**Unformatted text preview: **THE DEFINITE INTEGRAL
PLANE AREA
(POLAR COORDINATE)
EM 1202
Lecture by: Engr. Stephanie Y. Canete
Summer 2019 POLAR COORDINATE SYSTEM
INTRODUCTION
• Graph of a polar equation
Application
• Solving for Plane Areas ion: 3. • SEATWORK: r cos 3sin V. Sketch the graph of the polar equation:
1. r 2(1 sin )
2. r 2(1 cos )
3. r
2sin
4. r 1 2cos )
5. r 6. r 2 25cos 2
1
1 2sin No. 1 No. 2 No. 3 No. 4 No. 5 No. 6 Section 3-8 : Area With Polar Coordinates
In this section we are going to look at areas enclosed by polar curves. Note as well that we said
“enclosed by” instead of “under” as we typically have in these problems. These problems work a
little differently in polar coordinates. Here is a sketch of what the area that we’ll be ﬁnding in this
section looks like. Area With Polar Coordinates (Single Curve) We’ll be looking for the shaded area in the sketch above. The formula for ﬁnding this area is, Notice that we use in the integral instead of so make sure and substitute accordingly Let’s take a look at an example.
Example
Example
1: 1 Determine the area of the inner loop of
Calculus II - Area with Polar Coordinates 7/9/2019 . Calculus II - Area with Polar Coordinates Hide Solution For this problem we’ll also need to know the valuesCan
of where
you seethe
why we needed to know the values of where the curve goes through th
points
curve goes through the origin. We can get these byThese
setting
the deﬁne where the inner loop starts and ends and hence are also the limit
integration in the formula.
equation
equal
to
zero
and
solving.
We
graphed
this
function
back when we ﬁrst started looking at polar co
ketch of this curve with the inner loop shaded in.
So, the area is then, problem we’ll also Here
need
tosketch
know
thecurve
values
ofinnerwhere
thein.curve goes th
is the
of this
with the
loop shaded
can get these by setting the equation equal to zero and solving.
tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx 7/9/2019 Calculus II - Area with Polar Coordinates So, that’s how we determine areas that are enclosed by a single curve, but what about situations
like the following sketch where we want to ﬁnd the area between two curves. Area With Polar Coordinates (Between Two Curves) In this case we can use the above formula to ﬁnd the area enclosed by both and then the actual
area is the difference between the two. The formula for this is, example of this.
Let’s take a look at an example of this.
e the area that lies inside
and outside
.
Example 2 Determine the area that lies inside Example 2: and outside . Hide Solution
e region that we are after.
Here is a sketch of the region that we are after.
Find the points of intersection by equating the two:
7/9/2019 a, we’ll need to know the values of Calculus II - Area with Polar Coordinates Here is a sketch of the ﬁgure with these angles added. for which the two curves intersect. We Note as well
that we also acknowledged that another representation for the angle
Let’s take a look at an example
ofhere
this.
7/9/2019 . This is important for this problem. In order to use the formula above the area must Calculus II - Area with Polar Coordinates
Calculus II - Area with Polar Coordinates
enclosed as we increase from Example 2: So, if we use
Example 2 Determine the area that lies insidethe smaller to larger angle.
and
outside Hide Solution to
. we w
enclose the shaded area, instead we will enclose the bottom most of the three regions.
However, if we use the angles
to
we will enclose the area that we’re after. So, the area is then,
Here is a sketch
of the
with these
angles
added.
Here
is ﬁgure
a sketch
of the
region
that we
h of the ﬁgure with these angles added. are after. Note as well here that we also acknowledged that another representation for the angle
is
re that we also
that
representation
forthe
theformula
angle above
is the area must be
. Thisacknowledged
is important for
thisanother
problem.
In order to use
Let’sabove
work a the
slight
modiﬁcation
portant for this problem. In order to use the formula
area
must be of the previous example.
enclosed as we increase from the smaller to larger angle. So, if we use
to
we will not 7/9/2019 n This time we’re looking for the following region. Let’s work a slight modiﬁcation of the previous example.
Example
Example
3: 3 Determine the area of the region outside
Calculus II - Area with Polar Coordinates Calculus II - Area with Polar Coordinates and inside . Hide Solution
tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx This
we’re looking
e looking
fortime
the following
region.for the following region.
So, this is the region
we region
get by that
using
. The area
So, thisthat
is the
wethe
getlimits
by usingtothe limits
to for .this
Thereg
ar So, this is the region that we get by using the limits region that we get by using the limits to . The area for this region is, to . The area for this regio Notice
that
for“outer”
this
the the
“outer”
andfunction
“inner”
opposite!
