Multiple Integrals

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# Multiple Integrals - PowerPoint PPT Presentation

Multiple Integrals. Part II. Triple Integral. If f ( x , y , z ) is a function of 3 variables and B = [ a , b ]×[ c , d ]×[ r , s ] is a box within the domain of f , then the triple integral of f over the box B is. if this limit exists.

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## Multiple Integrals

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Presentation Transcript
Triple Integral

If f(x,y,z) is a function of 3 variables and B = [a,b]×[c,d]×[r,s] is a box within the domain of f , then the triple integral of f over the box B is

if this limit exists.

One Physical Interpretation of the Triple Integral

If B is a solid box whose density is not uniform but can be described by a function ρ(x,y,z), then

In particular, if the density is a constant k, then

2

1

0

x

1

-1

2

y

-2

z-simple regions

A bounded region E in 3D is said to be z-simple or of type I if it is bounded by the graphs of two continuous functions of x and y, i.e.

E = {(x,y,z): (x,y) D, u1(x,y) ≤ z ≤ u2(x,y)}

Example:

2

1

0

x

1

-1

2

y

-2

z-simple regions

In this case,

Example:

x-simple regions

A bounded region E in 3D is said to be x-simple or of type II if it is bounded by the graphs of two continuous functions of y and z, i.e.

E = {(x,y,z): (y,z) D, u1(y,z) ≤ x ≤ u2(y,z)}

y-simple regions

A bounded region E in 3D is said to be y-simple or of type III if it is bounded by the graphs of two continuous functions of x and z, i.e.

E = {(x,y,z): (x,z) D, u1(x,z) ≤ y ≤ u2(x,z)}

A deformed torus

8

4

0

-4

x

y

Find the total mass of the solid bounded by

if the density function is ρ(x, y, z) = 6 – z

10

8

6

4

2

0

-1

y

0

1

-2

-1

0

x

1

2

Find the total mass of the solid bounded by

if the density function is ρ(x, y, z) = y

2

1

z

0

-1

-2

-2

-2

-1

-1

0

x

0

y

1

1

2

2

Examples

Volume of a sphere with radius r

1

0.8

0.6

0.4

0.2

0

-0.4

-0.4

-0.2

-0.2

0

0

0.2

0.2

0.4

0.4

Examples

Volume of an ice-cream cone with radius 0.5and angle 

6

4

2

z

0

-2

-4

-4

-2

4

2

0

0

2

-2

4

y

-4

x

Find the volume of the solid bounded above by

and below by

6

4

2

z

0

-2

-4

-4

-2

4

2

0

0

2

-2

4

y

-4

x

Find the volume of the solid bounded above by

and below by

6

4

2

z

0

-2

-4

-4

-2

4

2

0

0

2

-2

4

y

-4

x

where can be found by solving for  from the equation

z

1

0.5

0

-0.5

-1

-1

-1

-0.5

-0.5

0

x

0

0.5

y

0.5

1

1

Bumpy Sphere

The general equation is of the form

In this picture,

m = 6 and n = 5.

The volume can be computed by integration in spherical coordinates.

y

R

x

Change of variables

If we want to evaluate

but the function f(x,y) is difficult to integrate, we can try to create a substitution u = g(x,y), v = h(x,y)

If this works, we will be integrating an easier function w(u.v) but the domain of integration will be completely changed.

v

S

u

Change of variables

On the other hand, we may want to evaluate

but the region R is difficult to work with, we can try to create a one-to-one transformation T(u,v) = (g(u,v), h(u,v)) and a rectangle S such that T(S) = R.

If this works, we will be integrating the function f(g(u,v), h(u,v)) over the rectangle S !

y

v

T

R

S

u

x

The idea on the previous page is nice because we can now integrate over a rectangle, which is very easy indeed.

However, when we use a transformation, dA = ΔxΔy will not be the same as ΔuΔv, hence we need to consider an additional factor which is introduced on the next page.

Jacobian

Let T: 2→2 be a transformation given by x = g(u,v) and y = h(u,v), then the Jacobian of T is defined to be

Change of Variables
• Suppose that T is a C1 transformation that maps a region S in the uv-plane onto a region R in the xy-plane. If
• T is one-to-one on S (except possibly on the boundary),
• fis continuous on R,
• the Jacobian of T is nonzero on S and
• R is either a x-simple or y-simple plane region, then

v

y

2

2

1

1

u

x

0

-1

1

0

-1

1

Example 1: Find the image of the square S = [0,1]×[0,1] under the transformation T given by

x = u2 – v2y = 2uv

Alsocompute the Jacobian for this transformation

Example 2: Compute

v

3

2

1

S

0

1

2

3

u

where u = x + y and

v = x - y

Remark

In practice, we usually introduce u as a function of x, y and v also as a function of x, y, hence it will be easier to compute

fortunately, we have the result that

hence we only have to re-express the reciprocal in term of u and v.

Example 3: Compute

v

5

4

s

3

2

1

0

1

2

3

4

u

put u = xy and v = x2 – y2

1

-1

0

1

2

3

x

0

R

-1

-2

-3

Example 4: Compute

y = x - 1

v

5

4

?

3

2

y = x - 2

1

0

1

2

3

4

u

put u = x + y and v = x – y

v

3

v = 2

v = 1

u

1

2

3

-3

-2

-1

0

-1