Got a question from the "Tangent Planes and Differentials" section of multivariable calculus.

Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height 12cm if the tin is 0.04cm thick.

The book gives and answer of 16cm^3 but I don't see how to do this one. Thanks!

Question

Got a question from the "Tangent Planes and Differentials" section of multivariable calculus.
 
 Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height 12cm if the tin is 0.04cm thick.

The book gives and answer of 16cm^3 but I don't see how to do this one.  Thanks!

anonymous · Answer

Volume of a cylinder(tin can) 

where r = radius = 4cm

h = height = 12cm

Applying the definition of the differential dz,

The differential of V is:

Suppose that the radius is decreased by 0.04 cm (dr = -0.04) and the height is decreased by 0,08 cm 
(dh = - 0.08). Then

dV = 96 Pi (-0.04) + 16 Pi (-0.08) = 16.08
the amount of tin is approximately 16cm^3

anonymous · Answer

or this solution:

dv = 2(pi)rhdr + (pi)r^2dh

dr = .04 & dh = .08

dv = 16.08 cm^3

anonymous · Answer

Where did the 0.08 come from?

anonymous · Answer

from the given diameter dh=.08

anonymous · Answer

Oh wait, I see now.  But dh should equal 0.04 as this is the thickness dimension.  Also, since the can has both a top and a bottom the expression should be: $$dV=2 \pi rhdr + 2 \pi r^2dh=2 \pi(4)(12)(0.04) + 2 \pi (4)^2(0.04)$$This gives,$$dV=12.063+4.021\approx16cm^3$$Thanks for your help!

anonymous · Answer

yeah..thanks also and glad to help:-)