triple integrals
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12.7 - 3
z-Simple solids (Type 1)
Definition: A solid region E
is said to be z-Simple if it is bounded by two surfaces z=z1(x,y) and z=z2(x,y)(z1 zz2)
x
y
z
12.7 - 4Iterated Triple Integrals over z-Simple solid E
When you project a z-Simple solid E onto the xy-plane you obtain a planar region D. 1st you integrate wrt z (the simple variable) from
z=z1(x,y) to z=z2(x,y). You obtain some function of x and y to integrate
over the region D in the xy-plane. If D is Type I you have y=y1(x) to y=y2(x) and you
integrate over y. Finally you integrate over the constant limits from x=x1 to
x=x2 and this integration is wrt x.
If D is Type II you have x=x1(y) to x=x2(y). Finally you integrate over the constant limits from y=y1 to
y=y2 and this integration is wrt y.
E
dV(x,y,z)
12.7 - 5
xy
xy
D
xy
xy
D
yxzz
yxzz
dAf(x,y)
dAdz(x,y,z)
dV(x,y,z)
),(
),(
E
2
1
Triple to Double Integral for z-Simple solid
12.7 - 6
xy
xy
D
xy
xy
D
yxzz
z
dA(x,y)z
dAdz
dV
2
),(
0
E
2
z-Simple solid: special case =1, z1=0, z2>0.
This gives the volume V over the region Dxy in the xy-plane of the surface z=z2(x,y)
12.7 - 7
y-Simple solids (Type 2)
Definition: A solid region E
is said to be y-Simple if it is bounded by two surfaces y=y1(x,z) and y=y2(x,y)(y1 y y2)
xy
z
12.7 - 8
Example: Paraboloids
Find the volume of the solid enclosed by the two surfacesy= 0.5(x2+z2) and y=16-x2-z2.
We need to define a region Dxz in the xz-plane.
xy
z
12.7 - 10 It may be helpful to recall
the single integral calculus method for finding area between two curves y=y1(x) and y=y2(x) -- just think of z as constant, e.g., on a horizontal trace (say z=0)
Next we evaluatethe integral -- dy goesfirst since it is y-simple.
332
12.7 - 11
1.2683
256
)5.116(2
5.15.116
332
0
2
22
16
)(5.0E
22
22
r
r
D
xz
yz
D
zxy
zxy
drrr
dAzx
dAdydV
yz
yz
Polar coordinates x=r cos ,z=r sin , wereused so that dAxz can be written asr dr d instead ofdx dz.Compare to a cylinder of radius and height 16 which has
double this volume (anyone know why?) and contains our solid E inside it.
332
12.7 - 12
xy-Simple, yz-simple, xz-simple Approach
Often a solid is simple in more than one variable.An alternate approach is to look for the one variable that it is not simple in, and make that the outer limitof integration. The inner limit is then a double integral.
This approach is also helpful in sketching the solid of integration, because as we will see the outer limit ofintegration corresponds to constant values on whichcontour regions in the simple plane lie
12.7 - 13
If not z-simple, try:
dz(x,y,z)dA
dV(x,y,z)
zz
zz zD
xy
xy
2
1 )(
E
where Dxy(z) is the trace of the solid (a trace of a solid is a region instead of a curve) in
the plane z=constant.
12.7 - 14
Example (Text, page 892#7)
Sketch the domain of integration of the triple integral
where
Then evaluate the integral.
E
dV yz
20,20,10|),,( zxzyzzyxE
12.7 - 15
Solution: Perhaps the easiest way to see the solid E which is
our domain of integration is to first consider z fixed. Then x varies from 0 to (z + 2)
and y varies from 0 to 2z. This defines a rectangle in the plane z units above the
xy-plane. Question: What are its vertices? Answer:
(0,0,z), (z + 2, 0, z), (0, 2z, z) and (z+2, 2z,z).
20,20,10|),,( zxzyzzyxE
12.7 - 16Horizontal trace of domain E (z=constant) Rectangle (0,0,z), (z + 2, 0, z),
(0, 2z, z), (z+2, 2z,z).
