Suppose I have an equation in polar form as, $$f(r,\theta) = \begin{cases} a & x\leq R\\ b\arcsin{\left(\frac{R}{r}\right)} & r>R \end{cases}$$ The equation is independent of $\theta$. I want to integrate the function over the sector of a circle $x^2+y^2=r^2$ with sector being considered from i.e. $\theta_1 \leq \theta \leq \theta_2$. The cartesian coordinates of the sector of circle are as,
$$\mathcal{D} = \{(x,y) \,|\, x=r\cos\theta, y=r\sin\theta; \,\forall\, r \in \mathbb{R}, \theta_1 \leq \theta \leq \theta_2\}$$
The integration is easy in polar cordinates as, $$I = \int_{\theta_1}^{\theta_2} \int_0^r f(r,\theta) r \,{\rm d}r \,{\rm d}\theta$$
This you can easily calculate, now what if I shift my circle i.e., $x^2 + (y-k)^2 = r^2$, the sector is also shifted but the arc angle remains same i.e., $\theta_1 \leq \theta \leq \theta_2$. I want to find the the same integral for a shifted sector, where only the origin of sector is shifted in y-axis and the arc angle remains same. I am blank how to approach?. Any help is appreciated
