if x is really really small ,,,, and ur talkin abt surface area
then its 3x/2
If the percentage error in measuring the surface area of a sphere is x%,then find the error in its volume.
if x is really really small ,,,, and ur talkin abt surface area
then its 3x/2
percentage error in measuring surface area is x%
CASE I: x is small
Then % error in measuring length is x/2%
So % error in volume is 3x/2%
CASE II: x is not small
Let the % error in measuring length be a
Then if the length is measured as l, the actual length is l(1±a/100)
Let actual length be l(1+a/100) then surface area = l2(1+a/50+a2/10000)
So %error = (a/50+a2/10000)*100 = 2a+a2/100 = x ...(i)
volume = lact3 = l3(1+3a/100 + 3a2/10000 + a3/106)
So % error = (3a/100 + 3a2/10000 + a3/106)*100 = 3a + 3a2/100 + a3/10000
u can use the value of a obtained in (i)