Triple Integrals — §12.5 33 Triple Integrals For a function f (x , y , z ) defined over a bounded region E in three dimensions, we can take the triple integral ZZZ E f (x , y , z ) dV
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ z=s
z=r
∫ y=d
y=c
∫ x=b
x=af (x , y , z) dx dy dz
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ z=s
z=r
∫ y=d
y=c
∫ x=b
x=af (x , y , z) dx dy dz
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ z=s
z=r
∫ y=d
y=c
∫ x=b
x=af (x , y , z) dx dy dz
And that you are allowed to choose the order of integration you wish.
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ y=d
y=c
∫ x=b
x=a
∫ z=s
z=rf (x , y , z) dz dx dy
And that you are allowed to choose the order of integration you wish.
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ y=d
y=c
∫ z=s
z=r
∫ x=b
x=af (x , y , z) dx dz dy
And that you are allowed to choose the order of integration you wish.
Triple Integrals — §12.5 33
Triple Integrals
For a function f (x , y , z) defined over a bounded region E in threedimensions, we can take the triple integral∫∫∫
Ef (x , y , z) dV
If f is continuous over a region that is a box
B = [a, b]× [c, d ]× [r , s],
Fubini’s theorem says that∫∫∫Bf (x , y , z) dV =
∫ x=b
x=a
∫ y=d
y=c
∫ z=s
z=rf (x , y , z) dz dy dx
And that you are allowed to choose the order of integration you wish.
Triple Integrals — §12.5 34
Example
Example. Find∫∫∫
B xyz2 dV for B = [0, 1]× [−1, 2]× [0, 3]
Solution.
The difficulties arise when
I Regions are not boxes. (Today)
I Regions are best defined in polar-like coordinates. (Next time)
Triple Integrals — §12.5 34
Example
Example. Find∫∫∫
B xyz2 dV for B = [0, 1]× [−1, 2]× [0, 3]
Solution.
The difficulties arise when
I Regions are not boxes. (Today)
I Regions are best defined in polar-like coordinates. (Next time)
Triple Integrals — §12.5 34
Example
Example. Find∫∫∫
B xyz2 dV for B = [0, 1]× [−1, 2]× [0, 3]
Solution.
The difficulties arise when
I Regions are not boxes. (Today)
I Regions are best defined in polar-like coordinates. (Next time)
Triple Integrals — §12.5 35
Setting up complicated triple integrals
I We only consider triple integrals over regions that can bedefined as being between two surfaces.
I This allows us to reduce our triple integral to a double integral.(Which may itself be complicated...)
Three types:∫∫∫Ef dV =
∫∫D
[∫ z=u2(x ,y)
z=u1(x ,y)f (x , y , z) dz
]dA
Typ
e1
∫∫∫Ef dV =
∫∫D
[∫ x=u2(y ,z)
x=u1(y ,z)f (x , y , z) dx
]dA
Typ
e2
∫∫∫Ef dV =
∫∫D
[∫ y=u2(x ,z)
y=u1(x ,z)f (x , y , z) dy
]dA
Typ
e3
Triple Integrals — §12.5 35
Setting up complicated triple integrals
I We only consider triple integrals over regions that can bedefined as being between two surfaces.
I This allows us to reduce our triple integral to a double integral.(Which may itself be complicated...)
Three types:∫∫∫Ef dV =
∫∫D
[∫ z=u2(x ,y)
z=u1(x ,y)f (x , y , z) dz
]dA
Typ
e1
∫∫∫Ef dV =
∫∫D
[∫ x=u2(y ,z)
x=u1(y ,z)f (x , y , z) dx
]dA
Typ
e2
∫∫∫Ef dV =
∫∫D
[∫ y=u2(x ,z)
y=u1(x ,z)f (x , y , z) dy
]dA
Typ
e3
Triple Integrals — §12.5 35
Setting up complicated triple integrals
I We only consider triple integrals over regions that can bedefined as being between two surfaces.
I This allows us to reduce our triple integral to a double integral.(Which may itself be complicated...)
Three types:∫∫∫Ef dV =
∫∫D
[∫ z=u2(x ,y)
z=u1(x ,y)f (x , y , z) dz
]dA
Typ
e1
∫∫∫Ef dV =
∫∫D
[∫ x=u2(y ,z)
x=u1(y ,z)f (x , y , z) dx
]dA
Typ
e2
∫∫∫Ef dV =
∫∫D
[∫ y=u2(x ,z)
y=u1(x ,z)f (x , y , z) dy
]dA
Typ
e3
Triple Integrals — §12.5 35
Setting up complicated triple integrals
I We only consider triple integrals over regions that can bedefined as being between two surfaces.
I This allows us to reduce our triple integral to a double integral.(Which may itself be complicated...)
Three types:∫∫∫Ef dV =
∫∫D
[∫ z=u2(x ,y)
z=u1(x ,y)f (x , y , z) dz
]dA
Typ
e1
∫∫∫Ef dV =
∫∫D
[∫ x=u2(y ,z)
x=u1(y ,z)f (x , y , z) dx
]dA
Typ
e2
∫∫∫Ef dV =
∫∫D
[∫ y=u2(x ,z)
y=u1(x ,z)f (x , y , z) dy
]dA
Typ
e3
Triple Integrals — §12.5 35
Setting up complicated triple integrals
I We only consider triple integrals over regions that can bedefined as being between two surfaces.
