Study Companion: S18 — Centroids of Lines, Areas and Volumes

S17 located the centre of mass from \bar{\mathbf r} = \int\mathbf r\,dm/m. When the material is uniform, the density is the same factor top and bottom and cancels, and what is left depends only on the shape: the centroid. This lesson computes centroids of the three kinds of shape a body can be modelled as, a line, an area and a volume. The physics is one sentence, the principle of moments; the craft, as Meriam says, is in choosing the element and setting up the integral, and that is what most of the lesson is about.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 5/3 ("Centroids of Lines, Areas, and Volumes"), with its Key Concepts box on the choice of element and Sample Problems 5/1–5/5. Marghitu, Dupac and Madsen, Statics with MATLAB, section 3.4 ("First Moment of an Area"), its table of common areas (Fig. 3.6) and the examples of section 3.7 on curves and plane regions (Examples 3.1–3.8). Axes: x to the right, y up, z toward the reader unless a figure says otherwise; Marghitu's centroid C and Meriam's C agree.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A plastic protractor is a semicircular plate of radius 100\,\mathrm{mm}. Bend a wire into a semicircle of the same radius. Each will balance on a knife edge laid parallel to its straight side, at some height above that side.

How high is each balance line? The plausible guess is "the same for both, about half the radius, 50\,\mathrm{mm}: they are the same semicircle". The answers are in Nugget 03 and Nugget 05.

Nugget 01: With Uniform Density, Only the Shape Is Left

Claim. When a body's density is uniform, its centre of mass is a property of its geometry alone, the centroid of a line, an area or a volume. A slender rod is modelled as a line, a thin plate or shell as an area, and anything else as a volume. The centroid need not lie on the line or surface.

The picture. A wire bent into an L, and a thin curved shell.

Both centroids are in the air: the wire's inside the corner of the L, the shell's on its axis below the shell.

Check with numbers. The L-wire's legs are 0.3\,\mathrm m along x and 0.4\,\mathrm m along y from the corner, total length 0.7\,\mathrm m; each leg's centroid is at its midpoint. With uniform cross-section the mass of each leg is \rho A times its length, and \rho A cancels:

\bar x = \frac{0.3(0.15) + 0.4(0)}{0.7} = 0.0643\,\mathrm m, \qquad \bar y = \frac{0.3(0) + 0.4(0.2)}{0.7} = 0.1143\,\mathrm m,

a point 64\,\mathrm{mm} from one leg and 114\,\mathrm{mm} from the other, on neither. A 5\,\mathrm{mm} steel plate cut in any outline has \rho t = 7850(0.005) = 39.25\,\mathrm{kg/m^2} everywhere, and that factor too cancels from its centroid.

The idea. Meriam (5/3): when \rho is uniform it cancels from the numerators and denominators of Eq. 5/3, and the remaining expressions describe a purely geometric property. The word centroid is used for a geometric shape, and centre of mass for a physical body; they coincide when the density is uniform, and in general not otherwise. Three categories:

Lines. A slender rod or wire of length L, cross-section A, density \rho: dm = \rho A\,dL. With \rho and A constant, $\bar x = \frac{\int x\,dL}{L}, \quad \bar y = \frac{\int y\,dL}{L}, \quad \bar z = \frac{\int z\,dL}{L} \qquad (5/4)In generalC$ is not on the line; a rod in one plane needs only two coordinates.
Areas. A plate or shell of small constant thickness t: dm = \rho t\,dA, and $\bar x = \frac{\int x\,dA}{A}, \quad \bar y = \frac{\int y\,dA}{A}, \quad \bar z = \frac{\int z\,dA}{A} \qquad (5/5)The numerators are the **first moments of area** (second moments come in S24). A curved shell needs all three coordinates and itsC$ is in general not on the surface; a flat area needs only the two in its plane.
Volumes. dm = \rho\,dV, \rho cancels if constant: $\bar x = \frac{\int x\,dV}{V}, \quad \bar y = \frac{\int y\,dV}{V}, \quad \bar z = \frac{\int z\,dV}{V} \qquad (5/6)$

Marghitu (3.3) writes the same three cases with \rho_l = m/L and \rho_s = m/A (Eqs. 3.5–3.6).

The equation (definition of the centroid, from S17's centre of mass with \rho cancelled):

\bar{\mathbf r} = \frac{\int\mathbf r\,dL}{L}, \qquad \bar{\mathbf r} = \frac{\int\mathbf r\,dA}{A}, \qquad \bar{\mathbf r} = \frac{\int\mathbf r\,dV}{V}

Assumptions: uniform density; for lines, constant cross-section; for areas, constant small thickness.

Real example (one you can see). A coat hanger made of wire hangs from its hook with its centroid straight below, in the empty triangle, not on the wire.

Plausible wrong answer. "The centroid of a wire is somewhere along the wire." For the L it is 64\,\mathrm{mm} and 114\,\mathrm{mm} from the two legs.

Check. A plate's density doubles across half of it. Is its centre of mass its centroid? (No: the density no longer cancels; use S17's Eq. 5/3.)

