Let C_1 solve 2xydydx=y^2-x^2, x0; let C_2 solve 2xy=x^2-y^2dy/dx(schematic).

Question

Let $C_1$ solve $2xy\dfrac{dy}{dx}=y^2-x^2$, $x>0$; let $C_2$ solve $2xy=\dfrac{x^2-y^2}{dy/dx}\cdot(	ext{schematic})$. Both pass through $(1,1)$. The area enclosed by $C_1,C_2$ is:

Accepted Answer

$C_1: x^2+y^2=2x$ and $C_2: x^2+y^2=2y$ (both circles of radius 1 through the origin, centers $(1,0)$ and $(0,1)$, both through $(1,1)$).

$$\text{Area} = 2\left(\frac{\pi\cdot1^2}{4}-\frac12\cdot1\cdot1\right) = \frac{\pi}{2}-1.$$

Solutions · 1