How do I find the area for the yellow section if the sides of the square are equal to 1?

Zectico
Apr 19, 2015

#1**+16 **

I'll attempt this one......

Labeling the lower left-hand corner of the square as (0,0), the upper left-hand corner as (0,1) and the lower right-had corner as (1,0), it * appears* that one of the curves that goes from (0,1) to (1,0) is just the quarter circle with a radius of 1 centered at the origin that has the equation x^2 + y^2 = 1. And the area of this quarter circle is just pi/4.

So, the area of the whole square less the area of this quarter circle gives us the region bounded by two sides of the square and this quarter circle and is equal to 1 - pi/4 ≈.2146. I'm going to call this "Region A." And we have a similar area (region) at the bottom left. So 2 x .2146 ≈ .4292

Now.....we need to find the areas of one of both "ends of a football" found at the top left and bottom right of the square. Both of these areas will be similar (equal) to the area bounded by the three circles whose equations are x^2 + y^2 + 1, (x-1)^2 + y^2 = 1 and x^2 + (y-1)^2 = 1. I'll label this bounded area as Region B.

To proceed, we can find the area of the small sub-region at the top left corner of the square that is bounded by the line y = 1 and the equation of the circle in the first paragraph from x = 0 to x = .5

So we have the following set up :

.5

∫ 1 - √(1 - x^2) dx

0

And....with a little help from Wolfram Alpha, this ≈ .0216943

And, using symmetry, that are four such "sub-regions" within Region A, so their total area is ≈ .086777

So, the area of Region B - one of the "ends of a football," =

[Area of Region A - Area of 4 "sub-regions"] =

[ .2146 - .086777]= 0.127823.....and we have another one of these Region B areas, as well...so....

2 x 0.127823 ≈ 0.255646

So...the central yellow region is given by [1 - 2(Area of Region A) - 2 (Area of Region B) ] =

[ 1 - .4292 - 0.255646] ≈ 0.315154 sq. units

There are probably easier methods to do this (Jacobians and change of Variables, maybe ??) but I don't know how to proceed any differently.

Also......somebody else check my reasoning and math.....I don't feel "100%" about this one....

CPhill
Apr 19, 2015

#3**+13 **

**Thanks Chris**

**Zetico** have you done calculus yet - were you expected to do it a differerent way?

I decided to do it independently of Chris.

I tried a variety of ordinary geometry methods but eventually I also resorted to calculus.

Our answers are the same but maybe our methods are a little different.

Let the bottom corner is the point (0,0)

Each of these curves is a quadrant. I am only interested in the quadrant centred at (0,0)

The equation is

$$\\x^2+y^2=1\\

y^2=1-x^2\\

y=\sqrt{1-x^2}\\\\

When \;\;x=0.5\;\;y=\sqrt{1-0.25}=\sqrt{0.75}\\\\

$The top intersection point is $(0.5,\sqrt{0.75})\\\\

$The right most intersection point is (\sqrt{0.75},0.5)\\\\

0.25* Area=\int_{0.5}^{\sqrt{0.75}}\;(\sqrt{1-x^2}-0.5 )\;\;dx\\\\

Area=4*\int_{0.5}^{\sqrt{0.75}}\;(\sqrt{1-x^2}-0.5 )\;\;dx\\\\

Area\approx 0.315147\;units^2$$

I also used wolfram alpha to do the calculation but maybe i could do it without wolfram|alpha if I had to :/

Melody
Apr 19, 2015

#4**+18 **

Best Answer

I thought it would be of interest to find a non-calculus solution to this problem; so here is my attempt:

.

Alan
Apr 20, 2015

#6**+5 **

Thank Alan, I just could not think of an explanation.

I have not looked at your properly yet but it doesn't look to easy.

Melody
Apr 21, 2015

#8**0 **

Is your study list as long as mine Chris :(

Still is is much better to have a list that is too long than a list that is too short :D

Melody
Apr 22, 2015

#9**0 **

**Alan**, can you find a way that we can bring up questions with the ? icon using the search feature?

If you could that would be really good. I ask you because you seem to be best at this type of thing. :)

Melody
Apr 22, 2015

#10**0 **

Well, if you right-click over the question mark icon and then choose Copy image URL you can paste the result (a URL) into the search box and it seems to find some questions.

You can also choose Open image in new tab and then copy the URL at the top.

Not sure if this is what you are after!

Is the question mark icon only available to moderators?

.

Alan
Apr 22, 2015

#11**0 **

Thanks Alan,

Yes, the question mark Icon is only available to moderators and I cannot find a way of accessing questions with it via the search function. :(

your ideas are good but they don't seem to work :(

I commented on these icons here

Melody
Apr 22, 2015

#12**0 **

Odd - I can't get it to work myself now!!

Oh! Yes I can - I just forgot to paste it into the search box.

So what I get from "Open image in new tab", going to that tab, then copying the URL that is at the top, is the following:

http://web2.0calc.com/img/icons/question.gif

If I then paste that into a search box and click the magnifying glass icon I get to the following page:

http://web2.0calc.com/questions/search/?q=http%3A%2F%2Fweb2.0calc.com%2Fimg%2Ficons%2Fquestion.gif

.

Alan
Apr 22, 2015

#13**0 **

Thanks Alan,

I did use it one time where it appeared to work but it included other stuff as well. Very strange :/

It would be good if we could have some traceable and easy way of tracking interesting questions. I do have my own private 'watch' threads but they are a litle painful to use. Not to mention the fact that they are mostly a disorganised mess.

I don't know if you are interested but you could do that too.

Initially your thread would be seen by the whole forum but it would very soon fall out of site and no-one but yourself would ever see it. you could set it up with several 'empty posts' so that when you wanted to use a new on it was already ancient history.

Do you have any interest in collecting old posts for your own interest Alan?

--------------

Of course other members could do this too :)

Melody
Apr 22, 2015