# m213multipleintegration by nuhman10

VIEWS: 4 PAGES: 10

• pg 1
```									Math 213 DS                      Notes on Multiple Integration

Double- Integration in Polar Coordinates

(Transformation Equations: x = rcos y=rsin     PolarCartesian
r = (x2 + y2)1/2  =arctan(y/x) CartesianPolar)

Recall the definition of the double integral (In Cartesian Coordinates)

     f (x,y)dxdy = limn,m     
n
i=1
m
j=1   f (xi ,yj) xi yj

Geometrically, this limit corresponds to shrinking unit rectangles to arbitrarily small size:

(Shrinking the unit rectangular area of a rectangle A= x  y to a rectangle: dA = dx dy

By the same token, an infinitesimal unit of area in polar coordinates consists of isolating a circular
sector: A  rr (since for small enough r the arclengths of the circular sectors r are equal to
arbitrary error.) Observe that this error shrinks to zero when we replace “ r” with “dr” so we may may
write for the infinitesimal measure of area (in polar coordinates) dA = rdrd  `

So for polar coordinates, the double integral is defined as:

     f (r,)rdrd = limn,m    
n
i=1
m
j=1   f (ri ,j) ri  i r j

When (or why) would it be useful to perform a double-integral in polar coordinates? It’s a
strategic device applicable when:

       The area of integration  is a segment of a circle
       The integrand is composed of the forms: (x2 + y2)n/2 (which as you know already from Calc
II, are hell to integrate, since they involve Trigonometric Substitutions, which are actually
nothing more than a Polar Coordinate transformation!)

Example: Derive the area formula for polar coordinates from Calc II: Answer: A=                rdrd =
1

/2 (r22 - r12)d

1
r2()
y=d

u(y)                                                       h(y)

y=c
=
=

x=a                  r1()   x=b

   Example: Gaussian Integrals

A “Gaussian” Integral is any integral involving the term: exp(u 2). Naturally the evaluation of exp(u2)udu   

is easy enough (just do a u-substitution). But what about just exp(u2)du ? This innocuous looking
integral resists all the fancy integration techniques you learned in Calc II. In truth, the only way to
intgrate this is via a double integral using polar coordinates. The text gives derivation for the improper
integral:

b                    2
lim b  exp( x )dx 
2
  exp( r
2
)rdrd = 
b                     0 0

b

 exp(x
2
Another interesting question is evaluating the definite integral:                               )dx
b
b                b     b b                               b    2 R

 exp(x )dx             exp( x  y )dxdy                exp( r             )rdrd =  (1  exp( R 2 )
2                         2   2                                        2
Then: :
b               b    b b                             b   0 0

…Of course the exact equality would be over the square in polar coordiantes:

2

4 b sec

 exp(r )rdrd  4                    exp(r       )rdrd 
2                                        2

squarebxb                                 0

4


4
1 4
2

 exp(b sec  )  1 d
2   2

4

…an integral which can only be evaluated by expanding exp(b 2sec2)
into a McClaurin Series!

Example of Application of the Double Integral: Calculating mean and standrad deviation (Bonus
question on Exam IV)

Consider two continuous independent variables (x,y) with joint probability density:
p(x,y)=  -1/2exp(-(x2+y2)) (p gives a measure of probability of location of ordered pair (x,y))

consider the quantity r=(x2+y2)1/2 (the length of the position vector locating (x,y)). Consider a dissk of
radius R in the x,y plane. The average value of r (defined as: r or sometimes also as: r ) is defined as:

2 R                           2      R

 rpdA         r exp(r )rdrd             d  exp(r )r dr
2                               2  2

r    r       disk
   0 0
2 R
   0       0

 pdA                                             (1  exp( R 2 ))
  exp(r       )rdrd
2

0 0

2    R                            R                  exp(r 2 )  R 1 R
                                   

d  exp(r integral 2be evaluated byrparts: 2 
 The upstairs )r dr can  exp(r ) dr                            r    exp(r 2 )dr  
2      2                      2         2
               
0    0                     0                        

2        0 2 0             

 ( exp( R 2 ) R    (1 - exp(-R2 )
(Observe what happens in the R limit…we get our “old friend” !)

Hence the average value becomes:

3
 ( exp( standard deviation: exp(-R the
An interesting quantity is the  R ) R    (1 -defined as ) square root to the variance ( 2) where:
2                          2

2 = r 2  (r ) 2
  (1- exp(-R ) 2
2

Where:
2 R                              2        R

 r pdA               r exp(r )rdrd                d  exp(r )r dr
2                     2      2                                  2  3

r2 r         disk
   0 0
2 R
   0         0

 pdA                                                     (1  exp( R 2 ))
  exp(r )rdrd
2

0 0

The upstairs is ironically easier to integrate than in the previous example! Let dv=exp(-r2)r and u = r2.
