Volumes with Square and Rectangle… — AP® Calculus AB Notes

Volume by slicing

You know how to find the volume of a box: area of the face times thickness. A loaf of bread is not a box, but each thin slice of it almost is. Its face has some area $A$ and its thickness is small, so the slice's volume is about $A$ times the thickness. Add up the slices and you have the loaf.

Calculus makes that exact. Put the solid along the $x$ -axis and slice it with planes perpendicular to the axis. The slice at position $x$ is a flat shape, a cross section, with area $A(x)$ . A slab of thickness $\Delta x$ there has volume about $A(x)\,\Delta x$ . Adding the slabs is a Riemann sum for $A$ , and letting them get thin gives

V = \int_a^b A(x)\,dx.

Everything in this topic, and in the next one, comes down to finding $A(x)$ .

Squares standing on a base

A common way to describe a solid: draw a region on the floor (the base), and stand a shape of a given kind on every line across it. In the figure, the base is the region under $y = 4 - x^2$ , and on each segment perpendicular to the $x$ -axis stands a square. The segment is one side of the square.

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. — **Figure 1.** Square slices on the region under $y = 4 - x^2$ . The square at $x$ has side $s = 4 - x^2$ ; the tallest one is at $x = 0$ . LevlPrep original diagram.

The side of each square is the width of the base across that slice. For a base between a top curve $y = f(x)$ and a bottom curve $y = g(x)$ , the side is top minus bottom, exactly the height you used for area between curves:

s(x) = f(x) - g(x), \qquad A(x) = s(x)^2 = \left(f(x) - g(x)\right)^2.

Worked example. The base of a solid is the region under $y = 4 - x^2$ and above the $x$ -axis. Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Limits. $4 - x^2 = 0$ at $x = \pm 2$ .

Side. $s$ $= (4 - x^2) - 0$ $= 4 - x^2$ , so $A(x)$ $= (4 - x^2)^2$ $= 16 - 8x^2 + x^4$ .

Volume.

V = \int_{-2}^{2} \left(16 - 8x^2 + x^4\right)\,dx = 2\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = 2\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{512}{15}.

The integrand is even, so the integral from $-2$ to $2$ is twice the integral from $0$ to $2$ .

Square the difference, not each curve

The most common mistake is writing $f(x)^2 - g(x)^2$ . That is not the area of anything here. The square's side is the gap between the curves, and you square the gap.

Worked example. The base is the region between $y = \sqrt{x}$ and $y = \frac{x}{2}$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Limits. $\sqrt{x} = \frac{x}{2}$ at $x = 0$ and $x = 4$ ; between them $\sqrt{x}$ is on top (at $x = 1$ : $1 > \frac{1}{2}$ ).

Area. $A(x)$ $= \left(\sqrt{x} - \frac{x}{2}\right)^2$ $= x - x^{3/2}$ $+ \frac{x^2}{4}$ .

Volume.

V = \int_0^4 \left(x - x^{3/2} + \frac{x^2}{4}\right)\,dx = \left[\frac{x^2}{2} - \frac{2}{5}x^{5/2} + \frac{x^3}{12}\right]_0^4 = 8 - \frac{64}{5} + \frac{16}{3} = \frac{8}{15}.

Squaring each curve first would give $\int_0^4 \left(x - \frac{x^2}{4}\right)\,dx$ $= \frac{8}{3}$ , five times too big.

Rectangles, other directions, and units

Slice shape	Side in the base	Area $A$
Square	$s$	$s^2$
Rectangle of height $h$ (a number)	$s$	$s \cdot h$
Rectangle whose height is $k$ times its base	$s$	$k s^2$

If the slices are perpendicular to the $y$ -axis, slice the base horizontally: the side is right minus left, written in $y$ , and the integral is in $dy$ between $y$ -values.

Units follow the formula: a side in feet gives an area in square feet, and integrating over feet gives cubic feet. There is no $\pi$ anywhere, because squares and rectangles have none in their area formulas, even when the base has curved edges.

With a calculator, find the base's crossings numerically and store them, write $\int_a^b \left(f(x) - g(x)\right)^2\,dx$ as your setup, and let the calculator do the integral. Report three decimal places.