# Am I doing this double integral right?

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?

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It looks fine, but at this stage $\iint_{R'} (u+2) \ \mathrm{d}A'$ you could just split the integral into $\iint_{R'} u\ \mathrm{d}A'+\iint_{R'} 2 \ \mathrm{d}A'$ . The first part is zero by symmetry, and the second part is 2 times the area of $A'$, that is, $2\pi$.