Math Core

You already know how to find area and volume with definite integrals from AP Calculus AB. This lesson is a fast, focused review of the three setups that show up on nearly every AP exam: area between curves, volumes with known cross sections, and volumes of revolution by disks and washers. The one habit that ties them together is the same: slice the region, write the size of one slice, and add up the slices with an integral.

Area between two curves

Slice the region into thin vertical rectangles of width $dx$ . Each rectangle stretches from the lower curve up to the upper curve, so its height is $\text{top} - \text{bottom}$ .

Area between curves

If $f(x) \ge g(x)$ on $[a, b]$ , the area between the curves is

$A = \int_a^b \big(f(x) - g(x)\big)\,dx = \int_a^b (\text{top} - \text{bottom})\,dx.$

If $x = f(y)$ lies to the right of $x = g(y)$ for $c \le y \le d$ , slice horizontally instead:

$A = \int_c^d \big(f(y) - g(y)\big)\,dy = \int_c^d (\text{right} - \text{left})\,dy.$

The limits of integration are usually the $x$ -values (or $y$ -values) where the curves intersect, so solve $f = g$ first. If the curves cross inside the interval, the "top" curve changes, and you must split the integral at each crossing.

The region between y = 4x − x² (top) and y = x (bottom) for 0 ≤ x ≤ 3.Open in grapher →

Worked example: Vertical slices

Find the area of the region bounded by $y = 4x - x^2$ and $y = x$ .

Intersections: $4x - x^2 = x$ gives $x^2 - 3x = 0$ , so $x = 0$ or $x = 3$ .

Top and bottom: at $x = 1$ , the parabola gives $3$ and the line gives $1$ , so the parabola is on top.

$A = \int_0^3 \big((4x - x^2) - x\big)\,dx = \int_0^3 (3x - x^2)\,dx = \left[\tfrac{3}{2}x^2 - \tfrac{1}{3}x^3\right]_0^3 = \tfrac{27}{2} - 9 = \tfrac{9}{2}.$

Worked example: Horizontal slices

Find the area of the region bounded by $x = y^2$ and $x = 2y$ .

The curves are already written as $x$ in terms of $y$ , so use horizontal slices. They meet where $y^2 = 2y$ , so $y = 0$ or $y = 2$ . At $y = 1$ , the line gives $x = 2$ and the parabola gives $x = 1$ , so the line is on the right.

$A = \int_0^2 (2y - y^2)\,dy = \left[y^2 - \tfrac{1}{3}y^3\right]_0^2 = 4 - \tfrac{8}{3} = \tfrac{4}{3}.$

Using vertical slices would have required writing both curves as functions of $x$ , which is more work here.

Volumes with known cross sections

Now picture the region as the base of a solid. Each slice perpendicular to the $x$ -axis is a flat shape (a square, a semicircle, a triangle) whose size depends on $x$ . If $A(x)$ is the area of the slice at $x$ , the slice is a thin slab of volume $A(x)\,dx$ .

Cross-section volume

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

where $A(x)$ is the area of the cross section at $x$ . Find the length $s$ of the slice across the base (usually $\text{top} - \text{bottom}$ ), then use the area formula for the shape.

The area formulas you need, in terms of the side or diameter $s$ :

cross section	area
square with side $s$	$s^2$
semicircle with diameter $s$	$\dfrac{\pi s^2}{8}$
equilateral triangle with side $s$	$\dfrac{\sqrt{3}}{4}s^2$
isosceles right triangle with leg $s$	$\dfrac{1}{2}s^2$

Worked example: Square cross sections

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

At each $x$ , the base slice runs from $y = 0$ up to $y = \sqrt{x}$ , so $s = \sqrt{x}$ and $A(x) = s^2 = x$ .

$V = \int_0^4 x\,dx = \left[\tfrac{1}{2}x^2\right]_0^4 = 8.$

Disks and washers

When a region spins around a line, each slice perpendicular to the axis of rotation sweeps out a disk, or a washer (a disk with a hole) if the region does not touch the axis.

Washer method

For rotation about a horizontal line, with slices perpendicular to the axis,

$V = \pi \int_a^b \big(R(x)^2 - r(x)^2\big)\,dx,$

where $R(x)$ is the outer radius (distance from the axis to the far curve) and $r(x)$ is the inner radius (distance from the axis to the near curve). If the region touches the axis, $r = 0$ and this is the disk method. For rotation about a vertical line, slice horizontally and integrate in $y$ .

A radius is always a distance to the axis. Rotating about $y = k$ , a curve $y = f(x)$ has distance $|f(x) - k|$ from the axis. Rotating about $x = h$ , a curve $x = g(y)$ has distance $|g(y) - h|$ .

The region between y = x and y = x², with the axis of rotation y = −1 dashed.Open in grapher →

Worked example: Washers about a line below the region

The region between $y = x$ and $y = x^2$ is revolved about the line $y = -1$ . Find the volume.

The curves meet at $x = 0$ and $x = 1$ , and $y = x$ is on top. The axis is below the region, so:

outer radius (to the far curve, $y = x$ ): $R = x - (-1) = x + 1$
inner radius (to the near curve, $y = x^2$ ): $r = x^2 + 1$

\begin{aligned} V &= \pi \int_0^1 \big((x + 1)^2 - (x^2 + 1)^2\big)\,dx \\ &= \pi \int_0^1 \big(x^2 + 2x + 1 - x^4 - 2x^2 - 1\big)\,dx \\ &= \pi \int_0^1 (-x^4 - x^2 + 2x)\,dx \\ &= \pi\left(-\tfrac{1}{5} - \tfrac{1}{3} + 1\right) = \tfrac{7\pi}{15}. \end{aligned}

Common mistake

Square each radius separately: $\pi\big(R^2 - r^2\big)$ , never $\pi(R - r)^2$ . The expression $(R - r)^2$ gives the square of the gap between the curves, which is not the area of a washer.

Tip

Before integrating, test one value of $x$ in your setup. Every radius, side length or height should be positive there. A negative radius or a "top" that is actually below the "bottom" means the setup is wrong.

Choosing a setup quickly

On the AP exam, most of the work is the setup, not the antiderivative. Ask three questions:

What are the limits? Solve for the intersections.
Which way do I slice? Perpendicular to the axis of rotation, or perpendicular to the axis named in a cross-section problem. If the curves are easier as $x = g(y)$ , slice horizontally.
What is one slice? A rectangle ( $\text{top} - \text{bottom}$ ), a cross section $A(x)$ , or a washer $\pi(R^2 - r^2)$ .

Calculator questions often have intersection points you cannot find by hand. Store them in your calculator at full precision, and round only the final answer to three decimal places.