Hello from MrBillDoesMath!
Answer: 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
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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
