Triple integrals in Cartesian coordinates (Sect. 15.4 ...

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´╗┐Triple integrals in Cartesian coordinates (Sect. 15.4)

Review: Triple integrals in arbitrary domains. Examples: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrary domains.

Review: Triple integrals in arbitrary domains.

Theorem

If f : D R3 R is continuous in the domain

D = x [x0, x1], y [h0(x), h1(x)], z [g0(x, y ), g1(x, y )] ,

where g0, g1 : R2 R and h0, h1 : R R are continuous, then the triple integral of the function f in the region D is given by

x1 h1(x) g1(x,y )

f dv =

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

D

x0 h0(x) g0(x,y )

Example

In the case that D is an ellipsoid, the figure represents the graph of functions g1, g0 and h1, h0.

z y = h0( x )

x1 x

z = g1 ( x, y )

x0

y = h1( x )

y

z = g ( x, y )

0

Triple integrals in Cartesian coordinates (Sect. 15.4)

Review: Triple integrals in arbitrary domains. Examples: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrary domains.

Changing the order of integration.

Example

Change

the

order

of

integration

in

the

triple

integral

1 3 1-x2 2 1-x2-(y /3)2

V=

dz dy dx.

-1 -3 1-x2 -2 1-x2-(y /3)2

Solution: First: Sketch the integration region. Start from the outer integration limits to the inner limits.

Limits in x: x [-1, 1].

Limits in y : |y | 3 1 - x2,

so,

x2

+

y2 32

1.

z

2

The limits in z:

|z |

2

1

-

x2

-

y2 32

,

so,

x2

+

y2 32

+

z2 22

1.

-3

x

-1 1

-2

3

y

Changing the order of integration.

Example

Change the order of integration in the triple integral

1 3 1-x2 2 1-x2-(y /3)2

V=

dz dy dx.

-1 -3 1-x2 -2 1-x2-(y /3)2

Solution:

Region:

x2

+

y2 32

+

z2 22

1. We conclude:

1 2 1-x2 3 1-x2-(z/2)2

V=

dy dz dx.

-1 -2 1-x2 -3 1-x2-(z/2)2

2

1-(z/2)2 3 1-x2-(z/2)2

V=

dy dx dz.

-2 - 1-(z/2)2 -3 1-x2-(z/2)2

2 3 1-(z/2)2

1-(y /3)2-(z/2)2

V=

dx dy dz.

-2 -3 1-(z/2)2 - 1-(y /3)2-(z/2)2

Changing the order of integration.

Example

2 1-x/2 3-3y -3x/2

Interchange the limits in V =

dz dy dx.

00

0

Solution: Sketch the integration region starting from the outer integration limits to the inner integration limits.

x [0, 2],

x

y 0, 1 - so the upper

2

x

limit is the line y = 1 - .

2

3x z 0, 3 - - 3y so the

2 upper limit is the plane

3x z = 3 - - 3y . This plane

2 contains the points (2, 0, 0),

(0, 1, 0) and (0, 0, 3).

z

z = 3 - 3 x / 2

3 z = (6 - 3x - 6y) / 2 z = 3 - 3 y

2 x

1

y

y = 1 - x / 2

Changing the order of integration.

Example

2 1-x/2 3-3y -3x/2

Interchange the limits in V =

dz dy dx.

00

0

Solution: The region: x We conclude:

z z = 3 - 3 x / 2

3 z = (6 - 3x - 6y) / 2 z = 3 - 3 y

2 x

1

y

y = 1 - x / 2

0, y 0, z 0 and 6 3x + 6y + 2z.

3 1-z/3 2-2y -2z/3

V=

dx dy dz.

00

0

1 3-3y 2-2y -2z/3

V=

dx dz dy .

00

0

2 3-3x/2 1-x/2-z/3

V=

dy dz dx.

00

0

3 2-2z/3 1-x/2-z/3

V=

dy dx dz.

