Chapter 15 – Multiple Integrals 15.1 Double Integrals over Rectangles 1 Objectives: Use double integrals to find volumes Use double integrals to find

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Chapter 15 – Multiple Integrals15.1 Double Integrals over Rectangles

15.1 Double Integrals over Rectangles

Objectives: Use double integrals to

find volumes

Use double integrals to find average values


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Definite Integral ReviewFirst, let’s recall the basic facts concerning

definite integrals of functions of a single variable:

If we divide an interval into subintervals of the same width, then we form the Riemann Sum

If we take the limit of this then we obtain the definite integral

*

1

( )n

ii

f x x

*

1

lim ( ) ( )n b

i ani

f x x f x dx


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Volumes In a similar manner, we consider a function f

of two variables defined on a closed rectangle

R = [a, b] x [c, d] = {(x, y) 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).


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VolumesLet S be the solid that lies above R and under the graph of f, that is,

S = {(x, y, z) R3 | 0 ≤ z ≤ f(x, y), (x, y) R}

Our goal is to find the volume of S.


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VolumesWe can approximate the part of S that lies above each Rij by a thin rectangular box (or “column”) with:

◦ Base Rij

◦ Height f (xij*, yij*)

The volume of thisbox is the height of the box times the area of the base rectangle:

f(xij *, yij *) ∆A


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VolumesWe follow this procedure for all

the rectangles and add the volumes of the corresponding boxes.

Thus, we get an approximation to the total volume of S:

* *

1 1

( , )m n

ij iji j

V f x y A


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VolumesOur intuition tells us that the

approximation becomes better as

m and n become larger.

So, we would expect that:

* *

,1 1

lim ( , )m n

ij ijm n

i j

V f x y A


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VolumesThis double sum means that:

◦ For each subrectangle, we evaluate f at the chosen point and multiply by the area of the subrectangle.

◦ Then, we add the results.


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Definition – Double Integral


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Double IntegralThe sample point (xij*, yij*) can be chosen to

be any point in the subrectangle Rij*.However, suppose we choose it to be the

upper right-hand corner of Rij [namely (xi, yj)].

Then, the expression for the double integral looks simpler:

,1 1

( , ) lim ( , )m n

i im n

i jR

f x y dA x y A


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VolumeIf 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

V f x y dA


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Example 1If R= [-1,3] x [0,2], use a

Riemann sum with m = 4, n=2 to estimate the value of .

Take the sample points to be the upper left corners of the squares.

2 22R

y x dA


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Mid Point Rule


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Average ValueWe define the average value of a

function f of two variables defined on a rectangle R to be

1( , )

( )avg

R

f f x y dAA R


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Example 2 – pg.982 #12Evaluate the double integral by

first identifying it as the volume of a solid.

(5 ) , , | 0 5, 0 3R

x dA R x y x y


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More Examples

The video examples below are from section 14.6 in your textbook. Please watch them on your own time for extra instruction. Each video is about 2 minutes in length. ◦Example 1◦Example 2◦Example 3

http://www.wadsworthmedia.com/math/stewart/solution_videos/6et_15_01_01.html



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Chapter 15 – Multiple Integrals 15.1 Double Integrals over Rectangles 1 Objectives: Use double integrals to find volumes Use double integrals to find