MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: moments, centres of mass and hydrostatic force (MATH 141)

This sheet is not a summary of section 8.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on moments, centroids and hydrostatic force in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the final, and every centroid quoted has been checked twice: by the integral, and by a symmetry, a bounding box or a second method. Forces are left as multiples of $\rho g$ , with $\rho g = 9800$ N/m $^3$ for water.

The thread of the chapter

Every quantity of the chapter is a sum of pieces times their ARM: slice, then ask each slice how much and how far. For a moment the arm is the distance to the AXIS, for a pressure it is the depth below the SURFACE, and a strip's arm is taken at its MIDDLE.

This chapter is part of MATH 141, Calculus 2 (McGill)

The essentials

Moments: a piece times its distance to the axis

• Point masses: $M_y = \sum m_i x_i$ (arm = distance to the $y$ -axis), $M_x = \sum m_i y_i$ (arm = distance to the $x$ -axis). Then $\bar{x} = \frac{M_y}{m}$ , $\bar{y} = \frac{M_x}{m}$ .
• Uniform plate between $f$ on top and $g$ below, on $[a, b]$ : a vertical strip has area $(f - g)\,dx$ , abscissa $x$ and MIDPOINT at height $\frac{f + g}{2}$ .
• Hence $A = \int_a^b (f - g)\,dx$ , $M_y = \int_a^b x(f - g)\,dx$ , $M_x = \int_a^b \frac{1}{2}(f^2 - g^2)\,dx$ , and $(\bar{x}, \bar{y}) = \left(\frac{M_y}{A}, \frac{M_x}{A}\right)$ . The density cancels.
• Symmetry principle: a uniform plate symmetric about a line has its centroid on that line. Composite plate: add the moments, subtract those of a hole, divide by the total area.
• Pappus: a region of area $A$ on ONE side of an axis, with its centroid at distance $d$ from the axis, sweeps a volume $V = 2\pi d A$ .

The strip's arm is its midpoint

M

, at height

\frac{f + g}{2}

; the region is symmetric about the dashed line

y = x

, so its centroid

C

lies on it.

Before any integral, write the strip with its two attributes: how much (its area) and how far (its arm). The formulas follow from that sentence, and so does every trap below.

Hydrostatic force: a pressure times an area, the depth from the surface

• At depth $x$ below the surface, $P = \rho g x$ : $\rho g = 9800$ N/m $^3$ for water in SI units, $\delta = 62.5$ lb/ft $^3$ in imperial units.
• Slice the plate into HORIZONTAL strips, each at a single depth. With $x$ the depth and $w(x)$ the width: $F = \int \rho g\,x\,w(x)\,dx$ .
• $\int x\,w(x)\,dx$ is the first moment of the plate about the surface line, so $F = \rho g\,\bar{d}\,A$ with $\bar{d}$ the depth of the CENTROID.
• Inclined plate at angle $\theta$ to the horizontal: the strip has area $w(x)\frac{dx}{\sin\theta}$ .
• The volume of water behind the plate never enters: only the depth and the shape of the wetted face. The resultant acts at the centre of pressure, $\frac{\int x^2 w\,dx}{\int x\,w\,dx}$ , deeper than the centroid.

Say in words, on the first line of the answer, what $x$ measures and in which direction. Half the errors in hydrostatics are a depth measured from the wrong end.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Piece, arm, contribution: the whole chapter in one grid

Read a line as: the piece of the first column, multiplied by the arm of the second, gives the contribution of the third. The examples use the region between $y = \sqrt{x}$ and $y = x^2$ , of area $\frac{1}{3}$ . The red lines are the rules students invent.

