Corrected exercises: moments, centres of mass and hydrostatic force (MATH 141)

The total mass is $m = 4 + 2 + 6 = 12$ kg. Each mass contributes its mass times its signed position: $M = 4(-3) + 2(1) + 6(3) = -12 + 2 + 18 = 8$ kg m. Then $\bar{x} = \frac{M}{m} = \frac{8}{12} = \frac{2}{3}$ m. The sign of each $x_i$ matters: the $4$ kg mass on the left pulls the moment DOWN by $12$, and dropping that minus sign would give $M = 32$ and $\bar{x} = \frac{8}{3}$, far too close to the right end for a system with $4$ kg pulling on the far left.

With a fourth mass of $4$ kg at $x = p$, the new moment is $8 + 4p$ and the new total mass is $16$. The centre of mass is at the origin exactly when the MOMENT is zero, so $8 + 4p = 0$ and $p = -2$. Check: $4(-3) + 2(1) + 6(3) + 4(-2) = 0$. The average $\frac{1}{3}$ treats the three masses as equal: it is the centre of mass of three EQUAL masses at those positions, not of these ones. The $6$ kg mass at $x = 3$ weighs three times as much as the $2$ kg mass, so it drags $\bar{x}$ to the right, from $\frac{1}{3}$ to $\frac{2}{3}$. The centre of mass is a WEIGHTED average: $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$, and the weights are the masses.

Total mass $m = 3 + 1 + 2 = 6$ kg. About the $y$-axis the arm is the $x$-coordinate: $M_y = 3(1) + 1(-2) + 2(3) = 3 - 2 + 6 = 7$. About the $x$-axis the arm is the $y$-coordinate: $M_x = 3(2) + 1(4) + 2(-1) = 6 + 4 - 2 = 8$. Then $\bar{x} = \frac{M_y}{m} = \frac{7}{6}$ and $\bar{y} = \frac{M_x}{m} = \frac{8}{6} = \frac{4}{3}$. The pairing to memorise is crossed: $\bar{x}$ comes from $M_y$ and $\bar{y}$ comes from $M_x$. Sanity check: $\frac{7}{6}$ lies between the smallest and largest $x_i$, $-2$ and $3$, and $\frac{4}{3}$ between $-1$ and $4$, as any weighted average must.

A moment measures the tendency to ROTATE about an axis, and the lever arm of a mass for a rotation about the $x$-axis is its distance to that axis, which is $|y|$, not $|x|$. A mass at $(5, 0)$ sits ON the $x$-axis and cannot turn anything about it, whatever its $x$-coordinate: its contribution to $M_x$ must be $0$, and it is, since $y = 0$. The subscript names the axis, never the coordinate that is multiplied. A student who swaps the moments writes $\bar{x} = \frac{8}{6}$ and $\bar{y} = \frac{7}{6}$, and reports $\left(\frac{4}{3}, \frac{7}{6}\right)$: the reflection of the true centre of mass in the line $y = x$. It costs the whole answer, and it is invisible unless you check against the picture.

With $\bar{x} = \frac{2}{3}$ the arms become $-3 - \frac{2}{3} = -\frac{11}{3}$, $1 - \frac{2}{3} = \frac{1}{3}$ and $3 - \frac{2}{3} = \frac{7}{3}$, so $\sum m_i (x_i - \bar{x}) = \frac{-44 + 2 + 42}{3} = 0$. This is not a coincidence: $\sum m_i (x_i - \bar{x}) = M - m\bar{x} = 0$ by the very definition of $\bar{x}$. That is what the centre of mass MEANS: a pivot placed there balances the rod. About a pivot at $x = 1$ the arms are $-4$, $0$ and $2$, and the moment is $4(-4) + 2(0) + 6(2) = -4 \neq 0$. The negative sign says the masses on the left win: the rod rotates so that its left end goes down. Consistent: the balance point $\frac{2}{3}$ is to the LEFT of the pivot.

$A = \int_0^4 x^{1/2}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^4 = \frac{2}{3} \cdot 8 = \frac{16}{3}$. Writing $\sqrt{x}$ as $x^{1/2}$ before integrating is what keeps the power rule safe: $4^{3/2} = (\sqrt{4})^3 = 8$.

The moment about the $y$-axis adds the strips, each with area $f(x)\,dx$ and arm $x$: $M_y = \int_0^4 x \cdot x^{1/2}\,dx = \int_0^4 x^{3/2}\,dx = \left[\frac{2}{5}x^{5/2}\right]_0^4 = \frac{2}{5} \cdot 32 = \frac{64}{5}$. So $\bar{x} = \frac{M_y}{A} = \frac{64}{5} \cdot \frac{3}{16} = \frac{12}{5}$. Sanity check: $\frac{12}{5} = 2.4$ lies to the right of the midpoint $2$ of $[0, 4]$, as it must, since the region is taller on the right. A centroid on the wrong side of the obvious midpoint is a sign error or a missing factor $x$.

