For the convex shape to be maximum, Im supposing that it is bounded by a cylindrical surface (the outer edge of the dry space), the base plane and a tangent plane to the sphere, which cuts the base circle in a chord, with mid-point M say. Let A be the acute angle which this plane makes with the horizontal, and let t be the perpendicular distance, MC, from the chord to the centre, C, of the base circle.

To find the volume of the shape, note that a typical horizontal section at height y above the base, will always be a segment of a circle, radius 1, with a chord at a distance x (say) from the vertical axis through the centre, O, of the sphere, where dy/dx = tan(A).

Area of this segment = area sector area triangle

= arccos(x)-(1/2)*x*sqrt(1 x^{2})

Volume of shape = Integral(x=t..1) of [arccos(x) (1/2)*x*sqrt(1 x^{2})] dy

= Integral(x=t..1) of [arccos(x) (1/2)*x*sqrt(1 x^{2})]tan(A) dx

= tan(A)[x*arccos(x) sqrt(1 x^{2}) + (1/6)(1 x^{2})^{3/2}] between t & 1

Apart from the root t = 1 (V indeterminate), equation (2) cant be solved explicitly for t. But the single root between 0 and 1 is shown numerically to be t = 0.46506..., which gives a maximum volume of 0.3134..