Find the area of $S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$ Let $S$ be the domain defined by $$S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$$
find the area of $S$
This is middle school problem,so I think it can be solved without integral methods?
$$(x-\sin^2{\theta})^2+(y-\cos^2{\theta})^2=1$$
 A: $S$ is defined as the set of points $N(x,y)$ such that $\vec{ON}=\vec{OM}+\vec{MN}$.
$M(x=\sin^2{\theta},y=\cos^2{\theta})$ is such that $x+y=1$ with $x,y \geq 0$, thus describes line segment $AB$ where $A(1,0)$ and $B(0,1)$.
For each position of $M$, point $\vec{MN} (\sin{\beta},\cos{\beta})$ describes a circle with radius 1 centered in $M$. The area swept by all these circles hen $M$ moves from $A$ to $B$ generates set $S$  similar to an horserace area, with medium axis $AB$ (see figure). Otherwise said, the union of all these circles generates a limit set like a machining tool would do by "digging" all the area $S$ (what is called a "dilation" of line segment $AB$).
Thus the area  $S$ can be split into 3 parts : two half disks (with radius 1) + a rectangular part DCEF ; this makes a whole disk (area $\pi$) +  rectangle with area equal to diameter $\times$ length(AB) $\  = 2 \times \sqrt{2}$. Grand total :
$$\pi+2\sqrt{2}$$ 

Remark: In order to have a fully rigorous proof, a "reciprocal" would be needed, but I have tried here to give a description of the solution set and its area that is accessible to a middle school student. 
A: From $(x - \sin^2\theta)^2 + (y - \cos^2\theta)^2 = 1$ we get that $S$ is a set of circles with radius $1$ and centers on the points of $(\sin^2\theta, \cos^2\theta)$, i.e. on the segmet from point $(1,0)$ to point $(0,1)$ (see figure below).
So the area of $S$ is sum of areas of two semicircles and area of rectngle with sides lengths $2$ and $\sqrt{2}$: 
$$
\frac{1}{2}\pi\cdot 1^2 + 2\cdot\sqrt{2} + \frac{1}{2}\pi\cdot 1^2 = \pi + 2\sqrt{2}.
$$

