Double Integrals over Rectangles

Math 55 - Elementary Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

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Recall

If f(x) is defined for a x b, we subdivide the interval [a, b]into subintervals of length x. Choose xi on each subinterval.

The definite integral of f from a to b is

baf(x) dx = lim

n!1

nXi=1

f(xi )x

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The Volume Problem

Let f(x, y) 0 and continuous on the rectangular regionR = [a, b] [c, d] = {(x, y)|a x b.c y d}. We want tofind the volume of of the solid under the surface z = f(x, y)over R.

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The Volume Problem

Partition R by dividing the interval [a, b] into m subintervals[xi1, xi] of equal width x and [c, d] into n subintervals[yj1, yj ] of equal width y.

Let Rij = [xi1, xi] [yj1, yj ] and let Aij be the area of Rij ,i.e., A = xy.

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The Volume Problem

Take an arbitrary point (xi , yj ) 2 Rij and construct rectangularprisms with bases Rij and height f(xi , yj ) and denote thevolume of the prism by Vij = f(xi , yj )A

The volume of the solid is approximatelymXi=1

nXj=1

f(xi , yj )A

As we increase the number of partitions,

V = limm,n!1

mXi=1

nXj=1

f(xi , yj )A

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Double Integrals over Rectangles

Definition

If f(x, y) is continuous on a rectangular region R, then thedouble integral of f over R is

R

f(x, y) dA = limm,n!1

mXi=1

nXj=1

f(xi , yi )A

provided this limit exists.

If f(x, y) 0 on R, then the double integral can be interpretedas the volume under the surface given by z = f(x, y) over therectangular region R.

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Iterated Integrals

Suppose f(x, y) is integrable on the rectangular regionR = [a, b] [c, d]. By

dcf(x, y) dy

we mean the definite integral from c to d of f(x, y) with respectto y where x is held fixed, called the partial integral with respectto y. Note that the result is a function of x, so let

I(x) =

dcf(x, y) dy. Hence,

baI(x) dx =

ba

dcf(x, y) dy

dx =

ba

dcf(x, y) dy dx

called an iterated (double) integral.

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Iterated Integrals

Example

Evaluate the iterated integrals:

a.

21

106xy2 dy dx b.

10

216xy2 dx dy

Solution.

a. 21

10

6xy2 dy dx

=

21

106xy2 dy

dx

=

212xy3

y=1y=0

dx

=

212x dx = x2

21

= 4 1 = 3

b.

10

21

6xy2 dx dy

=

10

216xy2 dx

dy

=

103x2y2

x=2x=1

dy

=

109y2 dy = 3y3

10

= 3

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Double Integrals over Rectangles

Theorem (Fubinis Theorem)

If f is continuous on the rectangular region R = [a, b] [c, d],then

R

f(x, y) dA =

ba

dcf(x, y) dy dx =

dc

baf(x, y) dx dy

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Double Integrals over Rectangles

Example

Evaluate

R

y sin(xy) dA, where R = [0, 2] [0,].

Solution. We choose to integrate first with respect to x:

R

y sin(xy) dA =

0

20y sin(xy) dx dy

=

0 cos(xy)

x=2x=0

dy

=

0

( cos 2y + 1) dy

= sin 2y2

+ y

0

=

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Double Integrals over Rectangles

Example

Find the volume of the solid S bounded by the surface4x3 + 3y2 + z = 20, the planes x = 1, y = 2 and the coordinateplanes.

Solution. S is the solid under the surface z = 20 4x3 3y2 abovethe region R = [0, 1] [0, 2]. The volume of S is

R

(20 4x3 3y2) dA = 20

10(20 4x3 3y2) dx dy

=

20

20x x4 3y2x x=1

x=0

dy

=

20

19 3y2 dy

= 19y y320

= 38 8 = 30

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Double Integrals over Rectangles

Suppose f(x, y) can be written as a product of a function of xand a function of y, i.e., f(x, y) = g(x)h(y). If f is continous onR = [a, b] [c, d], then by Fubinis Theorem,

R

f(x, y) dA =

ba

dcf(x, y) dy dx

=

ba

dcg(x)h(y) dy dx

=

ba

dcg(x)h(y) dy

dx

=

bag(x)

dch(y) dy

dx

=

bag(x) dx

dch(y) dy

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Double Integrals over Rectangles

Example

Evaluate

R

x2 sin y dA where R = [1, 2] [0, 2 ].

Solution. Since x2 sin y is a product of g(x) = x2 andh(y) = sin y, by the prevoius theorem,

R

x2 sin y =

21

2

0x2 sin y dy dx

=

21

x2 dx

2

0sin y dy

=

x3

3

21

! cos y

20

!

=

8

3+

1

3

(0 + 1) = 3

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Exercises

1 Calculate the iterated integral.

a.

20

10(2x+ y)8dx dy b.

41

21

x

y+

y

x

dy dx

2 Calculate the double integral.

a.

R

yexy dA, R = [0, 2] [0, 1]

b.

R

x

x2 + y2dA, R = {(x, y)|1 x 2, 0 y 1}

3 Find the volume of the solid in the first octant bounded bythe cylinder z = 16 x2 and the plane y = 5.

4 Sketch the solid bounded by the paraboloidx2 + y2 + z = 4, the planes 2x+ 2y + z = 6, x = 1, y = 1and the coordinate planes then find its volume.

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References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Leithold, L., The Calculus 7, Harper Collins College Div., 1995

3 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/

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