00

0

Triple integrals in Cartesian coordinates (Sect. 15.4)

Review: Triple integrals in arbitrary domains. Examples: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrary domains.

Average value of a function in a region in space.

Review: The average of a single variable function.

Definition

The average of a function f : [a, b] R on the interval [a, b], denoted by f , is given by

1

b

f=

f (x) dx.

(b - a) a

y f(x)

f

0101010101000111000000111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111000000000111111111

a

bx

Definition

The average of a function f : R R3 R on the region R with

volume V , denoted by f , is given by

1 f=

V

f dv .

R

Average value of a function in a region in space.

Example

Find the average of f (x, y , z) = xyz in the first octant bounded by the planes x = 1, y = 2, z = 3.

Solution: The volume of the rectangular integration region is

123

V=

dz dy dx V = 6.

000

The average of function f is:

1 123

11

f=

xyz dz dy dx =

x dx

60 0 0

60

2

y dy

0

3

z dz

0

1 x2 1 y2 2 z2 3 1 1 4 9

f=

=

.

6 20 20 20 6 2 2 2

We conclude: f = 1/4.

Triple integrals in Cartesian coordinates (Sect. 15.4)

Review: Triple integrals in arbitrary domains. Examples: Changing the order of integration. The average value of a function in a region in space. Triple integrals in arbitrary domains.

Triple integrals in arbitrary domains.

Example

Compute the triple integral of f (x, y , z) = z in the region bounded by x 0, z 0, y 3x, and 9 y 2 + z2.

Solution: Sketch the integration region.

The integration region is in the first octant.

It is inside the cylinder y 2 + z2 = 9.

It is on one side of the plane 3x - y = 0. The plane has normal vector n = 3, -1, 0 and contains (0, 0, 0).

z

3x - y = 0

3

-1

1

n 3 x

3

y

y 2 + z2 = 9

Triple integrals in arbitrary domains.

Example

Compute the triple integral of f (x, y , z) = z in the region bounded by x 0, z 0, y 3x, and 9 y 2 + z2.

Solution: We have found the region:

z 3

z = 9 - y2

The integration limits are:

Limits in z: 0 z 9 - y2.

Limits in x: 0 x y /3.

1

3y

Limits in y : 0 y 3.

x

y = 3x

3 y /3

9-y 2

We obtain I =

z dz dx dy .

00

0

Triple integrals in arbitrary domains.

Example

Compute the triple integral of f (x, y , z) = z in the region bounded by x 0, z 0, y 3x, and 9 y 2 + z2.

Solution: Recall:

3 y /3

9-y 2

z dz dx dy .

00

0

For practice purpose only, let us

change the integration order to

dz dy dx:

z 3

1

x

13

9-y 2

The result is: I =

z dz dy dx.

0 3x 0

z = 9 - y2

3y y = 3x

Triple integrals in arbitrary domains.

Example

Compute the triple integral of f (x, y , z) = z in the region bounded

by x 0, z 0, y 3x, and 9 y 2 + z2.

13

9-y 2

Solution: Recall I =

z dz dy dx.

0 3x 0

We now compute the integral:

1 3 z 2 9-y 2

f dv =

dy dx,

D

0 3x 2 0

=1

1

3

(9 - y 2)dy dx,

2 0 3x

11

3

y3 3

=

9y -

dx .

20

3x

3 3x

Triple integrals in arbitrary domains.

Example

Compute the triple integral of f (x, y , z) = z in the region bounded by x 0, z 0, y 3x, and 9 y 2 + z2.

Solution: Recall: Therefore,

11

3

y3 3

f dv =

9y -

dx .

D

20

3x

3 3x

1

f dv =

D

2

9 =

2

1

27(1 - x) - 9(1 - x)3 dx,

0

1

3(1 - x) - (1 - x)3 dx.

0

Substitute u = 1 - x, then du = -dx, so,

9 f dv =

1

(3u - u3)du.

D

20

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