Question

The integral
`\pi\int\limits^2_1 {(1-(lnx)^(2)) } \, dx` represents the volume of a solid obtained by rotating a region around y=-1. Evaluate.

Answer 1

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55

Explanation:

V = π ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² (1 − (ln x)²) dx

V/π = ∫₁² dx − ∫₁² (ln x)² dx

V/π = x |₁² − ∫₁² (ln x)² dx

V/π = 1 − ∫₁² (ln x)² dx

To evaluate the second integral, integrate by parts.

If u = (ln x)², then du = 2 (ln x) / x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= (ln x)² x − ∫ x (2 (ln x) / x) dx

= x (ln x)² − 2 ∫ ln x dx

Integrate by parts again.

If u = ln x, then du = 1/x dx.

If dv = dx, then v = x.

∫ u dv = uv − ∫ v du

= x ln x − ∫ x (1/x dx)

= x ln x − ∫ dx

= x ln x − x

Substitute:

∫ (ln x)² dx = x (ln x)² − 2 ∫ ln x dx

∫ (ln x)² dx = x (ln x)² − 2 (x ln x − x)

∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x

Substitute again:

V/π = 1 − ∫₁² (ln x)² dx

V/π = 1 − (x (ln x)² − 2x ln x + 2x) |₁²

V/π = 1 + (-x (ln x)² + 2x ln x − 2x) |₁²

V/π = 1 + (-2 (ln 2)² + 4 ln 2 − 4) − (-1 (ln 1)² + 2 ln 1 − 2)

V/π = 1 − 2 (ln 2)² + 4 ln 2 − 4 + 2

V/π = -2 (ln 2)² + 4 ln 2 − 1

V = π (-2 (ln 2)² + 4 ln 2 − 1)

V ≈ 2.55