Given $A=\{(x,y)\in\mathbb{R}^2\ |\ x^2+(y-1)^2 \leq 1, \ x+y \geq 2\}$, I must evaluate $$\iint_A xdxdy$$ I used a change of coordinates, letting $x=r \cos t$ and $y=r \sin t+1$, so the set of integration became $B=\{(r,t)\in\mathbb{R}^2\ | \ 0 \leq r \leq 1, \ r(\cos t + \sin t) \geq 1\}$ and then I have to evaluate $$\iint_B r^2 \cos t drdt$$ I have two main doubts:
- sketching $A$ it is obvious that $0 \leq x \leq 1$ and $1 \leq y \leq 2$ and so $t \in [0,\pi/2]$, but how can I find out with algebra that $0 \leq x \leq 1$ and $1 \leq y \leq 2$?
My attempt is this: Since $x^2+(y-1)^2 \leq 1 \implies x^2 \leq 1 \ \land \ (y-1)^2 \leq 1 \iff -1 \leq x \leq 1 \ \land \ 0 \leq y \leq 2$ it follows that $y \geq 0$ and that $-2 \leq -y \leq 0$, hence from $x+y \geq 2 \iff x \geq 2-y \geq 2-2=0 \implies x \geq 0$.
So combining $-1 \leq x \leq 1 \ \land x \geq 0$ it follows that $0 \leq x \leq 1$.
Hence it is $-1 \leq -x \leq 0$ and then $x+y \geq 2 \iff y \geq 2-x \geq 2-1=1 \implies y \geq 1$; so combining $y \geq 1 \ \land \ y \leq 2$ it is $1 \leq y \leq 2$. Can someone check if it is correct?
I would like to deduce the angle and the fact that we are in the first quadrant from polar coordinates too: so in $B$ it is both $0 \leq r \leq 1$ and $r(\cos t + \sin t) \geq 1$, now I would like to divide for $\cos t + \sin t$ but I don't know its sign and so I don't know how to proceed; can someone help me figure this out?
In this case, if I assume $\cos t +\sin t >0$ I can divide and get $\frac{1}{\cos t +\sin t} \leq r \leq 1$, but sometimes there are two lower bounds for $r$ and I don't know how to proceed. So the questions is: if we imagine (for the sake of a better comprehension for me) that we don't know if the sign of $\frac{1}{\cos t +\sin t}$ is always positive, I can't deduce $0 \leq \frac{1}{\cos t +\sin t} \leq r \leq 1$ just because $r \geq 0$ and it is $\frac{1}{\cos t+ \sin t} \leq r$ (it would be different if it was $\frac{1}{\cos t+ \sin t} \geq r$, because $\frac{1}{\cos t+ \sin t} \geq r \geq 0 \implies \frac{1}{\cos t+ \sin t} \geq 0$), so I have to consider the values of $t$ that makes $\frac{1}{\cos t +\sin t}<0$ too? Or ignore them? Thanks.
$A=\{(x,y)\in\mathbb{R}^2\ |\ x^2+(y-1)^2 \leq 1, \ x+y \geq 2\}$
$x + y \geq 2$ indicates your region is part of the circle above the line $x + y = 2$.
For your question $1$ on not going by the sketch, you can find the intersection points of the circle and the line.
$x^2+(y-1)^2 = 1$ and $x + y = 2 \implies y = 2 - x$
$x^2 + (1-x)^2 = 1 \implies 2x^2 - 2x = 0 \implies x = 0, 1$
Plugging in $y = 2 - x$, we get two intersection points $(1, 1), (0, 2)$.
Now on your question $2$, given the points you know $\theta = 0, \frac{\pi}{2}$ but if you want to plug into the polar equation and check -
$x = \cos \theta = 1, y = 1 + \sin \theta = 1 \implies \theta = 0^0 \,$ ($r = 1$ on the circle)
Similarly check for point $(0, 2)$.