26 triple integrals

Triple Integrals
We write the rectangular box B
{(x, y, z)| a ≤ x ≤ b, c ≤ y ≤ d and e ≤ z ≤ f }
as [a, b] x [c, d] x [e, f].
y
a
b
c d
x
e
f
B

Triple Integrals
as [a, b] x [c, d] x [e, f].
Let W = h(x, y, z) be a
continuous function in x, y & z.
For instance, B could be a
block of unevenly mixed alloy
and h(x, y, z) represents the
density at the point (x, y, z).
y
c d
x
e
f
B
a
b

Triple Integrals
as [a, b] x [c, d] x [e, f].
Partition B into sub-boxes, let Bi be a sub-box and (xi, yi, zi)
be a point in Bi.
y
a
b
c d
x
e
f
B
Bi(xi, yi, zi)

Triple Integrals
as [a, b] x [c, d] x [e, f].
be a point in Bi. Let Δx, Δy, Δz be the sides of Bi so the
volume of Bi is ΔVi= ΔxΔyΔz, then Mi = h(xi, yi, zi)ΔVi is
approximately the mass of Bi and the sum ΣMi approximates
the total mass
M of the solid B.
y
a
b
c d
x
e
f
B
Bi(xi, yi, zi)

Triple Integrals
as [a, b] x [c, d] x [e, f].
be a point in Bi. Let Δx, Δy, Δz be the sides of Bi so the
volume of Bi is ΔVi= ΔxΔyΔz, then Mi = h(xi, yi, zi)ΔVi is
approximately the mass of Bi and the sum ΣMi approximates
the total mass
M of the solid B. In particular, we defineΔx,Δy,Δz0 Δx,Δy,Δz0
y
a
b
c d
x
e
f
B
Bi(xi, yi, zi)

Triple Integrals
We write lim Σh(xi, yi, zi)ΔxΔyΔz, if it exists, as
Δx,Δy,Δz0
M = h(x, y, z) dV.∫ ∫∫
and it is called the triple integral of f over the domain B.
B

Triple Integrals
Δx,Δy,Δz0
M = h(x, y, z) dV.∫ ∫∫
However, if h(x, y, z) is continuous then the limit exists and
the triple integral may be convert into iterated integral:
B

Triple Integrals
Δx,Δy,Δz0
M = h(x, y, z) dV.∫ ∫∫
Theorem: Let h(x, y, z) be a continuous function over the
domain B = [a, b] x [c, d] x [e, f], then
h(x, y, z) dV =∫ ∫∫
B
h(x, y, z) dxdydz∫ ∫∫
x=ay=cz=e
bdf
B

Triple Integrals
Δx,Δy,Δz0
M = h(x, y, z) dV.∫ ∫∫
Theorem: Let h(x, y, z) be a continuous function over the
domain B = [a, b] x [c, d] x [e, f], then
B
or in any other order of integration (if B is a box). There are
five other versions: h dxdzdy,∫∫∫ h dydxdz,
h dydzdx, h dzdxdy, h dzdydx,
h(x, y, z) dV =∫ ∫∫
B
h(x, y, z) dxdydz∫ ∫∫
x=ay=cz=e
bdf
∫∫∫
∫∫∫ ∫∫∫ ∫∫∫

Triple Integrals
x
y
D
z = F(x,y)
z = G(x,y)
V
a
b
A (bounded) 3D solid V is said to be z-regular if it is
bounded between two functions z = G(x, y) and z = F(x, y)
over the same domain D (in the xy plane).

Triple Integrals
x
y
D
z = F(x,y)
z = G(x,y)
V Suppose D is type I and that
D = {(x, y)| a < x < b; g(x) < y < f(x)}
a
b
y = g(x)
y = f(x)

Triple Integrals
x
y
D
z = F(x,y)
z = G(x,y)
D = {(x, y)| a < x < b; g(x) < y < f(x)}
a
b
y = g(x)
y = f(x)
For any fixed (x*
, y*
) in D,
the point (x*
, y*
, z) in V must have
the z-coordinates satisfying
G(x*
, y*
) < z < F(x*
, y*
).
(x*,y*)
G(x*, y*)
F(x*, y*)

Triple Integrals
x
y
D
z = F(x,y)
z = G(x,y)
D = {(x, y)| a < x < b; g(x) < y < f(x)}
a
b
y = g(x)
y = f(x)
For any fixed (x*
, y*
) in D,
the point (x*
, y*
, z) in V must have
G(x*
, y*
) < z < F(x*
, y*
). Hence a
z–regular solid V with type I domains
D may be described as shown below.
Type I: {(x, y, z)| a < x < b; g(x) < y < f(x); G(x, y) < z < F(x,
y)}
(x*,y*)
F(x*, y*)
G(x*, y*)

