AP Physics Geometry Formulas: The Printed Sheet Box

All four AP Physics booklets print the same geometry box: fourteen formulas covering the areas of a rectangle, triangle and circle, circumference, arc length, the volume and surface area of a rectangular solid, cylinder and sphere, and the four right-triangle relations.

The GEOMETRY AND TRIGONOMETRY box is printed in all four AP Physics Course and Exam Descriptions effective Fall 2024, identically: Physics 1 appendix page 211, Physics 2 page 220, Physics C: Mechanics page 206, Physics C: Electricity and Magnetism page 181.

The Box as the Booklet Prints It

The box is titled GEOMETRY AND TRIGONOMETRY and it sits at the foot of the equation pages in the reference booklet. It holds fourteen formulas, arranged as three shapes in two dimensions, three solids, and one right triangle.

Shape	Printed formulas
Rectangle	$A = bh$
Triangle	$A = \frac{1}{2}bh$
Circle	$A = \pi r^2$ , $C = 2\pi r$ , $s = r\theta$
Rectangular Solid	$V = \ell wh$
Cylinder	$V = \pi r^2 \ell$ , $S = 2\pi r\ell + 2\pi r^2$
Sphere	$V = \frac{4}{3}\pi r^3$ , $S = 4\pi r^2$
Right Triangle	$a^2 + b^2 = c^2$ , $\sin\theta = \frac{a}{c}$ , $\cos\theta = \frac{b}{c}$ , $\tan\theta = \frac{a}{b}$

Count them and you get fourteen: one for the rectangle, one for the triangle, three for the circle, one for the rectangular solid, two for the cylinder, two for the sphere and four for the right triangle.

The box also carries two figures. One is a dashed circle with a radius $r$ , an arc $s$ and the angle $\theta$ between them, which is the picture for $s = r\theta$ . The other is a right triangle with $\theta$ at the left vertex, the right angle at the lower right, base $b$ , vertical side $a$ and hypotenuse $c$ . That labelling is what fixes $\sin\theta = a/c$ rather than $b/c$ .

Where it is printed:

AP Physics 1: Algebra-Based, appendix page 211
AP Physics 2: Algebra-Based, appendix page 220
AP Physics C: Mechanics, appendix page 206
AP Physics C: Electricity and Magnetism, appendix page 181

The four printings are identical. This box does not change between the algebra-based and the calculus-based courses, which surprises people, because the Physics C booklets do add separate boxes for vectors, calculus and identities that Physics 1 and Physics 2 never see.

The Symbol Key Printed Alongside It

The middle of the box is a legend, and it is worth reading once because two of the letters are not the ones you would guess.

Symbol	Meaning
$A$	area
$b$	base
$C$	circumference
$h$	height
$\ell$	length
$r$	radius
$s$	arc length
$S$	surface area
$V$	volume
$w$	width
$\theta$	angle

Eleven entries. The two that catch people:

$S$ is surface area, not distance and not entropy. Everywhere else in physics $s$ tends to mean a displacement, and in this box the lowercase $s$ does mean arc length while the capital $S$ means surface area. Two different quantities, one letter, distinguished only by case.

The cylinder uses $\ell$ for its length, not $h$ for its height. So the printed volume is $V = \pi r^2 \ell$ . If you have memorised $\pi r^2 h$ you are not wrong, but the sheet will not confirm it for you, and $\ell$ is also the letter the sheet uses for the length of a current-carrying wire in $F_B = I\ell B\sin\theta$ and for the length of a resistor in $R = \rho\ell/A$ . The reuse is deliberate: a wire is a cylinder.

What Each Formula Is Actually There For

This is the part no formula sheet tells you. Every one of the fourteen entries is there because a specific AP result needs it, and knowing which result turns a lookup into a shortcut.

