How do I know whether to use discs or washers?

Use a disc if the region touches the axis of rotation along the whole interval, and a washer if there's a gap between them.

Do cross-section problems always have pi in them?

No. Pi appears only when the shape of the slice needs it, such as a semicircle.

When should I integrate with respect to y?

When the curves are given as x in terms of y, when the region's boundaries are easier to describe that way, or when you revolve around a vertical line with discs or washers.

Are cylindrical shells on the AP Calculus AB exam?

No. The AB course uses discs, washers, and cross sections for volume.

Area and volume for AP Calculus AB

Updated November 9, 2026

Area and volume questions on the AP Calculus AB exam come down to writing the right integral. The integration itself is often the easy part, and on calculator questions Desmos does it for you.

This guide covers area between curves, volume by discs and washers, and volume with known cross sections, and it ends with one region that uses all of them.

Area between curves

In terms of $x$ : if $f(x) \ge g(x)$ on $[a, b]$ , the area is $\int_a^b (f(x) - g(x))\,dx$ . Use top minus bottom.
In terms of $y$ : if $p(y) \ge q(y)$ on $[c, d]$ , the area is $\int_c^d (p(y) - q(y))\,dy$ . Use right minus left.
If the curves cross on the interval, split the integral at the crossing, or integrate the absolute value of the difference.

Curves that cross inside the interval

Find the area of the region between

y = x^2

and

y = x + 2

from

x = 0

x = 3

Setting $x^2 = x + 2$ gives $x = 2$ in this interval. The line is on top on $[0, 2]$ , and the parabola is on top on $[2, 3]$ .

y = x^2

y = x + 2

Area $= \int_0^2 (x + 2 - x^2)\,dx + \int_2^3 (x^2 - x - 2)\,dx = \frac{10}{3} + \frac{11}{6} = \frac{31}{6}$ .

Answer: $\frac{31}{6}$

Integrating with respect to y

Find the area of the region enclosed by

x = y^2 - 2y

and

x = y + 4

Set them equal: $y^2 - 3y - 4 = 0$ , so $y = -1$ and $y = 4$ . On that interval, the line is on the right.

Area $= \int_{-1}^4 \left(y + 4 - (y^2 - 2y)\right)dy = \int_{-1}^4 (-y^2 + 3y + 4)\,dy = \frac{125}{6}$ .

Answer: $\frac{125}{6}$

Volume by discs and washers

When a region spins around a line, each slice perpendicular to that line is a disc or a washer. These formulas assume the axis of rotation doesn't pass through the inside of the region. The AB course uses only discs and washers for these solids.

Disc: $V = \pi\int_a^b R^2\,dx$ , when the region touches the axis along its whole length.
Washer: $V = \pi\int_a^b (R^2 - r^2)\,dx$ , when there's a gap between the region and the axis.
$R$ is the distance from the axis to the outer edge, and $r$ is the distance from the axis to the inner edge.
Revolving around a vertical line? Write $R$ and $r$ in terms of $y$ and integrate with respect to $y$ .

A disc

The region under

y = \sqrt{2x + 1}

from

x = 0

x = 4

, above the

x

-axis, is revolved around the

x

-axis. Find the volume.

$R = \sqrt{2x + 1}$ , so $V = \pi\int_0^4 (2x + 1)\,dx = \pi\left[x^2 + x\right]_0^4 = 20\pi$ .

Answer: $20\pi$

A washer around a line that isn't an axis

The region between

y = 3x

and

y = x^2

on the interval

[0, 3]

is revolved around the line

y = -2

. Find the volume.

The line $y = 3x$ is on top on this interval. Distance from $y = -2$ to each curve is the height plus $2$ , so $R = 3x + 2$ and $r = x^2 + 2$ .

$V = \pi\int_0^3 \left((3x + 2)^2 - (x^2 + 2)^2\right)dx = \pi\int_0^3 (5x^2 + 12x - x^4)\,dx = \frac{252\pi}{5}$ .

Answer: $\frac{252\pi}{5}$

Revolving around the y-axis

The region bounded by

x = 2\sqrt{y}

, the

y

-axis, and the line

y = 4

is revolved around the

y

-axis. Find the volume.

Slices are horizontal, and each one is a disc with $R = 2\sqrt{y}$ . So $V = \pi\int_0^4 4y\,dy = \pi\left[2y^2\right]_0^4 = 32\pi$ .

Answer: $32\pi$

Volume with known cross sections

Here a base region sits flat, and each slice perpendicular to an axis is a shape you know. The volume is the integral of the area of a slice. There's no automatic $\pi$ out front, since $\pi$ appears only if the shape needs it.

Let $s$ be the length of the slice across the base, usually top minus bottom. The area of each slice is:

Square with side $s$ : $s^2$ .
Semicircle with diameter $s$ : $\frac{\pi}{8}s^2$ .
Equilateral triangle with side $s$ : $\frac{\sqrt{3}}{4}s^2$ .
Isosceles right triangle with a leg on the base: $\frac{1}{2}s^2$ .
Isosceles right triangle with the hypotenuse on the base: $\frac{1}{4}s^2$ .
Rectangle with height $h$ : $sh$ .

Semicircle cross sections

The base of a solid is the region bounded by

y = 4 - \frac{x^2}{4}

and the

x

-axis. Cross sections perpendicular to the

x

-axis are semicircles with diameters in the base. Find the volume.

The parabola meets the $x$ -axis at $x = \pm 4$ , and $s = 4 - \frac{x^2}{4}$ .

$V = \frac{\pi}{8}\int_{-4}^4 \left(4 - \frac{x^2}{4}\right)^2 dx = \frac{\pi}{8} \cdot \frac{1024}{15} = \frac{128\pi}{15}$ .

Answer: $\frac{128\pi}{15}$

One region, every setup

With a calculator

Let

R

be the region in the first quadrant bounded by

y = 2\cos\left(\frac{x}{2}\right) + 1

y = \frac{x + 1}{2}

, and the

y

-axis. Find: (a) the area of

R

; (b) the volume when

R

is revolved around the

x

-axis; (c) the volume when

R

is revolved around

y = 4

; (d) the volume of the solid with base

R

and square cross sections perpendicular to the

x

-axis.

Graph both curves and click their crossing, at $x \approx 2.41724$ . Call that value $c$ in Desmos. Let $T(x)$ be the cosine curve and $B(x)$ be the line. $T$ is on top on $[0, c]$ .

T(x) = 2\cos(x/2) + 1

B(x) = (x + 1)/2

(a) $\int_0^c (T(x) - B(x))\,dx \approx 3.488$ .
(b) $\pi\int_0^c (T(x)^2 - B(x)^2)\,dx \approx 40.263$ . The cosine curve is farther from the $x$ -axis, so it's the outer radius.
(c) $\pi\int_0^c \left((4 - B(x))^2 - (4 - T(x))^2\right)dx \approx 47.410$ . Now the line is farther from $y = 4$ , so the outer and inner radii switch.
(d) $\int_0^c (T(x) - B(x))^2\,dx \approx 6.345$ .

Answer: (a) $3.488$ . (b) $40.263$ . (c) $47.410$ . (d) $6.345$ .

On free response, write the integral with its limits before you give the value. The AP Calculus AB free-response guide and the Desmos guide explain why.

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Practice on exam-style questions

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Questions

How do I know whether to use discs or washers?: Use a disc if the region touches the axis of rotation along the whole interval, and a washer if there's a gap between them.
Do cross-section problems always have pi in them?: No. Pi appears only when the shape of the slice needs it, such as a semicircle.
When should I integrate with respect to y?: When the curves are given as x in terms of y, when the region's boundaries are easier to describe that way, or when you revolve around a vertical line with discs or washers.
Are cylindrical shells on the AP Calculus AB exam?: No. The AB course uses discs, washers, and cross sections for volume.