I This allows us to reduce our triple integral to a double integral.(Which may itself be complicated...)
Three types:∫∫∫Ef dV =
∫∫D
[∫ z=u2(x ,y)
z=u1(x ,y)f (x , y , z) dz
]dA
Typ
e1
∫∫∫Ef dV =
∫∫D
[∫ x=u2(y ,z)
x=u1(y ,z)f (x , y , z) dx
]dA
Typ
e2
∫∫∫Ef dV =
∫∫D
[∫ y=u2(x ,z)
y=u1(x ,z)f (x , y , z) dy
]dA
Typ
e3
Triple Integrals — §12.5 36
Triple Integral Strategies
The hard part is figuring out the bounds of your integrals.
I Project your region E onto one of the xy -, yz-, or xz-planes, anduse the boundary of this projection to find bounds on domain D.
I Over this domain D, the region E is defined by some“higher function” and some “lower function”.These give the bounds on the innermost integral.
You may need to try multiple projections to find the easiest integralto integrate. Then use all the tools in your toolbox to integrate it.
Triple Integrals — §12.5 36
Triple Integral Strategies
The hard part is figuring out the bounds of your integrals.
I Project your region E onto one of the xy -, yz-, or xz-planes, anduse the boundary of this projection to find bounds on domain D.
I Over this domain D, the region E is defined by some“higher function” and some “lower function”.These give the bounds on the innermost integral.
You may need to try multiple projections to find the easiest integralto integrate. Then use all the tools in your toolbox to integrate it.
Triple Integrals — §12.5 36
Triple Integral Strategies
The hard part is figuring out the bounds of your integrals.
I Project your region E onto one of the xy -, yz-, or xz-planes, anduse the boundary of this projection to find bounds on domain D.
I Over this domain D, the region E is defined by some“higher function” and some “lower function”.These give the bounds on the innermost integral.
You may need to try multiple projections to find the easiest integralto integrate. Then use all the tools in your toolbox to integrate it.
Triple Integrals — §12.5 36
Triple Integral Strategies
The hard part is figuring out the bounds of your integrals.
I Project your region E onto one of the xy -, yz-, or xz-planes, anduse the boundary of this projection to find bounds on domain D.
I Over this domain D, the region E is defined by some“higher function” and some “lower function”.These give the bounds on the innermost integral.
You may need to try multiple projections to find the easiest integralto integrate. Then use all the tools in your toolbox to integrate it.
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)
dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)
dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)
dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)
dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)
dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)
dA
Triple Integrals — §12.5 37
Example
Example. Evaluate∫∫∫
E
√x2 + z2 dV where E is the region
bounded by the hyperboloid y = x2 + z2 and the plane y = 4.
Project onto xy -plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫ 2
−2
∫ 4
x2
(∫ √y−x2
−√
y−x2
√x2 + z2 dz
)dy dx
Project onto xz-plane
D is defined by
The higher and lower functions are
∫∫∫E
√x2 + z2 dV =∫∫
D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)dA
Triple Integrals — §12.5 38
Example, continued
Now calculate
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)dA
There is no y , so the innermost integral is easy:
=
∫∫D=circle
x2+z2=4
(√x2 + z2 · y
]4x2+z2 dA
=
∫∫D=circle
x2+z2=4
√x2 + z2 ·
(4− x2 − z2
)dA
This integral is easier to do using
Triple Integrals — §12.5 38
Example, continued
Now calculate
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)dA
There is no y , so the innermost integral is easy:
=
∫∫D=circle
x2+z2=4
(√x2 + z2 · y
]4x2+z2 dA
=
∫∫D=circle
x2+z2=4
√x2 + z2 ·
(4− x2 − z2
)dA
This integral is easier to do using
Triple Integrals — §12.5 38
Example, continued
Now calculate
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)dA
There is no y , so the innermost integral is easy:
=
∫∫D=circle
x2+z2=4
(√x2 + z2 · y
]4x2+z2 dA
=
∫∫D=circle
x2+z2=4
√x2 + z2 ·
(4− x2 − z2
)dA
This integral is easier to do using
Triple Integrals — §12.5 38
Example, continued
Now calculate
∫∫D=circle
x2+z2=4
(∫ 4
x2+z2
√x2 + z2 dy
)dA
There is no y , so the innermost integral is easy:
=
∫∫D=circle
x2+z2=4
(√x2 + z2 · y
]4x2+z2 dA
=
∫∫D=circle
x2+z2=4
√x2 + z2 ·
(4− x2 − z2
)dA
This integral is easier to do using
Triple Integrals — §12.5 39
Loose ends
Density in three dimensions
I Given a mass density function ρ(x , y , z) (mass per unit volume)
mass =
∫∫∫Eρ(x , y , z) dV .
Average value in three dimensions
I The average value of a function f (x , y , z) over a region E is
fave =1
V (E )
∫∫∫Ef (x , y , z) dV .