Quantity	Piece	Arm	Contribution
$M_y$ , point masses	$m_i$	$x_i$	$m_i x_i$
Example: $3$ kg at $(1, 2)$ contributes $3$ to $M_y$ and $6$ to $M_x$ .
$M_y$ , plate	$(f - g)\,dx$	$x$	$x(f - g)\,dx$
Example: $\int_0^1 x(\sqrt{x} - x^2)\,dx = \frac{2}{5} - \frac{1}{4} = \frac{3}{20}$ , so $\bar{x} = \frac{9}{20}$ .
$M_x$ , vertical strip	$(f - g)\,dx$	$\frac{f + g}{2}$	$\frac{1}{2}(f^2 - g^2)\,dx$
Example: $\frac{1}{2}\int_0^1 (x - x^4)\,dx = \frac{3}{20}$ , so $\bar{y} = \frac{9}{20}$ .
$M_x$ , horizontal strip	$(\sqrt{y} - y^2)\,dy$	$y$	$y(\sqrt{y} - y^2)\,dy$
Example: $\int_0^1 (y^{3/2} - y^3)\,dy = \frac{2}{5} - \frac{1}{4} = \frac{3}{20}$ again, with no factor $\frac{1}{2}$ .
Force on a plate	$w(x)\,dx$	depth $x$	$\rho g\,x\,w(x)\,dx$
Example: Triangle, base $4$ on the surface, height $3$ : $\rho g\int_0^3 x \cdot \frac{4}{3}(3 - x)\,dx = 6\rho g$ .
Volume, Pappus	$A$	$2\pi d$	$2\pi d A$
Example: About the $x$ -axis: $2\pi \cdot \frac{9}{20} \cdot \frac{1}{3} = \frac{3\pi}{10}$ .
$M_x$ with $(f - g)^2$	$(f - g)\,dx$	$\frac{f - g}{2}$	$\frac{1}{2}(f - g)^2\,dx$ no such rule
Example: $\frac{1}{2}\int_0^1 (\sqrt{x} - x^2)^2\,dx = \frac{9}{140}$ gives $\bar{y} = \frac{27}{140}$ , not $\frac{9}{20}$ . What to do: The arm is the MIDPOINT height $\frac{f + g}{2}$ , not half the length: $(f - g) \cdot \frac{f + g}{2} = \frac{1}{2}(f^2 - g^2)$ .
Composite plate	each piece	its centroid	average of centroids no such rule
Example: Rectangle $6 \times 4$ plus half-disc of radius $3$ : the average gives $3.64$ , the true $\bar{y}$ is $3.21$ . What to do: Add the moments $A_k\bar{y}_k$ and divide by the total area $\sum A_k$ .

Every blue line is (piece) $\times$ (arm), and the arm is always measured from the axis or from the surface, at the centre of the piece. The red lines are what you get by guessing the arm.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Building M_x from the x-coordinates

the whole centroid, reflected in the line y = x

What not to write

“ $3$ kg at $(1, 2)$ and $1$ kg at $(-2, 4)$ : $M_x = 3(1) + 1(-2) = 1$ .”

What to write

“ $M_x = \sum m_i y_i = 3(2) + 1(4) = 10$ and $M_y = \sum m_i x_i = 3(1) + 1(-2) = 1$ , so $(\bar{x}, \bar{y}) = \left(\frac{1}{4}, \frac{5}{2}\right)$ .”

Why: A moment about the $x$ -axis measures a rotation ABOUT that axis, and the arm is the distance to it, $|y|$ . A mass sitting on the $x$ -axis turns nothing about it, whatever its $x$ . The subscript names the axis, then $\bar{x} = \frac{M_y}{m}$ and $\bar{y} = \frac{M_x}{m}$ cross over.

2. Squaring the difference instead of taking the difference of the squares

3 to 4 marks, the whole of the y-coordinate

What not to write

“Between $y = \sqrt{x}$ and $y = x^2$ : $M_x = \frac{1}{2}\int_0^1 (\sqrt{x} - x^2)^2\,dx = \frac{9}{140}$ , so $\bar{y} = \frac{27}{140}$ .”

What to write

“ $M_x = \frac{1}{2}\int_0^1 \left((\sqrt{x})^2 - (x^2)^2\right)dx = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{5}\right) = \frac{3}{20}$ , so $\bar{y} = \frac{3/20}{1/3} = \frac{9}{20}$ .”

Why: The strip has length $f - g$ and its MIDPOINT is at height $\frac{f + g}{2}$ : the product is $\frac{1}{2}(f - g)(f + g) = \frac{1}{2}(f^2 - g^2)$ . Squaring $f - g$ slides every strip down to the $x$ -axis. Here $\bar{y} = \frac{27}{140} \approx 0.19$ would also break the symmetry $\bar{x} = \bar{y}$ of this region.

3. Using the top of the strip as its arm

the whole y-coordinate, and it goes unnoticed

What not to write

“ $\bar{y} = \frac{1}{A}\int_a^b f(x)\,dx$ , the height of the curve averaged over the region.”

What to write

“ $\bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}[f(x)]^2\,dx$ : the strip of area $f(x)\,dx$ has its own centroid at height $\frac{1}{2}f(x)$ .”

