If the volume of a sphere is triples, then its surface area will grow by a factor of what?

Question

can you please explain how?

Answer 1

The volume is ~ r^3. So if the volume triples, it means r increase by a factor of 3^(1/3).

Now the surface area is ~ r^2. So that increase by a factor of
[3^(1/3)]^2 = 3^(2/3)

Answer 2

V = 4/3 pi * r^3
SA = 4 * r^2

V1 = 3 V0
V1/r1^3 = 3*V0/r0^3

V1*r0^3 = 3V0*r1^3
r1^3 = 3 * r0^3
r1 = 3^(1/3) * r0
r1 = 1.442 r0
You thus calculate the increase of the radius and plug this into the surface area equation

SA1/SA0 = 4 pi * (1.442r0)^2/ 4 pi* r0^2
SA1/SA0 = 1.442^2 = 2.08

The Surface area will roughly increase by a factor of 2.

Edit: The top two answers have it right I just simply put it in numerical form without the powers.

Answer 3

v=4/3pir^3 is the original volume

v=4/3piR^3 is the new volume

(4/3piR^3)/(4/3pir^3)=3

R^3/r^3=3

R/r=3^1/3

area of original sphere = 4pir^2
area of new sphere =4piR^2

the ratio = 4piR^2/4pir^2 =(R/r)^2

but we know that the ration of R/r is 3^1/3

Hence, the surface area will increase by a factor of (3^1/3)^2 or 3^2/3 of 9^1/3

Answer 4

Doctor risk said it very well. But for gory long winded details.

Volume of a sphere is (4/3)*pi*r^3 and call this V
The surface area is 4*pi*r^2 and call this A

So V = (4*pi*r^2)*(r/3) = A*(r/3)

You want to compare two volumes:
V1 = A1*(r1/3) and V2 = A2*(r2/3)
Divide these and get: V2/V1 = (A2/A1)*(r2/r1)
Since we want the new area switch this around to:
(A2/A1) = (V2/V1)*(r1/r2) - [EQ 1]

From above V1= (4/3)*pi*r1^3 and V2= (4/3)*pi*r2^3
Divide V1 by V2 giving: V1/V2= (r1/r2)^3
and (r1/r2) = CubeRoot(V1/V2) - [EQ 2]

Substitude EQ 2 into EQ 1 and get:
(A2/A1) = (V2/V1)*CubeRoot(V1/V2)
and finally: (A2/A1) = (V2/V1)^(2/3)

So for the case of V2/V1 = 3 ..... A2/A1 = (3)^(2/3) or the cuberoot of 9

Answer 5

3^(2/3)