Multivariable-Calculus · Unit 15 · Video 2 · Interactive Practice

Area, Mass, Averages and Center of Mass by Double Integrals

IKey Formulas

Formula	Name	What you add up
$\text{Area}(R) = \iint_R 1\,dA$	Area	$1$ on every piece
$M = \iint_R \delta\,dA$	Mass	Density $\delta(x, y)$ , mass per unit area
$\bar f = \dfrac{1}{\text{Area}(R)}\iint_R f\,dA$	Average value (uniform)	$f$ , divided by the area
$\bar x = \dfrac{1}{M}\iint_R x\,\delta\,dA,\quad \bar y = \dfrac{1}{M}\iint_R y\,\delta\,dA$	Center of mass	$x\,\delta$ and $y\,\delta$ , divided by the mass

Key Insight: $\delta\,dA$ is the mass of a small piece, so it is the weight in a weighted average $\frac{1}{M}\iint_R f\,\delta\,dA$ . With $\delta = 1$ the mass equals the area and every weighted average becomes the ordinary average.

IIVisualization 1 — Adding Up Pieces: Area or Mass

On the plate $0 \le x \le 2$ , $0 \le y \le 1$ with $\delta = 1 + x^2$ , summing $\delta\,\Delta A$ over shrinking pieces converges to $M$ .

Add up Pieces per unit n 2

IIIVisualization 2 — Where Does the Plate Balance?

An L-shaped plate balances at $(\bar x, \bar y)$ , the mass-weighted average of $x$ and $y$ ; a heavier arm pulls it over.

Density of the right arm δ₂ 3

IVVisualization 3 — Averaging r and θ Is Not the Center of Mass

For the uniform unit disk, $\bar r = \tfrac{2}{3}$ and $\bar\theta$ depends on the chosen $\theta$ -interval, yet $(\bar x, \bar y) = (0, 0)$ .

θ interval starts at α 0

💡 Polar coordinates are still the right way to evaluate $\iint_R x\,dA$ and $\iint_R y\,dA$ on a round plate; what must be averaged is $x = r\cos\theta$ and $y = r\sin\theta$ , not $r$ and $\theta$ .

VQuiz Questions

Problem 1 · Mass of a Plate

Given: a plate occupies $R = [0, 2] \times [0, 3]$ with density $\delta(x, y) = y$ . Find its mass $M$ .

✅ Correct!

M = \iint_R y\,dA = \int_0^2 \tfrac{9}{2}\,dx = 9

❌ That is the area.

6 = \iint_R 1\,dA

; the mass adds up

\delta\,dA = y\,dA

, not

1\,dA

❌ Only the inner integral.

\int_0^3 y\,dy = \tfrac{9}{2}

is the mass of a strip of unit width; the plate is

2

units wide, so integrate over

0 \le x \le 2

as well.

❌ Check the antiderivative.

\int_0^3 y\,dy = \tfrac{y^2}{2}\Big|_0^3 = \tfrac{9}{2}

, not

9

— the factor

\tfrac12

was dropped.

Show solution

The mass is the double integral of the density:

M = \iint_R \delta\,dA = \int_0^2 \int_0^3 y\,dy\,dx = \int_0^2 \frac{y^2}{2}\Big|_0^3\,dx = \int_0^2 \frac{9}{2}\,dx = 9.

For comparison, the area is $\iint_R 1\,dA = 6$ , so the average density is $M / \text{Area}(R) = \tfrac{9}{6} = \tfrac{3}{2}$ , the average of $y$ over $[0, 3]$ .

Problem 2 · The Polar Averaging Trap

Given: the uniform upper half-disk $0 \le r \le 1$ , $0 \le \theta \le \pi$ (take $\delta = 1$ ). A student averages $r$ and $\theta$ and gets $\bar r = \tfrac{2}{3}$ , $\bar\theta = \tfrac{\pi}{2}$ , i.e. the point $(0, \tfrac23)$ . Find the true $\bar y$ .

✅ Correct!

\bar y = \tfrac{2/3}{\pi/2} = \tfrac{4}{3\pi} \approx 0.424

, well below the student's

\tfrac23

❌ That is the student's $\bar r$ . Averaging

r

is not averaging

y

\bar y

is the average of

y = r\sin\theta

, and points near the base of the half-disk have small

y

even when

r

is large.

❌ The area element is missing its $r$ . In polar coordinates

dA = r\,dr\,d\theta

, so the integrand is

r\sin\theta \cdot r = r^2\sin\theta

, not

r\sin\theta

❌ Wrong mass. The half-disk has area

M = \tfrac{\pi}{2}

, not

\pi

; dividing

\tfrac23

\pi

halves the answer.

