Area of an annulus through integration So, I have an annulus with inner-hole radius $r_1=1$ and outer-hole radius $r_2=5$.
To be exact, I have the area of only half of the annulus in this assignment, from $\theta_1=-\frac{3\pi}{4}$ and $\theta_2=\frac{\pi}{4}$.
I have a problem in transforming the integral function into coordinate form, and here's what I'm doing:
$$\iint_A y\,dx\,dy=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\text{stuff})\,dr\,d\theta.$$
The it's $(\text{stuff})$ I have a problem with.
I'm thinking something along the lines of (outer circle) $x^2+y^2=5^2$ (inner circle) $x^2+y^2=1.$
$$y=\sqrt{25-x^2}\implies \int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\sqrt{25-r^2} \,dr\,d\theta,$$
but this here is not the correct answer. Also, it is very possible to solve this without integral, but that would defeat the point.
 A: Using polar coordinate you would get 
$$
\int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} 1\ \color{red}{r} dr d\theta
$$
Where we have the additional $\color{red}{r}$ coming from the Jacobian of the change of coordinate.

However if you want to stick with classical cartesian coordinate, one could first try to rotate the annulus to have it in the position of an upper semi-circle and then use the function definition of the upper semi-circle : $y = \sqrt{R^2 - x^2}$ giving us :
$$
\int_{-r_1}^{r_1} dx \int_{\sqrt{r_1^2 -x^2}}^{\sqrt{r_2^2 - x^2}} dy\ 1 + 2 \cdot\int_{r_1}^{r_2} dx \int_0^{\sqrt{r_2^2 - x^2}} dy \ 1 \overset{(*)}{=} 12 \pi
$$
$(*)$ Thanks Mathematica because the computations are not that easy, especially the simplification to $12 \pi$. That's why I would strongly suggest using polar coordinate.
A: If you want the area of the whole annulus, $$\iint\limits_{1 \leq x^2 + y^2 \leq 25}1\,dx\,dy = \int_{1}^{5}\int_{0}^{2\pi}1\cdot r\, d\theta \, dr.$$
If $\theta_1 \leq \theta \leq \theta_2$, then this reduces to
$$\int_{r=1}^5 \, \int_{-\frac 34\pi}^{\frac \pi 4}1\cdot r\, d\theta \, dr = \left(\frac 14 + \frac 3 4\right)\pi\cdot\frac 12 (5^2 - 1^2).  $$
Az @Zubzub said, the $r$ comes from the absolute value of the determinant of the Jacobian of the map
$$(x,y) \mapsto (r\cos\theta, r\sin\theta).$$
The Jacobian is $$\pmatrix{ \cos\theta & \sin\theta \\ -r\sin\theta &r\cos\theta}, $$ so its determinant is $r(\cos^2\theta + \sin^2\theta) = r$.