Narration spine. A steel L-bracket; its density label fades, its thickness label fades, and only the outline remains. The same happens to a wire L and a curved shell. A dot labelled C appears for each, in the empty space.

Bridge. For an area there is a second way to say where the centroid is: the first moment about any axis through it is zero.

Nugget 02: About an Axis Through the Centroid, the First Moment Is Zero

Claim. The first moment of an area about the x-axis is M_x = \int y\,dA, about the y-axis M_y = \int x\,dA; the centroid is \bar x = M_y/A, \bar y = M_x/A. About any axis through the centroid, the first moment is zero. An axis of symmetry is such an axis, so it contains the centroid.

The picture. Marghitu's Example 3.6, the triangle under y = bx/a, with axes through its centroid.

Across the dashed axis x' the strips of area above and below have moments that cancel; so do the strips left and right of y'. That cancelling is what makes C the centroid.

Check with numbers. Marghitu integrates with the second-order element dA = dx\,dy, inner integral over y from 0 to bx/a:

A = \int_0^a\frac{bx}{a}\,dx = \frac{ab}{2}, \quad M_y = \int_0^a x\frac{bx}{a}\,dx = \frac{ba^2}{3}, \quad M_x = \int_0^a\frac12\Big(\frac{bx}{a}\Big)^2dx = \frac{b^2a}{6},

so \bar x = M_y/A = 2a/3 and \bar y = M_x/A = b/3; with a = b = 1\,\mathrm m, C = (0.667, 0.333)\,\mathrm m. About the vertical axis through C: \int(x - 2a/3)\,dA = M_y - \frac{2a}3A = \frac{ba^2}{3} - \frac{2a}{3}\cdot\frac{ab}2 = 0.

The idea. Marghitu (3.4) defines the first moments of a plane area, M_x = \int_A y\,dA (3.9) and M_y = \int_A x\,dA (3.10), and the centroid x_C = M_y/A, y_C = M_x/A (3.11–3.12). Note the naming: M_x is the moment about the x-axis, so its arm is y. The first moment carries information on the area's shape, size and orientation; the centroid's location does not depend on the axes used. Axes with their origin at the centroid are centroidal axes, and the first moments about them are zero. An area with an axis of symmetry has its centroid on it, since the first moment about it is zero; with two orthogonal axes of symmetry the centroid is at their intersection. Circles, rectangles and triangles can be done by inspection. Marghitu's Fig. 3.6 tabulates the common shapes, as Meriam's Tables D/3 and D/4 do (triangle h/3; semicircle 4r/3\pi; quarter circle 4r/3\pi from each edge; semiparabola 3a/8, 3h/5; parabolic spandrel 3a/4, 3h/10; semi-ellipse 4b/3\pi).

The equation (definition of the first moments; mathematical identity for a centroidal axis):

M_x = \int_A y\,dA = A\bar y, \qquad M_y = \int_A x\,dA = A\bar x, \qquad \int_A (x - \bar x)\,dA = 0

Worked number. A rectangular cover plate 0.6\,\mathrm m \times 0.4\,\mathrm m has two axes of symmetry: its centroid is at its centre, (0.3, 0.2)\,\mathrm m from a corner, with M_x = 0.24(0.2) = 0.048\,\mathrm{m^3} about the bottom edge, and zero about the centre lines.

Real example. The neutral axis of a beam in bending passes through the centroid of its cross-section, the axis about which the first moment of the section is zero; that is where a beam designer starts (S24 builds on it).

Plausible wrong answer. "M_x uses x as the arm." It is the moment about the x-axis, and the distance from the x-axis is y.

Check. An area has one axis of symmetry. How many centroid coordinates must still be computed? (One, along that axis.)

Narration spine. The triangle fills in, strip by strip. A dashed axis slides across it, the strips on each side turning blue and orange with moment bars that grow and shrink; where they balance, the axis locks: it passes through C. Then the horizontal axis does the same.

Bridge. For curved boundaries the coordinates matter. A circular arc is hopeless in x and y, and simple in angle.

Nugget 03: Polar Coordinates Turn a Circular Arc Into One Integral

Claim. Choose coordinates that match the boundary. For an arc of radius r and total angle 2\alpha, the element dL = r\,d\theta at x = r\cos\theta gives \bar x = r\sin\alpha/\alpha from the centre, along the axis of symmetry.

The picture. Meriam's Sample Problem 5/1.

The x-axis is placed on the axis of symmetry, so \bar y = 0 with no work. Each element of arc has the same length r\,d\theta; its moment arm about the y-axis is r\cos\theta.

Check with numbers. By the first of Eqs. 5/4, with L = 2\alpha r:

(2\alpha r)\bar x = \int_{-\alpha}^{\alpha}(r\cos\theta)\,r\,d\theta = 2r^2\sin\alpha \;\Rightarrow\; \bar x = \frac{r\sin\alpha}{\alpha}.