The strip at $x$ is a thin rectangle of height $f(x)$ and width $dx$. Its own centroid is at its MIDDLE, the point $M$ of the figure, at height $\frac{1}{2}f(x)$. Its moment about the $x$-axis is therefore (area) $\times$ (arm) $= f(x)\,dx \cdot \frac{1}{2}f(x) = \frac{1}{2}[f(x)]^2\,dx$. That is the whole origin of the formula. Here $M_x = \int_0^4 \frac{1}{2}(\sqrt{x})^2\,dx = \frac{1}{2}\int_0^4 x\,dx = \frac{1}{2} \cdot 8 = 4$, and $\bar{y} = \frac{M_x}{A} = 4 \cdot \frac{3}{16} = \frac{3}{4}$. Check: the region never rises above $2$, and most of its area is low, so $\bar{y}$ must be well below $1$: $0.75$ is plausible.

For any region under $f \ge 0$, $\int_a^b f(x)\,dx$ IS the area $A$, so $\frac{1}{A}\int_a^b f(x)\,dx = \frac{A}{A} = 1$, whatever $f$, $a$ and $b$ are. A formula that gives $\bar{y} = 1$ for a region of height $0.01$ and for a region of height $1000$ cannot be a centroid formula. The mistake is to use $y = f(x)$, the TOP of the strip, as the arm: the arm is the height of the strip's own centroid, $\frac{1}{2}f(x)$, and the area of the strip multiplies it once more, hence $\frac{1}{2}f^2$. Note also that $\bar{y}$ is not the average value of $f$: here $f_{\text{ave}} = \frac{1}{4} \cdot \frac{16}{3} = \frac{4}{3}$, far from $\frac{3}{4}$.

$A = \int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 1 + 1 = 2$. The region is symmetric about the vertical line $x = \frac{\pi}{2}$, since $\sin(\pi - x) = \sin x$, and a uniform plate symmetric about a line has its centroid ON that line: $\bar{x} = \frac{\pi}{2}$ with no integral (integrating $x \sin x$ by parts gives $M_y = \pi$ and confirms it). For $\bar{y}$ there is no symmetry to use: $M_x = \frac{1}{2}\int_0^\pi \sin^2 x\,dx = \frac{1}{2}\int_0^\pi \frac{1 - \cos 2x}{2}\,dx = \frac{1}{4}\left[x - \frac{\sin 2x}{2}\right]_0^\pi = \frac{\pi}{4}$, so $\bar{y} = \frac{\pi/4}{2} = \frac{\pi}{8}$. With $\pi \approx 3.14$, $\bar{y} \approx 0.39$: below half the maximum height, as for every region that is wider at the bottom than at the top.

$2x = x^2$ gives $x(x - 2) = 0$, so $x = 0$ or $x = 2$, at the points $(0, 0)$ and $(2, 4)$. At $x = 1$, $2x = 2 > 1 = x^2$: the line is on top on the whole of $[0, 2]$. Then $A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.

$M_y = \int_0^2 x(2x - x^2)\,dx = \int_0^2 (2x^2 - x^3)\,dx = \left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2 = \frac{16}{3} - 4 = \frac{4}{3}$, so $\bar{x} = \frac{4/3}{4/3} = 1$. The region is NOT symmetric about $x = 1$, its top edge is a line and its bottom edge a parabola. But $\bar{x}$ only depends on HOW LONG each vertical strip is, not on where it sits vertically: sliding strips up or down never changes their $x$. The length $L(x) = 2x - x^2 = 1 - (x - 1)^2$ is symmetric about $x = 1$, so the strips balance about that line. Useful reflex: before computing $\bar{x}$, look at $f - g$, not at the picture.

The strip at $x$ runs from $g(x) = x^2$ up to $f(x) = 2x$. Its length is $f - g$, its midpoint $M$ is at height $\frac{f + g}{2}$, so its moment about the $x$-axis is (area) $\times$ (arm) $= (f - g)\,dx \cdot \frac{f + g}{2} = \frac{1}{2}(f^2 - g^2)\,dx$: a difference of squares, not a square of a difference. Here $M_x = \frac{1}{2}\int_0^2 (4x^2 - x^4)\,dx = \frac{1}{2}\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^2 = \frac{1}{2}\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{1}{2} \cdot \frac{64}{15} = \frac{32}{15}$, and $\bar{y} = \frac{32}{15} \cdot \frac{3}{4} = \frac{8}{5}$. The centroid is $\left(1, \frac{8}{5}\right)$; at $x = 1$ the region runs from $y = 1$ to $y = 2$, and $1.6$ is inside, as expected for this convex region.