Triple Integrals
x
y
D
z = F(x,y)
z = G(x,y)
D = {(x, y)| a < x < b; g(x) < y < f(x)}
a
b
y = g(x)
y = f(x)
For any fixed (x*
, y*
) in D,
the point (x*
, y*
, z) in V must have
G(x*
, y*
) < z < F(x*
, y*
). Hence a
z–regular solid V with type I domains
D may be described as shown below.
Type I: {(x, y, z)| a < x < b; g(x) < y < f(x); G(x, y) < z < F(x,
y)}
(x*,y*)
The description for a solid V with
type II domain D is also given below.
Type II: {(x, y, z)| c < y < d; g(y) < x < f(y); G(x, y) < z < F(x,
F(x*, y*)
G(x*, y*)

Triple Integrals
Example A. Write V in set notation with D as type I and type II.
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
Solve for the intercepts and get
x = 2, y = 3, & z = 6.
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
x = 2, y = 3, & z = 6. Set z = 0, the
equation for the line in the xy-plane is
3x + 2y = 6.
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
x = 2, y = 3, & z = 6. Set z = 0, the
3x + 2y = 6.
Treat D as type I domain, then
D = {(x, y)| 0 < x < 2; 0 < y < 3 – x}3
2
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
x = 2, y = 3, & z = 6. Set z = 0, the
3x + 2y = 6.
D = {(x, y)| 0 < x < 2; 0 < y < 3 – x}3
2
So V = {(x, y, z)| 0 < x < 2; 0 < y < 3 – 3x/2; 0 < z < 6 – 3x –
2y}
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
x = 2, y = 3, & z = 6. Set z = 0, the
3x + 2y = 6.
D = {(x, y)| 0 < x < 2; 0 < y < 3 – x}3
2
So V = {(x, y, z)| 0 < x < 2; 0 < y < 3 – 3x/2; 0 < z < 6 – 3x –
2y}
Treat D as type II, then D = {(x, y)| 0 < y < 3; 0 < x < 2 – y}
2
3
x
y
z + 3x + 2y = 6
V
D
2
3
6
3x + 2y = 6

Triple Integrals
x
y
z + 3x + 2y = 6
V
D
x = 2, y = 3, & z = 6. Set z = 0, the
3x + 2y = 6.
2
3
6
3x + 2y = 6
D = {(x, y)| 0 < x < 2; 0 < y < 3 – x}3
2
So V = {(x, y, z)| 0 < x < 2; 0 < y < 3 – 3x/2; 0 < z < 6 – 3x –
2y}
Treat D as type II, then D = {(x, y)| 0 < y < 3; 0 < x < 2 – y}
2
3
So V = {(x, y, z)| 0 < y < 3; 0 < x < 2 – 2y/3; 0 < z < 6 – 3x – 2y}

Triple Integrals
Example B. Write V in set notation with D as type I and type
II. V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
II.
Solve for the x&y intercepts for D.
Set z = 2, the equation for the line in
the xy-plane is 4 – 4x – y = 0.
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
II.
Hence the intercepts are x = 1, y = 4.
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
II.
D = {(x, y)| 0 < x < 1; 0 < y < 4 – 4x}
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
V
x
y
2z – 4x – y = 0
2
D
II.
1
4
4 – 4x – y = 0 Treat D as type I domain, then
D = {(x, y)| 0 < x < 1; 0 < y < 4 – 4x}
So V = {(x, y, z)| 0 < x < 1; 0 < y < 4 – 4x; 2x + y/2 < z < 2}

Triple Integrals
II.
D = {(x, y)| 0 < x < 1; 0 < y < 4 – 4x}
So V = {(x, y, z)| 0 < x < 1; 0 < y < 4 – 4x; 2x + y/2 < z < 2}
Treat D as type II, then D = {(x, y)| 0 < y < 4; 0 < x < 1 – y/4}
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
II.
D = {(x, y)| 0 < x < 1; 0 < y < 4 – 4x}
So V = {(x, y, z)| 0 < x < 1; 0 < y < 4 – 4x; 2x + y/2 < z < 2}
Treat D as type II, then D = {(x, y)| 0 < y < 4; 0 < x < 1 – y/4}
So V = {(x, y, z)| 0 < y < 4; 0 < x < 1 – y/4; 2x + y/2 < z < 2}
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Theorem (Triple integral over a z–regular solid).
Let w = h(x, y, z) be defined over a z–regular solid
x
y
z = F(x,y)
z = G(x,y)
V
D
V = {(x, y, z)| a < x < b; g(x) < y < f(x); G(x, y) < z < F(x, y)},
Triple Integrals
A z–regular solid V defined
over a type I domain D.
y = g(x) y = f(x)
a
b

