Center of Mass and Centroid: Formulas and Examples

What Is Center Of Mass And Centroid?

The center of mass and centroid are two versions of the same idea: the single point where an object would balance. The centroid is the geometric center of a shape and depends only on the shape. The center of mass is the balance point of the actual object, so it also depends on how the mass is spread through the shape.

Both are averages. The centroid averages position over area, and the center of mass averages position over mass, which makes it a weighted average. Calculus turns those averages into definite integrals.

For a region under the graph of $y = f(x)$, with $f$ continuous and $f(x) \geq 0$ on $[a, b]$, the centroid $(\bar{x}, \bar{y})$ is

$$\bar{x} = \frac{1}{A}\int_{a}^{b} x,f(x),dx, \qquad \bar{y} = \frac{1}{A}\int_{a}^{b} \frac{1}{2}\left[f(x)\right]^2 dx, \qquad A = \int_{a}^{b} f(x),dx$$

• $A$ is the area of the region.

• $\bar{x}$ is the average $x$-position of the area, and $\bar{y}$ is the average height.

• $x,f(x),dx$ is the moment of a thin vertical strip about the $y$-axis, and $\frac{1}{2}[f(x)]^2,dx$ is its moment about the $x$-axis.

The numerators are the moments $M_y$ and $M_x$. The moments page builds those from point masses on a seesaw, so this page starts where it stops: rods, regions between curves, uneven density, symmetry and Pappus's theorem.

Is The Centroid The Same As The Center Of Mass?

The centroid and the center of mass are the same point only when the density is uniform. Cut a shape from one sheet of cardboard and the two coincide. Glue a coin to one corner and the center of mass slides toward the coin, while the centroid stays exactly where it was, because the shape has not changed.

Where Do The Centroid Formulas Come From?

The centroid formulas come from slicing the region into thin vertical strips and averaging their positions. A strip at position $x$ with width $dx$ has height $f(x)$ and area $f(x),dx$.

1. Horizontal position. The whole strip sits at distance $x$ from the $y$-axis, so its moment about that axis is $x \cdot f(x),dx$.

2. Vertical position. The strip is a thin rectangle, so its own balance point is halfway up, at height $\frac{f(x)}{2}$. Its moment about the $x$-axis is $\frac{f(x)}{2} \cdot f(x),dx = \frac{1}{2}[f(x)]^2,dx$.

3. Average. Add the moments of all strips (integrate) and divide by the total area. That is the same move as averaging test scores weighted by how many students got each one.

The half in $\bar{y}$ is the part students most often drop, and it is not a decoration. It comes straight from step 2: a strip's weight acts at its middle, not at its top.

How Do You Find The Center Of Mass Of A Rod?

You find the center of mass of a thin rod by weighting each point by the rod's density there. Lay the rod along the $x$-axis from $x = a$ to $x = b$ with linear density $\rho(x)$ in kilograms per metre. A short piece of length $dx$ has mass $\rho(x),dx$, so

$$m = \int_{a}^{b} \rho(x),dx, \qquad \bar{x} = \frac{1}{m}\int_{a}^{b} x,\rho(x),dx$$

When $\rho$ is constant, $\rho$ cancels and $\bar{x}$ lands at the midpoint. When the rod gets heavier toward one end, $\bar{x}$ shifts toward that end, which Example 1 shows.

How Do You Find The Centroid Of A Region Between Two Curves?

You find the centroid of a region between two curves by using the difference of the curves as each strip's height. If $f(x) \geq g(x)$ on $[a, b]$, with $f$ on top and $g$ underneath, then the area between two curves is $A = \int_{a}^{b} \left[f(x) - g(x)\right] dx$ and

$$\bar{x} = \frac{1}{A}\int_{a}^{b} x\left[f(x) - g(x)\right] dx, \qquad \bar{y} = \frac{1}{A}\int_{a}^{b} \frac{1}{2}\left(\left[f(x)\right]^2 - \left[g(x)\right]^2\right) dx$$

The $\bar{y}$ numerator subtracts the squares, $f^2 - g^2$, not the square of the difference, $(f - g)^2$. Here is why. A strip running from $g$ up to $f$ has its middle at $\frac{f + g}{2}$, and its moment is $\frac{f + g}{2}(f - g),dx = \frac{1}{2}(f^2 - g^2),dx$.