$\frac{1}{2}\int_0^2 (2x - x^2)^2\,dx = \frac{1}{2}\int_0^2 (4x^2 - 4x^3 + x^4)\,dx = \frac{1}{2}\left(\frac{32}{3} - 16 + \frac{32}{5}\right) = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15}$, and dividing by $A$ gives $\frac{8}{15} \cdot \frac{3}{4} = \frac{2}{5}$. The point $\left(1, \frac{2}{5}\right)$ lies BELOW the parabola, whose height at $x = 1$ is $1$: it is not in $R$ at all. What went wrong: $\frac{1}{2}(f - g)^2$ is the moment of a strip that starts on the $x$-axis, as if every strip had been slid down to $y = 0$. It is correct only when $g = 0$, which is exactly Exercise 2. Checking that $(\bar{x}, \bar{y})$ lands inside a convex region is a five-second test that catches this error every time.

Horizontal strips: for $0 \le y \le 4$ the region runs from the line, $x = \frac{y}{2}$, to the parabola, $x = \sqrt{y}$ (the parabola is on the RIGHT for horizontal strips). A strip at height $y$ has area $\left(\sqrt{y} - \frac{y}{2}\right)dy$ and arm $y$ for a rotation about the $x$-axis, with no factor $\frac{1}{2}$ this time, since the whole strip is at height $y$. So $M_x = \int_0^4 \left(y^{3/2} - \frac{y^2}{2}\right)dy = \left[\frac{2}{5}y^{5/2} - \frac{y^3}{6}\right]_0^4 = \frac{64}{5} - \frac{32}{3} = \frac{32}{15}$, the same value. The factor $\frac{1}{2}$ belongs to strips PERPENDICULAR to the axis; strips parallel to it use their coordinate directly.

The half-disc is symmetric about the $y$-axis, so $\bar{x} = 0$ with no computation. For $\bar{y}$, use vertical strips under $f(x) = \sqrt{r^2 - x^2}$ on $[-r, r]$: $M_x = \int_{-r}^{r} \frac{1}{2}(r^2 - x^2)\,dx = \frac{1}{2}\left[r^2 x - \frac{x^3}{3}\right]_{-r}^{r} = \frac{1}{2}\left(2r^3 - \frac{2r^3}{3}\right) = \frac{2r^3}{3}$. The square root disappears because the formula uses $f^2$: no trigonometric substitution is needed. The area is half a disc, $A = \frac{\pi r^2}{2}$, quoted from geometry. So $\bar{y} = \frac{2r^3}{3} \cdot \frac{2}{\pi r^2} = \frac{4r}{3\pi}$. With $\pi \approx 3.14$, $\frac{4}{3\pi} \approx 0.42$: the centroid sits at $42$ percent of the radius, below the middle height, because the half-disc is widest at the bottom.

Symmetry about the $y$-axis gives $\bar{x} = 0$. For $\bar{y}$, add the MOMENTS of the two pieces, each being (area) $\times$ (height of its own centroid). Rectangle: area $24$, centroid height $2$, moment $48$. Half-disc: area $\frac{9\pi}{2}$, centroid height $4 + \frac{4 \cdot 3}{3\pi} = 4 + \frac{4}{\pi}$ (measured from $y = 0$, not from its own base), moment $\frac{9\pi}{2}\left(4 + \frac{4}{\pi}\right) = 18\pi + 18$. Total moment $66 + 18\pi$, total area $24 + \frac{9\pi}{2} = \frac{48 + 9\pi}{2}$, so $\bar{y} = \frac{2(66 + 18\pi)}{48 + 9\pi} = \frac{132 + 36\pi}{48 + 9\pi} = \frac{4(11 + 3\pi)}{16 + 3\pi}$. Order of magnitude: $\frac{44 + 37.7}{16 + 9.4} \approx 3.21$.

He writes $\bar{y} = \frac{1}{2}\left(2 + 4 + \frac{4}{\pi}\right) = 3 + \frac{2}{\pi} \approx 3.64$. That is the centroid of two EQUAL masses placed at the two centroids. But the rectangle has area $24$ and the half-disc only $\frac{9\pi}{2} \approx 14.1$: the rectangle weighs more and must pull the centroid down, to about $3.21$. The correct combination is the weighted average $\bar{y} = \frac{A_1\bar{y}_1 + A_2\bar{y}_2}{A_1 + A_2}$, which is exactly the sum of the moments divided by the total area. It is the plate version of Exercise 1: moments add, centroids do not.

