Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area of region inside a square

  1. Aug 11, 2013 #1
    Hi,
    This question is killing me (please note that it's not homework, this is from self study):
    The shaded region inside a square of side "a" consists of all points that are closer to the centre of the square than any of its edges (emphasis on any of its edges--the resulting region is like a square with inflated edges). I know the answer involves integration, in fact I have the entire answer. It begins:
    "By symmetry, we consider only the portion of the region between the lines of y=0 and y=x, and then multiply the resulting area by 8. The distance from the origin to point (x, y) on the boundary of the region equals the distance from (x, y) to the line x=a/2, that is, such points satisfy √(x^2 + y^2)=√(x-a/2)^2..." then it solves for x and integrates. The final answer it gives is a^2(4√2 -5)/3 . I'm sorry to be slow, but I'm just not getting it. Why x-a/2? And how is x-a/2 a representative for distance all distances between the centre and the boundary of the region? Any help understanding the intuition behind this answer/a different logical answer would be massively appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Aug 11, 2013 #2
    The distance between a point and a line is found by finding the shortest distance between the given point and the line. This shortest distance is found by drawing a perpendicular segment from the point to the line. Look at the picture you have. The center of the square is at the origin (0,0).

    - What is the equation of the vertical line which forms the right side of the square given that the origin is at the center of the square?
    - How far is (x,y) from that line given that you have to draw a perpendicular segment to the line to find the distance?

    Junaid Mansuri
     
  4. Aug 12, 2013 #3
    That was the perfect push in the right direction, thanks so much!
     
  5. Aug 12, 2013 #4
    You're welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook