Area within a section of a circle

[n00blar]

Hi,

I am stuck on a problem here. =( I would appreciate it if anyone could help me out with this. The problem asks to find the area within the circle $\displaystyle x^2 + y^2 = 4$ and above the line $\displaystyle y = 1$. All I have been able to do so far is come up with this integral:

$\displaystyle \int_{0}^{2}{\sqrt {4 - x^2} dx}$

Am I on the right track?

 

[stapel]

...find the area within the circle $\displaystyle x^2 + y^2 = 4$ and above the line $\displaystyle y = 1$. All I have been able to do so far is come up with this integral:

$\displaystyle \int_{0}^{2}{\sqrt {4 - x^2} dx}$

How did you arrive at your limits of integration? And where do you account for the fact that you're only measuring the area above the line y = 1? Looking at your graph of the circle and the horizontal line, how does this relate to your integral?

Please be complete. Thank you!

Eliz.

 

[skeeter]

Hi,

I am stuck on a problem here. =( I would appreciate it if anyone could help me out with this. The problem asks to find the area within the circle $\displaystyle x^2 + y^2 = 4$ and above the line $\displaystyle y = 1$. All I have been able to do so far is come up with this integral:

$\displaystyle \int_{0}^{2}{\sqrt {4 - x^2} dx}$

Am I on the right track?

lower limit of integration should be 1 ... your integral would yield the area of the semicircle.

 

[stapel]

lower limit of integration should be 1...

Really? In my drawing, I have the circle with radius 2 and centered at the origin, which is intersected by the horizontal line through y = 1. The intersection points of the two curves do not appear, to me, to be x = 1 and x = 2:

      ^ y
      |
      |
____*___*______ y=1
  *   |   *
-*----+----*---> x
  *   |   * <=circle
    * | *
      |

...and the area between the two curves does not appear, to me, to be between the x-axis and the top of the circle. Where am I going wrong...?

Thank you!

Eliz.

 

[galactus]

The intersections points of the line and circle are $\displaystyle \sqrt{2^{2}-1^{2}}=\pm\sqrt{3}$

The area of our circle segment can be gotten from the integral:

$\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}[\sqrt{4-x^{2}}-1]dx$

Or without calc by using the area of the segment of a circle formula.

$\displaystyle \frac{1}{2}(2)^{2}(\frac{2\pi}{3}-sin(\frac{2\pi}{3}))$

 

[n00blar]

Ah okay, I see it now. Thank you. =)