# Area problem

1. Feb 19, 2012

### naaa00

1. The problem statement, all variables and given/known data
Well, the problem is to find the area of:

{(x,y), -(pi/2) <= x <= Pi/2, 1/2< y <=cosx}

3. The attempt at a solution

Well, I know that for an angle of 60 degrees, the cosine is 1/2. So I guess that the limits of my integral will be from 0 to Pi/3.

But I'm getting confused with the domain of x: for 270, and 90 degrees... So on the graph of cosine, and considering the condition of 1/2< y <=cosx = 1, it seems to me that the only point of intersection between the condition and the cosine graph is just a small piece of the cosine graph. So after I evaluate the inegral I got sqrt(3)/2.

2. Feb 19, 2012

### Dick

If you drew a graph of y=cos(x) and y=1/2 for x between -pi/2 and pi/2 you should have found that the crossed twice. And what integral did you work out to get sqrt(3)/2?

3. Feb 19, 2012

### naaa00

Well, the integral of cos(x) from 0 to Pi/3. Did the graph really cross it twice? But 270 is on the negative side and the condition says that y is in between 1/2< y <= cosx = 1

4. Feb 19, 2012

### Dick

cos(-pi/3) is also equal to 1/2. And if you want to find the area between two curves you integrate the difference between the two curves. You want to find the area between y=cos(x) and y=1/2.

5. Feb 19, 2012

### naaa00

Ok, I understand now. I would by symmetry multiply the integral by 2, and get sqrt(3).

The problem now is that I cannot visualize where is (-pi/3). So I don't know if there is another segment which I should consider. Is 360 - 60 = 300? I think I'm wrong.

...

6. Feb 19, 2012

### Dick

-pi/3 is -60 degrees. And I'm trying saying you should integrate cos(x)-1/2. Not just cos(x). It's the difference between the two curves you should be integrating.

7. Feb 19, 2012

### naaa00

Hello! Ok.

Well, I got 2 - 3Pi/4. My limits where -Pi/2, Pi/2. And the integral was (cos(x) - 1/2), right?

8. Feb 19, 2012

### Dick

You want to integrate between the points where the two curves cross. cos(x) is only greater than 1/2 between -pi/3 and pi/3.

9. Feb 19, 2012

### naaa00

Hello there!

Well, I tried the integration with the new limits. My answer is 1 - Pi/3

Is this fine?

10. Feb 19, 2012

### Dick

No, the -pi/3 part is ok. The 1 isn't. Can you show your steps?

11. Feb 19, 2012

### naaa00

Sure.

Integral of (Cosx - 1/2) dx = Sinx - (1/2)x,

f(b) - f(a) = [Sin(Pi/3) - Pi/6] - [Sin(-Pi/3) + Pi/6] = 1/2 - Pi/6 + 1/2 - Pi/6 = 1 - Pi/3.

Sin(-Pi/3) = -1/2. Or?

12. Feb 19, 2012

### naaa00

Ohhh, no. My bad. I messed it up.

Yeah, the answer is - Pi/3.

But what does a negative area in this case means exactly?

13. Feb 20, 2012

### Dick

You are still messing it up. sin(pi/3) isn't 1/2. And the answer isn't negative. -pi/3 is only part of the answer.

14. Feb 20, 2012

### naaa00

Yes, indeed, I'm still messing it up...

f(b) - f(a) = [Sin(Pi/3) - Pi/6] - [Sin(-Pi/3) - (-Pi/6)] = sqrt(3)/2 - Pi/6 + sqrt(3)/2 - Pi/6 = sqrt(3) - Pi/3.

That should be fine!

15. Feb 20, 2012

### Dick

That looks better.

16. Feb 20, 2012

Thank you!
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}