A semicircle, 2\alpha = \pi: \bar x = 2r/\pi = 0.637r. The puzzle's wire balances at 2(100)/\pi = 63.7\,\mathrm{mm} above its straight side. The quarter-circle arc, measured from either bounding radius, has the same 2r/\pi by symmetry.

The idea. Meriam (5/3, guideline 4): choose the coordinate system that best matches the boundaries, rectangular for straight edges, polar for circular ones; "it should be perfectly evident" that polar coordinates suit an arc. Marghitu (Example 3.2) reaches the same results parametrically: with x = r\cos t, y = r\sin t, dl = \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt = r\,dt; the full circle has L = 2\pi r, the semicircle (0, 2r/\pi), the quarter circle (2r/\pi, 2r/\pi).

The equation (geometric relationship: centroid of a circular arc):

\bar x = \frac{r\sin\alpha}{\alpha} \quad \text{from the centre, on the bisector}; \qquad \text{semicircle: } \frac{2r}{\pi}

Worked number. Meriam's Problem 5/7: a rod bent into an arc of radius a from 45^\circ to 165^\circ (not symmetric about either axis). With \theta from the x-axis and dL = a\,d\theta:

\bar x = \frac{a(\sin165^\circ - \sin45^\circ)}{2\pi/3} = -0.214a, \qquad \bar y = \frac{a(\cos45^\circ - \cos165^\circ)}{2\pi/3} = 0.799a.

Real example (one you can see). A curved handrail or an arched pipe bridge is hung for transport from a point above its centroid, which is inside the arch, 0.637r from the chord for a half circle.

Plausible wrong answer. "Half the radius." That ignores that the whole wire is at distance r from the centre: 2r/\pi is larger, 0.637r.

Check. As \alpha \to 0 the arc shrinks to a point at distance r. Does the formula agree? (\sin\alpha/\alpha \to 1, so \bar x \to r. Yes.)

Narration spine. A semicircular wire. A small orange element runs along it, its arm r\cos\theta shown as a dropping line; a running balance point slides out from the centre and settles at 0.637r, while a "50\,\mathrm{mm}" guess fades.

Bridge. The arc needed a first-order element and one integral. For areas the choice of element decides whether it takes one integral or three.

Nugget 04: Pick the Element That Covers the Shape in One Sweep

Claim. Prefer a first-order element (a strip, a slice) to a second- or third-order one, so a single integration covers the figure; prefer an element whose formula does not change partway across; drop higher-order terms. For a triangle a strip parallel to the base gives \bar y = h/3 in one line.

The picture. Meriam's Figs. 5/8a and 5/9: a strip that sweeps the whole triangle, and strips that change formula partway.

In (a) every horizontal strip has width proportional to its distance below the apex: one formula from bottom to top. In (b) a vertical strip's height grows up to x_1, then shrinks: two formulas, two integrals.

Check with numbers (Meriam's Sample Problem 5/2). Base b, altitude h, strip dA = x\,dy at height y, with x = b(h - y)/h by similar triangles. By the second of Eqs. 5/5, the arm of a horizontal strip is simply y:

\frac{bh}{2}\bar y = \int_0^h y\,\frac{b(h - y)}{h}\,dy = \frac{bh^2}{6} \;\Rightarrow\; \bar y = \frac h3.

The same holds measured from each side as a base, so the centroid is a third of the way up every altitude: the point where the medians meet. For the triangle of (b), base 0.6\,\mathrm m and apex 0.45\,\mathrm m above x = 0.2\,\mathrm m, the two-integral route gives A = 0.135\,\mathrm{m^2}, \bar x = 0.267\,\mathrm m and \bar y = 0.150\,\mathrm m: the mean of the vertices' coordinates and h/3, as the one-integral route would.

The idea. Meriam's Key Concepts box: the concept of moments is simple; the difficult steps are the element and the integral. Five guidelines; the first three are here.

Order of element. A first-order strip dA = l\,dy needs one integration; the second-order dx\,dy needs two. For a solid cone, a disc dV = \pi r^2\,dy beats a third-order dx\,dy\,dz.
Continuity. Choose an element that is integrated in one continuous operation: the horizontal strip of Fig. 5/8a over the vertical strip of Fig. 5/9, whose height formula changes at x = x_1.
Discarding higher-order terms. The vertical strip under a curve is dA = y\,dx; the small triangle \frac12dx\,dy on top is second order and dropped (Art. 1/7). In the limit there is no error.

Second view, reconciled. Marghitu sets up nearly every area as the double integral \int\int dx\,dy (Examples 3.6–3.8), the second-order element Meriam advises against for hand work. In MATLAB it costs nothing, and it is the same calculation: the inner integral, \int_{y_1}^{y_2}dy = y_2 - y_1, is the height of Meriam's vertical strip, and \int_{y_1}^{y_2}y\,dy = \frac12(y_2^2 - y_1^2) is that strip's moment about the x-axis.

The equation (geometric relationship: centroid of a triangle):

\bar y = \frac h3 \text{ above any side taken as base}; \qquad C = \text{intersection of the medians} = \frac{\mathbf r_1 + \mathbf r_2 + \mathbf r_3}{3}

Worked number. A triangular gusset plate with base 0.6\,\mathrm m and altitude 0.45\,\mathrm m: \bar y = 0.15\,\mathrm m above the base, whatever the position of the apex.

