Question

Let R be the region enclosed by the curve \[y = 1 / (r^2 + 1)\] and r-axis from 0 to 1. Find the volume of the solid generated when the Region R is revolved about the r-axis

Answer 1

what's r axis?

Answer 2

its the question :)

Answer 3

whats the answers i also want to know too

Answer 4

i'm working on it :)
i don'T have the answer yet too

Answer 5

\[V = \int\limits_{0}^{1}\pi ( 1 / (r^2 + 1))^2dx \]

Answer 6

ok so the formula is
\[V=\pi \int\limits_{0}^{1}f(r)^2 dr\]
so
\[V=\pi \int\limits_{0}^{1}1/(r^2+1)^2 dr\]
using the substitution r=tan(theta) you find
\[dr=\sec ^2(\theta) d \theta\]
so \[V=\pi \int\limits\limits_{0}^{\pi/4}\sec^2\theta/(\tan ^2\theta+1)^2 d \theta\]
which is
\[\pi(2+\pi)/8\]

Answer 7

yeap i got the same answer too :)