Problem $9(10$ Marks$)-$ Use surface integral to find the area of the part of the surface $z=x^2+y^2,$ that is below the plane $z=4$ in the first octant.

Note ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{a^2+u^2}du=0\mathrm{.}5u\sqrt[\phantom{2}]{a^2+u^2}+0\mathrm{.}5a^{2}ln(u+\sqrt[\phantom{2}]{a^2+u^2}).$