Question

Find the volume of the solid obtained by rotating the region bounded by y= 1/x^5, y=0, x=2, x=9

about the -axis.

please help with the answer i have tried and i keep gettign wrong answer :(

Answer 1

whats your integral?

Answer 2

When x=2, y=1/32. When x=9, y approaches zero.
Approximately, \[A=\int\limits_{0}^{1/32}π[f ^{-1}(y)]^2dy\]\[=π \int\limits_{0}^{1/32}[y ^{-1/5}]^2dy\]\[=π[y ^{3/5}/(3/5)] from 0 1/32\]\[=\frac{ 3π}{ 5 }[0-(\frac{ 1 }{ 32 })^{3/5}]\]
Since we want the magnitude,
\[A = \frac{ 3π }{ 5\times32^{3/5} }\] = 0.236
Is it right?