$$ \int_{r=0}^{r=3/\sqrt2} \int_{\theta=0}^{\theta=2\pi} \left(\sqrt{9-r^2}-r\right) r dr $$
Then compute them to
$I_1 = 2\pi\int_{r=0}^{3/\sqrt2}\left(\sqrt{9-r^2}\right)rdr$ and $I_2$ which isn't relevant for this question.
When I then use the substitution method I would usually think I would use $u=9-r^2$ and get:
$$2\pi\int_{u=9}^{9/2}\sqrt u.du$$
But when I look In the solution it shows that it becomes
$$-\pi\int_{u=9}^{9/2} \sqrt u.du$$
Could someone explain why the $2\pi$ changes to $-\pi$? Would be really appreciated.
$u = 9-r^2, du = -2r \,dr, dr = -\frac{1}{2r}du$
$\displaystyle \int_{r=0}^{r=3/\sqrt2} \int_{\theta=0}^{\theta=2\pi} \sqrt{9-r^2} \, r \, dr \, d\theta = -\pi \int_{9}^{9/2} \sqrt{u} \, du = \pi \int_{9/2}^{9} \sqrt{u} \, du$