So my professor gave as the following question: Given the function u which is continuous in $\bar{D}$ , where $D = \left \{ (x,y) \in \mathbb{R}^{2} : x^{2}+y^{2}<1 \right \}$, and harmonic in $D$.
Find $u(0,0)$, if $u(x,y)=x^{2}y^{2}$ for $(x,y) \in \partial D$.
Now I know from the mean value theorem that the value of $u$ at the center of the disc is the mean of its values at the border of the disc $\partial D $. So we get $u(0,0)=\frac{1}{\left | \partial D \right |} \int_{(x,y) \in \partial D} u(x,y)ds$, but how do we calculate this? Please help me, thank you in advance.
$$u(0,0)= \frac{1}{2 \pi}\int_0^{2 \pi}u( \cos t, \sin t) dt=\frac{1}{2 \pi}\int_0^{2 \pi} \cos^2 t \sin^2 t dt.$$