Why: Since $\int_a^b f\,dx = A$ , the wrong formula returns $\frac{A}{A} = 1$ for EVERY region, a pebble or a mountain. A formula that cannot see the region is not a centroid formula. For $y = \sqrt{x}$ on $[0, 4]$ the true value is $\frac{3}{4}$ .

4. Averaging the centroids of the pieces of a composite plate

2 to 3 marks

What not to write

“Rectangle $6 \times 4$ with centroid at height $2$ , half-disc of radius $3$ on top with centroid at $4 + \frac{4}{\pi}$ : $\bar{y} = \frac{1}{2}\left(6 + \frac{4}{\pi}\right) \approx 3.64$ .”

What to write

“Moments add: $\bar{y} = \frac{24 \cdot 2 + \frac{9\pi}{2}\left(4 + \frac{4}{\pi}\right)}{24 + \frac{9\pi}{2}} = \frac{4(11 + 3\pi)}{16 + 3\pi} \approx 3.21$ .”

Why: A plain average treats the two pieces as equally heavy. The rectangle has area $24$ and the half-disc about $14.1$ : the weights are the AREAS (the masses, if the densities differ). A hole enters with a NEGATIVE area and a negative moment.

5. Measuring the depth from the bottom of the plate

the whole force, and here exactly the force on the triangle turned over

What not to write

“Triangle, base $4$ along the surface, vertex $3$ m down. With $y$ up from the vertex, width $\frac{4}{3}y$ : $F = \rho g\int_0^3 y \cdot \frac{4}{3}y\,dy = 12\rho g$ .”

What to write

“With $y$ up from the vertex, the DEPTH is $3 - y$ : $F = \rho g\int_0^3 (3 - y)\frac{4}{3}y\,dy = 6\rho g$ .”

Why: Pressure is $\rho g$ times the depth below the SURFACE. Any other variable is allowed, but the depth must then be rewritten in it. The safest choice is to let $x$ be the depth itself, measured downward from the surface, so the pressure is simply $\rho g x$ .

6. Taking the pressure at the deepest point, or at the top

the whole force

What not to write

“The triangle above reaches depth $3$ , so $F = \rho g \cdot 3 \cdot 6 = 18\rho g$ .”

What to write

“ $F = \rho g\,\bar{d}\,A$ with $\bar{d}$ the depth of the CENTROID: $\bar{d} = 1$ (one third of the height, from the base), so $F = \rho g \cdot 1 \cdot 6 = 6\rho g$ .”

Same triangle, same area

6

. Base on the surface: centroid at depth

1

F = 6\rho g

. Turned over: centroid at depth

2

F = 12\rho g

. Only the centroid depth changed.

Why: The pressure varies with depth, so no single point represents the plate, except the centroid: $\int (\text{depth})\,dA$ is the first moment about the surface line, which equals $\bar{d}A$ . The deepest point overestimates, the top underestimates, and the true value always lies between the two.

7. Believing the centroid must lie on the plate

1 to 2 marks, and a correct answer rejected as absurd

What not to write

“The centroid of the half-ring $2 \le r \le 3$ , $y \ge 0$ , is on the ring, so $2 \le \bar{y} \le 3$ .”

What to write

“ $M_x = \frac{2}{3}(27 - 8) = \frac{38}{3}$ and $A = \frac{\pi}{2}(9 - 4) = \frac{5\pi}{2}$ , so $\bar{y} = \frac{76}{15\pi} < 2$ , since $\pi > \frac{38}{15}$ : the centroid is in the hole.”

The half-ring between radii

2

and

3

: its centroid

C

, at height

\frac{76}{15\pi} \approx 1.61

, sits in the hole, on the symmetry axis.

Why: The centroid is the balance point, not a point of the plate. It lies on the plate when the plate is CONVEX; a ring, an L or a crescent can put it in the empty part. Reject a centroid outside the bounding box, never one outside a non-convex plate.

8. Measuring the Pappus distance from the origin, or through the region

the whole volume

What not to write

“The region between $\sqrt{x}$ and $x^2$ turns about $x = 2$ ; its centroid is at $x = \frac{9}{20}$ , so $V = 2\pi \cdot \frac{9}{20} \cdot \frac{1}{3} = \frac{3\pi}{10}$ .”

