cm high and 16 cm in diameter if the metal in the top and bottom is 0.2 cm thick, and the meta me 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 enclose differential for the volume is dr+ and dh = (be careful) approximate volume of material is 3 cm.
cm high and 16 cm in diameter if the metal in the top and bottom is 0.2 cm thick, and the meta me 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 enclose differential for the volume is dr+ and dh = (be careful) approximate volume of material is 3 cm.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
can you help me with this cal 3 homework problem.
![Use differentials to estimate the amount of material in a closed cylindrical can that is 40 cm high and 16 cm in diameter if the metal in the top and bottom is 0.2 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
dr
and dh = (be careful)
The approximate volume of material is
cm³.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff29743aa-20ba-4376-81ba-ec57f5b19078%2F592eaa90-1048-4612-8f77-4142fddeef90%2Fedfyo49_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Use differentials to estimate the amount of material in a closed cylindrical can that is 40 cm high and 16 cm in diameter if the metal in the top and bottom is 0.2 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
dr
and dh = (be careful)
The approximate volume of material is
cm³.
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