Then: du = 2rdr and v = -(1/2)exp(-r2)

So:

2                                                                                              R

                       
R                             R                                                             R

 d  exp(r )r dr  2  [r (exp(r )]r dr    r exp(r )                                          r exp(r 2 )dr
2     3                           2       2                 2         2

0          0                              0                                                    0       0

                     1             1
  1  R 2 exp( R 2 )  exp( R 2 )  
                     2             2

POST EXAM EXERCISE: Method of Iteration:

1     1
x3
(16.3,32)                     x4  y2
dxdy
0    x2

y=x                 y=x2
x=y

1 y                             1

4

0 y
(x4 + y2)-1/2 x3 dxdy=   
0
[1/4(y2 + y2)1/2 - 1/4(y4 + y2)1/2]dy

Triple Integration

3
Over Some Region  in R the triple integral is defined:

    f (x,y,z)dxdydz = limn,m,q           
n
i=1
m
j=1
q
k=1   f (xi ,yj,, zk) xi yjzk

*             Remark : If the region  is a rectangular solid: [a,b][c,d][e,f] =
{(x,y,z) : a< x <b, c< y <d, e< z <f } (constant upper/lower limits) and if the integrand is
separable: f(x,y,z) = g(x)h(y)u(z), the triple integral can be broken down into a simple product of
three definite integrals:

b           d        f

    f (x,y,z)dxdydz =  g(x)h(y)u(z)dxdydz =  g(x)dx h(y)dy u(z)dz
a           c            e

However, IF the region  is any other than a simple rectangular solid, then even though
separable terms of the integrand may be factored out, the integral must still be evaulated in
nested form, as in the case of the double integral.

Evaluating the Triple Integral Using Nested Functions (Method of Iterated Functions)

Prior to introducing the method, observe that the unit Cartesian volume element dV can be
expressed in 3!=6 different ways:

dV= dxdydz = dxdzdy = dydxdz = dydzdx = dzdydx = dzdxdy

Where the volume of any region  is:           dxdydz = dxdzdy = ...
                  

This tells us that there are logically six different possible ways to represent a triple integral. The 3D
region first of all must obviously be sketched. Important: whichever variable we choose to start the
procedure, the upper/lower limits of integration are functions of two variables (the remaining variables of
integration). The next (Second) variable shall be bounded above and below by two functions of one
variable (the reamining variable of integration). The last (3rd) variable of integration shall be bounded
above and below by its constant limits.

5
The best way to introduce the interated functions technique is by way of concrete example. Suppose I
wanted to integrate f(x,y,z) on the sphere: x2 +y2 +z2 = a2 in the first octant:
(a2 - y2 - x2)1/2

y
(a2 - x2)1/2

Integrating f(x,y,z) on the sphere: x2 +y2 +z2 = a2 : If one chooses dV = dzdydx, then the limits of z are
bound by the functions 0(bottom) and (a 2 - y2 - x2)1/2 (top) ( functions of two variables; the remaining
variables of integration, which are y,x). The limits of y are bound by the functions 0( left ) and
(a2 - x2)1/2 ( right ) ( functions of one variable; the remaining variable of integration, which is x) The limts
of of x are bound by the constants 0 and a. This can be visualized as planting a rectangular solid paralell
to the z-axis, bounded above accordingly by solving for z in terms of x and y. Then this solid can be
visualized as sliding along a rectangular "track" paralell to y bounded by (a 2 - x2)1/2 (notice how at this
point the integral becomes a double integral). Hence:
2 2 1/2   2 2 2 1/2
a       (a - x )  (a - x -z )

   f (x,y,z)dxdydz =      { {              f (x,y,z)dx} dy }dz
0         0               0

If f(x,y,z) = g(x)h(y)u(z), (if the integrand were separable), then we can factor out terms:
2 2 1/2         2 2 2 1/2
a              (a -x )         (a - x -z )

   f (x,y,z)dxdydz = u(z){ h(y){g(x)dx} dy }dz
0              0               0

Which is NOT the same thing as merely reducing the integral to a product of three itegrals! (Refer to
above remark)

6
Further Tips for Evaluation of Triple Integrals in Cartesian Coordinates:

     Pay close attention to symmetrical regions about origin (cubes, spheres, etc). If the integrand has
any terms which are odd in any (or every) of the variables (x,y,z) then the entire integral is zero.