What to write

“The distance to the AXIS is $2 - \frac{9}{20} = \frac{31}{20}$ , so $V = 2\pi \cdot \frac{31}{20} \cdot \frac{1}{3} = \frac{31\pi}{30}$ .”

Why: In $V = 2\pi dA$ , $d$ is the radius of the circle travelled by the centroid, measured to the axis of rotation. The theorem also requires the region to lie on ONE side of the axis: a disc turned about a diameter would give $V = 0$ . Shells confirm: $2\pi\int_0^1 (2 - x)(\sqrt{x} - x^2)\,dx = \frac{31\pi}{30}$ .

9. Forgetting the slant on an inclined plate

half the force

What not to write

“Plate $4$ m wide, inclined at $30^\circ$ to the horizontal between depths $1$ and $2$ : $F = \rho g\int_1^2 4x\,dx = 6\rho g$ .”

What to write

“A band between depths $x$ and $x + dx$ is $\frac{dx}{\sin 30^\circ} = 2\,dx$ long on the plate, so $F = \rho g\int_1^2 x \cdot 4 \cdot 2\,dx = 12\rho g$ .”

Why: The strips are still taken at constant depth, but their width ALONG the plate is $\frac{dx}{\sin\theta}$ , $\theta$ being the angle with the horizontal. Without that factor you compute the force on the shadow of the plate on a vertical wall.

Which method to choose

Hydrostatic force: which route, by the FORM of the plate

Look at the shape of the plate and at where it sits in the water before writing any integral

Last branch: the crest is dry, so

x = 0

moves down to the water line and the width becomes

56 - \frac{2}{3}x

[0, 24]

, not

60 - \frac{2}{3}x

If a rectangle, a triangle or a disc, vertical → $F = \rho g\,\bar{d}\,A$ with the known centroid: mid-height, one third from the base, the centre

Example: triangle with its vertex at the surface, base $4$ at depth $3$ : $\rho g \cdot 2 \cdot 6 = 12\rho g$
If a plate bounded by curves → horizontal strips, $x$ = depth from the surface, $w(x)$ read from the equations and checked at both ends

Example: gate $x^2 \le y \le 4$ : depth $4 - y$ , width $2\sqrt{y}$ , $F = \frac{256}{15}\rho g$
If a plate symmetric about a horizontal line → put the origin at its centre: the odd part of the integral vanishes

Example: porthole of radius $R$ , centre at depth $d$ : $F = \rho g\,d\,\pi R^2$
If a plate inclined at $\theta$ to the horizontal → same strips at constant depth, area $w(x)\frac{dx}{\sin\theta}$

Example: $4$ m wide, $30^\circ$ , depths $1$ to $2$ : $F = 12\rho g$
If the water level is below the top of the plate → integrate the wetted part only, depth 0 at the NEW surface, width rewritten from there

Example: dam $60$ to $40$ m wide, $30$ m high, water $6$ m below the crest: $w = 56 - \frac{2}{3}x$ on $[0, 24]$

Whatever the route, finish with the bracket check: $\rho g\,d_{\text{top}}A < F < \rho g\,d_{\text{bottom}}A$ .

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Setting up a hydrostatic force integral

When to use it: Any question that asks for the force exerted by a liquid on a plate, a window, a gate or a dam

1 Draw the plate, the surface, and ONE horizontal strip. State the variable in words: $x$ = depth below the surface, measured downward, in metres.
2 Give the bounds as depths: where the wetted plate starts and where it ends.
3 Write the width $w(x)$ from the geometry (similar triangles, the equation of a curve), and check it at both bounds.
4 Write $F = \int \rho g\,x\,w(x)\,dx$ and evaluate; keep $\rho g$ as a factor until the end.
5 Check with $\rho g\,\bar{d}\,A$ when the centroid is known, or with the bracket between top and bottom pressures, then give the units.

Concluding sentence

“Let $x$ be the depth below the surface. The strip at depth $x$ has width $w(x) = 6 - 2x$ and area $(6 - 2x)\,dx$ , and the pressure there is $\rho g x$ , so $F = \rho g\int_0^3 x(6 - 2x)\,dx = 9\rho g$ , that is $88\,200$ N.”

The trap: A width function taken from the drawing without checking its two ends, or a depth that silently becomes a height. The check of step 3 costs ten seconds and saves the question.