Real example. A triangular sail's area is pushed by the wind with its resultant through the sail's centroid, a third of the way up from the foot; sailmakers locate the "centre of effort" by drawing two medians.

Plausible wrong answer. "A triangle's centroid is halfway up." Half the area lies below a line at 0.293h from the base, and the area's moment puts C at h/3.

Check. Which strip would you choose for the area between y = x and y = x^2, and why? (Either: both boundaries run the whole way, so neither strip changes formula.)

Narration spine. A triangle; a horizontal strip sweeps up it, shrinking as it rises, and one integral sign writes itself. Medians draw in and cross at h/3. A second triangle with its apex off-centre: a vertical strip sweeps across, and at x_1 its top changes slope; a red marker splits the integral in two.

Bridge. The strip is chosen; now its moment arm. That is where most mistakes are made.

Nugget 05: The Moment Arm Is the Centroid of the Element

Claim. In \int x_c\,dA, the arm x_c is the coordinate of the element's own centroid, not the coordinate of its edge. A vertical strip of height y has its centroid at y/2; a horizontal strip from x to a at (a + x)/2; a thin triangle from the vertex at two-thirds of its length.

The picture. Meriam's Sample Problem 5/4: the area under x = ky^3, swept two ways.

For the \bar y of the vertical strip, the arm is y/2: the strip's moment about the x-axis is that of its own area acting at its middle. For the \bar x of the horizontal strip, the arm is (a + x)/2, the mean of its two ends.

Check with numbers. x = ky^3 with k = a/b^3, so y = b(x/a)^{1/3}; area A = \int_0^a y\,dx = 3ab/4.

Vertical strip, x_c = x: \frac{3ab}{4}\bar x = \int_0^a xy\,dx = \frac{3a^2b}{7}, so \bar x = \frac47a.
Vertical strip, y_c = y/2: \frac{3ab}{4}\bar y = \int_0^a\frac y2\,y\,dx = \frac{3ab^2}{10}, so \bar y = \frac25b.
Horizontal strip, x_c = (a + x)/2, y_c = y: \int_0^b\frac{a + x}{2}(a - x)\,dy and \int_0^b y(a - x)\,dy give the same \frac47a and \frac25b.

With a = 2\,\mathrm m, b = 1\,\mathrm m: C = (1.143, 0.400)\,\mathrm m. Using the edge y instead of y/2 would give \bar y = \frac45b, twice too high, outside the area's lower half where most of it lies.

The idea. Meriam's guideline 5: when a first- or second-order element is used, the moment arm must be the coordinate of the element's centroid; the subscript c is the reminder:

\bar x = \frac{\int x_c\,dA}{A}, \quad \bar y = \frac{\int y_c\,dA}{A}, \quad \bar z = \frac{\int z_c\,dA}{A} \quad (5/5a), \qquad \bar x = \frac{\int x_c\,dV}{V}, \ \ldots \quad (5/6a)

For a horizontal strip, y_c is in the limit the y of both its edges; x_c is not the x of either boundary. For the semicircular slice of a half cone (Fig. 5/12b) x_c is the slice's own centroid, 4\rho/3\pi from its straight edge, while z_c is the slice's z. Keep the principle of moments in view: the moment of the resultant equals the integral of the moments of the elements, each acting at its own centroid.

Sample Problem 5/3, the sector of radius r and half-angle \alpha, shows the choice is free once the arms are right. Solution I: rings of radius r_0, width dr_0, area dA = 2\alpha r_0\,dr_0, each an arc with x_c = r_0\sin\alpha/\alpha (SP 5/1 with r_0 for r):

(\alpha r^2)\bar x = \int_0^r\Big(\frac{r_0\sin\alpha}{\alpha}\Big)(2\alpha r_0\,dr_0) = \frac23r^3\sin\alpha \;\Rightarrow\; \bar x = \frac{2r\sin\alpha}{3\alpha}.

Solution II: thin triangles from the vertex, dA = \frac12r(r\,d\theta), each with its centroid at two-thirds of its length (SP 5/2), x_c = \frac23r\cos\theta: (\alpha r^2)\bar x = \int_{-\alpha}^{\alpha}\frac23r\cos\theta\cdot\frac12r^2\,d\theta = \frac23r^3\sin\alpha, the same. Meriam's hints: keep the variable r_0 distinct from the constant r, and do not use r_0 as the element's centroid. A second-order element r_0\,dr_0\,d\theta integrated first in \theta gives the ring, first in r_0 the triangle.

The equation (geometric relationship: centroid of a circular sector):

\bar x = \frac{2r\sin\alpha}{3\alpha} \text{ from the vertex}; \qquad \text{semicircle and quarter circle: } \frac{4r}{3\pi}

Worked number. The puzzle's protractor: a semicircular area, 2\alpha = \pi, \bar y = 4(100)/3\pi = 42.4\,\mathrm{mm} above its straight edge. The wire balances at 63.7\,\mathrm{mm}: 1.5 times higher, because all of the wire is at the full radius while much of the plate's area is near the centre.

