MULTIVARIABLE CALCULUS 06

Multiple Integrals

Multiple integrals extend ordinary integration to quantities spread over regions in two or more dimensions.

The main idea

An ordinary integral such as

\[ \int_a^b f(x)\,dx \]

accumulates contributions along a one-dimensional interval.

A double integral such as

\[ \boxed{ \iint_R f(x,y)\,dA } \]

accumulates contributions over a two-dimensional region \(R\).

1. From small intervals to small areas

In one variable, a small contribution has the form

\[ f(x)\,\Delta x. \]

In two variables, divide a region into small pieces of area

\[ \Delta A. \]

A small contribution is approximately

\[ f(x,y)\Delta A. \]

Adding many such pieces gives

\[ \sum f(x_i,y_i)\Delta A_i. \]

As the pieces become smaller, this approaches the double integral.

2. Double integral notation

The expression

\[ \iint_R f(x,y)\,dA \]

is read as

“the double integral of \(f(x,y)\) over the region \(R\)”.

The symbol

\[ dA \]

represents a small element of area.

3. Rectangular regions

Suppose

\[ R= \{(x,y):a\le x\le b,\;c\le y\le d\}. \]

Then

\[ dA=dy\,dx \]

\[ dA=dx\,dy. \]

The double integral can be calculated as an iterated integral:

\[ \boxed{ \iint_R f(x,y)\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dx } \]

4. Why two integrations?

The inner integral accumulates in one direction.

The outer integral then accumulates those results in the second direction.

\[ \boxed{ \text{integrate in }y \rightarrow \text{then integrate in }x } \]

This turns a two-dimensional accumulation problem into two one-dimensional integrations.

5. Simple example

Evaluate

\[ \int_0^2\int_0^1 (x+y)\,dy\,dx. \]

First integrate with respect to \(y\), treating \(x\) as constant:

\[ \int_0^1(x+y)\,dy. \]

This gives

\[ xy+\frac{y^2}{2} \Bigg|_0^1. \]

Therefore,

\[ x+\frac12. \]

6. Now integrate with respect to \(x\)

So the double integral becomes

\[ \int_0^2 \left( x+\frac12 \right)dx. \]

Therefore,

\[ \left[ \frac{x^2}{2}+\frac{x}{2} \right]_0^2. \]

\[ 2+1=3. \]

Hence,

\[ \boxed{ \int_0^2\int_0^1(x+y)\,dy\,dx=3 } \]

7. Geometric meaning

\[ f(x,y)\ge0, \]

then

\[ \iint_R f(x,y)\,dA \]

can be interpreted as the volume under the surface

\[ z=f(x,y) \]

and above the region \(R\).

8. Diagram

9. Integrating the constant function \(1\)

\[ f(x,y)=1, \]

then

\[ \iint_R1\,dA \]

gives the area of the region \(R\).

\[ \boxed{ \text{Area}(R) = \iint_R dA } \]

10. Example: area of a rectangle

Let

\[ 0\le x\le3, \qquad 0\le y\le2. \]

Then

\[ \text{Area} = \int_0^3\int_0^2 1\,dy\,dx. \]

The inner integral gives

\[ 2. \]

Then

\[ \int_0^3 2\,dx=6. \]

\[ \boxed{ \text{Area}=6 } \]

11. Non-rectangular regions

Suppose a region is bounded by

\[ a\le x\le b \]

and

\[ g_1(x)\le y\le g_2(x). \]

Then

\[ \boxed{ \iint_R f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx } \]

The inner limits may depend on the outer variable.

12. Example over a triangular region

Consider the region

\[ 0\le x\le1, \qquad 0\le y\le x. \]

Its area is

\[ \int_0^1\int_0^x1\,dy\,dx. \]

The inner integral gives

\[ x. \]

Therefore,

\[ \int_0^1x\,dx = \frac12. \]

Hence,

\[ \boxed{ \text{Area}=\frac12 } \]

13. Order of integration

We may sometimes integrate

\[ dy\,dx \]

or reverse the order and integrate

\[ dx\,dy. \]

For sufficiently well-behaved functions over suitable regions, both orders give the same result.

But the limits must be rewritten to describe the region correctly.

14. Why changing the order can help

One order of integration may produce a difficult integral while the other is simpler.

So understanding the geometry of the region is often as important as the integration itself.

15. Triple integrals

A triple integral accumulates over a three-dimensional region \(V\):

\[ \boxed{ \iiint_V f(x,y,z)\,dV } \]

Here

\[ dV \]

represents a small volume element.

16. Volume from a triple integral

\[ f(x,y,z)=1, \]

then

\[ \iiint_V1\,dV \]

gives the volume of \(V\).

\[ \boxed{ \text{Volume}(V) = \iiint_V dV } \]

17. Density and mass

Suppose a thin plate has density

\[ \rho(x,y). \]

Then a small piece of area \(dA\) has approximate mass

\[ \rho(x,y)\,dA. \]

The total mass is

\[ \boxed{ M = \iint_R \rho(x,y)\,dA } \]

18. Probability densities

For two continuous random variables with joint density

\[ p(x,y), \]

the probability of lying in a region \(R\) is

\[ \boxed{ P((X,Y)\in R) = \iint_R p(x,y)\,dA } \]

The total probability must satisfy

\[ \iint_{\mathbb{R}^2} p(x,y)\,dA = 1. \]

19. Average value over a region

The average value of \(f\) over a region \(R\) is

\[ \boxed{ f_{\text{avg}} = \frac{1}{\text{Area}(R)} \iint_R f(x,y)\,dA } \]

This is the two-dimensional version of the one-variable average-value formula.

20. The pattern from one to several dimensions

\[ \boxed{ \int f(x)\,dx = \text{accumulation along a line} } \] \[ \boxed{ \iint f(x,y)\,dA = \text{accumulation over an area} } \] \[ \boxed{ \iiint f(x,y,z)\,dV = \text{accumulation through a volume} } \]

21. The central idea

Multiple integrals extend the same accumulation principle used in ordinary integration.

\[ \boxed{ \text{small contribution} = \text{density or height} \times \text{small region} } \] \[ \downarrow \] \[ \boxed{ \text{sum over all small regions} } \] \[ \downarrow \] \[ \boxed{ \text{multiple integral} } \]

The dimension changes, but the underlying idea of accumulation remains the same.