Marking: Typically 1 mark for the variable and bounds, 1 for the width, 1 for the integral, 1 for the evaluation; the check protects the other four.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Symmetry and bounding box

The centroid lies on every line of symmetry of a uniform plate, and inside the smallest rectangle containing the plate. A convex plate must contain it.

Region between $y = 2x$ and $y = x^2$ : $\left(1, \frac{8}{5}\right)$ is inside; $\left(1, \frac{2}{5}\right)$ is below the parabola, so it is wrong.
The heavy side wins

The centroid moves toward the side where the plate is wider or heavier; if yours moved the other way, look for a sign or a missing factor $x$ .

Under $y = \sqrt{x}$ on $[0, 4]$ the plate is taller on the right, and $\bar{x} = \frac{12}{5} > 2$ .
The pressure bracket

The force lies between the area times the pressure at the top and the area times the pressure at the bottom of the wetted plate.

Rectangle $4 \times 3$ from depth $2$ to $5$ : $24\rho g < 42\rho g < 60\rho g$ .
Pappus against washers

About a coordinate axis, Pappus and the washer or shell integral are the same computation. Doing both takes one extra line and catches the wrong distance.

$2\pi \cdot \frac{9}{20} \cdot \frac{1}{3} = \frac{3\pi}{10} = \pi\int_0^1 (x - x^4)\,dx$ .

The typical problem, taken apart

The centroid of a region between two curves, checked by symmetry, then used by Pappus

Let $R$ be the region between $y = \sqrt{x}$ and $y = x^2$ . Find its centroid, then the volume of the solid obtained by rotating $R$ about the $x$ -axis.

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

$\sqrt{x} = x^2$ gives $x = x^4$ , so $x = 0$ or $x = 1$ . At $x = \frac{1}{4}$ , $\sqrt{x} = \frac{1}{2} > \frac{1}{16}$ : $\sqrt{x}$ is on top. $A = \int_0^1 (x^{1/2} - x^2)\,dx = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ .

Why

Which curve is on top decides every sign that follows. Testing one point between the intersections is faster and safer than reading a sketch.

Step 2

$M_y = \int_0^1 x(x^{1/2} - x^2)\,dx = \int_0^1 (x^{3/2} - x^3)\,dx = \frac{2}{5} - \frac{1}{4} = \frac{3}{20}$ , so $\bar{x} = \frac{3}{20} \cdot 3 = \frac{9}{20}$ .

Why

The arm of a vertical strip for $M_y$ is its abscissa $x$ . Rewriting $\sqrt{x}$ as $x^{1/2}$ before multiplying by $x$ keeps the power rule clean.

Step 3

$M_x = \int_0^1 \frac{1}{2}\left((\sqrt{x})^2 - (x^2)^2\right)dx = \frac{1}{2}\int_0^1 (x - x^4)\,dx = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{5}\right) = \frac{3}{20}$ , so $\bar{y} = \frac{9}{20}$ .

Why

Difference of the squares, from the midpoint $\frac{f + g}{2}$ of the strip. This is the step where the marks are, and where $\frac{1}{2}(f - g)^2$ would give $\frac{27}{140}$ .

Step 4

Check: $y = \sqrt{x}$ and $y = x^2$ are inverse functions on $[0, 1]$ , so $R$ is symmetric about $y = x$ and $\bar{x} = \bar{y}$ is FORCED. Our $\frac{9}{20} = \frac{9}{20}$ agrees.

Why

A symmetry check costs one sentence and verifies two independent integrals at once. It also tells you, before computing, that one of them could have been skipped.

Step 5

Pappus: $R$ lies above the $x$ -axis, at distance $\bar{y} = \frac{9}{20}$ , so $V = 2\pi \cdot \frac{9}{20} \cdot \frac{1}{3} = \frac{3\pi}{10}$ . Washers: $\pi\int_0^1 (x - x^4)\,dx = \frac{3\pi}{10}$ .

Why

About the $x$ -axis, $V = 2\pi M_x$ exactly, so the two methods share one integral. Pappus pays off most when the axis is a line like $y = -1$ or $x = 2$ .

The conclusion, written out

“The centroid of $R$ is $\left(\frac{9}{20}, \frac{9}{20}\right)$ , on the line of symmetry $y = x$ , and the solid obtained by rotating $R$ about the $x$ -axis has volume $\frac{3\pi}{10}$ .”

The classic mistake on this problem: Writing $M_x = \frac{1}{2}\int_0^1 (\sqrt{x} - x^2)^2\,dx$ , which gives $\bar{y} = \frac{27}{140}$ and breaks the symmetry; or dividing $M_y$ by $M_x$ instead of by $A$ .

