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Limits when finding area of polar curves.

by apples
Tags: curves, limits, polar
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apples
#1
May5-08, 06:55 AM
P: 167
The problem is related to polar curves. most of the topics i need to do are easy (finding the slope, finding the area etc.)
What i'm facing problems with is that when I find the area, I don't know how to find the limits.

1. The problem statement, all variables and given/known data
Sample problem:
Find the area of the region in the plane enclosed by the cardioid r=4+(4sinθ)


2. Relevant equations
b
A = (1/2)(r^2) dθ
a

3. The attempt at a solution

Graphing the curve is no biggie, I use my calculator. The problem is when I use the equation to find the Area, I don't know what the interval is I don't know 'b' and 'a' are.
I have a calculus book, an AP calculus book, and a pre-calculus book. None of the books tell you how to find the interval. In their solved questions, they tell you that in the solution. The problem I wrote above is an example also.
They tell you the interval is (for the q above) 0 to 2π(pi)
but how do i know that what about other questions.

Another example:
Find the area inside the smaller loop of the limacon r=1+(2cosθ)

Here they say the limits are (2π(pi))/(3) to (4π(pi))/3


The exact explanation to this in the book is:
(for first question) "Because r swoops out the region as θ goes from 0 to 2π(pi), these are our limits of integration.
(For 2nd q) Because in the inner loop, r sweeps out the region as θ goes from (2π(pi))/(3) to (4π(pi))/3, these are our limits of integration.

But why!? What does r swoops mean. How do I know the limits!?

please help!
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Hootenanny
#2
May5-08, 08:13 AM
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Okay let's take your first example. So looking at your sketch you should be able to see that the cardioid is a closed path, which means that there are no breaks in the line. Now imagine that you are stood on the point where the cardioid intersects the x-axis and you have hold of a piece of string that is anchored at the origin. Now imagine walking around the cardioid whilst keeping hold of the string, every time you take a step you measure the angle between the x-axis and the string and write it down. Once you get back to where you started, what will be the range of the angles that you measured?

Once you've got your head around the angular limits we'll deal with the radial limits.
apples
#3
May5-08, 10:09 AM
P: 167
Thank you for the help

I think it would be 0-2π(pi) in this case.

Can we make this a bit quick. If I explain the reason, you people might be disgusted, but...
...I am serious this is very urgent.
I need to have mastered this by today night at most.

Hootenanny
#4
May5-08, 10:44 AM
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Limits when finding area of polar curves.

Quote Quote by apples View Post
Thank you for the help

I think it would be 0-2π(pi) in this case.

Can we make this a bit quick. If I explain the reason, you people might be disgusted, but...
...I am serious this is very urgent.
I need to have mastered this by today night at most.
I'm guessing that you've got your final tomorrow .

You're correct, [itex]\theta[/itex] can vary between zero and [itex]2\pi[/itex], this is the angle swept out. Hence, your angular limits are zero and [itex]2\pi[/itex]. Do you follow?

Now for the radial limits. The question states that the area is bounded by the cardioid. So now imagine that the cardioid is a fence and you are inside it, you're allowed to walk anywhere inside the fence, but you can't jump over it. Now, the radius is simply the distance from the origin, so given the boundary, what is the minimum distance you can be from the origin?
apples
#5
May5-08, 10:58 AM
P: 167
No, I have my AP Calculus BC exam day after tomorrow.
I follow the first part. The angle "sweeps out" creating an interval. (i know this might be incorrect terminology).

The second part, I don't fully understand. I understand that the area is bounded/enclosed by the cardioid.
The radius is the distance from the origin, so the minimum distance can be zero? right? because the cardioid intersects at the origin?

OK, so radius=distance, what's the interval?
Hootenanny
#6
May5-08, 11:06 AM
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Quote Quote by apples View Post
I follow the first part. The angle "sweeps out" creating an interval. (i know this might be incorrect terminology).
Sounds good to me.
Quote Quote by apples View Post
The radius is the distance from the origin, so the minimum distance can be zero? right? because the cardioid intersects at the origin?

OK, so radius=distance, what's the interval?
Correct! So we now know that the lower limit is r=0. What about the upper limit? What's the further distance that you can go away from the origin?
apples
#7
May5-08, 11:10 AM
P: 167
At the y-intercept? Other than the origin?

