Area between two curves (x = cos(y) and y = cos (x))

[igorrn]

Homework Statement

Give the area marked in the graph (graph as jpg file)

Relevant Equations

x = cos y
y = cos x
x E [0,1] and y E [0,1]

I tried this:
X = cos(y) → y = arccos(x) for x E(-1,1) and y E (0,2)
Then:
There's a point I(Xi,Yi) in which:
Cos(Xi) =Arccos(Xi)
Then I said area1 (file: A1)
A1 = ∫cosx dx definite in 0, Xi
And A2 (file:A2):
A2 = ∫cosy dy definite in 0, Yi
And the overlapping area as A3 (file: A3):
A3 = ∫Yi dx definite in 0, Xi

And total area, then, is:
A = A1 + A2 - A3

I had trouble finding the value of Xi though. The best Approach I could find is 3/4, but I had not found a method further narrow the aprroach answer. I think Xi is an irrational number, I'd want to know if it has a name and definition to it like Pi or Euler's Number to find it.
I'd want to know if there's another method to calculating this area also.

 

[pasmith]

By symmetry, the area you want is <br /> 2 \int_0^{X} \cos x - x \,dx where X = \cos X. Unfortunately that can only be solved numerically.

 

[igorrn]

By symmetry, the area you want is <br /> 2 \int_0^{X} \cos x - x \,dx where X = \cos X. Unfortunately that can only be solved numerically.

Thank you very much, but isn't it?
\int_0^{X} \ 2cos x - X \,dx
where X = \cos X.
To discount the overlapping area?
And could you explain me what is "numerically'?
(The int should be the integral symbol. Really don't know how to use it)

 

[LCKurtz]

No. There is no overlapping area if your first .jpg is the correct area. The integral$$
A = 2\int_0^p \cos x - x~dx$$is correct (here $p$ is the value where $\cos p = p$). What he means by having to do it numerically is the fact that even though you can integrate to get $$A =2\sin(p) - p^2$$you still have to find $p$ numerically. (About 0.7390851332).

 

[vela]

Thank you very much, but isn't it?
int_0^{X} \ 2cos x - X \,dx
where X = \cos X.
To discount the overlapping area?

Note that $X \ne x$. Either method will get you the same answer.

 

[Mark44]

(The int should be the integral symbol. Really don't know how to use it)

It's this: \int
As a definite integral, \int_{a}^{b}. Note that you don't need the braces for a limit of integration that is one character, but you do need them for two or more characters. I.e., \int_0^{2 \pi}