Centroid
The centroid of a geometric object is the point that corresponds to the arithmetic mean position of all the points in the shape, serving as its geometric center or balance point. For objects with uniform density, such as a planar lamina or solid body, the centroid coincides with the center of mass, representing the point where the object would balance if suspended.[1] [2] In two-dimensional geometry, the centroid (\bar{x}, \bar{y}) of a plane region R with area A is calculated using the formulas \bar{x} = \frac{1}{A} \iint_R x \, dA and \bar{y} = \frac{1}{A} \iint_R y \, dA, which integrate the first moments of the area about the coordinate axes.[3][4] For specific shapes like triangles, the centroid is the intersection point of the three medians (lines from each vertex to the midpoint of the opposite side) and can be found as the average of the vertices' coordinates: if the vertices are (x_1, y_1), (x_2, y_2), and (x_3, y_3), then \bar{x} = \frac{x_1 + x_2 + x_3}{3} and \bar{y} = \frac{y_1 + y_2 + y_3}{3}.[5] This point divides each median in a 2:1 ratio, with the longer segment (two-thirds of the median length) from the vertex to the centroid.[6] For three-dimensional solids with uniform density and volume V, the centroid (\bar{x}, \bar{y}, \bar{z}) extends this concept via \bar{x} = \frac{1}{V} \iiint_E x \, dV, and similarly for \bar{y} and \bar{z}, where E is the solid region.[2] These properties make the centroid essential in applications such as statics, where it determines balance and stability, and in computational geometry for shape analysis. The concept traces back to ancient geometry, particularly in the study of triangle centers by Greek mathematicians, though its integral formulations developed with the advent of calculus.[7]Definition
Geometric interpretation
The centroid of a geometric figure represents the arithmetic mean position of all the points comprising the figure, serving as its balance point under the assumption of uniform density across the shape. In two dimensions, this is conceptualized as the "center of area," the point about which the figure would balance if constructed from a thin, uniform sheet of material. For three-dimensional objects, it is the "center of volume," the analogous balance point for a solid of uniform density.[1] A particularly clear illustration of the centroid occurs in the case of a triangle, where it coincides with the average of the coordinates of its three vertices. Visually, the centroid is the unique point of intersection of the triangle's three medians—each median being the line segment joining a vertex to the midpoint of the opposite side. This concurrency arises geometrically because the medians divide the triangle into six smaller triangles of equal area, ensuring their common intersection balances the overall figure; the centroid divides each median in the ratio 2:1, with the longer portion directed toward the vertex.[5][8] To distinguish the centroid from other notable triangle centers, consider that the circumcenter is the intersection of the perpendicular bisectors of the sides and serves as the center of the circle passing through all three vertices, while the incenter is the intersection of the angle bisectors and the center of the circle tangent to all three sides. In contrast, the centroid emphasizes the uniform averaging of positional data via the medians. For clarity, one may visualize a triangle with its medians drawn, highlighting their convergence at the centroid, distinct from the locations of the circumcenter and incenter.[9][10] For figures of uniform density, the geometric centroid aligns with the physical center of mass, providing a foundational link to mechanics without considering variable mass distributions.[1]Relation to center of mass