Find the limits first by solving $f(x) = g(x)$. When the curves are easier to write as $x$ in terms of $y$ (sideways parabolas, for instance), use horizontal strips instead and swap the roles of $x$ and $y$ in every formula.

How Do You Find The Center Of Mass Of A Lamina With Variable Density?

A lamina is a thin flat plate. When its density changes only with $x$, written $\delta(x)$, every strip still has a single density, and you weight each strip's area by it:

$$m = \int_{a}^{b} \delta(x)\left[f(x) - g(x)\right] dx$$

$$M_y = \int_{a}^{b} x,\delta(x)\left[f(x) - g(x)\right] dx, \qquad M_x = \int_{a}^{b} \delta(x),\frac{1}{2}\left(\left[f(x)\right]^2 - \left[g(x)\right]^2\right) dx$$

$$\bar{x} = \frac{M_y}{m}, \qquad \bar{y} = \frac{M_x}{m}$$

Divide by the mass $m$ here, not by the area. A lamina whose density changes with both $x$ and $y$ needs a double integral, which belongs to multivariable calculus.

How Can You Use Symmetry To Find A Centroid Faster?

The symmetry principle says that if a uniform region has a line of symmetry, its centroid lies on that line. A shape symmetric about the $y$-axis has $\bar{x} = 0$ with no integration at all, and a shape with two lines of symmetry has its centroid where they cross.

One habit saves the most marks here. Before integrating, sketch the region and mark a rough guess for the centroid: toward the wide or tall part, on any line of symmetry, inside the region if it is convex. If the computed answer lands far from the guess, an integral has gone wrong.

For composite shapes, add moments, not centroids. Find each piece's area and centroid, multiply to get each piece's moment, add the moments, and divide by the total area.

These standard centroids come up often enough to know by sight:

The triangle row matches the vertex-average formula on the centroid of a triangle page. The last row is a curve rather than a region, so its centroid sits on the arc's own average position, farther out than the region's.

What Is The Theorem Of Pappus?

The Theorem of Pappus says that when a plane region of area $A$ is revolved about a line in its plane that does not cross the region's interior, the solid it sweeps out has volume

$$V = 2\pi,\bar{d},A,$$

where $\bar{d}$ is the distance from the region's centroid to the axis. In words: volume equals area times the distance the centroid travels. A companion version gives surface area: a curve of length $L$ revolved the same way sweeps a surface of area $S = 2\pi,\bar{d},L$, with $\bar{d}$ now the distance from the curve's centroid to the axis.

A quick check builds trust in both. A semicircular region of radius $r$ spun about its diameter makes a sphere, and Pappus gives $\frac{\pi r^2}{2} \cdot 2\pi \cdot \frac{4r}{3\pi} = \frac{4}{3}\pi r^3$. The semicircular arc spun the same way gives $\pi r \cdot 2\pi \cdot \frac{2r}{\pi} = 4\pi r^2$, the sphere's surface area.

Pappus is a shortcut for many volumes of solids of revolution, especially rings and doughnut shapes, where the disk or shell setup gets messy and the centroid is already known.

Examples Of Center Of Mass And Centroid

Example 1

A 4 m rod lies along the x-axis from x = 0 to x = 4 with density ρ(x) = 1 + x kilograms per metre. Find its mass and center of mass.

$$m = \int_{0}^{4} (1 + x),dx = \left[x + \frac{x^2}{2}\right]_{0}^{4} = 12,\text{kg}$$

$$M = \int_{0}^{4} x(1 + x),dx = \left[\frac{x^2}{2} + \frac{x^3}{3}\right]_{0}^{4} = 8 + \frac{64}{3} = \frac{88}{3}$$

$$\bar{x} = \frac{88/3}{12} = \frac{22}{9}$$

$$\bar{x} \approx 2.4444,\text{m}$$

The rod is heavier on the right, so its balance point sits right of the midpoint at 2 m.

Final answer: $m = 12,\text{kg}$ and $\bar{x} = \frac{22}{9} \approx 2.4444,\text{m}$.