(See problems #42, #43, 16.7). Also, If the integrand has any terms which are even in any (or every)
of the variables (x,y,z) then we can evaluate the integral in just one or two or four of the octants,
and multiply answer by 2 or 4 or 8 depending on extent of symmetry. For example:

1) Suppose region is Unit sphere, and suppose:                  f(-x,y,z) = f(x,y,z) (even in x), then:

    f (x,y,z)dxdydz = 2      /2   f (x,y,z)dxdydz

where: /2 is the unit hemisphere. (Formed by slicing original sphere along the yz plane)

2) Suppose region is Unit sphere, and suppose:                  f(-x,-y,z) = f(x,y,z) (even in x,y), then:

    f (x,y,z)dxdydz =4      /4   f (x,y,z)dxdydz

where: /4 is the unit quarter-sphere. (Formed by slicing original sphere along the yz and zx
planes)

3) Suppose region is Unit sphere, and suppose:                      f(-x,-y,-z) = f(x,y,z) (even in x,y,z), then:

    f (x,y,z)dxdydz =4      /8   f (x,y,z)dxdydz

where: /8 is the unit eighth-sphere. (Formed by slicing original sphere along the yz and zx and xy
planes)

     Much of the difficulty lies in originally setting up the region, getting clear the type of region of
integration to perform. Naturally this can be most difficult when all you're told is two regions.
Suggestion: Set the equation equal to one another: this give a space curve describing where two
such surfaces meet such that this curve may be used as a projection giving information of what the
"shadows"of the 3D surfaces look like when projected on any of the principal planes.

In the case of two intersecting cylindrical perpendicular to one another this may involve splitting the
integrands into different zones of applicability (See solution to problem 32, 16.7)

Last of all, if just given a region, and not told the order of variables by which to perform the integration,
it can appear to be real problem to judge ahead of time which out of the six possilities minimizes the
computation. Naturally one should keep in mind to minimize the involvement of radical
n/2
binomial/trinomial terms ( (k + x2)          , (k + x2 + z2)m/2 , etc.) . For example, suppose the xy
projection of a region  was parabolic : xy = { 0 < x < 1, 0 < y < x +1 } then it's easier to
2

integrate with respect to y first and then x, than vice versa, to avoid the use of radical expressions.
If it's however inevitable that that your expression contains radicals, once expressed as a double
integralfree free to use polar coordinates, or in the last resort, a trignonometric substitution.


7
Example (#49, 16.9) Find the center of mass of the solid in 35 if the density varies with 1+y

: { 2-x < z < 4-x2, 0< y <3 } To derive limits on x observe that: xy is a rectangular region
bounded by: 0< y <3. To find x simply solve: 2-x = 4-x2  x2 -x -2 =0  (x-2)(x+1)=0

Thus: : xy = {0< y <3, -1< x <2}

2   3    4-x2                   2       3

M=        dV =   k(1+y)dzdydx = k (1+y)[ (4-x ) - (2-x)]dydx
-1       0   2-x                -1       0
2

3       2                                 3                                     2

0
        
= k (1+y)dy [ (4-x2 ) - (2-x)]dx = k[y+ y2/2]
-1
[2x+ x /2 -x /3]  = 135k/4
0
2       3

-1

To calculate center of mass (X,Y,Z), note: MX =           xdV

2   3     4-x2                          2       3

8
MX =      xdV =   k(1+y)xdzdydx = k (1+y)[ (4-x ) - (2-x)]xdydx
-1   0   2-x                       -1       0
2

3      2                                        3                                     2

0
       
= k (1+y)dy [ (4x-x3) - (2x-x2)]dx = k[y+ y2/2]
-1
[ x + x /3 -x /4]  = 135k/8
0
2           3       4

-1

Hence: X=    MX/M = 0.5

2   3     4-x2                     2       3

MY =      ydV =   k(1+y)ydzdydx = k (1+y)y[ (4-x ) - (2-x)]dydx
-1   0   2-x                       -1       0
2

3           2                                        3                                            2

       -1

= k (1+y)ydy [ (4 -x2) - (2 -x)]dx = k[y2/2+ y3/3]
0
[2 x + x /2 -x /3]  = 351k/4
0
2       3

-1

9
Cylindrical Coordinate System:

Replace the cartesian specification with the polar. Hence:

(x,y,z)  (rcos, rsin,z)

By explicit construction, observe

dV =rdrddz

Hence:  f (x,y,z)dxdydz =       
   f (rcos, rsin,z) rdrddz

   Use Cylindrical system whenever regions are cylindrical (obviously -- but more specifically whenver
region is constant in z-direction). The system expecially comes in handy when cross secion of
cylinder is a cricular arc or parabolic

Example:

10

```
To top