The centroid of a geometric figure coincides with the center of mass of a physical object of uniform density that occupies the same figure.[11] For a continuous body with position-dependent density \rho(\mathbf{r}), the center of mass \mathbf{G} is defined as \mathbf{G} = \frac{1}{M} \iiint_V \mathbf{r} \, \rho(\mathbf{r}) \, dV, where M = \iiint_V \rho(\mathbf{r}) \, dV is the total mass and the integral extends over the volume V.[4] When \rho is constant (uniform density), M = \rho V and the formula simplifies to the centroid \mathbf{G} = \frac{1}{V} \iiint_V \mathbf{r} \, dV, bridging the geometric average position with physical mass distribution.[4] The center of mass represents the point where an object achieves torque equilibrium under its own weight, assuming uniform gravitational acceleration. At this point, the total torque due to gravity vanishes because the first moment of the mass distribution about \mathbf{G} is zero: for a discrete system of particles, \sum_i m_i (\mathbf{r}_i - \mathbf{G}) = \mathbf{0}, with the continuous analog \iiint_V (\mathbf{r} - \mathbf{G}) \rho(\mathbf{r}) \, dV = \mathbf{0}.[12] This property ensures the object balances perfectly when supported at \mathbf{G}, as the gravitational forces produce no net rotation.[13] In cases of non-uniform density, the centroid retains its geometric definition independent of material properties, while the center of mass shifts toward denser regions. For instance, consider a rod with uniform shape but varying density—increasing toward one end; the centroid lies at the geometric midpoint, but the center of mass moves closer to the heavier end.[14] For computation in Cartesian coordinates (x, y, z), the center of mass components separate as G_x = \frac{1}{M} \iiint_V x \, \rho \, dV, \quad G_y = \frac{1}{M} \iiint_V y \, \rho \, dV, \quad G_z = \frac{1}{M} \iiint_V z \, \rho \, dV, with analogous expressions for planar or linear cases by reducing the dimensionality.[15]Properties
Invariance under affine transformations
The centroid of a finite set of points in Euclidean space exhibits invariance under affine transformations, meaning that applying an affine map to the points results in the transformed centroid being the image of the original centroid under the same map. Consider an affine transformation T(\mathbf{x}) = A\mathbf{x} + \mathbf{b}, where A is an invertible linear transformation and \mathbf{b} is a translation vector. If G = \frac{1}{n} \sum_{i=1}^n \mathbf{x}_i is the centroid of points \{\mathbf{x}_1, \dots, \mathbf{x}_n\}, then the centroid G' of the transformed points \{T(\mathbf{x}_1), \dots, T(\mathbf{x}_n)\} satisfies G' = T(G) = AG + \mathbf{b}. This property arises because the centroid is defined as the arithmetic mean, and affine transformations preserve affine combinations, including the uniform average. To see this explicitly, substitute the definition: T(G) = A \left( \frac{1}{n} \sum_{i=1}^n \mathbf{x}_i \right) + \mathbf{b} = \frac{1}{n} \sum_{i=1}^n (A \mathbf{x}_i + \mathbf{b}) = \frac{1}{n} \sum_{i=1}^n T(\mathbf{x}_i) = G'. This linearity ensures the average property is maintained, making the centroid a natural representative point robust to shearing, scaling, rotation, and translation. For similarity transformations, which are affine maps composed of uniform scaling, rotation, and translation, the centroid transforms predictably while preserving relative positions. In particular, under a pure scaling by factor k > 0 centered at the origin (i.e., T(\mathbf{x}) = k \mathbf{x}), the centroid scales accordingly: if G is the original centroid, then G' = k G. For example, scaling a set of points forming an equilateral triangle with centroid at (0,0) by k=2 yields a new centroid at (0,0), while the side lengths double, demonstrating how the centroid anchors the shape's "center" invariantly under proportional enlargement. This behavior extends to general similarities, where G' = k R G + \mathbf{b} for rotation matrix R, underscoring the centroid's role in shape analysis under geometric distortions.