Example 2

Find the centroid of the region under y = x³ between the x-axis and x = 2.

$$A = \int_{0}^{2} x^3,dx = \left[\frac{x^4}{4}\right]_{0}^{2} = 4$$

$$\bar{x} = \frac{1}{4}\int_{0}^{2} x \cdot x^3,dx = \frac{1}{4}\left[\frac{x^5}{5}\right]_{0}^{2} = \frac{1}{4} \cdot \frac{32}{5} = \frac{8}{5}$$

For $\bar{y}$, the tempting move is to treat it like $\bar{x}$ and integrate the strip's height times its "position", writing $\frac{1}{A}\int_{0}^{2} \left[f(x)\right]^2 dx$:

$$\frac{1}{4}\int_{0}^{2} x^6,dx = \frac{1}{4} \cdot \frac{128}{7} = \frac{32}{7} \approx 4.5714$$

Try that same formula on a plain rectangle of height $h$. It gives $\frac{1}{bh}\int_{0}^{b} h^2,dx = h$, the top edge of the rectangle, but a rectangle balances halfway up. The formula places each strip's weight at its top instead of its middle, so every answer comes out twice too high.

Rescue: each strip's balance point sits at height $\frac{f(x)}{2}$, which puts the one-half back.

$$\bar{y} = \frac{1}{4}\int_{0}^{2} \frac{1}{2}\left(x^3\right)^2 dx$$

$$= \frac{1}{8}\left[\frac{x^7}{7}\right]_{0}^{2}$$

$$= \frac{1}{8} \cdot \frac{128}{7} = \frac{16}{7}$$

$$\approx 2.2857$$

Final answer: $(\bar{x}, \bar{y}) = \left(\frac{8}{5}, \frac{16}{7}\right) \approx (1.6, 2.2857)$, toward the tall right edge as the sketch suggests.

Example 3

Find the centroid of the region bounded by y = 4x − x² and y = x.

Set the curves equal: $4x - x^2 = x$ gives $x(3 - x) = 0$, so $x = 0$ and $x = 3$. On $[0, 3]$ the parabola is on top, and $f - g = 3x - x^2$.

$$A = \int_{0}^{3} \left(3x - x^2\right)dx = \frac{27}{2} - 9 = \frac{9}{2}$$

$$\bar{x} = \frac{2}{9}\int_{0}^{3} x\left(3x - x^2\right)dx = \frac{2}{9}\left(27 - \frac{81}{4}\right) = \frac{2}{9} \cdot \frac{27}{4} = \frac{3}{2}$$

$$\bar{y} = \frac{2}{9}\int_{0}^{3} \frac{1}{2}\left[\left(4x - x^2\right)^2 - x^2\right]dx$$

$$= \frac{2}{9} \cdot \frac{54}{5}$$

$$= \frac{12}{5}$$

The middle integral works out as $\frac{1}{2}\int_{0}^{3}\left(x^4 - 8x^3 + 15x^2\right)dx = \frac{1}{2}\left(\frac{243}{5} - 162 + 135\right) = \frac{54}{5}$.

Final answer: $(\bar{x}, \bar{y}) = \left(\frac{3}{2}, \frac{12}{5}\right) = (1.5, 2.4)$.

Example 4

Find the centroid of a semicircular region of radius 3 m with its diameter on the x-axis.

The region is symmetric about the $y$-axis, so $\bar{x} = 0$ with no integral. The top boundary is $f(x) = \sqrt{9 - x^2}$, and the area is $\frac{9\pi}{2}$.

$$\bar{y} = \frac{2}{9\pi}\int_{-3}^{3} \frac{1}{2}\left(9 - x^2\right)dx$$

$$= \frac{1}{9\pi}\left[9x - \frac{x^3}{3}\right]_{-3}^{3}$$

$$= \frac{1}{9\pi} \cdot 36 = \frac{4}{\pi}$$

$$\approx 1.2732,\text{m}$$

This matches the table entry $\frac{4r}{3\pi}$ with $r = 3$ for a semicircle.

Final answer: $(\bar{x}, \bar{y}) = \left(0, \frac{4}{\pi}\right) \approx (0, 1.2732)$ m.

Example 5

A rectangular plate covers 0 ≤ x ≤ 3 and 0 ≤ y ≤ 2 (in metres). Its density is δ(x) = 1 + x kilograms per square metre. Find its center of mass and compare it with the centroid.

