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.
Best Answer
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.