Real example (one you can see). Balance a protractor on a pencil laid parallel to its straight edge: the balance line sits well below half the radius, a little over four tenths of it, and a wire semicircle balances visibly higher.

Plausible wrong answer. "For a vertical strip, \bar y uses the height y of the strip." The strip's area acts at its middle, y/2; using y doubles \bar y.

Check. For a horizontal strip in SP 5/4, what is y_c, and why does it need no halving? (It is y: the strip is thin in y, so its centroid is at its own height.)

Narration spine. The area under x = ky^3; a vertical strip lights and a dot drops to its middle, labelled y/2; a ghost dot at its top, labelled "edge", turns red and fades. A horizontal strip replaces it, its dot at the middle of its length. Then the protractor beside the wire: two balance lines at 42.4 and 63.7\,\mathrm{mm}.

Bridge. Volumes follow the same rules, with slices and shells in place of strips.

Nugget 06: Solids Slice Into Discs or Peel Into Shells

Claim. For a solid of revolution, a disc slice perpendicular to the axis or a thin cylindrical shell around it is a first-order element that needs one integration. For a hemisphere of radius r both give \bar y = 3r/8 above the base; for a cone, \frac34 of the height from the vertex.

The picture. Meriam's Sample Problem 5/5, in section.

A disc at height y has radius \sqrt{r^2 - y^2} and its centroid at y. A shell of radius z has height y = \sqrt{r^2 - z^2} and its centroid at half that height.

Check with numbers. Volume V = \frac23\pi r^3. Solution I (discs, dV = \pi(r^2 - y^2)\,dy, y_c = y):

\frac23\pi r^3\,\bar y = \int_0^r y\,\pi(r^2 - y^2)\,dy = \frac14\pi r^4 \;\Rightarrow\; \bar y = \frac38r.

Solution II (shells, dV = 2\pi z\,y\,dz, y_c = y/2): \frac23\pi r^3\bar y = \int_0^r\frac{\sqrt{r^2 - z^2}}{2}\,2\pi z\sqrt{r^2 - z^2}\,dz = \pi\int_0^r(r^2z - z^3)\,dz = \frac{\pi r^4}{4}, the same \frac38r. Solution III uses the angle \theta from 0 to \pi/2 as the variable: the radius is r\sin\theta, the disc thickness dy = r\sin\theta\,d\theta and the shell thickness dz = r\cos\theta\,d\theta, with the shell's length y = r\cos\theta. All three agree. A dome of radius 0.5\,\mathrm m has its centroid 0.1875\,\mathrm m above its base.

The idea. Meriam (5/3): Eqs. 5/6 and 5/6a with an element of simple shape; Solutions I and II are equally useful, since each needs one integration. For a solid cone (Fig. 5/8b) the disc dV = \pi r^2\,dy is the element; for the half cone (Fig. 5/12b) the semicircular slice, whose own centroid sets x_c. If the density is not constant but a function of position, it does not cancel, and the mass-centre form (Eq. 5/3) must be used instead.

The equation (geometric relationship: centroids of the hemisphere and the cone):

\text{hemisphere: } \bar y = \frac38r \text{ above the base}; \qquad \text{cone: } \bar z = \frac34h \text{ from the vertex} = \frac h4 \text{ above the base}

Worked number. A solid cone of height h and base radius r: discs of radius rz/h at distance z from the vertex, dV = \pi(rz/h)^2\,dz: \bar z = \int_0^h z\cdot\pi r^2z^2/h^2\,dz \big/ \int_0^h\pi r^2z^2/h^2\,dz = (h^4/4)/(h^3/3) = \frac34h. A 0.7\,\mathrm m solid cone's centroid is 0.175\,\mathrm m above its base. The half cone, with semicircular slices whose centroids are 4(rz/h)/3\pi from the cut face, has its centroid r/\pi from that face.

Real example (one you can see). A solid rubber traffic cone (or a pile of gravel, a cone of material) has its mass centre only a quarter of the way up, which is why it is hard to knock over.

Plausible wrong answer. "A cone's centroid is a third of the way up, like a triangle's." The triangle is the cone's section; the cone's discs are larger near the base, which pulls C down to a quarter.

Check. A paraboloid of revolution (Meriam's Problem 5/9) has disc area proportional to the distance z from its vertex. Where is its centroid? (\int z\cdot z\,dz/\int z\,dz = \frac23h from the vertex.)

Narration spine. A hemisphere in section; discs stack up from the base, each shrinking, a balance point rising and settling at 3r/8. Then shells peel from the axis outward, each with a dot at mid-height, and the balance point settles at the same mark.

Bridge. Some curves and regions have no neat element at all. Marghitu hands those to the computer.

