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158 lessons

Essential Calculus for ML

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Essential Calculus for ML · 158 lessons

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What a double integral measures

Reading what an integral measures

Explanation

The double integral $\displaystyle\iint_R f(x,y)\,dA$ is defined as a signed volume, yet it also gives the area of $R$ when $f(x,y)=1$ . How can both be true?

To understand what an integral, double or otherwise, represents we need to make clear some basic terminology.

The piece and the integrand

Definition

Every integral is the limit of sums over small pieces. One integral sign chops an interval into small lengths $dx$ ; two chop a region into small areas $dA$ .

The integrand is what each piece is multiplied by before the pieces are added. When nothing is written, it is $1$ and each piece counts its own size.

Since a piece can never be negative, the sign of what is measured comes from the integrand alone.

The diagram below builds this understanding in five steps, one integral at a time. Work through them with the buttons, moving the piece as you go.

0,0

az = 3.14

el = 1.57

bank = 0.00

x

y

z

a

b

Δ x

height

f (x)

height

f (x)

R

Δ A

Δ x

Δ y

height

f (x, y)

A (x)

Drag the scene to rotate it.

Blue: the piece · Red: what the integrand multiplies by

Move the piece along

x

x = 0.75

Move the piece along

y

y = 0.75

Step 1: $\int_{a}^{b} d x$ , a length

One integral sign chops the interval from $a$ to $b$ into small lengths, one of them shown in blue. The picture shows a finite chop, and the integral is the limit as the pieces shrink, which is the value given below.

piece

Δ x = 0.25

integrand

1

this piece adds

0.25

integral

b - a = 2

Piece a lengthIntegrand

1

Measures a length

Set out by piece and by integrand, four of the five integrals in the diagram fill a grid.

	Integrand $1$	Integrand a height
One integral sign, piece a length $dx$	$\displaystyle\int_a^b dx$ , the length $b-a$	$\displaystyle\int_a^b f(x)\,dx$ , a signed area
Two integral signs, piece an area $dA$	$\displaystyle\iint_R dA$ , the area of $R$	$\displaystyle\iint_R f(x,y)\,dA$ , a signed volume

Down a column the piece changes and across a row the integrand changes. The left column is never signed, because the integrand $1$ is never negative. The right column is signed, because a height can be.

The fifth has a third kind of integrand. In $\displaystyle\int_a^b A(x)\,dx$ it is a cross-sectional area, so the integral sits in the top row with a length piece, and a length times an area is a signed volume.

Practice questions

6 questions

Suppose that

$R$ is a region of the $xy$ -plane.
$a<b$ .
$A(x)$ is the area of the cross-section at $x$ of a solid standing over $R$ .
$f(x,y)$ takes both positive and negative values on $R$ .

Which of these integrals can be negative?

Select the correct answer:

$\displaystyle\iint_R dA$

$\displaystyle\int_a^b dx$

$\displaystyle\iint_R f(x,y)\,dA$

$\displaystyle\int_a^b A(x)\,dx$

+ 5 more questions

Reading a density integral

Explanation

So far the integrand has been a pure number, a height, or a cross-sectional area, each turning a piece of the region into a length, an area, or a volume. It can instead be an amount per unit area, such as people per square kilometre in a district.

Density

Definition

A density on $R$ is a function $\delta(x,y)$ giving an amount per unit area at each point of $R$ .

A piece of area $\Delta A$ holds about $\delta(x,y)\,\Delta A$ of the amount, so adding over all the pieces of $R$ gives the total,

\iint_R \delta(x,y)\,dA .

The board below shows a district occupying a region $R$ , with one piece of area $\Delta A$ picked out and the number of people $\delta(x,y)\,\Delta A$ it holds marked beside it.

0,0

Δ A

y = 0

x = 0

y = 4 - x

R

x

y

Shown is a district occupying the triangle

R

bounded by

x = 0

y = 0

and

y = 4 - x

. The shaded piece of

R

has area

Δ A

, and where the population density on it is

δ (x, y)

, the number of people it holds is

δ (x, y) Δ A

Practice questions

4 questions

A district occupies the rectangle $R$ bounded by $x=0$ , $x=4$ , $y=0$ and $y=3$ , with $x$ and $y$ in kilometres. The region $R$ is shown below.

0,0

y = 0

y = 3

x = 0

x = 4

R

x

y

The population density at a point $(x,y)$ is $\delta(x,y)=100x$ people per square kilometre. Which integral gives the number of people in the district?

Select the correct answer:

$\displaystyle\int_0^4 100x\,dx$

$100x\times 12$

$\displaystyle\iint_R 100x\,dA$

$\displaystyle 100\iint_R dA$

+ 3 more questions

Part of

Double integrals

See all 10 lessons, what comes first, and what this unlocks.

Master the fundamentals

What a double integral measures

Explanation

Step 1: ∫abdx, a length

Explanation

Double integrals

Step 1: $\int_{a}^{b} d x$ , a length