Question

the error in the measurement of radius of sphere is.4 percentage find the error in surface area​

Answer 1

Answer:

the area of a sphere is 4*pi*r^2

where Pi is a constant and r = radius of sphere

So, if we imagine that we have a posibble error of 4% on the measurement of the radius of the sphere, that error could be separated on two situations:

A) The 4% of error gives an excess on the radius of the sphere

b) The 4% of error gives a reduction on the radius of the sphere

so case A the area would be equal to 4*pi*(1.04*r)^2 = 4.3264*pi*r^2

so case B the area would be equal to 4*pi*(0.96*r)^2 = *pi*r^2

now the definition of error

error % =100*( value calculated - value of reference)/value of reference

error from case A = 100*(3.6864*pi*r^2-4*pi*r^2)/4*pi*r^2 eliminating common terms we would get error from case A = 100*(-0.3136/4) = -7.84 %

error from case B = 100*(4.3264*pi*r^2-4*pi*r^2)/4*pi*r^2 applying the same procedure as above we would get error from case B = 100*(0.3264/4) = 8.16%

resuming the information,

if we have an error of 4% on the measurement of a the radius of a sphere

there will be an error on the surface area of the sphere that ranges from

-7.84 % to 8.16 %

Before to start to answer this question I thought that the change on the surface area of the sphere would behave in the same way for the increments and reductions but I noticed that it is not true so I used excel to made an easy calculus sheet to evidence what i noticed so:

If you introduce a change on the radius of a sphere you will get a different error based on what kind of change are you introducing, if you reduce the radius of the sphere on 1% you will get a reduction of about -1.99% on the other hand if you introduce an increment of 1% you will get an excess of 2.01 %.