Nugget 07: When the Integral Is Awkward, Let the Computer Evaluate It

Claim. The integrals of Eqs. 5/4–5/6 can be set up for any curve or region and evaluated symbolically or numerically. For a curve, the element of length is dl = \sqrt{1 + (dy/dx)^2}\,dx, or \sqrt{\dot x^2 + \dot y^2 + \dot z^2}\,dt for a parametric curve in space.

The picture. Marghitu's Example 3.1, a wire along y = 2x^{3/2}, and Example 3.8, the region between two curves.

The element of the curve is the hypotenuse of dx and dy. The region's boundaries cross at two points, which set the limits.

Check with numbers. Example 3.1: y = 2x^{3/2} for 0 \le x \le 1\,\mathrm m, dy/dx = 3x^{1/2}, dl = \sqrt{1 + 9x}\,dx:

L = \int_0^1\sqrt{1 + 9x}\,dx = 2.268\,\mathrm m, \qquad x_C = \frac{\int x\,dl}{L} = 0.575\,\mathrm m, \qquad y_C = \frac{\int y\,dl}{L} = 0.952\,\mathrm m,

a point off the curve, on its concave side. Example 3.8: the region between y_1 = x/4 and y_2 = \sqrt{2(x - 3)}, which cross at x_1 = 3.351\,\mathrm m and x_2 = 28.649\,\mathrm m: A = \int(y_2 - y_1)\,dx = 21.082\,\mathrm{m^2}, x_C = \int x(y_2 - y_1)\,dx/A = 14.000\,\mathrm m, y_C = \int\frac12(y_2^2 - y_1^2)\,dx/A = 4.000\,\mathrm m.

The idea. Marghitu (3.7) evaluates every case with MATLAB's int (symbolic) or numerically:

Plane curves (Ex. 3.1): dl = \sqrt{1 + (dy/dx)^2}\,dx.
Parametric curves (Ex. 3.2, 3.3): the quarter astroid x = a\cos^3t, y = a\sin^3t, a = 1\,\mathrm m, has dl = \frac32\lvert\sin2t\rvert\,dt, L = 1.5\,\mathrm m and C = (0.4, 0.4)\,\mathrm m.
Space curves (Ex. 3.4, 3.5): the helix x = a\cos t, y = a\sin t, z = ht with a = h = 1\,\mathrm m, over three turns (0 \le t \le 6\pi), has dl = \sqrt{a^2 + h^2}\,dt, L = 6\pi\sqrt2 = 26.657\,\mathrm m and C = (0, 0, 9.425)\,\mathrm m; the spiral x = 2e^t\cos t, y = 2e^t\sin t, z = 2e^t, 0 \le t \le 3, has L = 66.114\,\mathrm m and C = (-15.590, 10.778, 21.086)\,\mathrm m.
Plane regions (Ex. 3.6–3.8): \int\int dx\,dy; the region under y = 1.5\sin(0.75x) from 0 to 3\pi/(4k) has C = (1.854, 0.627)\,\mathrm m.

The computer does not choose the element or the arms; those are still the setup of Nuggets 04–05, and the results should be checked against symmetry and common sense (C inside the convex hull of the shape).

The equation (mathematical identity: arc length element):

dl = \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2}\,dx = \sqrt{\dot x^2 + \dot y^2 + \dot z^2}\,dt

Worked number. The helix's z_C = 9.425\,\mathrm m is exactly half its height 6\pi(1) = 18.85\,\mathrm m, as its uniform rise demands; x_C = y_C = 0 because three whole turns are symmetric about the axis.

Real example. Coil springs, cable runs and pipe routings in CAD are curves of this kind; the software reports their length and centroid by exactly this numerical integration.

Plausible wrong answer. "Use dx as the element of a curve." That measures the curve's shadow on the x-axis, not its length; for y = 2x^{3/2} the length is 2.268\,\mathrm m, not 1\,\mathrm m.

Check. A helix of two whole turns: what are x_C and y_C? (Zero: whole turns are symmetric about the axis.)

Narration spine. The curve y = 2x^{3/2} draws itself; a small right triangle dx, dy rides along it with its hypotenuse glowing; a counter for L runs to 2.268\,\mathrm m and a dot settles off the curve. Then the region between two curves fills with vertical strips, and its centroid lands at (14, 4).

Bridge. Every shape so far was one piece. Real parts are made of several.

Synthesis

The protractor balances 42.4\,\mathrm{mm} above its straight edge and the wire 63.7\,\mathrm{mm}, not both at 50\,\mathrm{mm}. Each answer took one integral: the wire's element was an arc r\,d\theta, all at the full radius; the plate's was a ring or a thin triangle, whose own centroid set the arm. The same semicircle has two centroids, because a line and an area are different shapes.

That is the content of the whole lesson. Uniform density cancels and leaves a shape; the shape's centroid comes from the principle of moments; the work is in the element, chosen to be first order, continuous and in coordinates that fit, with its own centroid as the arm. Marghitu's first moment says the same thing from the other side: the centroid is where the first moment of the shape about any axis through it vanishes. A centroid is the point about which a shape's first moments vanish; find it with one well-chosen element and its own centroid as the moment arm.

