A region is bounded by $y = ln x$ , the $x$ ‑axis, $x = 1$ and $x = e$ . Find the volume of the solid generated when this region is revolved about the $y$ ‑axis using the method of cylindrical shells.

hard 4 marks Class 12 · Mathematics · Application of Integrals

Answer

\frac{π}{2} (e^{2} + 1)

The correct answer is $\frac{π}{2} (e^{2} + 1)$ — Multiply by $2 π$ : $V = 2 π \times \frac{e ^{2} + 1}{4} = \frac{π}{2} (e^{2} + 1)$ .

Step-by-step solution

For shells about the $y$ ‑axis, radius $r = x$ , height $h = ln x$ , thickness $d x$ .
Shell volume element: $d V = 2 π r h d x = 2 π x ln x d x$ .
Integrate from $x = 1$ to $x = e$ : $V = 2 π \int_{1}^{e} x ln x d x$ .
Use integration by parts: let $u = ln x$ , $d v = x d x$ .
Then $d u = \frac{1}{x} d x$ , $v = \frac{x ^{2}}{2}$ .
Integral $\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \int \frac{x}{2} d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + C$ .
Evaluate between $1$ and $e$ : at $e$ , $\frac{e ^{2}}{2} \cdot 1 - \frac{e ^{2}}{4} = \frac{e ^{2}}{4}$ ; at $1$ , $0 - \frac{1}{4} = - \frac{1}{4}$ .
Difference: $\frac{e ^{2}}{4} + \frac{1}{4} = \frac{e ^{2} + 1}{4}$ .
Multiply by $2 π$ : $V = 2 π \times \frac{e ^{2} + 1}{4} = \frac{π}{2} (e^{2} + 1)$ .

Common mistake: Using the washer formula instead of shells for rotation about the

y

‑axis

Concept tested: Volume by shell method