Can you help me with this exercise? I did it like this but I'm not sure if until now it is correct:
Integral of $f(x,y) = x$ on $\{x^2+(y-1)^2\geq1; (x-1)^2+y^2 \leq 1\}$
I used polar coordinates $x = ρ\cosθ$, $y= ρ\sinθ$
With this substitution I found the integration extremes are now: $2\cosθ\leqρ\leq2\sinθ$ and $θ$ between $0$ and $\pi/2$.
The integral instead became $ρ^2\cosθ.$
Is that correct?
On
Circle A: $(x-1)^2 + y^2 = 1 \implies x^2 + y^2 = 2x$. Also note the circle is in fourth and first quadrant. So in polar coordinates, $r = 2 \cos\theta, -\pi/2 \leq \theta \leq \pi/2$
Circle B: $x^2 + (y-1)^2 = 1 \implies x^2 + y^2 = 2y$. The circle is in first and second quadrant. So in polar coordinates, $r = 2 \sin\theta, 0 \leq \theta \leq \pi$
Now given the inequalities, the region is inside circle $A$ and outside circle $B$ which is the shaded region in the diagram.
Also note that at the intersection of both circle, $r = 2 \cos\theta = 2\sin\theta \implies \theta = \pi/4$.
So set it up as split integrals where
(i) for $- \pi/2 \leq \theta \leq 0, 0 \leq r \leq 2 \cos\theta$
(ii) $0 \leq \theta \leq \pi/4, 2\sin\theta \leq r \leq 2\cos\theta$.
Not quite. You have$$x^2+(y-1)^2\geqslant0\iff\rho^2\geqslant2\rho\sin\theta\iff\rho\geqslant 2\sin\theta$$and$$(x-1)^2+y^2\leqslant0\iff\rho^2\leqslant2\rho\cos\theta\iff\rho\leqslant2\cos\theta.$$If $\theta\in\left[-\frac\pi2,0\right]$, then $\rho\geqslant2\sin\theta$ becomes $\rho\geqslant0$. So, your integral should be$$\int_{-\pi/2}^0\int_0^{2\cos\theta}\rho^2\cos\theta\,\mathrm d\rho\,\mathrm d\theta+\int_0^{\pi/4}\int_{2\sin\theta}^{2\cos\theta}\rho^2\cos\theta\,\mathrm dr\,\mathrm d\theta.$$