# Determining two sets of boundary conditions for a double integral prob

1. Dec 5, 2013

### ainster31

1. The problem statement, all variables and given/known data

Determining two sets of boundary conditions for a double integral problem in the polar coordinate system. Is the below correct?

2. Relevant equations

3. The attempt at a solution

There are two sets of boundary conditions that you can use to solve this problem in the polar coordinate system.

Set 1:

$$\theta=0\quad to\quad \theta=\frac{\pi}{2}\\r=0\quad to\quad r=2sin2\theta$$

Set 2:

$$\theta=0\quad to\quad \theta=\frac{sin^{-1}\frac{r}{2}}{2}\\r=0\quad to\quad r=2$$

2. Dec 5, 2013

### LCKurtz

Your second one isn't correct. Think about the range of $\theta$ if you let $r$ go from $0$ to $2$, looking at the picture.

3. Dec 5, 2013

### ainster31

$$a\le \theta\le b$$

Is that what you're asking for?

4. Dec 5, 2013

### LCKurtz

No. You have$$\theta = \frac{\arcsin(\frac r 2)} 2$$ and $r$ going from $0$ to $2$. What range of $\theta$ does that give you as you vary $r$?

5. Dec 5, 2013

### ainster31

$\theta=0$ to $\frac{\pi}{4}$

Hmm... how do I fix this?

6. Dec 5, 2013

### LCKurtz

There is a very simple fix using symmetry. Do you see it?

7. Dec 5, 2013

### ainster31

Yes. You just use the boundaries in set 2 and multiply the double integral by 2 because it is symmetric.

But what if it wasn't symmetric? How would I solve this? Would I have to split the graph into two parts?

Or I could just define a new inverse sine function on a different domain?

8. Dec 5, 2013

### LCKurtz

Yes, you would have to modify it somehow to make it work. I wouldn't lose a lot of sleep over this if I were you. When you are given $r = f(\theta)$, nobody in his right mind would do the integral by integrating $\theta$ first then $r$. In fact it isn't even clear from the statement of your original problem that that is what is wanted. Perhaps for the second method you could have just used the symmetry and integrate as in the first answer.

9. Dec 5, 2013

### ainster31

Well, in this example, it just makes things harder, but take a look at problem 30 I am currently trying to solve:

A graph can be shown from the overlap of the blue and the red below:

where red represents $0\le y\le 1$ and blue represents $0\le x\le \sqrt{2y-y^2}$

It seems like the only want to solve this without breaking the integral into multiple parts is with the following set:

$$\theta=0\quad to\quad \theta=sin^{-1}\frac{r}{2}\\r=0\quad to\quad r=1$$

Edit: never mind. Again with the range problem. :( How would I solve this problem? I can't even think of where to split the shape.

Last edited: Dec 5, 2013
10. Dec 5, 2013

### LCKurtz

You break it up into two parts. The outer boundary for $r$ is $r = 2\sin\theta$ up to $\theta = \frac \pi 4$. Then what is the outer boundary for $r$ in terms of $\theta$ for the greater values of $\theta$?

11. Dec 5, 2013

### ainster31

Like below?

$$\theta=0\quad to\quad \theta=\frac{\pi}{4}\\r=0\quad to\quad r=2sin\theta$$
$$\theta=\frac{\pi}{4}\quad to\quad \theta=\frac{\pi}{2}\\r=0\quad to\quad r=csc\theta$$

12. Dec 5, 2013

### LCKurtz

Yes. By the way, I understand your limits when you write them that way, but it actually is ambiguous to do so. It would be much better to write a double integral with limits on them to make clear the order of integration and which variables and limits go on the inner and outer integrals. Certainly you would want to do so on an exam.