1. No category

MULTIPLE INTEGRATION
MATH23
MULTIVARIABLE CALCULUS
GENERAL OBJECTIVE
At the end of the lesson the students are expected to:
• Determine the geometric interpretation of partial
derivatives and its derivation
14.1.1 (p. 1001)
The Volume Problem
Figure 14.1.2
Figure 14.1.3
Definition 14.1.2 (p. 1002)
Definition: Partial Definite Integrals

b
a
f ( x, y)dx
Integration with x as the variable of integration with the domain
a<x<b

d
c
f ( x, y)dy
Integration with y as the variable of integration with the domain
c<y<d
• Evaluation of Multiple Integrals:
Double Integral:
2! Possible orders of integration: f(x,y)
b
g2 ( x)
a
g1 ( x )

d
h2 ( y )
c
h1 ( y )

f ( x, y )dydx
f ( x, y )dxdy
Triple Integral:
3! Possible orders of integration: f(x,y,z)
b

y2 ( x )
a
y1 ( x )
d

x2 ( y )
c
x1 ( y )
f
y2 ( z )

e
y1 ( z )

z2 ( x , y )

z2 ( x , y )

x2 ( y , z )
z1 ( x , y )
z1 ( x , y )
x1 ( y , z )
f ( x, y, z )dzdydx
f ( x, y, z )dzdxdy
f ( x, y, z )dxdydz
b

z2 ( x )

y2 ( x , z )
a
z1 ( x )
d

z2 ( y )

x2 ( y , z )
c
z1 ( y )
f
x2 ( z )
 
y2 ( x , z )
x1 ( z )
y1 ( x , z )
e
y1 ( x , z )
x1 ( y , z )
f ( x, y, z )dydzdx
f ( x, y, z )dxdzdy
f ( x, y, z )dydxdz
Example:
Evaluate

2 1
1.
2
3
y
  cos xdydx.
0 0

2.
4 sec
  r dr d .
0
0
2 2x
3.

2
3
3x

8y

dy dx .
1 x


4.
5.
Applications of Multiple Integration
• Areas by double integration
• Volume by double Integration
– Rectangular Base
– Base bounded by given Curves
– Solids bounded by two surfaces
• Volume by triple integration
Area by Double Integral
If a region R is bounded below by y = g1(x) and above
by y = g2(x), and by a < x < b, then the area is given by
A
b
a

g2 ( x)
g1 ( x )
dydx
Consequently, if a region R is bounded on the left x= h1(y)
and to the right by x = h2(y), and by c < y < d, then the
area is given by
A
d
c

h2 ( y )
h1 ( y )
dxdy
Example
Set up the double integral that gives the area
between y = x2 and y = x3.
Find the area of the region bounded by
x = y2 and y = x.
Double Integral for Volumes
Let R be a region in the xy-plane and T be the
solid bounded below by R and bounded above
by the surface z = f(x,y).
Then the volume of T is found by
V= ∫∫ f(x,y) dxdy
Evaluation of Multiple Integral (Double Integral)
A. Rectangular Based Solids
y
d
c
a
b
x
• Domain
a<x<b
c<y<d
Example
(a)
 (40  2 xy)dA
R
Where R is the rectangle:
1≤ x ≤ 3 ; 2 ≤ y ≤ 4
1. Determine the volume above the xy plane
and below the Surface:
z = 5 – x2 - y 2
and bounded by the Domain
-1 < x < 1
0<y<1
Volumes by Multiple Integration
• Solids Bounded by curves at the base.
(b,d)
y = g2 (x)
x = h1 (x)
x = h2 (x)
y = g1 (x)
(a,c)
b
g2 ( x)
a
g1 ( x )

f ( x, y )dydx
d
h2 ( y )
c
h1 ( y )

f ( x, y )dxdy
Example
Find the double integral of f(x,y) = 6x2 + 2y
over the region enclosed by y = x2 and y = 4.
Evaluate the integral of
2
3
0
3y

x2
e dxdy
Example: Double Integral for Volumes
Set up the integral to find the volume of the
solid that lies below the cone
z = 4 – (x2 + y2)1/2
and above the xy-plane
Example
Setup and determine the volume above the xy – plane
and below the paraboloid
z = x2 + y2
And bounded by y = 2x and y = x2.
Bounded by two surfaces
• The volume bounded by two surfaces can be acquired as:
V 
b
 
a
y2 ( x )
y1 ( x )
( zupper  zlower ) dydx
• Where the limits of integration are obtained from the cylinder
that contains intersection between the surfaces as projected
against the xy-plane.
Example:
Bounded by two surfaces
Set up the double integral that gives the volume of
the solid that lies below the upper part of the
sphere x2 + y2 + z2 = 6 and above the paraboloid z =
x2 + y2 . DO NOT evaluate the integral.
Example:
Set up the integral of f(x,y,z) over G, the solid “ice
cream cone” bounded by the cone z = (x2 + y2)1/2
and the sphere
z = (1 – x2 – y2)1/2.
Definition: Triple Integral
Let f(x,y,z) be the density
of some 3-dimensional
solid W.
Objective: Define the
triple integral of f
over W.
Example:
Set-up the integral to find the volume of
the solid enclosed between the paraboloids
z = 5x2 + 5y2 and z = 6 – 7x2 – y2
Example:
Definition: Riemann Sum Value
Definition
Example:
Set up the integral of f(x,y,z) over G, the solid “ice
cream cone” bounded by the cone z = (x2 + y2)1/2
and the sphere
z = (1 – x2 – y2)1/2.
Example:
Assignment:!!!!!!!!!!!!!!!!!
Use triple integral to find the volume of the
solid within the cylinder x2 + y2 = 9 and
between the plane z = 1 and x + z = 5.
Example:
Set-up the integral to find the volume of
the solid enclosed between the paraboloids
z = 5x2 + 5y2 and z = 6 – 7x2 – y2
Example: