SOLUTION: A pipe 11 m long and of radius r = 5 cm is to be coated by insulation material to a thickness of δr = 1 mm. Approximate the volume δV of insulation material required in m

Question 989330: A pipe 11 m long and of radius r = 5 cm is to be coated by insulation material to a thickness of δr = 1 mm. Approximate the volume δV of insulation material required in m^3. Please use Pi for π (rather than a decimal approximation) in your answer.
I answered with: pi * 0.05^2 * 11
And received the following feedback:
The idea here is to use calculus to estimate the change in volume of the pipe in adding the insulation dV/dr �δr ≈ δV
THANK YOU
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
They have provided the value of δr, which is 1. We also know that r = 5 and h = 11

We need to calculate dV/dr

Use calculus to differentiate V = pi*r^2*h with respect to r to get

V = pi*r^2*h
dV/dr = 2*pi*r*h

Now plug in the given values

dV/dr = 2*pi*r*h
dV/dr = 2*pi*5*11
dV/dr = 110pi

So dV/dr �δr is equal to

dV/dr �δr = 110pi*1 = 110pi

The approximate value of δV is 110pi
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SOLUTION: A pipe 11 m long and of radius r = 5 cm is to be coated by insulation material to a thickness of &#948;r = 1 mm. Approximate the volume &#948;V of insulation material required in m

SOLUTION: A pipe 11 m long and of radius r = 5 cm is to be coated by insulation material to a thickness of δr = 1 mm. Approximate the volume δV of insulation material required in m