06 Double Integrals Over Rectangles - Handout
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Transcript of 06 Double Integrals Over Rectangles - Handout
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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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