[tex]\dfrac{\pi}{4}[/tex]

Explicação passo a passo:

Primeiro vamos mudar a região de coordenadas cartesianas para uma região de coordenadas polares:

[tex]R=\{(r,\;\theta)|0\leq r\leq1,\;0\leq \theta\leq 2\pi \} \;\;\;\;x=r\;\cos\theta\;\;\;\;\;y=r\;\sin \theta[/tex]

Agora montamos nossa integral:

[tex]\displaystyle\iint(x^{2} +y)\;dA=\displaystyle\int_0^{2\pi}\int_0^{1}(r^{2}\cos^2\theta+r\sin\theta )r\;dr\;d\theta=[/tex]

Calculando a integral temos:

[tex]=\displaystyle\int_0^{2\pi}\int_0^{1}(r^{3}\cos^2\theta+r^{2}\sin\theta )\;dr\;d\theta\\\\\\=\displaystyle\int_0^{2\pi}\left[\frac{r^4}{4} \cos^2\theta+\frac{r^3}{3} \sin\theta \right]_0^{1}\;d\theta\\\\\\=\displaystyle\int_0^{2\pi}\frac{\cos^2\theta}{4} +\frac{\sin\theta}{3} \;d\theta\\\\\\=\displaystyle\int_0^{2\pi}\frac{1}{8}\left( 1+\cos2\theta\right) +\frac{\sin\theta}{3} \;d\theta\\\\\\=\left[\frac{\theta}{8} +\frac{\sin2\theta}{8} -\frac{\cos\theta}{3} \right]_0^{2\pi}=\frac{\pi}{4}[/tex]