# Homework Help: Double integrals of polar equations help!

1. Mar 20, 2010

### JHans

Hey, everyone. I'm new to the forums and am hoping that someone can help me with a tricky problem from my multivariable calculus class. We're covering double integrals in polar coordinate systems, and there's a problem from my problem set for homework that I can't seem to get the grasp of.

The question asks: Use a double integral to find the area of the region within both of the circles $$r=cos( \theta )$$ and $$r=sin( \theta )$$

I'm having a lot of difficulty figuring out where to go with this, though. It seems to me that the region is traced by $$r=sin( \theta )$$ as $$\theta$$ goes from $$0$$ to $$\pi /4$$ and by $$r=cos( \theta )$$ as $$\theta$$ goes from $$\pi /4$$ to $$\pi /2$$. Other than that, I'm clueless as to how to solve this using a double integral. Can someone please help me?

2. Mar 20, 2010

### Dustinsfl

Do you know what the answer is for the cos integral?

3. Mar 20, 2010

### JHans

Do you just mean the antiderivative? Sure. The antiderivative of cos is sin.

4. Mar 20, 2010

### Dustinsfl

No the numerical answer. I have an answer but I am not 100% on it so before I make any suggestions I need to know if it is correct.

5. Mar 20, 2010

### JHans

Well, I don't know if this helps, but the answer given by the textbook is (1/8)(pi - 2). I'm not so much looking for the answer as I am looking for the method. I'm completely clueless about how to set up this problem.

6. Mar 20, 2010

### Dustinsfl

I am hesitant to say how I did since I have +2 not minus 2

7. Mar 20, 2010

### Dustinsfl

I just checked my answer in maple and it is plus 2.

8. Mar 20, 2010

### Dustinsfl

Why do you say cos goes from pi/4 to pi/2?
If you graph the function, it goes from 0 to 1 on the r axis and pi/4 to negative pi/4.

9. Mar 20, 2010

### JHans

That's weird... I was looking at a website earlier today that did the problem by a different method involving two single integrals, and it also came upon the answer (1/8)(pi - 2).

10. Mar 20, 2010

### Dustinsfl

Now we can integrate the whole function or use symmetry. How would you prefer to do it?

11. Mar 20, 2010

### JHans

Because the region bounded by sin(theta) and cos(theta) is the cats-eye shape shared by the two circles. The lower boundary of this region is drawn by sin(theta) as it goes from theta=0 to theta=pi/4. The upper boundary of this region is drawn by cos(theta) as it goes from theta=pi/4 to theta=pi/2.

12. Mar 20, 2010

### Dustinsfl

It was unclear to me you wanted to integrate their intersection. I will look it at it again now hold on.

13. Mar 21, 2010

### Dustinsfl

I have the answer but I don't like how I obtained it.

14. Mar 21, 2010

### Dustinsfl

Ok ok ok. So when looking at the intersection we notice, by looking at half it, that we can track sin from 0 to pi/4 to arrive at the intersection. Do you agree?

15. Mar 21, 2010

### Dustinsfl

Now we can start establishing the bounds of integration. We have already arrived at 0 to pi/4 for d theta. What do you think the bounds for r are?

16. Mar 21, 2010

### JHans

Let me make sure I'm thinking about this correctly... the integration of sin(theta) from 0 to pi/4 would be equal to exactly half of the area of the region, correct?

17. Mar 21, 2010

### Dustinsfl

Yes, but sin(theta) 0 to pi/4 is the bounds not the integration. You will be integrating r dr d(theta). What do you think the r bounds are or should be?

18. Mar 21, 2010

### ideasrule

It seems that the intersected area is just twice the area occupied by r=sin(theta) from theta=0 to theta=45 degrees. Why not find that integral and multiply it by 2?

EDIT: oops, somebody beat me to it.

19. Mar 21, 2010

### JHans

I recognize that the lower limit of integration with respect to r is 0, out of common sense. I'm tempted to say that the upper limit of integration with respect to r is sin(theta). That way, the integral allows a tracing from 0 to pi/4, and it allows the radius to be as large as sin(pi/4)=sqrt(2)/2, but no larger.

Am I correct in thinking that the setup would be:

integral(theta=0, theta=pi/4) integral(r=0,r=sin(theta)) r dr d(theta)?

20. Mar 21, 2010

### Dustinsfl

Times 2 since that is only half of if.