Printed formula	The AP problem it is printed for
$A = bh$	Area under a flat graph. A constant-velocity segment of a $v$ against $t$ graph gives displacement; a constant-force segment of an $F$ against $x$ graph gives work; a horizontal segment of a PV diagram gives $W = -P\Delta V$ .
$A = \frac{1}{2}bh$	Area under a straight sloping graph. The triangle under a linear $v$ against $t$ graph, and the triangle under $F = k\Delta x$ that integrates to $U_s = \frac{1}{2}k(\Delta x)^2$ .
$A = \pi r^2$	Flux through a circular loop in $\Phi_B = \vec{B} \cdot \vec{A}$ ; the cross-sectional area in $R = \rho\ell/A$ and in the continuity equation $A_1v_1 = A_2v_2$ ; the area in $P = F_{\perp}/A$ .
$C = 2\pi r$	The path length in Ampere's law, where $\oint \vec{B} \cdot d\vec{\ell}$ around a circle of radius $r$ becomes $B(2\pi r)$ . Also the distance covered per revolution in orbital and circular-motion problems.
$s = r\theta$	The bridge between linear and rotational motion, and the parent of $\Delta x_{cm} = r\Delta\theta$ for rolling without slipping, which is itself printed on the mechanics sheet.
$V = \ell wh$	Density $\rho = m/V$ for a rectangular block, and the container volume in $PV = nRT$ .
$V = \pi r^2 \ell$	The mass of a cylindrical rod from its density, which is where a rotational-inertia integral starts. Also the volume of fluid in a length of pipe.
$S = 2\pi r\ell + 2\pi r^2$	The Gaussian cylinder. The curved side, area $2\pi r\ell$ , carries all the flux from a line or cylinder of charge, and the two end caps, area $2\pi r^2$ together, carry none.
$V = \frac{4}{3}\pi r^3$	Uniform density in a sphere. It is what makes the enclosed charge inside a uniformly charged sphere scale as $r^3/R^3$ , and it converts a planet's radius and density into its mass.
$S = 4\pi r^2$	The Gaussian sphere. $E(4\pi r^2) = q_{enc}/\varepsilon_0$ collapses straight back into Coulomb's law, and the same $4\pi r^2$ is why every point-source field falls off as $1/r^2$ .
$a^2 + b^2 = c^2$	The magnitude of a vector from perpendicular components, and the resultant of two perpendicular forces.
$\sin\theta = \frac{a}{c}$	The opposite component: $mg\sin\theta$ down an incline, $v_0\sin\theta$ for a launch, the $\sin\theta$ in $\tau = rF\sin\theta$ .
$\cos\theta = \frac{b}{c}$	The adjacent component: $mg\cos\theta$ into a ramp, the $\cos\theta$ in $W = Fd\cos\theta$ .
$\tan\theta = \frac{a}{b}$	Recovering an angle from two components. Also the angle of repose, where $\tan\theta = \mu_s$ , and the ideal banking angle, where $\tan\theta = v^2/(rg)$ .

The two surface areas are the ones worth internalising. If you can see immediately that a Gaussian sphere contributes $4\pi r^2$ and a Gaussian cylinder contributes $2\pi r\ell$ , most of Gauss's law is arithmetic.

Arc Length, and the Radian Trap

The relation $s = r\theta$ is printed in the geometry box, under Circle, alongside the area and the circumference. It is not printed in the mechanics equations, and it is not in the rotational-kinematics group, which is where students go looking for it.

That placement matters, because $s = r\theta$ is the origin of the whole linear-to-rotational dictionary. Differentiate it once with respect to time and you get $v = r\omega$ ; differentiate again and you get $a_T = r\alpha$ . Both of those derived relations are printed in the mechanics equations, and their parent is sitting in the geometry box two boxes away.

The trap: $s = r\theta$ requires $\theta$ in radians. The booklet does not say so. Nothing in the box states a unit for $\theta$ , and the trig values table on the same page lists its angles in degrees only, which makes the omission worse. Put 120 into $s = r\theta$ instead of 2.09 and your arc length is too big by a factor of $180/\pi$ , roughly 57.

So does $v = r\omega$ , which is why angular velocity on an AP exam is in radians per second and never in degrees per second or revolutions per second without conversion. If a problem hands you revolutions per minute, the first move is to convert, because the sheet's relations assume radians throughout.

Reading the Right-Triangle Figure Correctly

The three trig ratios in the box are stated against a specific picture, and reading them off a different picture is how a sign or a sine becomes a cosine.

In the printed figure the angle $\theta$ sits at the left vertex. The right angle is at the lower right. The base along the bottom is labelled $b$ , the vertical side on the right is labelled $a$ , and the hypotenuse running from $\theta$ up to the top is labelled $c$ . Against that figure:

$\sin\theta = a/c$ , opposite over hypotenuse, where $a$ is the side across the triangle from $\theta$ .
$\cos\theta = b/c$ , adjacent over hypotenuse, where $b$ is the side touching $\theta$ .
$\tan\theta = a/b$ , opposite over adjacent.

The physics version of that: the component you get depends on where you measured the angle from, not on whether the component is horizontal or vertical. An incline problem measures $\theta$ from the horizontal and the down-slope weight component is $mg\sin\theta$ . A projectile launch measures $\theta$ from the horizontal and the vertical velocity is $v_0\sin\theta$ . But a problem that gives you the angle from the vertical, which some rope and pendulum questions do, flips both. Draw the triangle, mark the angle, and read the ratio off the printed figure rather than off memory.

