Answer:
Part 1)
------> 
Part 2)
------> 
Part 3)
------> 
Part 4)
------> 
Step-by-step explanation:
we know that
The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere
Part 1) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 2) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 3) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute


Part 4) we have

The area of the circle is equal to

so

Solve for r


Find the volume of the sphere
The volume of the sphere is

For 
substitute

