10.1-Double-integrals-definition-2.power point

In this Chapter:
 1 Double Integrals over Rectangles
 2 Double Integrals over General Regions
 3 Double Integrals in Polar Coordinates
 4 Applications of Double Integrals
 5 Triple Integrals
 6 Triple Integrals in Cylindrical Coordinates
 7 Triple Integrals in Spherical Coordinates
 8 Change of Variables in Multiple Integrals

Circumscribed rectangles
are all above the curve:
0
1
2
3
1 2 3 4
0
1
2
3
1 2 3 4
Inscribed rectangles are
all below the curve:

By taking the limit as n (the number of
rectangles) approaches infinity, the
actual area is approached.
n®¥
limsL
= A (actual area) =
n®¥
limSU

Riemann Sum
The area under the curve is subdivided
into n subintervals. Each subinterval is
treated as a rectangle. The area of all
subintervals are added to determine the
area under the curve.
There are several variations of Riemann
sum as applied to composite integration.

In Left Riemann
sum, the left-
side sample of
the function is
used as the
height of the
individual
rectangle.
a b
Left Riemann Sum
y
x
x


In Right Riemann
sum, the right-side
sample of the
function is used as
the height of the
individual rectangle.
a b
Right Riemann Sum
y
x
x


In the Midpoint
Rule, the sample at
the middle of the
subinterval is used
as the height of the
individual rectangle.
a b
Midpoint Rule
y
x
x


Definition:
A sum such as the one below is called a
Riemann sum.
f (x1
)+ f (x2
)+ f (x3
)+...+ f (xn
)
é
ë
ù
ûDx where Dx =
b-a
n
Riemann Sum

Note that the very small change in x
becomes dx.
lim
n®¥
f xk
( )Dx
k=1
n
å = f x
( )dx
a
b
ò
lim
n®¥
f xk
( )Dx
k=1
n
å

Multiple Integrals
• In this chapter, we extend the idea
of a definite integral to double and triple
integrals of functions of two or three
variables.
• These ideas are then used to compute
volumes, masses, and centroids of more general
regions , probabilities when two random variables
are involved.

• We will see that polar coordinates
are useful in computing double integrals over
some types of regions.
Triple Integrals:
– Cylindrical coordinates
– Spherical coordinates

Double Integrals
• Just as our attempt to solve the area
problem led to the definition of a definite
integral, we now seek to find the
volume of a solid.
• In the process, we arrive at the definition
of a double integral.

- With functions of one variable, we integrated
over interval (one-dimensional space)
- How about functions of two-variables?

• Let’s consider a function f of two
variables defined on a closed rectangle
• R = [a, b] x [c, d]
= {(x, y) in R2 | a ≤ x ≤ b, c ≤ y ≤ d
• and we first suppose that f(x, y) ≥ 0.
– The graph of f is a surface with equation
z = f(x, y).

Let’s “S” be the graph of f(x,y) over the rectangle R

How can we estimate/find the volume of the solid
bounded by the surface S and the xy-plane?

• The first step is to divide the rectangle R into
subrectangles.
– We divide the interval [a, b] into m subintervals
[xi–1, xi] of equal width ∆x = (b – a)/m.
– Then, we divide [c, d] into n subintervals
[yj–1, yj] of equal width ∆y = (d – c)/n.

R is divided into a series of smaller rectangles
From each of these rectangles, let’s choose a point (xi*, yi*)
in each Rij.

Over each of these smaller rectangles, let’s construct
a box whose height: f (xi*, yi*)

• What is the area of each rectangle?
• ∆A = ∆x ∆y
• How about its height?
• f (xi*, yj*)
So volume of the box ?
f(xi *, yj *) ∆A

Volume of the solid- double sum?
V » f (xi
*
, yj
*
)DA
j=1
m
å
i=1
n
å

• Our intuition tells us that the
approximation given in
becomes better as m and n become larger
and larger.
• So, we would expect that:
V = lim
n,m®¥
f (xi
*
, yj
*
)DA
j=1
m
å
i=1
n
å
V » f (xi
*
, yj
*
)DA
j=1
m
å
i=1
n
å

Double Integral
• The double integral of f over the rectangle R is:
if this limit exists.
f (x, y)dA =
R
òò lim
m,n®¥
f (xi
*
, yj
*
)DA
j=1
m
å
i=1
n
å

One Interpretation of the Double Integral:
If f (x, y) ≥ 0, then the volume V of the solid that lies
above the rectangle R and below the surface z=f (x, y)
is


R
dA
y
x
f
v )
,
(

How to calculate
this double Reimann sum
Or
Double Integral?

Midpoint rule for Double Integrals
where xi is the midpoint of [xi-1, xi] and yj is the
midpoint of [yj-1, yj ].
f (x, y)dA
R
òò » f (
j=1
m
å
i=1
n
å xi, yj )DA

Estimate the volume of the solid that lies above
the square R = [0, 2] x [0, 2] and
below the elliptic paraboloid z = 16 – x2 – 2y2.
– Divide R into four equal squares (m = n= ?) and
choose
the sample point to be the upper right corner
of each square Rij.
Example:

• The squares are shown here.
– The paraboloid is the graph
of f(x, y) = 16 – x2 – 2y2
– The area of each
square is 1.

• Approximating the volume by the Riemann sum with
m = n = 2, we have:
V » f (xi
, yj
)
j=1
2
å
i=1
2
å DA
= f (1,1)DA+ f (1,2)DA+ f (2,1)DA+ f (2,2)DA
=13(1)+7(1)+10(1)+4(1)
= 34

•That is
the volume of
the approximating
rectangular boxes
shown here.

•The following figure shows how, when we use 16,
64, and 256 squares,

Don’t be confused with Double sum!!!

10.1-Double-integrals-definition-2.power point

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