Each strip has height 2, so its area is $2,dx$ and its mass is $2(1 + x),dx$.

$$m = \int_{0}^{3} 2(1 + x),dx = 15,\text{kg}$$

$$M_y = \int_{0}^{3} 2x(1 + x),dx = 9 + 18 = 27$$

$$M_x = \int_{0}^{3} (1 + x) \cdot \frac{1}{2}\left(2^2\right)dx = 15$$

$$\bar{x} = \frac{27}{15} = \frac{9}{5}, \qquad \bar{y} = \frac{15}{15} = 1$$

The centroid of the rectangle is $(1.5, 1)$. The center of mass moves right to $(1.8, 1)$ because the plate is denser on the right, and stays at height 1 because the density does not change with $y$.

Final answer: center of mass $\left(\frac{9}{5}, 1\right)$; centroid $\left(\frac{3}{2}, 1\right)$.

Example 6

The region from Example 3 is revolved about the vertical line x = −1. Find the volume of the solid using the Theorem of Pappus, then check it with shells.

The centroid sits at $\bar{x} = \frac{3}{2}$, so its distance from the axis is $\bar{d} = \frac{3}{2} + 1 = \frac{5}{2}$. The axis lies outside the region, so Pappus applies.

$$V = 2\pi \cdot \frac{5}{2} \cdot \frac{9}{2}$$

$$= \frac{45\pi}{2}$$

$$\approx 70.6858$$

Check with the cylindrical shell method: a shell at $x$ has radius $x + 1$ and height $3x - x^2$.

$$V = \int_{0}^{3} 2\pi(x + 1)\left(3x - x^2\right)dx = 2\pi \cdot \frac{45}{4} = \frac{45\pi}{2}$$

Final answer: $V = \frac{45\pi}{2} \approx 70.6858$ cubic units, by both methods.

Why Does The Balance Point Of A Shape Matter?

It matters because one point can stand in for the whole object. Gravity pulls on every grain of a beam, a ladder or an aircraft, but for balance and tipping you can treat the entire weight as acting at the center of mass. That swap turns a problem about millions of particles into a problem about one point.

The centroid earns its place in three distinct jobs:

• Balance and tipping: an object stays upright while the vertical line through its center of mass falls inside its base, so designers move heavy parts low and central.

• Bending: a uniform beam bends about an axis through the centroid of its cross-section, so the centroid decides where the material does the most good.

• Volume and area shortcuts: through Pappus's theorem, a known centroid gives the volume of a revolved solid in one line.

Physics courses lean on the same simplification for moving objects, treating a car, a football or a falling cat as if all its mass sat at one point, as the OpenStax University Physics section on center of mass shows.

Where Are Center Of Mass And Centroid Used In Real Life?

Wherever weight, balance or bending matters, someone has computed a center of mass or a centroid:

• Aircraft loading: before flight, crews check that the loaded center of gravity (the point where the total weight acts, equal to the center of mass in uniform gravity) sits inside the aircraft's certified limits.

• Vehicle design: a lower center of mass makes a car or truck harder to roll over in a sharp turn.

• Structural engineering: beam and column sizes depend on the centroid of the cross-section, the axis about which bending happens.

• Shipbuilding: naval architects track a ship's center of gravity against its center of buoyancy, the centroid of the displaced water, to keep it stable.

• Robotics and animation: walking robots and game characters stay upright by keeping the center of mass over the feet.

• Manufacturing: Pappus's theorem gives the volume of rings, gaskets and doughnut-shaped tanks from a cross-section and its centroid.

Common Errors When Working With Center Of Mass And Centroid

Mistake 1: Dropping The One-Half In The Y-Bar Formula

Where it slips in: Computing $\bar{y}$ right after $\bar{x}$, when the pattern of $\bar{x}$ is still in your head.

Don't do this: Write $\bar{y} = \frac{1}{A}\int f^2,dx$. In Example 2 it gives $\frac{32}{7}$, double the true value.

The correct way: Remember that each strip balances at its middle: $\bar{y} = \frac{1}{A}\int \frac{1}{2}f^2,dx$.

The memorizer who learned "$\bar{x}$ is $x$ times $f$" tends to guess "$\bar{y}$ is $f$ times $f$" by analogy. Deriving the half once from a single strip, as in the formulas section, fixes it for good.

Mistake 2: Squaring The Difference Instead Of Subtracting The Squares

Where it slips in: Regions between two curves, like Example 3.

Don't do this: Use $\frac{1}{2}(f - g)^2$ for the strip's moment. In Example 3 that gives $\bar{y} = \frac{9}{10}$, a point below the line $y = x$ at $x = 1.5$ and outside the region.

