How to Find the Area of a Polar Equation Excluding Overlapping Part?

[jun9008]

It's my first time here so I guess I have to introduce myself first.

I am a junior in high school taking Calculus BC and Physics B (taking Physics C next year).
Currently, as of September 29, 2008, my class is learning of Polar equations.

We just went over basics but I am not perfectly grasping the concept on using polars.

One of the question I received during class was:

r=1/2 + sin(\theta

Now, without using calculator, how do I find the area of the equation which excludes the overlaping part.

I know that

A=1/2\intr^2 d\theta
but here we would subtract the inner area.
No point in using calculator;;; I cannot check my answer either since my teacher said the right answer choice is not shown (it is a multiple choice question).

ps. I don't know how to use the "tex" thing so...

 

[HallsofIvy]

It's my first time here so I guess I have to introduce myself first.

I am a junior in high school taking Calculus BC and Physics B (taking Physics C next year).
Currently, as of September 29, 2008, my class is learning of Polar equations.

We just went over basics but I am not perfectly grasping the concept on using polars.

One of the question I received during class was:

r=1/2 + sin(\theta

Now, without using calculator, how do I find the area of the equation which excludes the overlaping part.

I know that

A=1/2\intr^2 d\theta
but here we would subtract the inner area.
No point in using calculator;;; I cannot check my answer either since my teacher said the right answer choice is not shown (it is a multiple choice question).

ps. I don't know how to use the "tex" thing so...

Well, what have you done? You have given the formula. What do you get when you do the integration? One thing you can do is a quick graph of the region to see what "overlapping" part they are talking about. It's not that difficult even without a calculator! But I'll give you a hint: for \theta= 0 to \pi, the graph looks about like an ellipse. For \theta= \pi to 2\pi, it is a smaller ellipse inside the first one.