways, and each one changes $A (x)$ :

a fixed height, such as 3, gives $A (x) = 3 (f (x) - g (x))$ ;
a multiple of the base, such as half of it, gives $A (x) = \frac{1}{2} (f (x) - g (x))^{2}$ ;
a height that changes with $x$ comes with its own formula, and you multiply it by the base.

On the picture's base at $x = 2$ the slice is 4 long. A rectangle of height 3 standing there has area $4 \cdot 3 = 12$ , and one whose height is half its base is 4 by 2, with area 8. A square is the rectangle whose height equals its base.

Then integrate the area across the base, exactly as before: $V = \int_{a}^{b} A (x) d x$ . Before you integrate, read the problem's sentence about the height once more. Writing a rectangle's area as if it were a square's is an easy slip to make.

Quick check · Calculate

No calculator. The base of a solid is the triangle bounded by $y = x$ , $y = - x$ and $x = 3$ . Cross sections perpendicular to the $x$ -axis are rectangles whose height is twice the length of their base. What is the volume of the solid?

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Squaring each boundary and subtracting, $(f (x))^{2} - (g (x))^{2}$ , for a square cross section.	The side of each square is the one length $f (x) - g (x)$ , and its area is that whole length squared. Squaring each curve on its own measures no cross section of the solid.	Write the side $s (x) = f (x) - g (x)$ first, then square it: $A (x) = (s (x))^{2}$ .
Integrating the side length instead of the area.	$\int_{a}^{b} (f (x) - g (x)) d x$ adds up the lengths of the slices, which is the area of the base, not the volume of the solid.	Write $A (x)$ , the area of one cross section, and integrate that.
Writing a rectangle's area as if it were a square's.	A rectangle's height is whatever the problem says: a fixed number, a multiple of the base or a formula in $x$ . Only when the height equals the base is the area the base squared.	Write the area as base times height, $A (x) = s (x) \cdot h (x)$ , with the height taken from the problem's own sentence.

Part 3 Watch it done Squares on the region between a line and a parabola

Each step opens in turn. The last one is yours.

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 squares. Find the volume of the solid.

Step 1 · Find the ends of the base
$x^{2} = x + 2$ gives $x^{2} - x - 2 = 0$ , so $(x - 2) (x + 1) = 0$ and the curves meet at $x = - 1$ and $x = 2$ .
Step 2 · Measure the side
At $x = 0$ the line is at 2 and the parabola is at 0, so the line is on top on $[- 1, 2]$ . The slice at $x$ has length $s (x) = (x + 2) - x^{2}$ .
Step 3 · Set up the integral
$A (x) = {(x + 2 - x^{2})}^{2}$ , so $V = \int_{- 1}^{2} {(x + 2 - x^{2})}^{2} d x$ .
Why this step?
Why square $x + 2 - x^{2}$ instead of writing $(x + 2)^{2} - {(x^{2})}^{2}$ ?
Each cross section is one square, and its side is the whole slice, from the parabola up to the line. Its area is that one length squared. $(x + 2)^{2} - {(x^{2})}^{2}$ is a big square minus a small one, which is not the area of any cross section of this solid.
Step 4 · Expand and find an antiderivative
${(x + 2 - x^{2})}^{2} = x^{4} - 2 x^{3} - 3 x^{2} + 4 x + 4$ , which has antiderivative $F (x) = \frac{x^{5}}{5} - \frac{x^{4}}{2} - x^{3} + 2 x^{2} + 4 x$ . So $V = F (2) - F (- 1)$ , with $F (2) = 6.4$ and $F (- 1) = - 1.7$ .
Your step · Calculate

No calculator. Finish it. For the solid with square cross sections on the region between $y = x + 2$ and $y = x^{2}$ , $V = F (2) - F (- 1)$ , where $F (2) = 6.4$ and $F (- 1) = - 1.7$ . What is the volume? Give an exact decimal.

Your answer

Part 4 Try it Turn the slices, then scale them

Try it · Calculate

No calculator. The base of a solid is the region in the first quadrant bounded by $y = x^{2}$ , the line $y = 4$ and the $y$ -axis. Cross sections perpendicular to the $y$ -axis are squares. What is the volume of the solid?

Your answer

Try it · Decide and justify

No calculator. Solid $S$ has square cross sections perpendicular to the $x$ -axis. Solid $T$ also has square cross sections perpendicular to the $x$ -axis over the same interval of $x$ , but at every $x$ its slice across the base is twice as long as the slice for $S$ . A classmate says the volume of $T$ is twice the volume of $S$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, so work this one on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A sculptor casts a concrete piece whose 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. Cross sections of the piece perpendicular to the $x$ -axis are squares.

(a) Write an integral expression for the volume of the piece, and find the volume. Show the work that leads to your answer, and use correct units.

(b) A second piece has the same base $R$ , but its cross sections perpendicular to the $x$ -axis are rectangles whose height is half the length of their base. Find the volume of the second piece. Indicate units of measure.