Show solution

With $\delta = 1$ the mass is the area: $M = \tfrac12 \pi (1)^2 = \tfrac{\pi}{2}$ . Evaluate $\iint_R y\,dA$ in polar coordinates, with $y = r\sin\theta$ and $dA = r\,dr\,d\theta$ :

\iint_R y\,dA = \int_0^{\pi}\int_0^1 r^2 \sin\theta\,dr\,d\theta = \frac13 \int_0^{\pi} \sin\theta\,d\theta = \frac13 \cdot 2 = \frac23.

\bar y = \frac{1}{M}\iint_R y\,dA = \frac{2/3}{\pi/2} = \frac{4}{3\pi} \approx 0.424.

By symmetry $\bar x = 0$ . The polar-average point $(0, \tfrac23)$ sits too high: $\bar r$ weights the far rim heavily, but the points near the rim at the base, with $\theta$ near $0$ or $\pi$ , have small $y$ .

Problem 3 · Center of Mass with Variable Density

Given: the triangular plate $0 \le y \le x \le 2$ with density $\delta(x, y) = y$ . Find its center of mass $(\bar x, \bar y)$ .

What is $\bar x$ ?

What is $\bar y$ ?

✅ Correct!

(\bar x, \bar y) = (\tfrac32, 1)

: the density

\delta = y

pulls the balance point up and right of the uniform centroid

(\tfrac43, \tfrac23)

❌ That ignores the density.

\tfrac43

is the centroid of the uniform triangle; here each piece counts with weight

\delta = y

❌ Divide by the mass.

\iint_R x\,\delta\,dA = 2

; then

\bar x = 2 / M

with

M = \tfrac43

❌ Divided by the area instead of the mass.

\tfrac{2}{2} = 1

uses

\text{Area}(R) = 2

; a weighted average divides by the total weight

M = \iint_R \delta\,dA = \tfrac43

❌ That is the uniform centroid's $\bar y$ (it also comes from dividing

\tfrac43

by the area

2

). Divide

\iint_R y\,\delta\,dA = \tfrac43

by the mass

M = \tfrac43

❌ Divide by the mass.

\tfrac43 = \iint_R y\,\delta\,dA

is the moment, not the average;

\bar y = \tfrac{4/3}{4/3}

❌ That is $\bar x$ . For

\bar y

the integrand is

y\,\delta = y^2

, not

x\,\delta

Show solution

Integrate $y$ first, from $0$ to $x$ , then $x$ from $0$ to $2$ .

Mass:

M = \int_0^2\int_0^x y\,dy\,dx = \int_0^2 \frac{x^2}{2}\,dx = \frac{8}{6} = \frac43.

Moments:

\iint_R x\,\delta\,dA = \int_0^2\int_0^x xy\,dy\,dx = \int_0^2 \frac{x^3}{2}\,dx = 2,

\iint_R y\,\delta\,dA = \int_0^2\int_0^x y^2\,dy\,dx = \int_0^2 \frac{x^3}{3}\,dx = \frac43.

Center of mass:

\bar x = \frac{2}{4/3} = \frac32, \qquad \bar y = \frac{4/3}{4/3} = 1.

Problem 4 · Average Value over a Disk

Given: $f(x, y) = x^2 + y^2$ on the unit disk $x^2 + y^2 \le 1$ . Find the average value $\bar f$ .

✅ Correct!

\bar f = \tfrac{1}{\pi}\cdot\tfrac{\pi}{2} = \tfrac12

: the disk's area crowds toward the rim (half of it lies beyond

r = \tfrac{1}{\sqrt2}

), so

r^2

averages more than the

\tfrac13

it would average if every

r

[0, 1]

were equally likely.

❌ The area element is missing its $r$ .

dA = r\,dr\,d\theta

, so integrate

r^2 \cdot r = r^3

, not

r^2

❌ Divide by the area.

\tfrac{\pi}{2} = \iint_R f\,dA

is the sum of the values; the average divides by

\text{Area}(R) = \pi

❌ Wrong area.

2\pi

is the length of the

\theta

-interval, not the area of the disk;

\text{Area}(R) = \pi

Show solution

In polar coordinates $f = r^2$ and $dA = r\,dr\,d\theta$ :

\iint_R f\,dA = \int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta = 2\pi \cdot \frac14 = \frac{\pi}{2}.

The area of the unit disk is $\pi$ , so

\bar f = \frac{1}{\text{Area}(R)}\iint_R f\,dA = \frac{\pi/2}{\pi} = \frac12.

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