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?

Question

Bosco

Physics

21 July, 04:16

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?

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Deshawn Mcneil · Answer 1

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.