I'll work with a side of the rhombus.       (0,y0) |\

|  \ (x2,y2)

of segments corrisponding to that side of                    |__b\a___

the rhombuses    (x1,0)

a angle side/axe x  b=pi-a

Equation of the bundle is. y=y0+x*tan(a) with "a" variable and y0=2*sin(b)       [=2*sin(a)],    so: 

A point P (x2,y2) will be in the curve that envelopes the rhombuses if it belongs to the side of one rhombus (1) and its distance to the origen is bigger than the corrisponding point of any other rhombus (2). So that: 

(1): y2=2*sen(a)+x2*tan(a)

(2): d² =x2²+y2²  (will be maximum in realtion to variable "a")

d² =x2²+[2*sin(a)+x2*tan(a)]² =

=4*sin²(a)+x2²*tan²(a)+4*x2*sin(a) *tan(a)+x2² (call u this) 

d=sqr (u)

To get the maximum d'(u)=0 => u'/[2*sqr(u)]=0 =>u'=0

then:

 8*cos⁴(a)+4*x2*cos³(a)+4*x2*cos(a)+2*x2²=0

It's possible to express this as: 

One of both should be=0

Then our x2=-2*cos³(a)=2*cos³(b) 

(the other term=0 is the singularity of the curve in point 1 (x1,0).  

From there we get y2

y2=2*sin³(b)  (this are the coordinates of the point P (x2,y2) throught which each rhombus contributes to the curve. 

As we can see the coordinates depends on the angle b (or a, it's the same thing)

The requested area will be the area of the curve y2=f(x2). Then:

A=Integral y2 d(x2).  To calculate this we express d(x2) in terms of d(a)

d(x2)= 6 cos²(b)*sin(b)*d(a)  so that:

Area= Integral (2*sin³(b)*6 cos²(b)*sin(b)*d(a) [between pi and pi/2]

This integral should be done by recurrence [w: cos²(b)=1-sin²(b)] leading to: 

[Between pi and pi/2 all terms with sin*cos are 0. So that:

Area=3/4*pi-3/8*pi=3/8*pi

But this is the area of one of the four sectors of the rhombus. The total area, will be, then: