Surface integral over an annulus

Hi, would just like to know if what I have done is correct.

Find the area integral of the surface $\displaystyle z=y^2+2xy-x^2+2$ in polar form lying over the annulus $\displaystyle \frac{3}{8}\leq x^2+y^2\leq1$

The equation in polar form is $\displaystyle r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2$

$\displaystyle \displaystyle{\int^{\pi}_{-\pi}}\int^1_{\sqrt{\frac{3}{8}}}(r^2\sin^2\theta+2 r^2\cos\theta\sin\theta-r^2\cos^2\theta+2)\, r\, dr\, d\theta$

Integrating with respect to $\displaystyle r$

$\displaystyle {\frac {55}{256}}\, \left( \sin \left( \theta \right) \right) ^{2}+{

\frac {55}{128}}\,\cos \left( \theta \right) \sin \left( \theta

\right) -{\frac {55}{256}}\, \left( \cos \left( \theta \right)

\right) ^{2}+5/8

$

and with respect to $\displaystyle \theta$, I get the area as

$\displaystyle \frac{5}{4}\pi$

Does this look ok?