This rectangle lies on a plane which is located z units above the xy-plane.
Let’s graph the family of horizontal traces, for several values of z between 0 and 1.
z
xy
2zx
0x0y
zy 2
12.7 - 17
Domain E is not z-simple
Next, as z increases, the rectangles become larger (and higher)
If we stack them, one above the other, we get the solid domain of integration
Let’s assume (x,y,z)=1, so that the triple integral will give us a volume instead of a mass.
z
xy
zy 2
0y 0x2zx
12.7 - 18Cross-section Rxy (keeping z constant)
We integrate (x,y,z)=1 over Rxy
treating z as constant to get the area of Rxy
Its horizontal traces define regions R=Rxy(z) above the xy-plane. Each of these planar regions should be Type I (or Type II) so that the areas of the cross-sections can be evaluated as double integrals
Between z and z+dz the total volume is dV=Axydz. Sum from z=0 to 1 to get the total volume: that sum converges to the integral over z.
z
xy
12.7 - 19
Evaluation of the triple integral
31
0
1
0
2
0
1
0
2
0
2
0
1
0E
3
8)2(2
)2(
mdzzz
dzdyz
dzdydx
dzdAdV
z
z
z
z
z
z
z
zz
z
z D
xy
xy
kgdzzz
dzdyzyz
dzdydxyz
dzdAyzdVyz
z
z
z
z
z
z
z
zz
z
z D
xy
xy
5
7)2(2
)2(
1
0
3
1
0
2
0
1
0
2
0
2
0
1
0E
12.7 - 20
Challenge
Rewrite
as
or
using the fact that E is y-simple (i.e., dy on the inside)
2
0
2
0
1
0
zz
dzdydxyz
?
?
?
?
?
?
dzdxdyyz
?
?
?
?
?
?
dxdzdyyz
12.7 - 21
Answers: dy dx dz is easy
but dy dz dx has to be split into 2.
z zz
dzdxzdzdxdyyz2
0
2
0
31
0
2
0
1
0 5
72
5
2122
1
2
33
2
1
0
32
0
x
dxdzzdxdzz
12.7 - 22
Example: Tetrahedron(x,y, and z-simple :-) Evaluate the triple integral
where E is the solid tetrahedron with vertices (0,0,0), (1,0,0), (0,2,0), and (0,0,3).
E
dVxy
12.7 - 23
Example: Step 1
Visualize the solid. You need to get equations of the 4 planar sides of the tetrahedron.
Consider P(0,0,0), Q(1,0,0), R(0,2,0)Convince yourself -- or show using
that the equation of the plane through these 3 points is z=0.
0,,, zyx PzPyPxnPRPQn
12.7 - 24
Consider P(0,0,0), Q(1,0,0), S(0,0,3)Similarly the equation of the plane
through these 3 points is y=0. Consider P(0,0,0), R(0,2,0), S(0,0,3)
This corresponds the plan x=0. Now consider Q(1,0,0), R(0,2,0),
S(0,0,3)
06236,,2,3,6
2,3,63,0,10,2,1
zyxQzQyQx
QSQRn
zyx
12.7 - 25
Example: Step 2
The region is described by 0 x, 0 y, 0 z, and 6x+3y+2z 6.
This solid is x-simple, y-simple and z-simple. To describe it as z-simple, we let z1=0 and
z2=3-3x-1.5y. Equating z=z1=z2 we obtain the region Dxy
described by 3x+1.5y=3 in the xy-plane. Since 0 y 2-2x for 0 x 1 ...
12.7 - 26
Example: Step 3
We now are in the position to set up the triple integral with limits.
dxdyxyz
dxdydzxy
dVxyI
yxzz
x
yxx
E
5.1330
22
0
1
0
5.133
0
22
0
1
0
12.7 - 27
…continued...
1
0
22
0
32222
22
0
221
0
22
0
1
0
2
1
2
3
2
3
)5.133(
)5.133(
dxyxyxxy
dxdyxyyxxy
dxdyyxxyI
xy
y
x
x
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