Calculus/Multiple integration

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Multiple integration

Double integrationEdit

 
Illustration of partitioning interval

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  .

 
Illustration of Cartesian product by an example.

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

 
of  -tuples (or vectors).

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.

 
Illustration of partitioning rectangle in  .

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

 
exists. In that case, we denote this limit by
 
(  is a mnemonic of area), and call it the 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 integralsEdit

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

 
and
 

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.

1 Choose the correct expression(s) for the integral  .

 
 
 
 

2 Choose the correct expression(s) for the integral  .

 
 
 
 
 


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 R2Edit

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

 
if the latter integral exists.

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

 
(ii) If   in which functions   and   are continuous, then
 
in which   are real numbers satisfying the above conditions.

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,

 
The result follows. (We say that whether equality holds in the above inequalities does not matter, because the definite integral over a point equals zero anyway, so it does not affect the result.)

Part (ii): ( )

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

 
The result follows.  

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.
 
Illustration of approach 1
 
Illustration of approach 2

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

 
Approach 2: 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.
 
Illustration of finding volume of tetrahedron by double integration

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

 
Therefore, the equation of   is
 
The desired volume is the volume under the graph of   over a region  , and the region   is the projection of the tetrahedron on the xy-plane, which is the triangle with vertices   in  . Because the line passing through   and   has equation  , the bound for   is  , and the bound for   is   given a fixed  . Thus, the desired volume is
 
On the other hand,
 
which equals the desired volume.  

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

Proof.
 
Illustration of the bounds of the region for this integral (region ABC).

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

 
if the integrals exist.

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

 
if the integrals exist.

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

 
The result follows.  

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

 
by proposition about volume given by double integration (because   for each   )  

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 integrationEdit

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

 
exists. In that case, we denote this limit by
 
(  is a mnemonic of volume), and call it the 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

 
(ii) if  , then
 
(iii) if  , then
 
(iv) if  , then
 
(v) if  , then
 
(vi) if  , then
 
in which each function involved is continuous. That is, we can use either one of all   possible integration orders for iterated integrals to compute triple integrals, with suitable bounds.

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  .


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Multiple integration