Consider a sphere of radius r, surface area a and volume v. suppose you double the radius to 2r. how does the new surface area anew and the new volume vnew compare to the old values?

For a sphere, we know that the volume is equal to V = (4/3) * pie*r^3 and the surface area is equal to SA=4*pie*r^2. In those equations, r is the radius. If the radius is doubled to 2r, the new expression for V becomes V = (4/3) * pie * (2r) ^3 = (4/3) * pie*8*r^3 = (32/3) * pie*r^3. The new Volume is 8 times larger than the old volume. For the surface area, the new expression for SA becomes SA=4*pie * (2r) ^2=4*pie*4*r^2=16*pie*r^2. The new surface area is 4 times larger than the old surface area.