[16] The centroid also relates closely to barycentric coordinates, where it serves as the barycenter (weighted average) of the points with equal weights \frac{1}{n} for each of the n vertices of a simplex or point set. In barycentric terms, the coordinates of the centroid with respect to the points are \left( \frac{1}{n}, \frac{1}{n}, \dots, \frac{1}{n} \right), reflecting its position as the balance point under uniform mass distribution. This equal-weight formulation highlights the centroid's affine invariance, as barycentric coordinates are inherently preserved under such maps. Furthermore, the centroid is the unique point in the space that minimizes the sum of squared Euclidean distances to all points in the set. For points \{\mathbf{x}_1, \dots, \mathbf{x}_n\}, the objective function f(\mathbf{y}) = \sum_{i=1}^n \|\mathbf{y} - \mathbf{x}_i\|^2 achieves its minimum at \mathbf{y} = G, with the value f(G) = \sum_{i=1}^n \|\mathbf{x}_i\|^2 - n \|G\|^2. This uniqueness follows from the strict convexity of the quadratic function, whose Hessian is n I (positive definite for n \geq 1), ensuring a single global minimum.[17]Composition for composite shapes
The centroid of a composite shape, formed by combining multiple simpler geometric components, is calculated as the weighted average of the individual centroids, where the weights are the areas of the components for two-dimensional figures or the volumes for three-dimensional solids. This approach leverages the additivity of first moments, ensuring the overall balance point reflects the contributions of each part proportionally to its size. For a two-dimensional composite area, the coordinates of the centroid (\bar{x}, \bar{y}) are given by \bar{x} = \frac{\sum_{i=1}^{n} A_i \bar{x}_i}{\sum_{i=1}^{n} A_i}, \quad \bar{y} = \frac{\sum_{i=1}^{n} A_i \bar{y}_i}{\sum_{i=1}^{n} A_i}, where A_i is the area of the i-th component and (\bar{x}_i, \bar{y}_i) is the centroid of that component.[18] Similarly, for a three-dimensional composite solid, the centroid coordinates are \bar{x} = \frac{\sum_{i=1}^{n} V_i \bar{x}_i}{\sum_{i=1}^{n} V_i}, \quad \bar{y} = \frac{\sum_{i=1}^{n} V_i \bar{y}_i}{\sum_{i=1}^{n} V_i}, \quad \bar{z} = \frac{\sum_{i=1}^{n} V_i \bar{z}_i}{\sum_{i=1}^{n} V_i}, with V_i denoting the volume of the i-th component.[19] These formulas assume uniform density across the shape; if density varies, the weighting should use mass instead of area or volume, calculated as m_i = \rho_i A_i or m_i = \rho_i V_i.[20] To apply this method, follow a systematic process: first, decompose the composite shape into non-overlapping simpler parts whose centroids and measures (areas or volumes) can be readily determined, such as rectangles, triangles, or cylinders. Next, establish a common coordinate system and compute the centroid coordinates and measure for each part relative to this system. Then, sum the products of each part's measure and its centroid coordinates, and divide by the total measure to obtain the overall centroid. This decomposition simplifies analysis for irregular shapes that lack closed-form expressions.[21] Consider an L-shaped lamina in the xy-plane, formed by combining two rectangles of uniform thickness: a horizontal rectangle from x=0 to x=4 units and y=0 to y=2 units (area A_1 = 8 square units, centroid at (\bar{x}_1 = 2, \bar{y}_1 = 1)), and a vertical rectangle from x=2 to x=4 units and y=2 to y=6 units (area A_2 = 8 square units, centroid at (\bar{x}_2 = 3, \bar{y}_2 = 4)). The total area is \sum A_i = 16 square units. The x-coordinate of the composite centroid is \bar{x} = \frac{8 \cdot 2 + 8 \cdot 3}{16} = 2.5 units, and the y-coordinate is \bar{y} = \frac{8 \cdot 1 + 8 \cdot 4}{16} = 2.5 units. Note that the overlapping region where the rectangles join must be excluded in the decomposition to avoid double-counting area; in this configuration, the vertical rectangle starts at y=2, avoiding overlap with the horizontal part to form the intended L-shape.[22] Inaccurate results can arise from inconsistent density assumptions; for instance, treating a non-uniform material as uniform shifts the centroid toward denser regions, potentially leading to errors in stability analyses or load distribution calculations. Always verify the decomposition covers the entire shape without gaps or overlaps, and use consistent units for all measures to maintain precision.[18]Computation for Discrete Cases
Finite set of points
