Finding area of $D =\{(x,y): x^2+y^2 \geq 1 , y \geq x-1 , y \leq 1, x \geq 0\}$

Question

I want to calculate $$\iint_{D} dx \,dy$$ given that $$ D =\left\{(x,y): x^2+y^2 \geq 1 , y \geq x-1 , y \leq 1, x \geq 0\right\}$$

My attempt : I used polar cordinates $r,\theta$ such that $x =r \cos \theta , y= r \sin \theta$ and from $x^2+y^2 \geq 1$ we derive that $r \geq 1$ also from $ y \leq 1$ we derive that $ r \leq \frac{1}{\sin \theta}$ which implies that $ 0 \leq \theta \leq \pi$ with the additional condition that $ x \geq 0 $ we get that $ 0 \leq \theta \leq \frac{\pi}{2}$ but i don't know what to do with $y \geq x-1$ ??

Answer 1

HINT

Draw a nice plot to see the area required and calculate $$\int_0^1\int_{\sqrt{1-y^2}}^{y+1}dxdy$$

Answer 2

No calculus needed: the region is the trapezoid with vertices $\{(0,0),(0,1),(2,1),(1,0)\}$ with the quarter disk $x^2+y^2\leq 1, 0\leq x,y$ removed. You can find the area using geometry.

Answer 3

Solution in polar coordinates:

Line $y \geqslant x-1$ transforms to $r \sin \phi \geqslant r \cos \phi -1$ and to $r\geqslant \frac{1}{\cos \phi -\sin \phi }=\frac{1}{\sqrt{2} \sin(\frac{\pi}{4} - \phi)}=f_1(\phi)$

from $y \leqslant 1$ we obtain $r \leqslant \frac{1}{\sin \phi} = f_2(\phi)$

Functions $f_1(\phi)$ and $f_2(\phi)$ have intersection in $\tan(\phi)= \frac{1}{2}$, so we obtain: $$\left( \int\limits_{0}^{\arctan\left(\frac{1}{2} \right)}\int\limits_{1}^{\frac{1}{\cos \phi -\sin \phi }} + \int\limits_{\arctan\left(\frac{1}{2} \right)}^{\frac{\pi}{2}}\int\limits_{1}^{\frac{1}{\sin \phi }} \right)rdrd\phi$$ Its a beautiful drawing in polar coordinates, but I yet do not study how insert picture here.