The problem can be stated as a maximization/minimization.

Find maximum/minimum of

#S(x,y) = 2pi x y+2 pi x^2#

subjected to

#x^2+y^2=r^2#

#y > 0#

Introducing a slack variable #s# to transform the inequality restriction into an equality equivalent restriction we have

#g_1(x,y) = x^2+y^2-r^2 = 0#

#g_2(x,y) = y-s^2=0#

The lagrangian is

#L(x,y,lambda_1,lambda_2,s) = S(x,y)+lambda_1 g_1(x,y) + lambda_2 g_2(x,y,s)#

being analytic, the stationary conditions are given by

#grad L(x,y,lambda_1,lambda_2,s)=vec 0#

giving

#
{
(2 lambda_1 x + 4 pi x + 2 pi y = 0),
(lambda_2 + 2 pi x + 2 lambda_1 y = 0),
( x^2 + y^2 -r^2= 0),
(y -s^2= 0),
(2 lambda_2 s = 0)
:}
#

Solving for #x,y,lambda_1,lambda_2,s# we obtain

#
(x= -r, y= 0, lambda_1= -2pi, lambda_2= 2pi r, s= 0),
(x = r, y = 0, lambda_1= -2pi, lambda_2 = -2pi r, s = 0),
(x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0,
s ne 0),
(x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0,
s ne 0)
#

As we kow the active restriction is #g_1(x,y)# so, the stationary points qualification is made over

#f_{g_1}(x)=2 pi x^2 + 2 pi x sqrt[r^2 - x^2]#

having a minimum #(d^2f_{g_1})/(dx^2)>0# at

#
(x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0,
s ne 0),
#

and maximum #(d^2f_{g_1})/(dx^2)<0# at

#
(x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0,
s ne 0)
#

The maximum value being

#S(r/2 sqrt[2 + sqrt[2]], r/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi r^2#

If #r = 1# then #S(1/2 sqrt[2 + sqrt[2]], 1/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi#