# Calculus/Multiple integration

## Double integration[edit | edit source]

For (Riemann) integrals, we consider the Riemann sum.
Recall in the one-variable case, we partition an interval into more and more subintervals with smaller and smaller width, and we are integrating over the interval
by summing the areas of corresponding rectangles for each subinterval.
For the multivariable case, we need to do something similar, but the problem arises when we need to partition 'interval'
in or in general. (Actually, we only have the term *interval* in .)

In multivariable case, we need to consider not just 'interval' itself (which is undefined in multivariable case), but *Cartesian product*
over intervals for , and more generally *n-ary Cartesian product* over intervals for .

**Definition.** (*n*-ary Cartesian product)
The *n-ary Cartesian product* over sets is the set

**Remark.**

- recall that
*interval*is essentially a set, e.g. . - 2-ary Cartesian product is simply called Cartesian product, e.g. is
*Cartesian product*over two intervals, and also a*rectangle*with side lengths 1 and 2 geometrically. *special case*: is called*n-ary Cartesian power*and is denoted by , e.g. .

Area (for ), volume (for ) or measure (for each positive number ) of geometric objects (e.g. rectangles in and cubes in ) in is the product of the lengths of all its sides (in different dimensions).

**Example.**

- is the Cartesian product over two intervals (It is square with side length in geometrically)
- is the 3-ary Cartesian product over three intervals (It is a rectangular cuboid in geometrically)
- is the 4-ary Cartesian product over four intervals, or 4-ary Cartesian power (it is denoted as ) (It is a 4-dimensional cube in geometrically)

Now, we are ready to define multiple integral in an analogous way compared with single integral. For simplicity, let us first discuss double integral, and then generalize it to multiple integral in an analogous way.

**Definition.** (Double integrals)
Let be a function defined on a rectangle in .
Consider a partition of into small rectangles with areas
respectively. Choose an *arbitrary* point in the th rectangle. The function
is *integrable* over if

*double integral*of over .

**Remark.**

- the limit is the Riemann sum over with the partition of into small rectangles
- as we partition more and more small rectangles (with smaller and smaller area) from the rectangle ,
*which arbitrary*point we choose in each rectangle becomes less and less important, because different points in each small rectangle become 'closer and closer' together, and thus the position of different points are more and more 'similar' - we may assume that nice functions (functions given in the questions here are nice, unless stated otherwise) to be integrable, so we do not need to check integrability of every function we encounter here (Checking of integrability involves the formal definition of limit, and is out of scope here)
- the process of computing
*double integral*is called*double integration*

A physical meaning of double integration is computing volume.

**Proposition.** (Volume given by double integration)
Let be an *integrable* function defined on a rectangle in . Suppose
for each . Then, the volume under the graph of over is

**Remark.**

- if is
*not*always , the region below the -plane has negative volume, so this negative volume will cancel out the positive volume, which may or may not be desired - if is always negative, then the volume computed by this formula is negative, and we usually take its absolute value to get the volume, because volume is usually defined to be nonnegative

Let's also introduce some properties of double integral to ease computation of double integral.

**Proposition.** (Properties of double integral)
Let and be *integrable* functions defined on a rectangle in .
Then, the following properties hold.

- (Linearity) is also integrable over for each real number and , and
- (Monotonicity) if for each , then
- (Triangle inequality) is integrable over and

**Proposition.** (Continuity implies integrability)
Let be a *continuous* function defined on a rectangle in . Then, the function is integrable.

**Remark.**
Because nice functions are often continuous, most nice functions are integrable.

### Iterated integrals[edit | edit source]

Thankfully, we need not always work with Riemann sums every time we want to calculate an integral in more than one variable. There are some results that make life a bit easier for us.
Before stating the result, we need to define *iterated integral*, which is used in the results.

**Definition.** (Iterated integral)
Let be a continuous function defined on a rectangle in .
The iterated integral is defined by

**Remark.**
e.g., for the integral , we first compute the definite integral with respect to by treating as a *constant*. Then, we compute the remaining integral (in terms of ) with respect to .

Computation of iterated integrals is generally much easier than computing the double integral directly using Riemann sum. So, it will be nice if we have some relationships between iterated integral and double integral for us to compute double integral with the help of iterated integral. It is indeed the case and the following theorem is the bridge between iterated integral and double integral.

**Theorem.** (Fubini's theorem)
Let be a continuous function defined on a rectangle in . Then,

**Remark.**

- i.e., we can use iterated integral in
*either*order to compute the corresponding double integral - we should notice the change of bounds for each integral after changing the integration order

**Example.** (Volume of rectangular cuboid)

(i) Prove that the volume under the graph of over the rectangle is .

(ii) Hence, prove that volume of rectangular cuboid with length , width and height is .

**Proof.**

(i) Because for each , by the proposition of volume given by double integration, the volume is . By Fubini's theorem, this equals

**Remark.** Geometrically, the graph is a rectangular cuboid, and its volume is given by product of each side length, namely , which matches with our answer.

(ii) The desired volume is given by (double integral of a constant function over the rectangle with length and width ). (We may also express the integral as without affecting the result.) Then, by Fubini's theorem,

### Double integrals over more general regions in **R**^{2}[edit | edit source]

We have defined double integrals over rectangles in . However, we often want to compute double integral over regions with shape other than rectangle, e.g. circle, triangle, etc. Therefore, we will discuss an approach to compute double integrals over more general regions reasonably, without altering the definition of double integrals.