Learn by heart

• $M_x = \sum m_i y_i$ , $M_y = \sum m_i x_i$ ; $\bar{x} = \frac{M_y}{m}$ , $\bar{y} = \frac{M_x}{m}$ . The subscript names the AXIS.
• Plate: $M_y = \int x(f - g)\,dx$ , $M_x = \int \frac{1}{2}(f^2 - g^2)\,dx$ , divide by $A = \int (f - g)\,dx$ .
• Symmetry line of a uniform plate: the centroid is on it. Composite plate: add moments, never centroids.
• Pappus: $V = 2\pi d A$ , $d$ to the AXIS, region on one side of it.
• $P = \rho g x$ with $x$ the depth below the SURFACE; $\rho g = 9800$ N/m $^3$ , $\delta = 62.5$ lb/ft $^3$ .
• $F = \int \rho g\,x\,w(x)\,dx = \rho g\,\bar{d}\,A$ . Inclined plate: $\frac{dx}{\sin\theta}$ .
• Triangle: centroid one third of the height from the base. Half-disc: $\frac{4r}{3\pi}$ from the diameter.

Frequently asked questions

How do I find the centroid of a region between two curves?

Find where the curves meet and which one is on top. The area is the integral of top minus bottom. The x-coordinate is the integral of x times top minus bottom, divided by the area. The y-coordinate is the integral of one half of top squared minus bottom squared, divided by the area. Then check that the point lies inside the region and on any line of symmetry.

Why is there a one half in the formula for the y-coordinate of the centroid?

A thin vertical strip has its own centroid at its midpoint, halfway between the bottom curve and the top curve. Its moment about the x-axis is its area, top minus bottom times dx, multiplied by that midpoint height, top plus bottom over two. The product is one half of top squared minus bottom squared. The one half comes from the midpoint, not from anything else.

How do I calculate the hydrostatic force on a dam or a vertical plate?

Measure the depth x downward from the water surface. Cut the plate into thin horizontal strips, each at one depth. A strip of width w(x) and thickness dx feels the pressure rho g x, so the force is the integral of rho g times x times w(x). For water, rho g is 9800 newtons per cubic metre. The answer equals rho g times the depth of the centroid times the area.

When can I use the theorem of Pappus?

Whenever a plane region is rotated about a line in its plane that does not cut through the region. The volume is two pi times the distance from the centroid to the axis times the area. It is fastest when the centroid is known from symmetry, like a disc making a torus, and when the axis is not a coordinate axis. The distance is always measured to the axis, never to the origin.

Does the length of a lake change the force on its dam?

No. The pressure at a point depends only on its depth below the surface, not on how much water lies behind it. The force on the dam depends only on the depth of the water and the shape of the wetted face. A dam holding back a small pond and a dam holding back a long lake feel the same force if the depth and the face are the same.

Practise it

Corrected exercises: Moments, centres of mass and hydrostatic force, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: moments, centres of mass and hydrostatic force (MATH 141)

The essentials

Moments: a piece times its distance to the axis

Hydrostatic force: a pressure times an area, the depth from the surface

The rules in table form

Piece, arm, contribution: the whole chapter in one grid

The mistakes that cost marks

1. Building M_x from the x-coordinates

2. Squaring the difference instead of taking the difference of the squares

3. Using the top of the strip as its arm

4. Averaging the centroids of the pieces of a composite plate

5. Measuring the depth from the bottom of the plate

6. Taking the pressure at the deepest point, or at the top

7. Believing the centroid must lie on the plate

8. Measuring the Pappus distance from the origin, or through the region

9. Forgetting the slant on an inclined plate

Which method to choose

Hydrostatic force: which route, by the FORM of the plate

How the answer is expected to be written

Setting up a hydrostatic force integral

Check before you hand in

The typical problem, taken apart

The centroid of a region between two curves, checked by symmetry, then used by Pappus

Learn by heart

Frequently asked questions

How do I find the centroid of a region between two curves?

Why is there a one half in the formula for the y-coordinate of the centroid?

How do I calculate the hydrostatic force on a dam or a vertical plate?

When can I use the theorem of Pappus?

Does the length of a lake change the force on its dam?

Corrected exercises: Moments, centres of mass and hydrostatic force, MATH 141 at McGill

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