The centroid of a finite set of n points in Euclidean space is the arithmetic mean of their position vectors.[23] For points with coordinates (x_i, y_i, z_i) where i = 1, 2, \dots, n, the centroid \mathbf{G} = (G_x, G_y, G_z) is given by G_x = \frac{1}{n} \sum_{i=1}^n x_i, \quad G_y = \frac{1}{n} \sum_{i=1}^n y_i, \quad G_z = \frac{1}{n} \sum_{i=1}^n z_i. In vector notation, this is expressed as \mathbf{G} = \frac{1}{n} \sum_{i=1}^n \mathbf{r}_i, where \mathbf{r}_i is the position vector of the i-th point.[24] This formulation derives from the geometric property that the centroid minimizes the sum of squared Euclidean distances from the points to the candidate location, specifically solving \min_{\mathbf{r}} \sum_{i=1}^n \| \mathbf{r} - \mathbf{r}_i \|^2.[23] As an illustrative example, the vertices of a unit square located at (0,0), (1,0), (1,1), and (0,1) have centroid coordinates (0.5, 0.5).[25] The same averaging principle applies in n-dimensional space, where each coordinate of the centroid is the arithmetic mean of the corresponding coordinates across all points.[26]Weighted points
In the case of a finite set of points where each point has an associated weight or mass m_i rather than uniform weighting, the centroid, also known as the center of mass, is computed as a weighted average of the position vectors \vec{r}_i. This extends the unweighted discrete case by accounting for varying influences from each point.[27] The position of the weighted centroid \vec{G} in two or three dimensions is given by the formula \vec{G} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i}, where the sums are taken over all points, and \sum_i m_i = M is the total mass or weight. This expression represents the balance point of the system under gravitational forces, equivalent to concentrating the entire mass M at \vec{G}.[27]/02%3A_Applications_of_Integration/2.03%3A_Centre_of_Mass_and_Torque) This formula derives from the principle of torque balance in static equilibrium. Consider the system supported at a potential pivot point \vec{G}; for balance, the total torque due to the weights m_i g (where g is gravitational acceleration) about \vec{G} must be zero. The torque from each particle is m_i g (\vec{r}_i - \vec{G}) \times \hat{k} (in the plane, assuming vertical forces), leading to \sum_i m_i (\vec{r}_i - \vec{G}) = 0. Solving for \vec{G} yields the weighted average formula above./02%3A_Applications_of_Integration/2.03%3A_Centre_of_Mass_and_Torque)[28] For example, consider three points in the plane: one at (0,0) with mass 1, one at (1,0) with mass 2, and one at (0,1) with mass 3. The total mass is 6. The x-coordinate is (1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0)/6 = 2/6 = 1/3, and the y-coordinate is (1 \cdot 0 + 2 \cdot 0 + 3 \cdot 1)/6 = 3/6 = 1/2, so \vec{G} = (1/3, 1/2). Such weighted centroids find application in determining molecular centers of mass, where atomic masses serve as weights, and in statistics as the multivariate weighted mean for data analysis.[23]Computation for Continuous Cases
Integral formulas for plane figures
For a plane figure defined by a bounded region D in the xy-plane with uniform density, the centroid G = (G_x, G_y) is given by the integral formulas G_x = \frac{1}{A} \iint_D x \, dA, \quad G_y = \frac{1}{A} \iint_D y \, dA, where A = \iint_D dA is the area of the region.[4][29] These formulas arise as the limiting case of the discrete summation for the centroid of a finite set of weighted points, where the summation \sum x_i \Delta A_i / \sum \Delta A_i is replaced by the continuous integral as the partition of D into subregions of area \Delta A_i becomes infinitely fine.[30][31] This derivation aligns with the physical interpretation of moment balance, where the first moments about the axes divided by the total area yield the coordinates that ensure equilibrium.[4] In polar coordinates, which are particularly useful for regions exhibiting circular symmetry, the formulas adapt by substituting x = r \cos \theta, y = r \sin \theta, and dA = r \, dr \, d\theta, yielding G_x = \frac{1}{A} \iint_D r^2 \cos \theta \, dr \, d\theta, \quad G_y = \frac{1}{A} \iint_D r^2 \sin \theta \, dr \, d\theta, with the area A = \iint_D r \, dr \, d\theta.[15] The integrals are well-defined for bounded regions D with piecewise smooth boundaries, ensuring the existence of the double integrals over the domain.[4]Integral formulas for solids
