Triple integrals in Cartesian coordinates (Sect. 15.4 ...
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.
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 1x2 2 1x2(y /3)2
V=
dz dy dx.
1 3 1x2 2 1x2(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 1x2 2 1x2(y /3)2
V=
dz dy dx.
1 3 1x2 2 1x2(y /3)2
Solution:
Region:
x2
+
y2 32
+
z2 22
1. We conclude:
1 2 1x2 3 1x2(z/2)2
V=
dy dz dx.
1 2 1x2 3 1x2(z/2)2
2
1(z/2)2 3 1x2(z/2)2
V=
dy dx dz.
2  1(z/2)2 3 1x2(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 1x/2 33y 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 1x/2 33y 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 1z/3 22y 2z/3
V=
dx dy dz.
00
0
1 33y 22y 2z/3
V=
dx dz dy .
00
0
2 33x/2 1x/2z/3
V=
dy dz dx.
00
0
3 22z/3 1x/2z/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
9y 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
9y 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
9y 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
9y 2
Solution: Recall I =
z dz dy dx.
0 3x 0
We now compute the integral:
1 3 z 2 9y 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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