Multiple Integration - Iterated Integrals and Area in the Plane
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Transcript Multiple Integration - Iterated Integrals and Area in the Plane
Iterated Integrals and
Area in the Plane
By Dr. Julia Arnold
Courtesy of a CDPD grant
Objectives of this section:
1. Evaluate an Iterated Integral
2. Use an iterated integral to find the area of a plane
region
Objective 1
Evaluate an Iterated Integral
Definition of an Iterated Integral
Just as we can take partial derivative by considering only one of
the variables a true variable and holding the rest of the
variables constant, we can take a "partial integral". We indicate
which is the true variable by writing "dx", "dy", etc. Also as with
partial derivatives, we can take two "partial integrals" taking one
variable at a time. In practice, we will either take x first then y
or y first then x.
We call this an iterated integral or a double integral.
Notation:
Let f(x,y) be a function of two variables defined on a
region R bounded below and above by
y = g1(x)
and
y = g2(x)
and to the left and right by
x = a
and
x = b
then the double integral (or iterated integral) of f(x,y)
over R is defined by
The first integration gives us a function in x
second gives us a numerical value.
Let’s look at an example
while the
Example 1: Evaluate the iterated integral
3 2
2
x
y dydx
The order the dx
dy is in determines
which you do first.
0 1
3
2
0
1
2
[
x
y dy] dx
x 2 y 2 2
x 2 22 x 212 3 x 2
0 2 1 dx and since 2 2 2
3
3x 2
0 2
3
3 x 3 x 3 3 27
27
dx
0
6
2 0 2
2
We integrate with
respect to y
holding x term like
a constant.
Evaluate it at its
limits.
Then we integrate
with respect to x
and evaluate it at
its limits.
Example 2: Evaluate the iterated integral
4
x
2 ye x dydx
This is the solution to the first integral:
1 1
x
x
1 1 2 ye dy dx
4
4
xe
x
x
2 ye
x
dy y e
1
e x dx using parts
0
xe
x
2 x
4
1 4
4
1
4e e 4
1
e e
2
x
x e x 12 e x xe x e x
1
Hint: start with u = e-x and dv=(x – 1)
Objective 2
Use an iterated integral to find
the area of a plane region
Let’s begin by finding the area of a rectangular
y
region.
d
R
c
a
b
x
y
If we integrate with
respect to y first we
would go from c to d.
d
Then integrate with
respect to x and we
would go from a to b.
R
c
a
b
x
b d
b
a c
a
dydx d c dx (d c)(b a)
Which is the same as length times width.
Example 2: Use an integral to find the area of the region.
y
Y goes from 0 to
x2 y 2 4
4 x2
x goes from 0 to 2
2
4 x 2
0
dydx
Using a table of integrals
0
2
4 x 2
1
x2
2
2
dy
dx
4
x
dx
x
4
x
4
arcsin
0 0 0
2
2 0
1
2 1
0
2
2
4
4
4
arcsin
0
4
0
4
arcsin
2
2 2
2
2
x
2 arcsin1 2 arcsin 0 2
2
20
Since we know this is ¼ of a circle we can verify
by using the traditional formula.
Exercise: Use an iterated integral to find the area of the
region bounded by the graphs of the equations.
xy 9, y x, y 0, x 9
First let’s sketch
the bounded area.
y
It looks like we might
need to divide this
into two problems.
Since the
left area is
of a right
triangle we
could save
time and use
the formula.
x
Exercise: Use an iterated integral to find the area of the
region bounded by the graphs of the equations.
xy 9, y x, y 0, x 9
Triangle area
1
9
3 3 and
2
2
3 x
x2 3 9
0 0 dydx 2 0 2
area on right
9x
9
9
9
9
dy
dx
dx
ln
x
9
ln
9
9
ln
3
9
ln
9 ln 3
3 0 3 x
3
3
9
Total Area +9ln3
2
9
y
x
You may be wondering if you
integration. The answer is
be easier than the other.
Exercise: Sketch the region
order of integration.
0
can switch the order of
yes. However, one way may
R of integration and switch the
4 2
y
f ( x, y )dydx
y
y x 2 and 0 y 4
y x2 4
x
Exercise: Sketch the region
order of integration.
R of integration and switch the
y
Switched
4 2
0
f ( x, y )dxdy
y
y x 2 and 0 y 4
y x2 4
2
x2
0
0
f ( x, y)dydx
x
Last Exercise
Sketch the region R whose area is given by the iterated
integral. Then switch the order of integration and show that
both orders yield the same area.
0 x 4 y 2 and 2 y 2
2 4 y 2
2
2
dxdy
2
0
2
y3 2
4 y dy 4 y
3 2
2
3
2
4 2
4 2
3
3
3
8 8 8 8 16 16 32
3
3
3
3
Last Exercise
Sketch the region R whose area is given by the iterated
integral. Then switch the order of integration and show that
both orders yield the same area.
0 x 4 y 2 and 2 y 2
x 4 y2
y2 4 x
y 4 x
We will need
two double
integrals in
this order.
4
4 x
0
4
0
4
dydx
0
dydx
0 4 x
4
4
0
0
1
2
4 xdx 4 xdx 2 4 x (1)dx
0
4
1
2
2 4 x (1)dx
4 4 x
3
0
4 4 4
3
3
2
4 4 0
3
3
2
0
3
2
4
0
32 32
3
3
For comments on this
presentation you may email the
author
Dr. Julia Arnold at
[email protected].