Volumes with Square and Rectangle Cross…

Part 1 · Hook

Why this matters

Slice a loaf of bread and every slice is a flat shape with a tiny thickness. Know the area of each slice and you know the loaf: add up area times thickness. Calculus does exactly that, for solids built by standing squares on a region drawn on the floor.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A box has a square face of side 3 cm and a depth of 0.5 cm. What is its volume?

$4.5$ cubic centimeters
$1.5$ cubic centimeters
$9$ cubic centimeters
$6$ cubic centimeters

Show the answer

Volume = area of the face times thickness: $3^2 \cdot 0.5 = 4.5$ .

Correct: $4.5$ cubic centimeters:
$1.5$ cubic centimeters:
$9$ cubic centimeters:
$6$ cubic centimeters:

2. The region between $y = x + 2$ (top) and $y = x^2$ (bottom) has what height at $x = 1$ ?

$2$
$4$
$3$
$1$

Show the answer

Top minus bottom: $(1 + 2) - 1^2 = 2$ .

Correct: $2$ :
$4$ :
$3$ :
$1$ :

Part 4 · See it

See it first

A 3D view of a solid built on the region under y = 4 − x² from x = −2 to 2, lying flat on the floor. Square slices stand on the base, each perpendicular to the x-axis with one side across the base, so the slice at x has side s = 4 − x². One thin slice is highlighted in blue with its side s marked; dashed curves trace the top edges of the solid. A side panel shows a flat square of side s, its area A(x) = (4 − x²)², and the volume, the integral from −2 to 2 of A(x) dx = 512/15. — Square slices standing on the region under $y = 4 - x^2$ . The slice at $x$ has side $s = 4 - x^2$ and area $s^2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Slice the solid perpendicular to the $x$ -axiseach slice is a thin slab: a cross section of area $A(x)$ with thickness $\Delta x$
A slab's volume is about $A(x)\,\Delta x$ adding the slabs is a Riemann sum for $A$
Let the slabs get thinthe volume is $V$ $= \int_a^b A(x)\,dx$
For a square cross section, the side $s(x)$ is the length of the base across the slice $A(x)$ $= s(x)^2$ $= \left(f(x) - g(x)\right)^2$

Part 6 · Key ideas

Key ideas

Volume by cross sections: $V$ $= \int_a^b A(x)\,dx$ , where $A(x)$ is the area of the slice at $x$ .
For squares standing on the base, the side is top minus bottom of the base: $s = f(x) - g(x)$ , and $A(x) = s^2$ .
Square the whole difference: $(f - g)^2$ is not $f^2 - g^2$ .
Rectangles work the same way: $A = s \cdot h$ , with $h$ given (a number, or a multiple of $s$ ).
Slices perpendicular to the $y$ -axis use right minus left and $dy$ .

Part 7 · Misconception

A common mistake

The wrong idea: To get the square's area, square the top function and the bottom function and subtract: $f(x)^2 - g(x)^2$ .

What actually happens: The square's side is the gap between the curves, $f(x) - g(x)$ , so its area is $\left(f(x) - g(x)\right)^2$ . Squaring each curve first gives a different, wrong number.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The base: the region between $y = x + 2$ and $y = x^2$

1. The shaded region is the base of a solid. Cross sections perpendicular to the $x$ -axis are squares. Write the area $A(x)$ of the cross section at $x$ .

Type an expression for $A(x) =$ .

Show the answer

The side is the height of the base at $x$ : $s = (x + 2) - x^2$ . So $A(x)$ $= s^2$ $= (x + 2 - x^2)^2$ .

Answer: $A(x) =$ $(x + 2 - x^{2})^{2}$

2. Which integral gives the volume of the solid with this base and square cross sections perpendicular to the $x$ -axis?

$\int_{-1}^{2} \left(x + 2 - x^2\right)^2\,dx$
$\int_{-1}^{2} \left[(x + 2)^2 - (x^2)^2\right]\,dx$
$\int_{-1}^{2} \left(x + 2 - x^2\right)\,dx$
$\pi \int_{-1}^{2} \left(x + 2 - x^2\right)^2\,dx$

Show the answer

Volume is the integral of cross-sectional area: $\int_{-1}^{2} A(x)\,dx$ with $A(x) = s^2$ and $s = (x + 2) - x^2$ .