h(x, y, z)dV = ∫ h(x, y, z) dz dy dx
y=g(x)
y=f(x)
∫z=G(x,y)
z=F(x, y)
V
x
y
z = F(x,y)
z = G(x,y)
V
D
V = {(x, y, z)| a < x < b; g(x) < y < f(x); G(x, y) < z < F(x, y)}, then
∫
x=a
b
Triple Integrals
∫∫∫
y = g(x) y = f(x)
a
b

y=g(x)
y=f(x)
∫z=G(x,y)
z=F(x, y)
V
x
y
z = F(x,y)
z = G(x,y)
V
D
∫
x=a
b
For type II domain D, V is
{c < y < d; g(y) < x < f(y); G(x, y) < z < F(x, y)}
then the iterated integral is
∫ h(x, y, z) dz dx dy.
x=g(y)
f(y)
∫z=G(x,y)
F(x, y)
∫
d
y=c
Triple Integrals
∫∫∫
y = g(x) y = f(x)
a
b

y=g(x)
y=f(x)
∫z=G(x,y)
z=F(x, y)
V
x
y
z = F(x,y)
z = G(x,y)
V
D
∫
x=a
b
For type II domain D, V is
{c < y < d; g(y) < x < f(y); G(x, y) < z < F(x, y)}
then the iterated integral is
∫ h(x, y, z) dz dx dy.
x=g(y)
f(y)
∫z=G(x,y)
F(x, y)
∫
d
y=c
Triple Integrals
Analogous to double integrals,
1 dV = volume of V.
V
∫∫∫
∫∫∫
y = g(x) y = f(x)
a
b
we've that

Triple Integrals
Example C. Find
x
yD
V = {0 < x < 1; 0 < y < 1 – x; 0 < z < √1 – x2
}
2z dV =
V
∫ 2z dz dy dx
y=0
1–x
∫z=0
√1–x2
∫
1
x=0
= ∫ z2
| dy dx
y=0
1–x
∫
1
x=0
=
2z dV where V is as shown.
V
∫∫∫
V
x2
+z2
=1
x+y=1
z=0
√1–x2
∫
y=0
1–x
∫
1
1 – x2
dy dx
=
y=0
1–x
∫
1
(1 – x2
)y | dx
x=0
x=0
= ∫
1
(1 – x2
)(1 – x) dx
x=0
= ∫
1
1 – x – x2
+ x3
dx =
x=0
5
12
∫∫∫
Viewing V as z–regular and D as type I region,

Triple Integrals
x
y
z + x + y = 2
V
D
A.
Example B. Find the following volume.

Triple Integrals
x
y
z + x + y = 2
V
D
A. V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x – y}

Triple Integrals
x
y
z + x + y = 2
V
D
A. V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x – y}
Hence the volume is
∫ 1 dV =∫
V
∫ ∫ 1 dz dy dx
y=0
2–x
∫
z=0
2–x–y
∫
2
x=0

Triple Integrals
x
y
z + x + y = 2
V
D
A. V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x – y}
Hence the volume is
∫ 1 dV =∫
V
∫ ∫ 1 dz dy dx
y=0
2–x
∫
z=0
2–x–y
∫
2
x=0
∫ 2 – x – y dy dx
y=0
2–x
∫
2
x=0
=

Triple Integrals
x
y
z + x + y = 2
V
D
A. V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x – y}
Hence the volume is
∫ 1 dV =∫
V
∫ ∫ 1 dz dy dx
y=0
2–x
∫
z=0
2–x–y
∫
2
x=0
∫ 2y – xy – y2
/2 | dx
y=0
2–x2
x=0
∫ 2 – x – y dy dx
y=0
2–x
∫
2
x=0
=
=

Triple Integrals
x
y
z + x + y = 2
V
D
A. V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x – y}
Hence the volume is
∫ 1 dV =∫
V
∫ ∫ 1 dz dy dx
y=0
2–x
∫
z=0
2–x–y
∫
2
x=0
=
∫ 2y – xy – y2
/2 | dx
y=0
2–x2
x=0
∫ 2 – x – y dy dx
y=0
2–x
∫
2
x=0
=
∫ x2
– 4x + 4 dx
2
x=0
= 1/2

