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let f(x) = e^x and let R be the region bounded by the graph of f and the x axis for x < or equal to -1. Does the solid generated by revolving R about the x-axis have a finite volume? If so, find the volume. If not, show why not
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Yes|dw:1361431440425:dw|
is it pi/2 e^-2 ?
\[A=\pi (e^x)^2\]\[V=\pi \int_{-1}^{0}e^{2x}\]
thanks
I got \[\frac{\pi}{2}(1-e^{-2})\]
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