I want to calculate $\iint_R x \ \mathrm{d}A$, where $R$ is the unit disc centered at $(2, 0)$.
First, I made the following substitution: $$x' = x-2$$ $$\mathrm{d}x' = \mathrm{d}x$$ $$ \mathrm{d}A' = \mathrm{d}u\ \mathrm{d}y $$
And got this:
$$\iint_{R'} (u+2) \ \mathrm{d}A' $$
Since now my region $R'$ is centered at the origin, I can switch to polar coordinates:
$$x' = r \cos \theta$$ $$y = r \sin \theta$$ $$\mathrm{d}A' = r\ \mathrm{d}r\ \mathrm{d}\theta$$
And now my integral can be set up like this:
$$\begin{align*} \int_0^{2\pi} \int_0^1(r \cos\theta + 2)r \ \mathrm{d}r\ \mathrm{d}\theta &= \int_0^{2\pi} \int_0^1 (r^2 \cos\theta + 2r)\ \mathrm{d}r\ \mathrm{d}\theta \\ &=\int_0^{2\pi}(\frac1{3}\cos \theta +1) \ \mathrm{d}\theta \\ &= 2\pi \end{align*}$$
Is this correct?