Use differentials to estimate the amount of material in a closed cylindrical can

Question

Use differentials to estimate the amount of material in a closed cylindrical can that is 50 cm high and 20 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.1 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses.
The differential for the volume is
dV=________dr+________________dh(enter your answer in therms of r and h )
dr= ________ and dh=_________
the approximate volume of material is __________ cm^3

Explanation / Answer

V is a function of two variables, r and h, we have: V= pr²h The total differential of V is dV: dV = (?V/?r).dr + (?V/?h).dh ?V/?r = 2prh (differentiate V with respect to r, keeping h constant) ?V/?h = pr² (differentiate V with respect to h, keeping r constant) So, we now have: dV = (?V/?r).dr + (?V/?h).dh = 2prh.dr + pr².dh h= 50cm r=d/2 =10cm dr= 0.1cm dh= 2x0.1cm (Thickness of metal in top and bottom) dV= 2p(10)(50)(0.1) + p(10)²(2x0.1) =(120)p cm³

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