jmc

The solution set of the first inequality is a closed disk whose center is (4,0) and whose radius is 4. Each of the second two inequalities describes a line ($y=x-4$ and $y=-\frac{1}{3}x$, respectively), as well as the region above it. The intersection of these three sets is the shaded region shown. This region consists of a triangle, marked 1, along with a sector of a circle, marked 2. The vertices of the triangle are (0,0), (4,0) and the intersection point of $y=x-4$ and $y=-\frac{1}{3}x$. Setting the right-hand sides of these two equations equal to each other, we find $x=3$ and $y=-1$. Therefore, the height of the triangle is 1 unit and the base of the triangle is 4 units, so the area of the triangle is $\frac{1}{2}(1)(4)=2$ square units. The central angle of the shaded sector is $180^{\circ }-45^{\circ }=135^{\circ }$, so its area is $\frac{135}{360}\pi (4{)}^2=6\pi$ square units. In total, the area of the shaded region is $\boxed{6\pi +2}$ square units.