# Thread: Area inside a square/closest to the center

1. ## 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.

2. Hi DeuceJ! I'm a little tired, so forgive me if I make a mistake, but I think I can offer some help.

Originally Posted by DeuceJ
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 $\displaystyle y - k = \frac1{4p}(x - h)^2$, where $\displaystyle (h,\;k)$ is the vertex and $\displaystyle p$ is the distance between the vertex and the focus or directrix; for the horizontal parabolas just interchange $\displaystyle x$ and $\displaystyle y$.

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 $\displaystyle x$, 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.

3. thanks a lot Reckoner, that help me a lot!