Double integral in polar coordination between two circles Use polar coordinates in $\Bbb{R^2}$ to evaluate
$$\iint_{R} \frac{x^2}{x^2 +y^2} \,dx\,dy$$ where R is the region between the concentric circles of equations $x^2 +y^2=a$ and $x^2 +y^2=b$ with $a<b$ and $(x,y)$ are Cartesian coordinates in $\Bbb{R^2}$
So I know the region is the area between
the smaller circle $$x^2 +y^2=a$$ and the bigger circle
$$x^2 +y^2=b$$ but i'm not sure how to get limits and be able to evaluate it when I don't have any values for a and b which is throwing me off.
 A: In polar coordinates, the integration region is parametrized as
$$
\left\{x = r\cos\left(\theta\right), y = r\sin\left(\theta\right)\right\}, \quad 0\leq\theta\leq 2\pi, a\leq \rho \leq b.
$$
Changing the variables gives
$$
dxdy = rdrd\theta.
$$
$$
\iint_R\frac{x^2}{x^2+y^2}dxdy = \int_0^{2\pi}\int_a^b\frac{r^2\cos^2\left(\theta\right)}{r^2\cos^2\left(\theta\right) + r^2\sin^2\left(\theta\right)}rdrd\theta = \int_a^brdr\int_0^{2\pi}\cos^2\left(\theta\right)d\theta
$$
A: 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 $\frac{\pi}{2}(b^2-a^2)$
