se differentials to estimate the amount of material in a closed cylindrical can that is 30 cm high and 12 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.05 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 (in terms of r,hr,h) is dV=dV= dr+dr+ dhdh The differentials of rr and hh are: dr=dr= and dh=dh= (be careful) The approximate volume of material is cm3cm3. HINT. The metal fills in the space between two cylinders - the inner one and the outer one. Therefore, the volume of metal equals the increment in the volume VV of the cylinder whose measurements increase from rr and hh to r+drr+dr and h+dhh+dh, respectively. This increment ΔVΔV can be approximated by the differential dVdV.
Use differentials to estimate the amount of material in a closed cylindrical can that is 30 cm high and 12 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.05 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 (in terms of r,hr,h) is
dV=dV= dr+dr+ dhdh
The differentials of rr and hh are:
dr=dr= and dh=dh= (be careful)
The approximate volume of material is cm3cm3.
HINT. The metal fills in the space between two cylinders - the inner one and the outer one. Therefore, the volume of metal equals the increment in the volume VV of the cylinder whose measurements increase from rr and hh to r+drr+dr and h+dhh+dh, respectively. This increment ΔVΔV can be approximated by the differential dVdV.
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