The centroid of a three-dimensional solid is determined using volume integrals that extend the principles applied to two-dimensional figures. For a solid with volume V, the coordinates of the centroid (\bar{x}, \bar{y}, \bar{z}) are given by the first moments of the volume divided by the total volume. Specifically, \bar{x} = \frac{1}{V} \iiint_V x \, dV, \quad \bar{y} = \frac{1}{V} \iiint_V y \, dV, \quad \bar{z} = \frac{1}{V} \iiint_V z \, dV, where the volume is V = \iiint_V dV.[32][33] These formulas arise from considering the solid as composed of infinitesimal volume elements dV, analogous to the double integrals used for plane figures, but now requiring triple integrals over the three-dimensional region. The derivation parallels the two-dimensional case by balancing the moments about the coordinate planes, ensuring the centroid represents the average position of the volume elements.[32][13] Symmetry in the solid's geometry simplifies these calculations significantly. For instance, if the solid exhibits axial symmetry about the z-axis, such as a solid hemisphere of radius R with its flat base in the xy-plane, the integrals for \bar{x} and \bar{y} evaluate to zero due to the odd symmetry of the integrands x and y over the symmetric domain, leaving only \bar{z} nonzero along the axis of symmetry. In this case, \bar{z} = \frac{3R}{8}.[14][34] For solids with complex boundaries, such as cylinders or spheres, alternative coordinate systems facilitate evaluation of the triple integrals. Cylindrical coordinates (r, \theta, z) are particularly useful for solids of revolution around the z-axis, where the volume element becomes dV = r \, dr \, d\theta \, dz, simplifying expressions involving radial symmetry. Similarly, spherical coordinates (\rho, \theta, \phi) with dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi are advantageous for spherical or conical solids, aligning the integration limits with the natural geometry.[35][36][37]Specific Geometric Shapes
Triangles and polygons
The centroid of a triangle, assuming uniform density, is located at the average of its vertex coordinates. For a triangle with vertices at (x_1, y_1), (x_2, y_2), and (x_3, y_3), the coordinates of the centroid G are given by G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right). [38] This point also coincides with the intersection of the triangle's medians, where each median—a line segment from a vertex to the midpoint of the opposite side—is divided in the ratio 2:1, with the longer segment (two-thirds of the median's length) extending from the vertex to the centroid.[8] For a general polygon, the centroid can be computed by decomposing the shape into non-overlapping triangles and applying the composite centroid formula for areas. This involves calculating the area and centroid of each triangular component, then finding the weighted average based on their areas. The formula for the x-coordinate (and similarly for y) is \bar{x} = \frac{\sum (A_i \bar{x}_i)}{\sum A_i}, where A_i is the area of the i-th triangle and \bar{x}_i is the x-coordinate of its centroid.[21] To decompose, select a vertex and connect it to all non-adjacent vertices, forming n-2 triangles for an n-sided polygon, ensuring no overlaps and coverage of the entire area. Consider a convex quadrilateral with vertices A(0,0), B(4,0), C(3,3), and D(1,3), ordered counterclockwise. Decompose it into two triangles: ABC and ACD.- For triangle ABC: Vertices (0,0), (4,0), (3,3). Centroid \bar{x}_{ABC} = (0+4+3)/3 = 7/3, \bar{y}_{ABC} = (0+0+3)/3 = 1. Area A_{ABC} = (1/2)| (0(0-3) + 4(3-0) + 3(0-0)) | = 6.
- For triangle ACD: Vertices (0,0), (3,3), (1,3). Centroid \bar{x}_{ACD} = (0+3+1)/3 = 4/3, \bar{y}_{ACD} = (0+3+3)/3 = 2. Area A_{ACD} = (1/2)| (0(3-3) + 3(3-0) + 1(0-3)) | = 3.