Correct: $\int_{-1}^{2} \left(x + 2 - x^2\right)^2\,dx$ : Square the side, then integrate from $-1$ to $2$ .
$\int_{-1}^{2} \left[(x + 2)^2 - (x^2)^2\right]\,dx$ : This squares each curve separately. $(a - b)^2$ is not $a^2 - b^2$ .
$\int_{-1}^{2} \left(x + 2 - x^2\right)\,dx$ : This is the area of the base, not the volume.
$\pi \int_{-1}^{2} \left(x + 2 - x^2\right)^2\,dx$ : A square's area is $s^2$ , with no $\pi$ .

3. Find the volume of the solid with this base and square cross sections perpendicular to the $x$ -axis.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$(x + 2 - x^2)^2$ $= x^4 - 2x^3 - 3x^2$ $+ 4x$ $+ 4$ , so $V$ $= \left[\frac{x^5}{5} - \frac{x^4}{2} - x^3 + 2x^2 + 4x\right]_{-1}^{2}$ $= \frac{32}{5} - \left(-\frac{17}{10}\right)$ $= \frac{81}{10}$ .

Answer: $\frac{81}{10}$ ≈ 8.100

4. What is the area of the square cross section at $x = \frac{1}{2}$ ?

$\frac{81}{16}$
$\frac{9}{4}$
$\frac{99}{16}$
$\frac{25}{4}$

Show the answer

At $x = \frac{1}{2}$ the side is $\frac{5}{2} - \frac{1}{4} = \frac{9}{4}$ , so the area is $\left(\frac{9}{4}\right)^2 = \frac{81}{16}$ .

Correct: $\frac{81}{16}$ : Side $\frac{9}{4}$ , squared.
$\frac{9}{4}$ : $\frac{9}{4}$ is the side, a length; the area is its square.
$\frac{99}{16}$ : $\left(\frac{5}{2}\right)^2 - \left(\frac{1}{4}\right)^2 = \frac{99}{16}$ squares each height separately instead of the gap.
$\frac{25}{4}$ : $\left(\frac{5}{2}\right)^2$ uses the height of the line above the $x$ -axis; the base starts at the parabola, not the axis.

5. The base of a solid is the disk $x^2 + y^2 \le 4$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $x$ , the disk runs from $y = -\sqrt{4 - x^2}$ to $y = \sqrt{4 - x^2}$ , so $s = 2\sqrt{4 - x^2}$ and $A(x) = 4(4 - x^2)$ . $V$ $= \int_{-2}^{2} 4(4 - x^2)\,dx$ $= 4 \cdot \frac{32}{3}$ $= \frac{128}{3}$ .

Answer: $\frac{128}{3}$ ≈ 42.667

6. The volume of a solid with square cross sections has no $\pi$ , even when its base has a curved edge. Why?

Each slice is a square with area $s^2$ , and $s^2$ has no $\pi$ in it
$\pi$ cancels out when the integral is evaluated
$\pi$ is left out because the base lies flat in the $xy$ -plane
$\pi$ is needed when the base region has a curved edge

Show the answer

The volume is $\int A(x)\,dx$ , and $A$ is whatever the shape of a slice gives. A square's area is $s^2$ . A curved edge only changes how $s$ depends on $x$ .

Correct: Each slice is a square with area $s^2$ , and $s^2$ has no $\pi$ in it: The slice's area formula decides; for a square it is $s^2$ .
$\pi$ cancels out when the integral is evaluated: There is no $\pi$ to cancel: it never enters.
$\pi$ is left out because the base lies flat in the $xy$ -plane: Every base lies in the $xy$ -plane; that does not decide whether $\pi$ appears.
$\pi$ is needed when the base region has a curved edge: A curved edge changes $s(x)$ , not the area formula of a square.

Part 9 · Summary

Summary

A solid with known cross sections has volume

\int_a^b A(x)\,dx

. When the slices are squares standing on a base, each side is the base's width across the slice,

f(x) - g(x)

, and

A(x) = \left(f(x) - g(x)\right)^2

. Square the difference, not each curve, and leave out

\pi

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The base: the region between y=x+2y = x + 2 and y=x2y = x^2

Summary

What comes next

Connections

The base: the region between $y = x + 2$ and $y = x^2$