Consider a function in which is a general region.
To apply the definition of double integrals, we need to transform the general region to a rectangle (say ).
An approach is finding a rectangle *containing* (i.e., ), and let for each
lying *outside* (i.e., for each ). Because the value of the function is zero
outside the region we are integrating over, this *does not* change the volume under the graph of over , so
this way is a good way to define such double integrals. Let's define such double integrals formally in the following.

**Definition.** (Double integrals over general regions)
Let be a function defined on a region (of arbitrary shape) . Then, we define
for each lying outside (i.e., for each ), and define the *double integral* of function
over the region by

**Remark.**
Then, we may compute double integrals over general regions by computing the corresponding Riemann sum for the latter integral.

However, this way of computation (by computing Riemann sum) is generally very difficult, and usually we use a generalized version of Fubini's theorem to compute such integrals. It will be discussed in the following.

**Theorem.** (Generalized Fubini's theorem)
Let be a continuous function defined on a region (with arbitrary shape) . Then, the following hold.

(i) If in which functions and are continuous, then

**Proof.**
We can prove this theorem by Fubini's theorem (ungeneralized version) and applying the definition of double integrals
over general regions. (We can use Fubini's theorem because we assume that functions and are continuous.)

Part (i): ()

Take arbitrary rectangle containing (i.e. and ). Then, define if . After that,

Part (ii): ()

Similarly, take arbitrary rectangle containing (i.e. and ). Then, define if . After that,

**Remark.**
Usually, finding the bounds for and is the most difficult step when we compute such double integrals.

**Example.**
Let be the triangle in with vertices and . Prove that .

**Proof.**

Approach 1: The bound for is . Given a fixed , the bound for is . Thus, the integral is

**Example.** (Volume of tetrahedron)
Consider a tetrahedron in with vertices and in which and are positive numbers. Define vectors and . Prove that volume of the tetrahedron is . (i.e., of the volume of parallelepiped spanned by the vectors and )

**Proof.**

Let the plane containing and be . To find the equation of , consider its normal vector. A normal vector of is

**Example.** (Switching integration order) Prove that is approximately . (correct to two decimal places)

**Proof.**

If we integrate in this order, the computation will be very tedious. Thus, we will interchange the integration order using generalized Fubini's theorem to ease the computation. Originally, the bounds are and (given a fixed ). If we integrate over first instead, the suitable expressions of bounds are and (given a fixed ). Therefore,

**Proposition.** (Union of regions for double integration)
Let be a continuous function defined on a bounded set in . If is
the union of and , in which the *overlapping set* of and (in ) has *zero area*, then

**Remark.**

- Example of analogous case in : (the overlapping set of and has
*zero length*) *Curves*and*points*have zero area in .

**Corollary.** (Subtraction of regions for double integration)
Let be a continuous function defined on a bounded set in . If is
the union of and , in which ,then

**Proof.**
First, the overlapping set of and is , which has zero area.
Next, because ,

**Proposition.** (Area given by double integration)
Let be a bounded region in . Then, the area of is

**Proof.**
The volume under the graph of the constant function over equals the base area (which is the area of ) times the height (the height is one in this case).
Therefore, the area of is the volume under the graph of over , which is

**Remark.**
Recall that area of a bounded region can also be found by single integration. In some cases, using this proposition instead is more convenient.

**Example.**
Let be a region bounded by the curves and , in which is a positive number. Prove that the area of bounded region is .

**Proof.**

Solving and , we get . Because , intersection points of these two curves are and . Therefore, the bound for is , and given a fixed , the bound for is . Thus, the desired area is

**Remark.**Geometrically, the bounded region is a disk (region in a plane bounded by a circle) of radius .

## Triple integration[edit | edit source]

The concepts in the section of double integrals apply to *triple integrals* (and also multiple integrals generally) analogously.
We will give several examples for triple integrals in this section.

**Definition.** (Triple integrals)
Let be a function defined on a rectangular box (or rectangular cuboid) in .
Consider a partition of into small boxes with volumes
respectively. Choose an *arbitrary* point in the th box. The function
is *integrable* over if

*triple integral*of over .

**Remark.**

- The process of computing
*triple integral*is called*triple integration*

**Theorem.** (Generalized Fubini's theorem (triple integrals version))
Let be a continuous function defined on a region (with arbitrary shape) . Then, the following hold.

(i) if , then

**Proposition.** (4-dimensional volume given by triple integration) Let be an *integrable* function defined on a rectangular box in . Suppose for each .
Then, the *4-dimensional volume* under the graph of over is

**Remark.**
It is just a theoretical result, and is difficult to be visualized.

**Proposition.** (Volume given by triple integration)
Let be a bounded region in . Then, the *volume* of is

**Example.**
Consider a region that is bounded by a sphere (with radius , which is positive) that lies in the
octant (+,+,+) (i.e., the octant in which and are positive).

(i) Prove that the volume of is .

(ii) Hence, prove that the volume of the region that is bounded by the sphere (in ) is .

**Proof.**

(i) The given bounds for is and . We can express the bounds as follows:

- given a fixed , aim: finding bounds for in the form of .

- Steps:

- given fixed and ( is selected from its bound, namely , and is selected from its bound, which depends on the fixed ), aim: finding bounds for in the form of .

- Steps:

Therefore, by generalized Fubini's theorem (triple integrals version) (i) and proposition about volume given by triple integration, the desired volume is

(ii) Because there are eight octants (in ), and each octant is symmetric to each other. There are seven more regions that are symmetric to in octants other than octant (+,+,+). Thus, the desired volume is

**Remark.**The region mentioned in (ii) is a ball (solid figure bounded by a sphere) of radius .