If the surface area of a sphere is doubled, how will the volume get affected

Hello from MrBillDoesMath!

the volume of the sphere increases by a factor of 2 * (2^(1/2)) or about 2.8

Discussion:

From Plane Geometry

Surface Area of a Sphere of radius "r" = 4 * Pi * r^2

Volume of a Sphere of radius "r" is (4/3)* Pi * r^3

Suppose the sphere with surface area "A" has radius "r1". Suppose further that the sphere with surface area "2A" has radius "r2". Then

4 * Pi * r1^2 = A     and

4 * Pi * r2^2 = 2A

Divide the bottom equation by the top one to get

( 4* Pi *  r2^2 )  /  (4 * Pi * r1^2 ) =  r2^2/ r1^2 = (r2/r1) ^2

This ratio also equals (2A)/A = 2. We conclude that

(r2/r1) ^ 2 =   2     or

r2/r1 =  2 ^ (1/2)                     (this is the square root of 2)

r2 = 2^ (1/2) * r1    

Now let's look at the volume

Consider the volume of the sphere with surface area 2A.

V =  

(4/3) * Pi * r2^3  =

(4/3) * Pi *  ( 2 ^ (1/2) * r1) ^3  =                (write r2 in terms of r1)      

(4/3) * Pi * (r1 ^3) * (2^ (1/2)) ^3  

As  (2^(1/2))^3 is the square root of 2 cubed (which equals 2 times the square root of 2), the last equation becomes

( (4/3) * Pi * r1^3) * (2 * 2 ^ (1/2) )

Note the boldfaced equation above is the volume of the original sphere of surface area A. So doubling the surface area causes the volume to increase by the factor 2 * 2^(1/2) which is about 2.8.

Regards, MrBill