Question

How do you find the volume of the solid obtained by rotating the region bounded by the curves `1/(1+x^2)`, `y=0`, `x=0`, and `x=2` rotated around the `x=2`?

Answer 1

`V_s = 2pi[2arctan(2)-1/2ln(5) ]`

Explanation:

Given: `1/(1+x^2)`, bound in the region, `R` over `y=0, x=0, and x=2` rotated around `x=2`
Required Volume?
Solution Strategy: Use the "Volume by Cylindrical Shells Method"
The "shell method" give the volume calculated in the region bound by `y = f (x), a ≤ x ≤ b` and the integral formula is:
`V_s=∫_a^b 2π ("shell radius") ("shell height") dx=2pi∫_a^br_sh_sdx`

From the figure `r_s(x) = (2-x) and h_s(x) = f(x)=1/(1+x^2)`
thus`=>V_s=2pi∫_a^b r_s(x)*f(x) dx`
`V_s=2pi∫_a^b (2-x)*1/(1+x^2) dx` Now simply integrate
Apply sum rule and write:
`V_s=2pi[∫_a^b 2/(1+x^2)dx - ∫_a^bx/(1+x^2)dx] =2pi[I_1-I2]`
`I_1 =∫_a^b 2/(1+x^2)= [2arctan(x)]_0^2`

`I_2= ∫_a^bx/(1+x^2)dx` let `u=1+x^2; du=2xdx; dx=1/(2x)du`
`I_2= ∫cancelx/u 1/(2cancelx)du=1/2ln(u)` substitute `u=1+x^2`
`I_2=1/2ln(1+x^2)`

`V_s= 2pi[I_1-I_2 ] = 2pi[2arctan(x)+1/2ln(1+x^2)]_0^2=`
`V_s = 2pi[2arctan(2)-1/2ln(5) ]`