Math Core

Area came from adding up thin rectangles. Volume comes from adding up thin slabs. If you can find the area of every cross section of a solid, the definite integral stacks those areas into a volume. This lesson handles solids built on a flat base, like a loaf of bread where every slice has a known shape.

Stacking slices

Imagine a solid lying along the $x$ -axis from $x = a$ to $x = b$ . Cut it with a plane perpendicular to the $x$ -axis at position $x$ , and call the area of that cross section $A(x)$ . A slab of thickness $\Delta x$ has volume about $A(x)\,\Delta x$ . Adding the slabs and letting $\Delta x \to 0$ gives the volume.

Volume by cross sections

If a solid's cross sections perpendicular to the $x$ -axis have area $A(x)$ for $a \le x \le b$ , then

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

If the cross sections are perpendicular to the $y$ -axis with area $A(y)$ for $c \le y \le d$ , then $V = \displaystyle\int_c^d A(y)\,dy$ .

Solids with a known base

The typical AP problem describes a base region $R$ in the $xy$ -plane and tells you the shape of the cross sections standing on it. Each cross section perpendicular to the $x$ -axis sits on a segment across $R$ . The length of that segment, call it $s$ , is

s = \text{top} - \text{bottom}

for slices perpendicular to the $x$ -axis, or $s = \text{right} - \text{left}$ for slices perpendicular to the $y$ -axis. It's exactly the height of the slice you used for area.

The base region between y = 2x and y = x². At each x, the cross section stands on a segment of length s = 2x − x², straight up out of the page.Open in grapher →

Then you use the area formula for the shape, written in terms of $s$ .

Cross section	Area in terms of $s$
Square with side $s$	$s^2$
Rectangle with base $s$ and height $h$	$sh$
Semicircle with diameter $s$	$\dfrac{\pi}{8}s^2$
Equilateral triangle with side $s$	$\dfrac{\sqrt{3}}{4}s^2$
Isosceles right triangle with a leg $s$	$\dfrac{1}{2}s^2$
Isosceles right triangle with hypotenuse $s$	$\dfrac{1}{4}s^2$

The semicircle formula comes from radius $\dfrac{s}{2}$ : area $= \dfrac{1}{2}\pi\left(\dfrac{s}{2}\right)^2 = \dfrac{\pi}{8}s^2$ .

Common mistake

For semicircles, $s$ is the diameter, not the radius. Using $\dfrac{1}{2}\pi s^2$ makes the answer four times too big. Always convert to the radius $\dfrac{s}{2}$ first, or use $\dfrac{\pi}{8}s^2$ directly.

A procedure you can repeat

Sketch the base region and draw one slice perpendicular to the given axis.
Write the length $s$ of that slice in terms of $x$ (or $y$ ).
Write the cross-section area $A$ in terms of $s$ , then in terms of $x$ (or $y$ ).
Integrate $A$ between the limits of the base region.

Worked examples

Worked example: Square cross sections

The base of a solid is the region under $y = \sqrt{x}$ and above the $x$ -axis for $0 \le x \le 4$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.

Solution. The side of each square is $s = \sqrt{x} - 0 = \sqrt{x}$ , so $A(x) = (\sqrt{x})^2 = x$ .

V = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = 8.

Worked example: Semicircle cross sections

The base of a solid is the region between $y = 2x$ and $y = x^2$ . Cross sections perpendicular to the $x$ -axis are semicircles. Find the volume.

Solution. The curves meet at $x = 0$ and $x = 2$ , with the line on top, so $s = 2x - x^2$ and $A(x) = \dfrac{\pi}{8}(2x - x^2)^2$ .

\begin{aligned} V &= \frac{\pi}{8}\int_0^2 \big(4x^2 - 4x^3 + x^4\big)\,dx = \frac{\pi}{8}\left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]_0^2 \\ &= \frac{\pi}{8}\left(\frac{32}{3} - 16 + \frac{32}{5}\right) = \frac{\pi}{8}\cdot\frac{16}{15} = \frac{2\pi}{15}. \end{aligned}

Worked example: Equilateral triangles on a disk

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

Solution. At position $x$ , the slice runs from $y = -\sqrt{4 - x^2}$ to $y = \sqrt{4 - x^2}$ , so $s = 2\sqrt{4 - x^2}$ and

A(x) = \frac{\sqrt{3}}{4}\big(2\sqrt{4 - x^2}\big)^2 = \sqrt{3}\,(4 - x^2).

V = \sqrt{3}\int_{-2}^{2} (4 - x^2)\,dx = \sqrt{3}\cdot\frac{32}{3} = \frac{32\sqrt{3}}{3}.

Tip

When $s^2$ involves a square root, squaring it first usually makes the integral easy, as in the last example: $(2\sqrt{4 - x^2})^2 = 4(4 - x^2)$ . Expand before you integrate; there is no need for $u$ -substitution.

Practice

Practice 1

The base of a solid is the region under $y = \sin x$ and above the $x$ -axis for $0 \le x \le \pi$ . Cross sections perpendicular to the $x$ -axis are squares. Find the exact volume.