If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

NOTE: IT SEEMS THAT r + delta r = 7 but answer comes only by taking initial value r = 7

V=(4pi/3)r³, dr=0.02m,,(r+dr)³=(r³+3r²dr+3r(dr)² +(dr)³)≈r³(1+3(dr/r)) dV=(4pi/3)(3dr/r)=(4pi/3)*3*0.02/7

as r= 7m   dr = 0.02m 

V = (4pi/3)r³

differentiating both sides

dV = (4pi)/3*3r²dr

=4*22/7*49*0.02

= 15.32m³ 

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Dear manoj jain,

Given that,

 r = 7m   dr  = 0.02m 

V = (4π/3)r³

Now, differentiating both sides

dV = (4π)/3*3r²dr

=4 * 3.14 * 49 * 0.02

= 15.32m³ 

Hope this helped you immensely...

All the Very Best & Good Luck to you dear..

Regards,

AskIITians Expert,

Godfrey Classic Prince

IIT-Madras

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according to question initial r is 6.98 or 7.02