Treat the hole as a piece of NEGATIVE area. Full square: area $16$, centroid $(2, 2)$, $M_y = 32$. Hole: area $\pi$, centroid $(3, 3)$, $M_y = 3\pi$. Plate: area $16 - \pi$ and $M_y = 32 - 3\pi$, so $\bar{x} = \frac{32 - 3\pi}{16 - \pi}$. The plate is symmetric about the diagonal $y = x$ (the square is, and the hole is centred on it), so $\bar{y} = \bar{x}$ without a second computation. Check the direction: removing material from the upper right must push the centroid down and to the left of $(2, 2)$, and indeed $\frac{32 - 3\pi}{16 - \pi} < 2 \iff 32 - 3\pi < 32 - 2\pi$, which is true. Numerically $\frac{32 - 9.42}{16 - 3.14} \approx 1.76$.

With two densities the weights are the MASSES, not the areas. Take the rectangle's density as $\rho$ and the glass as $\frac{\rho}{2}$. Rectangle: mass $24\rho$, moment $48\rho$. Half-disc: mass $\frac{\rho}{2} \cdot \frac{9\pi}{2} = \frac{9\pi\rho}{4}$, moment $\frac{\rho}{2}(18\pi + 18) = (9\pi + 9)\rho$. Then $\bar{y} = \frac{48 + 9\pi + 9}{24 + \frac{9\pi}{4}} = \frac{4(57 + 9\pi)}{96 + 9\pi} = \frac{4(19 + 3\pi)}{32 + 3\pi}$, about $\frac{4 \cdot 28.4}{41.4} \approx 2.75$. It moved DOWN from $3.21$: lightening the top leaves the heavy rectangle in charge. The density $\rho$ cancels only when it is the same everywhere; that is why parts a) to d) never mention it, and why the symmetry principle requires a uniform plate.

The disc has area $A = \pi \cdot 1^2 = \pi$ and, by symmetry, its centroid is its centre $(4, 0)$, at distance $d = 4$ from the $y$-axis. The disc lies entirely in $x \ge 3$, on one side of the axis, so Pappus applies: $V = A \cdot 2\pi d = \pi \cdot 8\pi = 8\pi^2$. Read the formula as a sentence: the centroid travels once around a circle of length $2\pi d = 8\pi$, and the area $\pi$ sweeps a tube of that length. No integral was needed; with shells it would take a trigonometric substitution.

The distance from the centroid to the $x$-axis is $\bar{y} = \frac{8}{5}$, so $V = \frac{4}{3} \cdot 2\pi \cdot \frac{8}{5} = \frac{64\pi}{15}$. With washers, the outer radius is $2x$, the inner radius $x^2$: $V = \pi\int_0^2 (4x^2 - x^4)\,dx = \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15}$. The agreement is no accident. The washer integral is $\pi\int (f^2 - g^2)\,dx$, and $M_x = \frac{1}{2}\int (f^2 - g^2)\,dx$: so $V = 2\pi M_x = 2\pi \bar{y} A$. Pappus about the $x$-axis IS the washer method, rewritten with the centroid.

The axis $x = 3$ lies to the right of the whole region ($0 \le x \le 2$), and the centroid is at $x = 1$, so the distance is $d = 3 - 1 = 2$, measured from the AXIS, not from the origin. $V = \frac{4}{3} \cdot 2\pi \cdot 2 = \frac{16\pi}{3}$. With shells about $x = 3$, the radius is $3 - x$ and the height $2x - x^2$: $V = 2\pi\int_0^2 (3 - x)(2x - x^2)\,dx = 2\pi\int_0^2 (6x - 5x^2 + x^3)\,dx = 2\pi\left(12 - \frac{40}{3} + 4\right) = 2\pi \cdot \frac{8}{3} = \frac{16\pi}{3}$. Using $d = 1$, the distance to the origin, would give $\frac{8\pi}{3}$, half the true volume.

The triangle has area $\frac{ah}{2}$ and lies in $x \ge 0$, on one side of the $y$-axis. Pappus gives $2\pi\bar{x} \cdot \frac{ah}{2} = \pi a h \bar{x}$, and this must equal the volume of the cone, $\frac{1}{3}\pi a^2 h$. Hence $\bar{x} = \frac{a}{3}$: the centroid of a right triangle is at one third of each leg from the right angle. Rotating about the $x$-axis instead gives a cone of radius $h$ and height $a$, and the same argument gives $\bar{y} = \frac{h}{3}$. Pappus is a two-way street: a known volume gives a centroid for free.