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#?

Question

Christopher Agresta · Accepted Answer

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

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#

Paisley Johnson · Answer

undefined To find the volume of the solid obtained by rotating the region bounded by the curves $ y = \frac{1}{1+x^2} $, $ y = 0 $, $ x = 0 $, and $ x = 2 $ about the line $ x = 2 $, you can use the method of cylindrical shells.

The volume $ V $ of the solid generated by rotating a curve $ y = f(x) $ from $ x = a $ to $ x = b $ about the line $ x = c $ is given by:

$$ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx $$

In this case:

1. The curve is $ y = \frac{1}{1+x^2} $.
2. The bounds of integration are $ x = 0 $ and $ x = 2 $.
3. The axis of rotation is $ x = 2 $.

Substitute these values into the formula:

$$ V = 2\pi \int_{0}^{2} x \cdot \frac{1}{1+x^2} \, dx $$

Now, compute the integral:

$$ V = 2\pi \int_{0}^{2} \frac{x}{1+x^2} \, dx $$

To solve this integral, perform a substitution:

Let $ u = 1 + x^2 $, then $ du = 2x \, dx $.

When $ x = 0 $, $ u = 1 $ and when $ x = 2 $, $ u = 5 $.

The integral becomes:

$$ V = 2\pi \int_{1}^{5} \frac{1}{u} \, du $$
$$ V = 2\pi [\ln|u|]_{1}^{5} $$
$$ V = 2\pi [\ln(5) - \ln(1)] $$
$$ V = 2\pi \ln(5) $$