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Math Help - Area inside a square/closest to the center

  1. #1
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    Area inside a square/closest to the center

    Region R consists of all points inside the square that are closer to the center of the square than to any of the sides. Find the area of R

    http://img.photobucket.com/albums/v3...epinoy/box.jpg

    Just to verify the length of the sides of the square is 2.

    Right now i'm just focusing on quadrant I and making a triangle. Then i make the outside point of Region R say, point P, from the right side 1-x and from the top 1-y. But then i don't know where to go from there. At first i thought it was a circle but its not.
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    Hi DeuceJ! I'm a little tired, so forgive me if I make a mistake, but I think I can offer some help.

    Quote Originally Posted by DeuceJ View Post
    Region R consists of all points inside the square that are closer to the center of the square than to any of the sides. Find the area of R

    http://img.photobucket.com/albums/v3...epinoy/box.jpg

    Just to verify the length of the sides of the square is 2.

    Right now i'm just focusing on quadrant I and making a triangle. Then i make the outside point of Region R say, point P, from the right side 1-x and from the top 1-y. But then i don't know where to go from there. At first i thought it was a circle but its not.
    Here's a big hint: a parabola is the set of all points in the plane equidistant from a fixed point (the focus) and a fixed line (the directrix). So, the locus of points equidistant from the center of the square and any one side will be a parabola. Considering all sides, you will have four intersecting parabolas which will, in the center, close off R (see graph).

    Finding the area of this region will involve some integration. To find equations for the parabolas, start with y - k = \frac1{4p}(x - h)^2, where (h,\;k) is the vertex and p is the distance between the vertex and the focus or directrix; for the horizontal parabolas just interchange x and y.

    Now, since the region is symmetric as you seem to have observed, consider one corner. The two parabolas that lie in that quadrant can be written as a function of x, as long as you only worry about one half of the horizontal one. Find their intersection, and integrate along the appropriate intervals. Then remember to multiply by 4 to get the total area.
    Attached Thumbnails Attached Thumbnails Area inside a square/closest to the center-area_problem.png  
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    thanks a lot Reckoner, that help me a lot!
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