Notice that for Notice
this
area
the
and
“inner”
were were
opposite!
that
for
thisarea
area
“outer”
and function
“inner”
function
were opposite So, this is the region that we get by using the limits to . The area for this region is, Let’s do one ﬁnal modiﬁcation of this example.
Example
Example
4:4 Determine the area that is inside both and . Hide Solution
Calculus II - Area with Polar Coordinates Here is the sketch for this example.
We are not going to be able to do this problem in the same fashion that we did the previous two.
There is no set of limits that will allow us to enclose this area as we increase from one to the
other. Remember that as we increase the area we’re after must be enclosed. However, the
only two ranges for that we can work with enclose the area from the previous two examples
and not this region. In this case however, that is not a major problem. There are two ways to do get the area in this
tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
problem. We’ll take a look at both of them.
Solution 1
In this case let’s notice that the circle is divided up into two portions and we’re after the upper
portion. Also notice that we found the area of the lower portion in Example 3. Therefore, the
area is, e able to do this problem in the same fashion that we did the previous two.
s that will allow us to enclose this area as we increase from one to the
as we increase the area we’re after must be enclosed. However, the Let’s do one ﬁnal only two ranges for that we can work with enclose the area from the previous two examples
and not this region.
modiﬁcation
of this example.
In this case however, that is not a major problem. There are two ways to do get the area in this
problem. We’ll take a look at both of them. Example
Example
4:4 Determine the area that is inside both
Hide Solution and . Solution 1
In this case let’s notice that the circle is divided up into two portions and we’re after the upper
portion. Also notice that we found the area of the lower portion in Example 3. Therefore, the
SOLUTION NO. 1
area is, Calculus II - Area with Polar Coordinates Here is the sketch for this example. Solution 2
SOLUTION NO. 2
In7/9/2019
this case we do pretty much the sameCalculus
thing
this
time we’ll think of the area as the
II - except
Area with Polar
Coordinates
other portion of the limacon than the portion that we were dealing with in Example 2. We’ll also
need to actually compute the area of the limacon in this case.
So, the area using this approach is then,
tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx e able to do this problem in the same fashion that we did the previous two.
s that will allow us to enclose this area as we increase from one to the
as we increase the area we’re after must be enclosed. However, the
A slightly longer approach, but sometimes we are forced to take this longer approach. = 2∫
5 + 18sin θ + 2 (1 − cos ( 2θ ) ) dθ
3π
−0.3398
2
9
1 107
Make sure you can do=the
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θ
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42.0376
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this section
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If you don’t recall how
The
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at the Integrals
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1. Find theinarea
thewill
loop
of
.
r
=
3
−
8cos
θ
= ∫involve
−
−
+
Ainner
4
2
cos
θ
6
2
cos
θ
integrals
thisinside
section
this
kind
of
manipulation.
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) (
) Ifdθyou don’t recall how t
2π 2
3
them
go
back
and
take
a
look
at the Integrals Involving Trig Functions section.
Ans.
A
=
15.2695
sq.
units.
Make sure you can do the trig
4 π manipulations required to do these integrals. Most of th
3
= ∫involve
−10 −this
θ dof
θ manipulation. If you don’t recall how
20 cos
integrals
thisinside
section
kind
2. Find theinarea
thewill
graph
2 π of r = 7 + 3cos θ and to the left of the y-axis.
Step 1
3
them
go
back
and
take
a
look
at the
the Integrals
Involving
Trig Functions
section.
Do not
getisA too
excited
about
all
the
second
step
above.
Just because
4 π inwe
Ans.
=a42.0376
sq. units.
First,
here
quick
sketch
of the
graphminus
of thesigns
region
are
interested
in.
3
−10θ
θ ) )not
= 13.6971
20sindoes
(
(
terms have minus signs in=front
of−them
2 π mean that we should get a negative va
3.
Find
the area that is inside r = 3 + 3sin θ and outside
r = 2.
3
Step
1
out integral!
is the “inner” graph. Problem SET # 8 Ans. is
A =a33.7074
sq. units.of the graph of the region we are interested in.
First, here
quick sketch
Do
not the
get area
too excited
aboutrall
minus
signs
the
integral.
Just because all the term
4. Find
that is inside
and
outside
= 2the
3sin
r =in3 +
θ. minus signs in front of them does not mean that we should get a negative value from ou
Ans. A = 3.8622 sq. units. 5. Find
Step
1 the area that is inside r = 4 − 2 cos θ and outside r = 6 + 2 cos θ .
Ans. is
A =a13.6971
sq. units.of the graph of the region we are interested in.
First, here
quick sketch
6. Find the area that is inside both r = 1 − sin θ and r = 2 + sin θ . Step 1
#$
Ans. A =
− 2 + 3 3 = 2.6578 . %
First, here is a quick
sketch of the graph of the region we are interested in. ...

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