Central idea: with uniform density, centre of mass = centroid of the line, area or volume.
Key relationship: \bar x = \int x_c\,dA/A (Eq. 5/5a), and its line and volume forms.
Most important assumption: uniform density (and constant cross-section or thickness); otherwise return to S17's Eq. 5/3.
Open question: a bracket made of a plate, a hole and a rod, or a table of shapes from Appendix D: how are known centroids combined without integrating again? (S19.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's programs (Section 3.7); the Python is a translation, and both print the same numbers.

Python (numpy, scipy). S18_centroids.py has one function for any parametric curve. It integrates |\mathbf r'(t)| and \mathbf r\,|\mathbf r'(t)| with quad, and runs it over Marghitu's wire, astroid, helix and spiral and Meriam's arc. It then does the areas, finding Example 3.8's limits with brentq, and the two cones.

Part 1: lines
  Ex 3.1 y = 2x^1.5: L = 2.268 m, C = [0.575 0.952] m
  Ex 3.2 astroid: L = 1.500 m, C = [0.4 0.4] m
  Ex 3.4 helix: L = 26.657 m, C = [0.    0.    9.425] m
  Ex 3.5 spiral: L = 66.114 m, C = [-15.59   10.778  21.086] m
  P 5/7 arc 45 to 165 deg: C = [-0.214  0.799] a
Part 2: areas
  semicircular plate r = 100 mm: y_bar = 42.4 mm; wire: 63.7 mm
  Ex 3.8: curves cross at 3.351 and 28.649 m, A = 21.082 m^2, C = (14.000, 4.000) m
  gusset, base 0.6 m, altitude 0.45 m: y_bar = 0.150 m
Part 3: volumes (h = 0.7 m, r = 1)
  solid cone: z_bar = 0.5250 m from the vertex, 0.175 m above the base
  half cone: 0.3183 r from the cut face (1/pi = 0.3183)

MATLAB (Octave-compatible). S18_centroids.m uses the numerical integral and fzero in place of Marghitu's symbolic int and solve. The numbers are the same.

Ex 3.1: L = 2.268 m, C = (0.575, 0.952) m
Ex 3.4: L = 26.657 m, z_C = 9.425 m
Ex 3.5: L = 66.114 m, C = (-15.590, 10.778, 21.086) m
P 5/7: C = (-0.214, 0.799) a
Ex 3.8: x = 3.351 to 28.649 m, A = 21.082 m^2, C = (14.000, 4.000) m
plate y_bar = 42.4 mm, wire 63.7 mm

Traps

Assuming the centroid lies on the wire or shell. It is usually in space. (From Nugget 01.)
Using the centroid formula when the density varies. Then \rho does not cancel. (From Nugget 01.)
Swapping M_x and M_y. M_x = \int y\,dA is the moment about the x-axis. (From Nugget 02.)
Fighting a circular boundary in x and y. Use polar coordinates. (From Nugget 03.)
A strip whose formula changes partway. Turn it the other way or split the integral. (From Nugget 04.)
Taking the edge of the element as its arm. Use the element's own centroid: y/2 for a vertical strip. (From Nugget 05.)
Using the variable r_0 as the ring's arm. The ring's centroid is at r_0\sin\alpha/\alpha. (From Nugget 05.)
Placing a cone's centroid at a third of its height. That is the triangle; the cone's is at a quarter above the base. (From Nugget 06.)
Integrating a curve over dx. The element is dl = \sqrt{1 + y'^2}\,dx. (From Nugget 07.)

Self-check

Meriam's Problem 5/7: a uniform rod bent into an arc of radius a from 45^\circ to 165^\circ measured from the x-axis. Find \bar x and \bar y.
Meriam's Problem 5/2: a circular sector of radius 200\,\mathrm{mm} whose straight sides rise at 30^\circ above the horizontal on each side of its vertex. How far above the vertex is its centroid?
Meriam's Problem 5/6: the area under y = a\sin(\pi x/b), 0 \le x \le b. Find \bar y.
Meriam's Problem 5/8: a trapezoid with a vertical left side of height h = 0.4\,\mathrm m, base b = 0.6\,\mathrm m and top a = 0.3\,\mathrm m. Find its centroid.
The parabolic spandrel under y = h(x/a)^2, 0 \le x \le a: find \bar x and \bar y, and check \bar y with a horizontal strip.
Marghitu's Problem 3.14: the region between y = x and y = \sqrt x, 0 \le x \le 1. Find its centroid.
Marghitu's Problem 3.15: the region between y = x^2 and y = \sqrt x. Find its centroid, and explain why \bar x = \bar y.
Meriam's Problem 5/10: the area under y = 1 + x^3/8 between x = 1 and x = 2 (lengths in metres). Find its centroid.
Meriam's Problem 5/9: a homogeneous paraboloid of revolution, height h, vertex down. Find \bar z above the vertex.
Meriam's Problem 5/17: show that a solid right circular cone's centroid is \frac34h from the vertex, and evaluate it for h = 0.6\,\mathrm m.
Marghitu's Example 3.4: the helix x = \cos t, y = \sin t, z = t (m), 0 \le t \le 6\pi. Find its length and centroid, using symmetry where you can.
A student finds \bar y = \frac45b for SP 5/4's area. What went wrong?