(c) A cut perpendicular to the $x$ -axis at $x = k$ divides the first piece into two parts, and the part between $x = 0$ and $x = k$ holds one fourth of the volume. Write an equation involving an integral that $k$ satisfies, and find $k$ .

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

Model answer

(a) The slice at $x$ runs from $y = 0$ to $y = 2 \sqrt{x}$ , so the side is $2 \sqrt{x}$ feet and $A (x) = {(2 \sqrt{x})}^{2} = 4 x$ square feet. $V = \int_{0}^{4} {(2 \sqrt{x})}^{2} d x = \int_{0}^{4} 4 x d x = [2 x^{2}]_{0}^{4} = 32$ cubic feet.

(b) Each rectangle has base $2 \sqrt{x}$ and height $\sqrt{x}$ , so $A (x) = 2 \sqrt{x} \cdot \sqrt{x} = 2 x$ . The volume is $\int_{0}^{4} 2 x d x = [x^{2}]_{0}^{4} = 16$ cubic feet.

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

(a) Writes $\int_{0}^{4} {(2 \sqrt{x})}^{2} d x$ , with the side squared and limits 0 and 4.
(a) Gets 32 cubic feet, with units.
(b) Writes the area as base times height, $2 \sqrt{x} \cdot \sqrt{x} = 2 x$ , and gets 16 cubic feet.
(c) Writes $\int_{0}^{k} 4 x d x = 8$ , or the same integral set equal to one fourth of the answer to (a).
(c) Solves to get $k = 2$ .

Part 6 Check yourself A tapered table leg

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

A woodworker shapes a square table leg 40 centimeters long. At a distance of $x$ centimeters from the top of the leg, its cross section perpendicular to the leg's length is a square with side $s (x)$ centimeters, where $s$ is continuous. Some values are in the table.

Side of the leg's square cross section
$x$ (centimeters)	$s (x)$ (centimeters)
0	6
10	5
20	4.5
30	4
40	5

Case question 1 · Multiple choice

No calculator. Which expression gives the volume of the leg, in cubic centimeters?

Case question 2 · Calculate

No calculator. Use a right Riemann sum with the four subintervals of the table to approximate the volume of the leg, in cubic centimeters.

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 bounded by $y = 4$ and $y = x^{2}$ . Cross sections perpendicular to the $x$ -axis are squares. A student finds the volume. Tap the first line with a mistake.

Check yourself · Match

No calculator. The base of a solid runs from $x = 0$ to $x = 3$ , and its slice perpendicular to the $x$ -axis at $x$ has length $6 - 2 x$ . Tap a description of the cross sections, then the area $A (x)$ of the cross section at $x$ .

Part 7 Keep it From earlier in the course

Two earlier ideas worth keeping: the area between two curves, which every base region here starts from, and changing the limits when you substitute. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. Find the area of the region bounded by $y = 3 x$ and $y = x^{2}$ . Give an exact decimal.

Your answer

Bring it back · Multiple choice

No calculator. With the substitution $u = x^{2} + 1$ , which integral is equal to $\int_{0}^{2} x {(x^{2} + 1)}^{3} d x$ ?

In three lines

A solid's volume is the integral of its cross-sectional area, $V = \int_{a}^{b} A (x) d x$ .
For squares, $A (x)$ is the side squared, and the side is the length of the slice across the base.
For rectangles, $A (x)$ is base times height, with the height taken from the problem.

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.

Write the area of one cross section, $A (x)$ , before the integral, then write the integral with both limits and its differential. Without a calculator, show the antiderivative and its values at both limits before the number, and give a volume in context in cubic units.

Questions students ask

Do I need to draw the solid?

No. Draw the base region in the $x y$ -plane and one slice across it. The slice gives you the side, and the shape's area formula does the rest.

What if the base dips below the

x

-axis?

Nothing changes: the side is still top minus bottom. A bottom boundary below the axis makes $f (x) - g (x)$ larger, as at $x = 2$ in the picture at the top, where the side is $3 - (- 1) = 4$ .

When do I integrate with respect to

y

When the cross sections are perpendicular to the $y$ -axis. Then each slice is horizontal, its length is right minus left written in terms of $y$ , and the limits are the lowest and highest $y$ in the base.

Is a square just a special rectangle here?

Yes. A square is the rectangle whose height equals its base, so $A (x) = s (x) \cdot s (x) = (s (x))^{2}$ .

Does the pyramid formula from geometry agree with the integral?

Yes. Slicing a pyramid with a square base into square cross sections and integrating their areas gives one third of the base area times the height, which is where that formula comes from.

Course alignment, for teachers

AP Calculus AB topic 8.7, Unit 8: Applications of Integration.

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

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first How much glass is in the paperweight?

Part 2 Learn it Slice, measure, integrate

7a. Volume is the integral of cross-sectional area

7b. The side is the slice across the base

7c. Rectangles: base times height

Part 3 Watch it done Squares on the region between a line and a parabola

Part 4 Try it Turn the slices, then scale them

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A tapered table leg

Part 7 Keep it From earlier in the course

Exam strategy for this topic

Questions students ask

Past exam questions on this topic