The correct way: Use $\frac{1}{2}\left(f^2 - g^2\right)$, which comes from the strip's middle height $\frac{f + g}{2}$ times its length $f - g$.

The rusher writes $(f - g)^2$ because the area integral used $f - g$. The quickest check is the sketch: a centroid that lands outside a convex region is always wrong.

Mistake 3: Dividing By Area When The Density Varies

Where it slips in: Laminas and rods whose density is a function, like Examples 1 and 5.

Don't do this: Compute the density-weighted moments, then divide by the area. In Example 5, $\frac{27}{6} = 4.5$ puts the center of mass outside the plate.

The correct way: Divide every moment by the total mass, $m = \int \delta \cdot (\text{strip area})$.

Students who first find the centroid, then say out loud whether the density pulls the answer left or right, catch this error immediately, because 4.5 m lies off a plate that ends at 3 m.

Mistake 4: Measuring Pappus Distances To The Wrong Point

Where it slips in: Pappus problems where the axis sits away from the region, as in Example 6.

Don't do this: Use the distance from the axis to the region's nearest edge or to the origin. Using $\bar{x} = \frac{3}{2}$ as the distance in Example 6 gives $\frac{27\pi}{2}$, the volume about the $y$-axis instead.

The correct way: Measure from the axis to the centroid: $\bar{d} = \frac{3}{2} + 1 = \frac{5}{2}$.

Getting a center of gravity wrong has cost more than marks. In January 2003, Air Midwest Flight 5481 lost pitch control on takeoff from Charlotte; the NTSB accident report found incorrect elevator rigging compounded by a center of gravity "substantially aft" of the certified limit, with the FAA's average weight assumptions in weight and balance guidance at the time listed as a contributing factor. Wrong masses fed into a weighted average give a wrong balance point, exactly as in Mistake 3.

Practice Problems On Center Of Mass And Centroid

1. Find the centroid of the region under $y = 9 - x^2$ above the $x$-axis.

2. A 2 m rod has density $\rho(x) = 3x^2$ kilograms per metre for $0 \leq x \leq 2$. Find its center of mass.

3. Find the centroid of a quarter-disc of radius 6 in the first quadrant.

4. Sketch the region bounded by $y = x^2$ and $y = 2x$, then find the centroid of its area.

5. A disc of radius 1, centered at $(3, 0)$, is revolved about the $y$-axis. Use Pappus's theorem to find the volume of the ring it sweeps out.

Answer to Problem 1: By symmetry $\bar{x} = 0$; with $A = 36$, $\bar{y} = \frac{1}{36}\int_{-3}^{3} \frac{1}{2}\left(9 - x^2\right)^2 dx = \frac{18}{5} = 3.6$.

Answer to Problem 2: $\bar{x} = \frac{\int_{0}^{2} 3x^3,dx}{\int_{0}^{2} 3x^2,dx} = \frac{12}{8} = \frac{3}{2},\text{m}$.

Answer to Problem 3: $\left(\frac{8}{\pi}, \frac{8}{\pi}\right) \approx (2.5465, 2.5465)$.

Answer to Problem 4: $A = \frac{4}{3}$, $\bar{x} = 1$, $\bar{y} = \frac{8}{5}$.

Answer to Problem 5: $V = 2\pi \cdot 3 \cdot \pi(1)^2 = 6\pi^2 \approx 59.2176$ cubic units.

Key Takeaways

• The center of mass and centroid are balance points, and they coincide only when density is uniform.

• For a region under $y = f(x)$, $\bar{x} = \frac{1}{A}\int x f,dx$ and $\bar{y} = \frac{1}{A}\int \frac{1}{2}f^2,dx$, with the half coming from each strip balancing at its middle.

• Between two curves, the $\bar{y}$ moment uses $f^2 - g^2$, never $(f - g)^2$.

• With variable density, every moment is divided by the mass, not the area.

• Symmetry gives one coordinate for free, and Pappus's theorem turns a known centroid into a volume.

Where To Go From Here

Work through Practice Problems 1 to 5; if Problem 4 trips you up, return to the section on regions between two curves and check that you subtracted the squares. Want a live Bhanzu trainer to walk through more centroid setups with you? Book a free demo class.

Read More

• Applications of integration

• Washer method

• Arc length of a curve

• Calculus in engineering

• Calculus in physics