Calculating Shaded Object Area | Integrating Inner Square

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For question 16, I am having trouble assessing the question. I set the sides equal to x and put the image in the middle of a coordinate axes hoping it would be easier to assess. I also set r2 = [(1/2)x]2 + y2 and tried to set up an integration but cannot come up with a relation for the inner square's area. Any help with this question would be great!
 
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The question says that the shaded region is the set of all points inside of the square that are closer to the origin than the edges.

Can you write down an inequality for this (Hint: Pythagorean theorem)? What if you were only to pick the outermost edges of the region, what would the equation for that be?
 
Further to Sentin3l's reply...
Simplify the shape by cutting it into a number of identical regions, each with a much simpler boundary.
Pick one of those pieces. If (x, y) is a point in that piece, how far is it from the origin, and how far is it from the square?
 
Ok, well my solution was:

Split the square into 4 quadrants and integrate 1 of them and then multiply that area by 4 to find the answer. So, I did:

x2 + y2 < 0.25 since it must not be closer to the outer edge of the square

So I rearranged and got:

y2 < 0.25 - x3/3

Now I integrated:

0<0.5∫y2dy < A

(1/3)y3l0<0.5 < A

(0.53)/3 > A since using 0.5 means it is the same distance b/w the square's edge and the actual shape

0.125/3 > A and so I multiply this by 4 and find 0.5/3 > A...

this isn't an exact answer and my textbook doesn't have the answer for this question, so any help would be great!

Also, is it mathematically improper to put inequalities as the upper or lower limits if you're just trying to assess the maximum and minimum possible area?
 
Also, I found this solution. http://mymathforum.com/viewtopic.php?f=15&t=33895

The problem I have is that he assumes that (x2 + y2) = (y-1)2 since the origin and square are equidistant from the edge of the shaded shape. Why can you assume they're equidistant? Doesn't the questions say that it is always closer to the origin?
 
MathewsMD said:
Ok, well my solution was:

Split the square into 4 quadrants
Do you mean the usual four quadrants, as bounded by the x and y axes? That won't help much because of the awkward bends in the boundary at NE, SE, SW and NW positions. Cut the shape so that those bends are at the boundaries of the cuts.
x2 + y2 < 0.25 since it must not be closer to the outer edge of the square
That is the arc of a circle centred at the origin, radius 1/2. It is not composed of arcs of circles, and it extends beyond such a circle.
You need an algebraic expression for the boundary of the shape (or rather, the boundary as it is within one of the regions you cut it into). Write an equation representing points which are equidistant from the centre and the square.
So I rearranged and got:

y2 < 0.25 - x3/3
Where did the cube come from?
Now I integrated:

0<0.5∫y2dy < A
Why are you integrating y2? Area is ∫y.dx
 
haruspex said:
Do you mean the usual four quadrants, as bounded by the x and y axes? That won't help much because of the awkward bends in the boundary at NE, SE, SW and NW positions. Cut the shape so that those bends are at the boundaries of the cuts.

That is the arc of a circle centred at the origin, radius 1/2. It is not composed of arcs of circles, and it extends beyond such a circle.
You need an algebraic expression for the boundary of the shape (or rather, the boundary as it is within one of the regions you cut it into). Write an equation representing points which are equidistant from the centre and the square.

Where did the cube come from?

Why are you integrating y2? Area is ∫y.dx

Ok, I initially thought the radius was constant for some absurd reason. I realize the solution I gave is incorrect. Why can we assume the area in each of the 8 quadrants is circular? Also, why do we consider the lines equidistant when the question says they square is farther from the shape compared to the origin at all points?
 
MathewsMD said:
Why can we assume the area in each of the 8 quadrants is circular?
It isn't circular, and I didn't say it was. A one eighth sector is an octant.
Also, why do we consider the lines equidistant when the question says they square is farther from the shape compared to the origin at all points?
If a region is defined by f(x,y) < c then its boundary is given by f(x,y)=c.
 
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