How to find the volume of the region bounded by three graphs?
By Daniel Johnston
The problem asks given the region bounded by the graphs of y=lnx, y=0, and x=e, find
a. the volume of the solid generated by revolving the region about the x-axis.
b. the volume of the solid generated by revolving the region about the y-axis.
For my first solution I first integrated with respect to the x-axis using the disc method and got 2.2565 as my answer
$$ (V) = \int_a^b \pi(f(x))^2dx $$
$$ (V) = \int_1^e \pi(\ln x)^2dx = \pi(e-2)\approx 2.2565 $$ Then for my second solution I integrated with respect to the y-axis using the shell method but instead of getting the same answer I got a different answer. Is there something wrong with my shell method setup?
$$ (V) = \int_a^b 2\pi x[f(x)-g(x)]dx $$
$$ (V) = \int_1^e (2\pi x*\ln x )dx = 1/2(1+e^2)\pi \approx 13.1775 $$
$\endgroup$1 Answer
$\begingroup$Your first integral is calculating the volume of the solid formed when the region is revolved around the $x$-axis. Your second integral is calculating the volume of the solid formed when the region is revolved around the $y$-axis. These two solids have different volumes, hence why you got different answers. For your second solution, if you want to integrate with respect to the $y$-axis, then your integrand should be in terms of $y$ not $x$.
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