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?