Note that the leftmost point of the area is at the intersection of #x=-y# and #x=2-y^2# at #(-2, 2)# and the rightmost is at the intersection of #x=2-y^2# and the #x#-axis at #(2, 0)#. Thus, we will consider integrals as #x# goes from #-2# to #2#.

As the bounds of the area change at #x=0#, we will split this problem into two integrals.

From #x=-2# to #x=0#, the area is bounded above by #x=2-y^2# and below by #x=-y#.

Solving for #y# gives us the equations #y=sqrt(2-x)# and #y=-x# (note that we want the positive square root for our upper bound). Thus, we can represent this part of the area by the integral

#int_(-2)^0(sqrt(2-x)-(-x))dx = int_(-2)^0(sqrt(2-x)+x)dx#

#=[-2/3(2-x)^(3/2)+x^2/2]_(-2)^0#

#=(-(4sqrt(2))/3+0)-(-16/3+2)#

#=(10-4sqrt(2))/3#

From #x=0# to #x=2#, the area is still bounded above by #x=2-y^2#, but below by #y=0#.

Once again rewriting #x=2-y^2# as #y = sqrt(2-x)#, this gives us the integral

#int_0^2(sqrt(2-x)-0)dx = int_0^2sqrt(2-x)dx#

#=[-2/3(2-x)^(3/2)]_0^2#

#=0-(-(4sqrt(2))/3)#

#=(4sqrt(2))/3#

Adding the two areas together, we get our total area:

#(10-4sqrt(2))/3 + (4sqrt(2))/3 = 10/3#