Triple Integrals
x
y
z + x + y = 2V
D
V = {0 < x < 2; 0 < y < 2 – x; 0 < z < 2 – x –
y}Hence the volume is
1 dV = ∫ 1 dz dy dx
y=0
2–x
∫z=0
2–x–y
∫
2
x=0
= ∫ 2y – xy – y2
/2 | dx
y=0
2–x2
x=0
∫ 2 – x – y dy dx
y=0
2–x
∫
2
x=0
=
∫ x2
– 4x + 4 dx
2
x=0
= 1/2 = 4
3
Example D. Find the following volume.
V
∫∫∫
Viewing V as z–regular and D as type I region,
Just as in R2
we have types I and II, we group solids in R3
into
different types depending on how the boundaries are defined.
We start out with the standard type, the z–regular solids.

Triple Integrals
A (bounded) 3D solid V is said to be y-regular if it is
bounded between two functions y = G(x, z) and y = F(x, z)
over the same domain D (in the xz plane).
x
y y = F(x,z)
y = G(x,z)
V
z
D
x = g(z)
x = f(z)
a
b

Triple Integrals
If D = {(z, x)| a < z < b; g(z) < x < f(z)}, then
V = {a < z < b; g(z) < x < f(z); G(x,z) < y < F(x, z)}.
x
y y = F(x,z)
y = G(x,z)
V
z
D
x = g(z)
x = f(z)
a
b

Triple Integrals
h(x, y, z)dV
∫ h(x, y, z) dy dx dz
x=g(z)
f(z)
∫
y=G(x,z
)
F(x, z)
∫
z=a
b
=
Theorem.
(Triple Integral over a y–regular V)
Let w = h(x, y, z) be defined over a
y–regular solid V as the above, then
the triple integral over V
x
y y = F(x,z)
y = G(x,z)
V
z
D
x = g(z)
x = f(z)
a
b
V
∫∫∫

Triple Integrals
A (bounded) 3D solid V is said to be x-regular if it is
bounded between two functions x = G(y, z) and x = F(y, z)
y
z x = F(y,z)x = G(y,z)
V
D
x

Triple Integrals
A (bounded) 3D solid V is said to be x-regular if it is
bounded between two functions x = G(y, z) and x = F(y, z)
y
z x = F(y,z)x = G(y,z)
V
D
x
h(x, y, z)dV
∫ h(x, y, z) dy dx dz
x=g(z)
f(z)
∫
y=G(x,z
)
F(x, z)
∫
z=a
b
=
Theorem.
(Triple Integral over a x–regular V)
Let w = h(x, y, z) be defined over a
y–regular solid V as the above, then
the triple integral over V
V
∫∫∫

Triple Integrals
Example 3: View V as x-regular to find the volume of V.
Viewing V as x-regular, the domain D
is in the yz plane.
V
x
y
2z – 4x – y = 0
2
D
1
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
The equation of the lower boundary
of D, in the plane x = 0, is 2z – y = 0.
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}
Solve 2z – 4x + y = 0 for x and get x = z/2 – y/4.
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}
So V is {0 < z < 2; 0 < y < 2z; 0 < x < z/2 – y/4}.
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}
So V is {0 < z < 2; 0 < y < 2z; 0 < x < z/2 – y/4}.
Hence the volume of V is ∫ 1 dx dy dz
y=0
2z
∫
x=0
z/2–y/4
∫
z=0
2
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}
So V is {0 < z < 2; 0 < y < 2z; 0 < x < z/2 – y/4}.
y=0
2z
∫
x=0
z/2–y/4
∫
z=0
2
z/2 – y/4 dy dz= ∫
y=0
2z
∫
z=0
2
2z – y = 0
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0

Triple Integrals
2z – y = 0
So V is {0 < z < 2; 0 < y < 2z; 0 < x < z/2 – y/4}.
y=0
2z
∫
x=0
z/2–y/4
∫
z=0
2
z/2 – y/4 dy dz= ∫
y=0
2z
∫
z=0
2
z2
/ 2 dz=∫
z=0
2
= 4/3
V
x
y
2z – 4x – y = 0
2
D
1
4
4 – 4x – y = 0
is in the yz plane.
Hence D = {0 < z < 2; 0 < y < 2z}

26 triple integrals

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26 triple integrals