Fully worked solutions

1. L = a(165^\circ - 45^\circ) = a(2\pi/3). \bar x = \int a\cos\theta\,a\,d\theta/L = a(\sin165^\circ - \sin45^\circ)/(2\pi/3) = a(0.2588 - 0.7071)/2.0944 = -0.214a. \bar y = \int a\sin\theta\,a\,d\theta/L = a(\cos45^\circ - \cos165^\circ)/(2\pi/3) = a(0.7071 + 0.9659)/2.0944 = 0.799a.

2. The sector spans 30^\circ to 150^\circ: 2\alpha = 120^\circ, \alpha = \pi/3, symmetric about the vertical. \bar y = 2r\sin\alpha/(3\alpha) = 2(200)(0.8660)/(3 \times 1.0472) = 110.3\,\mathrm{mm}.

3. Vertical strip, y_c = y/2: A = \int_0^b a\sin(\pi x/b)\,dx = 2ab/\pi; \int_0^b\frac12a^2\sin^2(\pi x/b)\,dx = a^2b/4. \bar y = (a^2b/4)/(2ab/\pi) = \pi a/8 = 0.393a. (\bar x = b/2 by symmetry.)

4. Horizontal strip at height y: width w = b + (a - b)y/h = 0.6 - 0.75y, from x = 0, so x_c = w/2. A = (a + b)h/2 = 0.18\,\mathrm{m^2}. \bar x = \int\frac{w^2}{2}\,dy/A = (a^2 + ab + b^2)/[3(a + b)] = (0.09 + 0.18 + 0.36)/2.7 = 0.2333\,\mathrm m; \bar y = \int yw\,dy/A = h(2a + b)/[3(a + b)] = 0.4(1.2)/2.7 = 0.1778\,\mathrm m.

5. A = \int_0^a h(x/a)^2\,dx = ah/3. \bar x = \int x\,y\,dx/A = (a^2h/4)/(ah/3) = \frac34a; \bar y = \int\frac{y^2}{2}\,dx/A = (ah^2/10)/(ah/3) = \frac{3}{10}h. Horizontal strip from x = a\sqrt{y/h} to a: \int_0^h y\,a(1 - \sqrt{y/h})\,dy = a(h^2/2 - \frac25h^2) = ah^2/10, so again \bar y = \frac3{10}h.

6. \sqrt x \ge x on [0, 1]. A = \int(\sqrt x - x)\,dx = \frac23 - \frac12 = \frac16. \bar x = \int x(\sqrt x - x)\,dx/A = (\frac25 - \frac13)/\frac16 = 0.4; \bar y = \int\frac12(x - x^2)\,dx/A = \frac12(\frac12 - \frac13)/\frac16 = 0.5. C = (0.4, 0.5).

7. A = \int(\sqrt x - x^2)\,dx = \frac23 - \frac13 = \frac13. \bar x = \int x(\sqrt x - x^2)\,dx/A = (\frac25 - \frac14)/\frac13 = 0.45; \bar y = \int\frac12(x - x^4)\,dx/A = \frac12(\frac12 - \frac15)/\frac13 = 0.45. The curves y = x^2 and y = \sqrt x are reflections of each other in the line y = x, so the region is symmetric about it and \bar x = \bar y.

8. A = \int_1^2(1 + x^3/8)\,dx = 1 + \frac{15}{32} = 1.469\,\mathrm{m^2}. \bar x = \int_1^2 x(1 + x^3/8)\,dx/A = (1.5 + 31/40)/1.469 = 1.549\,\mathrm m; \bar y = \int_1^2\frac12(1 + x^3/8)^2\,dx/A = 1.110/1.469 = 0.756\,\mathrm m.

9. Disc at height z: radius^2 \propto z, so dV = cz\,dz. \bar z = \int_0^h z\cdot z\,dz/\int_0^h z\,dz = (h^3/3)/(h^2/2) = \frac23h.

10. Disc radius rz/h at z from the vertex: \bar z = \int_0^h z^3\,dz/\int_0^h z^2\,dz = \frac34h. For h = 0.6\,\mathrm m: 0.45\,\mathrm m from the vertex, 0.15\,\mathrm m above the base.

11. dl = \sqrt{\sin^2t + \cos^2t + 1}\,dt = \sqrt2\,dt, L = 6\pi\sqrt2 = 26.657\,\mathrm m. Three whole turns are symmetric about the z-axis: x_C = y_C = 0. The rise is uniform, so z_C is half the height: \frac12(6\pi) = 3\pi = 9.425\,\mathrm m.

12. They used the strip's top, y, as its moment arm for \bar y: \int y\cdot y\,dx/A = (3ab^2/5)/(3ab/4) = \frac45b. The strip's area acts at its own centroid, y/2, which halves the integral and gives \frac25b (Nugget 05).