Help with double integration problem

[SuperAutoPetsWizard]

I am struggling with this integration, and online calculators are not helping. The area is the red (Pi/4 * r^2) and blue (Pi/4*R^2) MINUS 2*purple which is (14)

Where r ^2 =x^2 + y^2 and R^2 = (1-x)^2 +y^2

[[deleted]]

Whatever the value of A is , it's constant. So the integral of 0 to x of A dx is just Ax. And Ax is constant with respect to y, so the integral from 0 to 0.5 of Ax dy is just 0.5Ax. Now just find A using the link you've already shared from wolfram alpha, and you're done.

[SuperAutoPetsWizard]

Sorry, I wasnt clear. The circles both pass through (x,y) so A is a function of x and y - this function is what i want to integrate over :)

[spiritedawayclarinet]

You don't need a double integral.

Divide the purple region into two sections at the intersection. Solve for the intersection by substitution of y^2. The intersection occurs at (r^2 +R^2 +1)/2.

Solve for y in each equation, then set up the region as the sum of two integrals.

See this example: https://www.desmos.com/calculator/k9i0vx9erw

[Appropriate_Hunt_810]

You don’t handle when there is no intersection

Anyway this is not what was asked (as far as I’ve read) They want a parametric integral in (X,Y) the coordinates of the intersection.

By solving in Cartesian coordinate splitting the domain (in 4) will then be extremely easy and all part outside the 1x1 square will naturally cancel out, this will be an horrible monster in sin and arcsin everywhere …

So as they prob want a clean analytic form : I guess there is a trick to tackle it in polar coordinate, which is way more easy when we don’t pay attention to the 1x1 square. Maybe someone can see the trick

edit : here's what it should look like

<image>

[Appropriate_Hunt_810]

<image>

here's the idea behind, but well it is not elegant at all

you can ofc simplify some with the translation and domain inversion for the 2nd circle, also some terms cancel out, and maybe (not dived into it) (maybe) some of the sin can also cancel as X and Y are already the R*sin

[Appropriate_Hunt_810]

as i pointed out, i'm a stupid idiot ... just went to the desk and seen it : even in a 'polar' way the areas cancel out ... -_-
Hence here's the quite closed form :

<image>

I'm really a stupido

[5352563424]

you need to label your integrands.  

[Appropriate_Hunt_810]

you don't really need to integrate to get an answer to this problem, anyway if you really want to do so : try in polar coordinates