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

a) 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.

b) 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.

c) 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.

d) 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.

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

b) 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 $x$.

c) 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.

d) 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 $\frac{3}{4}$.

e) $A = \int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 1 + 1 = 2$. 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.

a) $2x = x^2$ 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 $A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$.

b) $M_y = \int_0^2 x(2x - x^2)\,dx = \int_0^2 (2x^2 - x^3)\,dx = \left[\frac{2x^3}{3} - \frac{x^4}{4}\right]_0^2 = \frac{16}{3} - 4 = \frac{4}{3}$, 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.

c) 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.

d) $\frac{1}{2}\int_0^2 (2x - x^2)^2\,dx = \frac{1}{2}\int_0^2 (4x^2 - 4x^3 + x^4)\,dx = \frac{1}{2}\left(\frac{32}{3} - 16 + \frac{32}{5}\right) = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15}$, 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.

e) 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.

a) 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.

b) 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: $\frac{44 + 37.7}{16 + 9.4} \approx 3.21$.

c) 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.

d) 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 $\frac{32 - 9.42}{16 - 3.14} \approx 1.76$.

e) 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.

a) 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.

b) 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.

c) 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.

d) 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.

e) The hypothesis of Pappus is that $R$ lies entirely on ONE side of the axis. The $y$-axis cuts the disc in half. When it turns, the left half sweeps the same ball as the right half, and in the formula the moments of the two halves cancel, which produces the absurd $V = 0$. The fix is to rotate only the right half-disc $x^2 + y^2 \le 1$, $x \ge 0$: it produces the whole ball, has area $\frac{\pi}{2}$, and its centroid is at distance $\frac{4 \cdot 1}{3\pi}$ from the $y$-axis (Exercise 4 a, turned on its side). So $V = \frac{\pi}{2} \cdot 2\pi \cdot \frac{4}{3\pi} = \frac{4\pi}{3}$, the familiar volume of the unit ball.

a) Choose the variable first and say it in words: $x$ is the DEPTH below the surface, measured downward. The plate occupies $2 \le x \le 5$, and at every depth its width is $4$. The strip between depths $x$ and $x + dx$ has area $4\,dx$ and feels the pressure $\rho g x$, so $F = \int_2^5 \rho g x \cdot 4\,dx = 2\rho g\left[x^2\right]_2^5 = 2\rho g(25 - 4) = 42\rho g = 42 \cdot 9800 = 411\,600$ N. The bounds are depths, $2$ and $5$, not $0$ and $3$: the plate does not start at the surface. Using $\int_0^3$ would give $18\rho g$, the force on the same plate touching the surface.

b) With $x$ the depth, the base is at $x = 0$ and the vertex at $x = 3$. The width shrinks linearly from $6$ to $0$, so $w(x) = 6 \cdot \frac{3 - x}{3} = 2(3 - x)$; check the two ends, $w(0) = 6$ and $w(3) = 0$. Then $F = \rho g\int_0^3 x \cdot 2(3 - x)\,dx = 2\rho g\left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = 2\rho g\left(\frac{27}{2} - 9\right) = 9\rho g = 88\,200$ N. The width function is where most marks are lost: write it from similar triangles, then CHECK it at both ends before integrating.

c) Now the vertex is at depth $0$ and the base at depth $3$, so $w(x) = 2x$, with $w(0) = 0$ and $w(3) = 6$. $F = \rho g\int_0^3 x \cdot 2x\,dx = \rho g\left[\frac{2x^3}{3}\right]_0^3 = 18\rho g = 176\,400$ N. Same triangle, same area, twice the force. A student who keeps the width function of b) because the triangle is the same one gets $9\rho g$ again and loses the question: the width must be read at each DEPTH, and turning the plate over changes which depths carry the wide part.