The seven angles you are most likely to be handed are tabulated on the trig values page, which is the box printed directly above this one in the booklet.

Areas Under Graphs, Which Is Why the Rectangle Is There

It looks odd that a physics reference sheet needs to tell you the area of a rectangle. It is there because AP Physics asks you to read areas off graphs constantly, and almost every such area decomposes into rectangles and triangles.

Graph	The area means	Shape you decompose it into
Velocity against time	Displacement	Rectangles under constant segments, triangles under sloping ones
Acceleration against time	Change in velocity	Same decomposition
Force against position	Work done	Rectangle for a constant force, triangle for a spring
Force against time	Impulse, so change in momentum	Rectangle for a constant force, triangle for a sharp collision spike
Pressure against volume	Work done on the gas, with a minus sign	Rectangle for an isobaric step, and the area is zero for an isochoric one
Current against time	Charge transferred	Rectangle for steady current

The spring case is the neatest. The force needed to stretch a spring is $F = k\Delta x$ , a straight line through the origin. The area under it from 0 to $\Delta x$ is a triangle of base $\Delta x$ and height $k\Delta x$ , so the area is $\frac{1}{2}(\Delta x)(k\Delta x) = \frac{1}{2}k(\Delta x)^2$ , which is exactly the elastic potential energy printed on every one of the four sheets. The geometry box and the mechanics box are telling you the same thing twice.

In the calculus-based courses that decomposition becomes an integral, and the relevant rules are on the calculus formulas page. The geometry still works, and on a multiple-choice question it is faster.

What the Box Does Not Give You

Knowing the boundary saves you from hunting during an exam. None of the following is printed in the geometry box, or anywhere else in any of the four booklets:

Any moment of inertia. No $\frac{1}{2}MR^2$ for a disc, no $\frac{2}{5}MR^2$ for a sphere, no $\frac{1}{12}ML^2$ for a rod. The mechanics sheets print the definitions $I = \sum m_i r_i^2$ and, for Physics C, $I = \int r^2\,dm$ , plus the parallel-axis theorem $I' = I_{cm} + Md^2$ , and nothing else. Physics C candidates are expected to do the integral or be given the result in the stem.
The cone. No volume, no surface area, no slant height.
The trapezoid. Which matters, because the area under a trapezoidal $v$ against $t$ segment is a common question. Split it into a rectangle plus a triangle, both of which are printed.
The ellipse. No area and no eccentricity, even though orbits are elliptical and Kepler's laws are on the C: Mechanics syllabus.
The law of sines or the law of cosines. Vector addition on the AP exam is done by components, not by solving oblique triangles.
Solid angle, and no steradian anywhere.
Any circle-segment or circle-sector area. The arc length $s = r\theta$ is printed; the sector area $\frac{1}{2}r^2\theta$ is not.

One caution about checking this yourself: a site or a revision guide that lists what is on the sheet often works from a transcription of the physics equations only, and those transcriptions routinely leave the geometry, trigonometry, vector, calculus and identity boxes out. Absence from a transcription is not absence from the booklet. Every formula on this page was read off a rendered image of the appendix page named above.

The Gaussian cylinder, where the surface-area formula pays off

An infinite line of charge carries a uniform linear charge density of $\lambda = 2.0 \times 10^{-6}$ C/m. Use a coaxial Gaussian cylinder of radius $r = 0.10$ m and length $\ell = 0.50$ m to find the electric field magnitude at that radius. Take $\varepsilon_0 = 8.85 \times 10^{-12}$ C $^2$ /(N m $^2$ ).

Split the printed cylinder surface area into its two printed pieces. The curved side has area $2\pi r\ell = 2\pi(0.10)(0.50) = 0.3142$ m $^2$ . The two end caps have combined area $2\pi r^2 = 2\pi(0.10)^2 = 0.0628$ m $^2$ .
By symmetry the field from an infinite line points radially outward, so it is parallel to the end caps and passes through none of them. The caps contribute zero flux, and only the $2\pi r\ell$ term survives. That is why the sheet prints the surface area as a sum rather than a single expression.
Charge enclosed by the cylinder: $q_{enc} = \lambda\ell = (2.0 \times 10^{-6})(0.50) = 1.0 \times 10^{-6}$ C.
Apply Gauss's law with $E$ constant over the curved side: $E(2\pi r\ell) = q_{enc}/\varepsilon_0$ , so $E = \frac{1.0 \times 10^{-6}}{(8.85 \times 10^{-12})(0.3142)} = \frac{1.0 \times 10^{-6}}{2.780 \times 10^{-12}} = 3.6 \times 10^5$ N/C.
Check it against the standard result $E = 2k\lambda/r$ with $k = 9.0 \times 10^9$ : $E = 2(9.0 \times 10^9)(2.0 \times 10^{-6})/0.10 = 3.6 \times 10^4/0.10 = 3.6 \times 10^5$ N/C. The two agree.
Notice the length cancelled. $q_{enc}$ carried a factor of $\ell$ and so did the area, which is why the answer does not depend on how long a cylinder you chose.

$E = 3.6 \times 10^5$ N/C, directed radially away from the line. The whole calculation ran on the printed $S = 2\pi r\ell + 2\pi r^2$ , with the second term contributing nothing.

Arc length in radians, and what degrees would have cost you

A wheel of radius 0.30 m rolls without slipping and turns through 120°. How far along the ground does its centre travel?

The printed relation is $s = r\theta$ , and rolling without slipping makes the centre's displacement equal that arc length, which the mechanics sheet states directly as $\Delta x_{cm} = r\Delta\theta$ .
Convert the angle to radians, because $s = r\theta$ assumes them: $\theta = 120° \times \frac{\pi}{180°} = \frac{2\pi}{3} = 2.0944$ rad.
Substitute: $s = (0.30)(2.0944) = 0.6283$ m, so 0.63 m to two significant figures.
For contrast, feeding degrees straight in gives $s = (0.30)(120) = 36$ m, a wheel of radius 30 cm travelling 36 metres in a third of a turn. The error factor is $180/\pi \approx 57.3$ , and it is large enough that the absurdity is your own check.

0.63 m. The angle must be in radians, a requirement the geometry box does not state anywhere.

Frequently asked questions

Are geometry formulas on the AP Physics equation sheet?

Yes. Every AP Physics reference booklet prints a box titled GEOMETRY AND TRIGONOMETRY holding fourteen formulas: areas of a rectangle, triangle and circle, the circumference of a circle, arc length, volumes of a rectangular solid, cylinder and sphere, surface areas of a cylinder and sphere, and the four right-triangle relations. It appears on appendix page 211 in Physics 1, 220 in Physics 2, 206 in C: Mechanics and 181 in C: Electricity and Magnetism, and the four printings are identical.

Is s = r theta on the AP Physics formula sheet?

Yes. It is printed in the GEOMETRY AND TRIGONOMETRY box under the heading Circle, alongside the area and circumference, in all four courses. It is not printed among the rotational-kinematics equations, which is where most students look for it. The angle must be in radians, a requirement the booklet does not state.

Is the surface area of a sphere on the AP Physics sheet?

Yes, S = 4 pi r squared is printed in the geometry box in all four courses, along with the sphere volume V = four thirds pi r cubed. It is there for Gauss's law: a Gaussian sphere of radius r has area 4 pi r squared, so E times 4 pi r squared equals the enclosed charge over epsilon zero, which rearranges into Coulomb's law.

Does the AP Physics sheet give moments of inertia for common shapes?

No. No rotational inertia for a disc, rod, hoop or sphere appears anywhere in any of the four booklets. The mechanics sheets print only the definition, I equals the sum of m times r squared, and for Physics C the integral form I equals the integral of r squared dm, plus the parallel-axis theorem. Any specific shape's rotational inertia will either be given in the question stem or must be derived.

Is the volume of a cone on the AP Physics formula sheet?

No. The geometry box prints only three solids: the rectangular solid, the cylinder and the sphere. There is no cone, no pyramid, no trapezoid area and no ellipse area in any of the four booklets. If a question needs the area under a trapezoidal graph, split it into a rectangle plus a triangle, both of which are printed.

Why does the AP Physics sheet write the cylinder volume with an l instead of an h?

The printed form is V equals pi r squared times script l, using the same letter the sheet uses for length elsewhere, such as the length of a current-carrying wire and the length of a resistor. The legend in the box defines script l as length and h as height. Both readings give the same volume, but the sheet will only confirm the script l version.

Is the geometry box different in AP Physics C than in AP Physics 1?

No. The GEOMETRY AND TRIGONOMETRY box is identical in all four courses. What differs is that the two Physics C booklets add three boxes the algebra-based booklets never print, covering vectors, calculus rules and a short list of identities. Those three sit directly beneath the geometry box on the same appendix page.