Triple Integrals In this section, we will learn about: Triple integrals.

Natural Science Department – Duy Tan University

Triple Integrals

In this section, we will learn about:
Triple integrals.

Lecturer: Ho Xuan Binh
Da Nang-02/2015


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS

Just as we defined single integrals for functions
of one variable and double integrals for
functions of two variables, so we can define
triple integrals for functions of three variables.

Triple Integrals  


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS.

Let’s first deal with the simplest case where f is
defined on a rectangular box:

B = { ( x , y , z ) a ≤ x ≤ b, c ≤ y ≤ d , r ≤ z ≤ s }

Triple Integrals s  


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS

The first step is
to divide B into
sub-boxes—by
dividing:
The interval [a, b] into l
subintervals [xi-1, xi]
of equal width Δx.
[c, d] into m subintervals of
width Δy.
[r, s] into n subintervals of
width Δz.

Triple Integrals s  


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS
The planes through the
endpoints of these
subintervals parallel to
the coordinate planes

divide the box B into
lmn sub-boxes

Bijk = [ xi −1 , xi ] ×  y j −1 , y j  × [ zk −1 , zk ]
Each sub-box has volume ΔV = Δx Δy Δz
Triple Integrals  


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS
Then, we form the triple Riemann
l
m n
sum

∑∑∑ f ( x
i =1 j =1 k =1

*
ijk

where the sample point
is in Bijk.

Triple Integrals  

*
ijk

*
ijk

) ∆V

*
ijk

*
ijk

,y ,z

(x

*
ijk

,y ,z

)


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS
The triple integral of f over the box B
is:

∫∫∫ f ( x, y, z ) dV

B

= lim

l , m , n →∞

l

m

n

∑∑∑ f ( x
i =1 j =1 k =1

*
ijk

*
ijk

*
ijk

,y ,z

if this limit exists
Again, the triple integral always exists if f
is continuous.
Triple Integrals in Cylindrical Coordinates  

) ∆V


Natural Science Department – Duy Tan University

TRIPLE INTEGRALS

We can choose the sample point to be any point in the
sub-box.
However, if we choose it to be the point
(xi, yj, zk) we get a simpler-looking expression:

∫∫∫ f ( x, y, z ) dV =
B

lim

l , m , n →∞

l

m

n

∑∑∑ f ( x , y , z ) ∆V
i =1 j =1 k =1

Triple Integrals 

i

j

k


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Thank you for your attention