EDIT: in angular limits, where do we start from, the x-axis? always?
Hootenanny
#8
May5-08, 11:12 AM
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Quote Quote by apples View Post
At the y-intercept? Other than the origin?
Forget the mathematics for the moment. Your in a fenced area, what is the furthest you can get away from the centre of the area? What stops you getting out?
apples
#9
May5-08, 11:14 AM
P: 167
I don't know. The farthest from the origin is the top pointy corner of the cardioid
Hootenanny
#10
May5-08, 11:17 AM
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Quote Quote by apples View Post
I don't know. The farthest from the origin is the top pointy corner of the cardioid
If the area is bounded by the curve [itex]r = 4\left(1+\sin\theta\right)[/itex] then you know that [itex]r < 4\left(1+\sin\theta\right)[/itex] so the upper bound is...
apples
#11
May5-08, 11:20 AM
P: 167
I don't understand.
The curve?
Hootenanny
#12
May5-08, 11:32 AM
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Quote Quote by apples View Post
I don't understand.
What exactly don't you understand?
Quote Quote by apples View Post
The curve?
Correct
apples
#13
May5-08, 11:33 AM
P: 167
now what are the limits?
Hootenanny
#14
May5-08, 11:34 AM
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Quote Quote by apples View Post
now what are the limits?
I've just told you,
Quote Quote by Hootenanny View Post
you know that [itex]r < 4\left(1+\sin\theta\right)[/itex] so the upper bound is...
Quote Quote by Hootenanny View Post
So we now know that the lower limit is r=0.
DavidWhitbeck
#15
May5-08, 11:50 AM
P: 352
Putting aside the multivariable calc ways of doing it ([tex]r dr d\theta[/tex]) which is what some people seem to be discussing here-- in AP calc you will generally see rose type curves and you set the bounds of integration by finding the values of [tex]\theta[/tex] where [tex]r=0[/tex] (because the curves you usually see loop back on itself at the origin) and integrating that gives you the area of one leaf, and you multiply by the number of leaves to get the total area.

The more general method is to find the points where the curve intersects itself, and use that to find the bounds of integration.
apples
#16
May5-08, 12:16 PM
P: 167
OK, I'm confused.
I understood angular and radial limits.

b
A = ∫ (1/2)(r^2) dθ
a


But which one do I use in the formula above? What happens to the other limit.

If I make r=0, and find the value of θ, which limit will i get?
How do I find the other one.
Could you please explain in a bit more detail, and tell me the process.

I just simply need to know one thing.


How to find the limits when finding the area of polar curves.
DavidWhitbeck
#17
May5-08, 01:02 PM
P: 352
If you find [tex]\theta[/tex] there will be multiple solutions. You take the smallest for a, and the second smallest for b so that [tex]a < b[/tex] and there are no other zeroes inbetween them. For example if you had [tex]r=a\sin4\theta[/tex] then the values are [tex]\theta=0,\pi/4,\pi/2,\hdots[/tex] and one loop would be from [tex]\theta=0[/tex] to [tex]\theta=\pi/4[/tex]. And there are eight such leaves for that rose.
Hootenanny
#18
May6-08, 03:33 AM
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Quote Quote by apples View Post
OK, I'm confused.
I understood angular and radial limits.

b
A = ∫ (1/2)(r^2) dθ
a


But which one do I use in the formula above? What happens to the other limit.

If I make r=0, and find the value of θ, which limit will i get?
How do I find the other one.
Could you please explain in a bit more detail, and tell me the process.

I just simply need to know one thing.


How to find the limits when finding the area of polar curves.
Your area integral isn't correct. To find the area enclosed by a curve one needs to evaluate,

[tex]A = \iint_A dA[/tex]

Notice that the integral is a double integral. So in cylindrical coordinates,

[tex]A = \iint_A rdrd\theta[/tex]

Now for the limits. Notice that the radial limits depend on [itex]\theta[/itex] and therefore, we must evaluate the radial integral first,

[tex]A = \int^{2\pi}_{0}\left[ \int^{4\left(1+\sin\theta\right)}_0 r dr\right]d\theta[/tex]

Does that make sense?


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