Area of Polar Curve: Find Outer Loop

[steel1]

Homework Statement

Find the area inside the larger loop and outside the smaller loop of the limacon r=.5+cosθ
Picture here http://www.wolframalpha.com/input/?i=r=.5+costheta

Homework Equations

Area = .5∫r^2

The Attempt at a Solution

To get the area of the outer loop, you just get the value of the entire area, and then just subtract the area of the inner loop. I use two integrals for this

To start, i get the area of the inner loop.
r=.5+cosθ
-.5=cosθ
θ=120 degrees or 2pi/3. and 240 degrees, or 1.33pi.
.5∫(.5+cosθ)^2 evaluated at 2pi/3(lower bound on integral) and 1.33pi(upper bound on integral).
I get .135879

Now, I have to get the area of the entire limacon. I use the bounds zero and 2pi

.5∫(.5+cos)^2 which 2pi as upper limit and 0 as lower limit.

I get 2.35619

Now, 2.35619-.135879 i get 2.2203.

This is the wrong answer though, the correct answer is 2.08. If i was to subtract another .135879 from 2.2203, that would give me the right answer. But why do i have to subtract another .135879? I should be getting the correct answer with the set up i have.

 

[haruspex]

The integral right around will include the area in the inner loop twice: once as part of the outer loop and again as the inner loop part of the range.

 

[steel1]

So your saying that it takes longer than 2pi to complete before it starts tracing over itself again? and idea how i can find the time it takes a polar curve to complete 1 cycle?

 

[SammyS]

Homework Statement

Find the area inside the larger loop and outside the smaller loop of the limacon r=.5+cosθ
Picture here http://www.wolframalpha.com/input/?i=r=.5+costheta

Homework Equations

Area = .5∫r^2

This integral needs to have dθ with it.

The Attempt at a Solution

To get the area of the outer loop, you just get the value of the entire area, and then just subtract the area of the inner loop. I use two integrals for this

To start, i get the area of the inner loop.
r=.5+cosθ
-.5=cosθ
θ=120 degrees or 2pi/3. and 240 degrees, or 1.33pi.
.5∫(.5+cosθ)^2 evaluated at 2pi/3(lower bound on integral) and 1.33pi(upper bound on integral).
I get .135879

Now, I have to get the area of the entire limacon. I use the bounds zero and 2pi

.5∫(.5+cos)^2 which 2pi as upper limit and 0 as lower limit.

I get 2.35619

Now, 2.35619-.135879 i get 2.2203.

This is the wrong answer though, the correct answer is 2.08. If i was to subtract another .135879 from 2.2203, that would give me the right answer. But why do i have to subtract another .135879? I should be getting the correct answer with the set up i have.

The inner loop is for the case where r is negative. That's for values of θ satisfying [itex]\displaystyle \ \frac{1}{2}+\cos(\theta)<0\ .[/itex]

 

[haruspex]

So your saying that it takes longer than 2pi to complete before it starts tracing over itself again? and idea how i can find the time it takes a polar curve to complete 1 cycle?

No, I'm not saying that. The outer loop range is θ from -2π/3 to 2π/3, say, and the inner loop from 2π/3 to 4π/3. Suppose the two loops have areas A and B. The outer loop contains the inner loop, so the area betwen them is A-B. You computed the integral from 0 to 2π, which is A+B. A-B = A+B -2B.

 

[steel1]

Hmm. you would think A+B-B, the B's just cancel and your left with A. but i guess not though

 

[haruspex]

Hmm. you would think A+B-B, the B's just cancel and your left with A. but i guess not though

But you don't want A. A is the area within the outer loop. That includes the area of the inner loop. you want A-B, the area between the two loops.

 

[steel1]

Hmm, ok thanks for explaining. Also, 1 small question. why is the inner loop at .5 at 180 degrees, when cos(pi)+.5 is actually -.5?

 

[haruspex]

When theta = pi, r = -.5 in the direction pi, which is the same as +.5 in the direction 0.