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Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ].PowerPoint Presentation

Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ].

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Suppose f(x,y,z) is bounded for all (x,y,z) in a box B = [a,b][c,d][p,q].

z

(a,c,q)

(a,d,q)

(b,c,q)

(b,d,q)

y

(a,c,p)

(a,d,p)

x

(b,c,p)

(b,d,p)

Just as in the development of the double integral, we can

partition [a,b] into n equal sub-intervals each with length x,

partition [c,d] into n equal sub-intervals each with length y,

partition [p,q] into n equal sub-intervals each with length z.

For each of the n3 sub-boxes, we may multiply a value of the function f(x,y,z) anywhere on the sub-box by the volume V = x y z.

If the limit as n exists for the sum of these products (which will be true for continuous functions), then we say that f is integrable over B, and this limit is called the triple integral

f(x,y,z) dx dy dz = f(x,y,z) dV =

B

B

q

q

d

b

b

d

f(x,y,z) dx dy dz = f(x,y,z) dy dx dz=

etc.

p

c

a

p

a

c

Example Integrate f(x,y,z) = (3x + 2y + z)2 over the box defined by [0,1/3][–1/2,0][0,1] using any two different orders of integration.

1

0

1/3

(3x + 2y + z)2dx dy dz =

1/12

0

–1/2

0

An elementary region in R3 is one which can be defined by restricting one of the variables to be between two functions of the remaining two variables, where the domains of these functions is an elementary region in R2.

Similar to the way the integral of a function f(x,y) over an elementary region in R2 is defined, we can define the integral of a function f(x,y,z) over an elementary region W in R3 to be

f(x,y,z) dx dy dz = f(x,y,z) dx dy dz

W

B

where B is a box containing the region W, and the function f(x,y,z) is defined to be equal to f(x,y,z) on the region W and equal to 0 (zero) outside the region W.

Note that the volume of an elementary region W in R3 can be found from

dx dy dz

which can be written as a double integral of the difference between two functions of two variables.

W

Example Find the volume inside the sphere of radius a defined by

x2 + y2 + z2 = a2 .

One way the region inside the sphere can be described is

– a x a

– a2 – x2 y a2 – x2

– a2 – x2 – y2 z a2 – x2 – y2

a

a2 – x2

a2 – x2 – y2

dz dy dx =

– a

– a2 – x2

– a2 – x2 – y2

a2 – x2

a2 – x2 – y2

z dy dx =

– a

– a2 – x2

z = – a2 – x2 – y2

a

a2 – x2

Note: One way of evaluating this integral is to use a table of integrals.

2 a2 – x2 – y2 dy dx

– a

– a2 – x2

b

Another way of evaluating this integral is to realize that

b2 – t2 dt

– b

is the area of a semicircle of radius b, which is equal to b2/2.

We see then (with b = a2 – x2 and t = y) that

a

a2 – x2

a

2 a2 – x2 – y2 dy dx =

(a2 – x2)dx =

– a

– a

– a2 – x2

a

4a3

——

3

(a2x – x3/3) =

x = – a

Integrate the function xyze over the region between the xy plane, the plane y = x, the plane x = 4, and the cone z2 = x2 + y2 in the region where x, y, and z are all non-negative.

z2

z

z2 = x2 + y2

First, we observe that this region can be described by

x

0

4

0

y

x

x

y

x = 4

0

z

x2 + y2

y = x

4

x

x2 + y2

x2 + y2

4

x

z2

z2

xyze dz dy dx =

xye

—— dy dx =

2

0

0

0

0

0

z = 0

x

4

x

x2 + y2

x2 + y2

xyexy

——— – — dy dx =

2 2

xexy2

——— – — dx =

4 4

0

0

0

y = 0

4

2x2

x2

xex3xe

——— – — – —— dx =

4 4 4

0

4

2x2

x2

ex4e

—— – — – —— =

16 16 8

x = 0

e32– 256 – 2e16– 1 + 2 e32– 2e16– 255

—————————— = ———————

16 16

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