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Thank you for your time!

I will work on this right away

Hm. I was thinking parabolas, too. I think that might be just what I am looking for.

Let the very bottom line be the directrix and let the very center be the focus...

Thank you so much!

No problem, it was a bit of a learning experience for me as well.

Would you agree with the points I drew in and labeled ?

Actually, let me think about it for a moment...

be the*

Wait now that I think about it..

Those numbers should be cut in half

|dw:1356849810908:dw|
I know it's not a circle but think of it this way.

You're absolutely right. Back to the drawing board for me.

|dw:1356849937491:dw|

|dw:1356850006505:dw|

In this case, it was just finding the right algebra.... lol

\[
\sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2}
\]

Right, sqrt(2) . Hm.

Yeah, well, just solve for \(x\)...

@LogicalApple Hmm?

I'm still trying to figure out why x + y = sqrt(1^2 +1^2) instead of x + y = sqrt(2^2 + 2^2)?

Because the total length is 2, but we're only doing halfway... which is 1.

Yeah scratch that

I am in agreement that x = 2 - sqrt 2

I will modify my equations

|dw:1356850703108:dw|

4x as in 4 times

I like that idea a lot actually

4(2 - sqrt 2)^2

We also already found the equation of the parabola, and we know its limits.

@LogicalApple Got an idea for finding the area under that parabola?

I don't know how you're getting that...

We could just ignore that it a parabola altogether and just try using the distance formula.

Should be a plus in there.

Now I think I'm getting a rational answer of 8/3 for the area of the region.

But notice how our slope is pretty much the same.

Just reflected because this time we took the concave down parabola at top.

I'm following.. and I'm curious. I am going to review this entire post when it's all said and done.

\[
\int_{-(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx
\]

Sorry, been having quite a few typos today.

Don't sweat it

Right but I can just use symmetry again. Plus there is wolfram alpha

plugging that pellet in that is gonna be a pain in the retrice..

\[\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} \frac{ \sqrt{2} - 1 }{ 4\sqrt{2} - 3} x^{2}dx -\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} (2 - \sqrt{2}) dx = 4\sqrt{2} - 16\]
The above integral is based on the attached picture (below)
It determines the area of one of the parabolic regions bounded by the inner square. Multiply this result by 4 and add the area of the square, we obtain 8/3 as the area of the region.

should be a dx in that second integral (sorry am still getting used to the equation creator)

That looks right... makes me wonder where I went wrong.

@LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

But didn't you calculate the distance from a point to the line y = 1 instead of to the line y = 2?

Of course you did. Hm.

Let me see what I obtain next.

Wow it's really bothering me how I messed up. I can't figure out why my trig method was wrong.

Ooooh! I figured out my error!

|dw:1356855949526:dw|
We found \(x\)! We wanted \(z\) in the diagram!

|dw:1356856038380:dw|

sorry I mean^

Modifying .. 1 sec

I did worry about it!

That looks correct.

I will integrate now.. and I have a feeling it will be irrational this time ;(

I now see why there was a discrepancy with my earlier parabolic equation.

I knew from the very start there was very little chance of a rational result.

My image is still wrong. One more modification ..

That's pretty much it.
It's an easy integral because it's just a 2 degree polynomial.

Final answer is ...
\[\frac{ 4 }{ 3 }(4\sqrt{2} - 5) \]

For just the parabola?

That is 4 times the parabola plus the area of the square.

It lists your answer as an alternate form.