Double Integral, cannot understand the region

Question

I need to calculate the area knowing the domain and I have a hard time finding the region. I know that that $1\leq x^2+y^2 \leq 2x , y\geq 0$. I need to transform $x,y$ in polar coords.

Answer 1

Making the usual substitutions,\begin{align*} 1 \leq r^2 &\leq 2r\cos \theta \\ r \sin \theta &\geq 0 \end{align*}

Since $r \geq 0$, the latter tells us $\sin \theta \geq 0$, so $0 \leq \theta \leq \pi$. (We could make other choices for the range of $\theta$s, but let's pick a range including the interval just found.) Similarly, $\cos \theta \leq 0$ for $\theta$ in $[\pi/2,\pi]$, which would violate $1 \leq 2 r \cos \theta$ for those angles. Therefore, $0 \leq \theta < \pi/2$.

Now $1 \leq r^2$ means all points of the region are on or exterior to the unit circle. (It may be better to see this as "$r \leq -1$ or $1 \leq r$" by taking square roots, then reject the negative radius.)

From $r^2 \leq 2 r \cos \theta$, we have $r=0$ or $r \leq 2 \cos \theta$. We have removed $r=0$ above. Recall that $\cos \theta$ starts at $(\theta, \cos \theta) = (0,1)$, so at angle $0$, our radii are bounded below by $1$ (by the previous paragraph) and above by $2 \cdot 1 = 2$. The upper bound on the radius decreases with cosine until that bound meets the lower bound, that is until $2 \cos \theta = 1$. Since $\cos \theta = 1/2$ only for $\theta = \pi/3$ for angles in $[0,\pi/2]$, we know where the region ends, at $(r,\theta) = (1,\pi/3)$.

As others have shown, this is the disk of radius $1$ centered at $(r,\theta) = (1,0)$ minus the disk of radius $1$ centered at $(r,\theta) = (0,0)$, restricted to the upper half plane.

Answer 2

If you write $x$ and $y$ as $r\cos\theta$ and $r\sin\theta$ respectively, then your conditions become:

1. $r\geqslant1$;
2. $r^2\leqslant2r\cos\theta(\iff r\leqslant2\cos\theta)$;
3. $\theta\in[0,\pi]$.

Can you take it from here?

Answer 3

Corresponding to the boundaries $1\leq x^2+y^2 \leq 2x ,\space y\geq 0$ in the $xy$-plane, the boundaries in the polar coordinates are

$$ 1 \le r\le2\cos{\theta}, \space \space 0 \le \theta \le \pi$$

But, given the configuration of the two circles, you only need to integrate over the region $0<\theta<\pi/3$, since the circles intersect at $\theta = \pi/3$.

Thus, the area integral is

$$ A = \int_0^{\pi/3} d\theta\int_1^{2\cos\theta}rdr =\frac{\pi}{6}+\frac{\sqrt{3}}{4}$$

Answer 4

The second inequality is $$x^2+y^2\leq 2x$$ That is equivalent to $$x^2-2x+1+y^2\leq 1$$ which may be written as $$(x-1)^2 + y^2\leq 1$$ So that is all points inside a circle of radius $1$ centered at $(1,0)$. Couple that with the first inequality, where you need to be outside the unit circle centered at the origin. The shape is a kind of crescent moon shape. (Well, the top half, once you consider the third inequality.) Polar coordinates will be a smooth way to set up limits of integration.