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surface area integration
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y=3x^1/3
rotate about the x axis
\[A=\int\limits_{0}^{1}2 \pi y ds\]
\[ds=\sqrt{1+(x^{-2/3})^{2}}dx\]
\[A=\int\limits_{0}^{1}\sqrt[3]{x}\sqrt{1+\frac{ 1 }{ x^{4/9} }}\]
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I am stuck at this point and I am not sure what to use
you did't say what the boundaries are.... but if you are confident that surface area you are looking for is from y=0 to y=1, then ok...
sorry yes the boundaries are from 0 to 1
from y=0 to y=1, are you sure? maybe it is from x=0 to x=1 ?
@recon14193 it might make things easier if you look at the integral in terms of y instead of x
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you mean in terms of x, than y ?
yes it is from x=0 to x=1
very nice.
what does that mean @SolomonZelman
i mean putting everything in terms of y
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