You can use polar coordinates, then you have
$$ 9\leq x^2+y^2\leq 25 \implies 3\leq r\leq 5, $$
and
$$ 0\leq \theta \leq \frac{\pi}{4},\quad \rm{since}\quad y=x. $$
$$ \displaystyle\iint\limits_{R}\frac{y^2}{x^2}\ dA = \int_{0}^{\pi/4}\int_{3}^{5}\frac{\sin^{2}(\theta)}{\cos^2(\theta)}rdrd\theta $$
For circle one, the equation is simply
$$C_1: r=2$$
For circle two $x^2+y^2=r^2,\quad x=r\cos\theta$, so
$$C_2: r^2=2r\cos\theta \implies r(r-2\cos\theta)=0 \implies r=0,r=2\cos\theta$$
Of these only the second solution is useful (the first just says $C_2$ passes through the origin), so the inner and outer radii for the integral are $2\cos\theta$ and $2$, respectively. And $\theta$ varies from $0$ to $\pi/2$ in the first quadrant.
Since $dA=r\,dr\,d\theta$, we get
$$A=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r}\,dr\,d\theta}$$
(The original integral for area should have been $A=\iint{dA}$.)
If $I=\iint{x\,dA}$ is required, then from $x=r\cos\theta$
$$I=\int_{0}^{\pi/2}{\int_{2\cos\theta}^{2}{r^2\cos\theta}\,dr\,d\theta}$$
Best Answer
The region you are integrating over can be written as $[a,b] \times [0, 2 \pi )$ when you are transforming to polar coordinates. The radius $r$ ranges between $a$ and $b$ while the angle $\phi$ goes around the whole circle, hence the interval from $0$ to $2\pi$.
Now you can transform $$(x,y)=(r \cos(\phi), r \sin(\phi))$$ and with the Jacobian determinant being $J=r$ for two-dimensional polar coordinates we get the integral $$\int_a^b \int_0^{2 \pi} r \cos^2(\phi) \ d \phi dr$$ which can be evaluated relatively easy when first integrating with respect to $\phi$ and then with respect to $r$. The antiderivative of $\cos^2(x)$ is $\frac{\cos\left(x\right)\sin\left(x\right)+x}{2}$. Plugging in $2\pi$ and $0$ gives us $\pi$ as the value for the first integral. Now we just evaluate $$\pi \int_a^b r \ dr$$ which is $\frac{\pi}{2}(b^2-a^2)$