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
 a b f ( x , y ) dx
Integration with x as the variable of integration with the domain a < x < b
 c d 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   a g
1 g
2
(
( x x
)
) f ( x , y ) dydx c d   h
2 h
1
(
( y y )
) f ( x , y ) dxdy
Triple Integral:
3! Possible orders of integration: f(x,y,z) a y
2 y
1
(
( x b    x )
) z
1 z
2
(
( x , x , y y
)
) f ( x , y , z ) dzdydx z z
1
2
(
( b    a x x
)
) y
1 y
2
(
( x , x , z z
)
) f ( x , y , z ) dydzdx c x
2 x
1
(
( y d    y )
) z
1 z
2
(
( x , x , y ) y ) f ( x , y , z ) dzdxdy e y
2 y
1
(
( z f    z )
) x
1 x
2
(
( y , y , z z
)
) f ( x , y , z ) dxdydz c z z
1
2
(
( y d    y )
) x
1 x
2
(
( y , y , z z
)
) f ( x , y , z ) dxdzdy e x
2 x
1
(
( z f    z )
) y
1 y
2
(
( x , x , z z
)
) f ( x , y , z ) dydxdz

Example:
Evaluate
1.
2.
3.
0
2

1
0

4 

2

1
0
3 sec 
 y
2 cos r dr d  x dy dx .
.

 x
0
2 x 
3 x 2  8 y 3
 dy dx .

4.

5.

• 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 = g
1
(x) and above by y = g
2
(x), and by a < x < b, then the area is given by
A
 b   a g
1 g
2
(
( x x
)
) dydx
Consequently, if a region R is bounded on the left x= h
1
(y) and to the right by x = h
2
(y), and by c < y < d, then the area is given by
A
 c d   h
1 h
2
(
( y y
)
) dxdy

Example
Set up the double integral that gives the area between y = x 2 and y = x 3 .
Find the area of the region bounded by x = y 2 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
• Domain a < x < b b x c < y < d

( a )
Example

( 40

R
2 xy ) dA
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 – x 2 - y 2 and bounded by the Domain
-1 < x < 1 0 < y < 1

• Solids Bounded by curves at the base.
(b,d) y = g
2
(x) x = h
1
(x) x = h
2 y = g
1
(x)
(x)
(a,c) b   a g
1 g
2
(
( x x
)
) f ( x , y ) dydx c d   h
1 h
2
(
( y ) y ) f ( x , y ) dxdy

Example
Find the double integral of f(x,y) = 6x 2 + 2y over the region enclosed by y = x 2 and y = 4.
Evaluate the integral of
2  
0 3
3 y e x
2 dxdy

Example: Double Integral for Volumes
Set up the integral to find the volume of the solid that lies below the cone z = 4 – (x 2 + y 2 ) 1/2 and above the xy-plane

Example
Setup and determine the volume above the xy – plane and below the paraboloid
2
2
And bounded by y = 2x and y = x 2.

• The volume bounded by two surfaces can be acquired as:
V
 b   a y
1 y
2
(
( x x
)
)
( z upper
 z lower
) 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 x 2 + y 2 + z 2 = 6 and above the paraboloid z = x 2 + y 2 . 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 = (x 2 + y 2 ) 1/2 and the sphere z = (1 – x 2 – y 2 ) 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 = 5x 2 + 5y 2 and z = 6 – 7x 2 – y 2

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 = (x 2 + y 2 ) 1/2 and the sphere z = (1 – x 2 – y 2 ) 1/2 .

Example:

Assignment:!!!!!!!!!!!!!!!!!
Use triple integral to find the volume of the solid within the cylinder x 2 + y 2 = 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 = 5x 2 + 5y 2 and z = 6 – 7x 2 – y 2

Example: