# 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?

Mar 26, 2016

${V}_{s} = 2 \pi \left[2 \arctan \left(2\right) - \frac{1}{2} \ln \left(5\right)\right]$

#### Explanation:

Given: $\frac{1}{1 + {x}^{2}}$, bound in the region, $R$ over $y = 0 , x = 0 , \mathmr{and} x = 2$ 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 ${r}_{s} \left(x\right) = \left(2 - x\right) \mathmr{and} {h}_{s} \left(x\right) = f \left(x\right) = \frac{1}{1 + {x}^{2}}$
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 $u = 1 + {x}^{2}$
${I}_{2} = \frac{1}{2} \ln \left(1 + {x}^{2}\right)$

${V}_{s} = 2 \pi \left[{I}_{1} - {I}_{2}\right] = 2 \pi {\left[2 \arctan \left(x\right) + \frac{1}{2} \ln \left(1 + {x}^{2}\right)\right]}_{0}^{2} =$
${V}_{s} = 2 \pi \left[2 \arctan \left(2\right) - \frac{1}{2} \ln \left(5\right)\right]$