For x=-√(3-y),

From y=-6 to y=3, remove a right triangle with an area of 3*6/2=9.

For x=√(3-y),

From y=0 to y=3, plus the area of the right triangle is 1*2/2= 1,

The area of the curved trapezoid on the left side of Y axis is:

-∫√(3-y)dy (-6→3)

=∫√(3-y)d(3-y ) ( 3-y)(0→9)

=[(3-y)^( 1/2+ 1)/( 1/2+ 1)(3 →- 6)

=2(3-y)^(3/2)/3 (3→-6)

= 18,

Y-axis left area = 18-9=9,

Similarly, the area of the curved trapezoid on the right side of the Y axis is ∫√(3-y)dy (0→3).

=-2(3-y)^(3/2)/3 (0→3)

=2/3,

The area on the right side of Y axis =2/3+ 1*2/2=5/3,

The area of the shaded part = 9+5/3=32/3,

So you should choose C.