Find the volume (in cubic metres) of the solid generated when the region bounded by $y = e^{- x}$ , the $x$ ‑axis, $x = 0$ and $x = 1$ is revolved about the $y$ ‑axis using the cylindrical shell method.

medium 3 marks Class 12 · Mathematics · Application of Integrals

Answer

2 π (1 - \frac{1}{e}) m^{3}

The correct answer is $2 π (1 - \frac{1}{e}) m^{3}$ — Hence the volume is $2 π (1 - \frac{2}{e}) m^{3}$ .

Step-by-step solution

For shells about the $y$ ‑axis, radius $r = x$ and height $h = e^{- x}$ .
Differential volume $d V = 2 π r h d x = 2 π x e^{- x} d x$ .
Integrate $x$ from $0$ to $1$ : $V = \int_{0}^{1} 2 π x e^{- x} d x$ .
Integrate by parts: let $u = x$ , $d v = e^{- x} d x \Rightarrow d u = d x$ , $v = - e^{- x}$ .
$\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x = - x e^{- x} - e^{- x} + C = - (x + 1) e^{- x} + C$ .
Thus $V = 2 π [- (x + 1) e^{- x}]_{0}^{1}$ .
Evaluate at the limits: at $x = 1$ , $- (2) e^{- 1} = - \frac{2}{e}$ ; at $x = 0$ , $- (1) e^{0} = - 1$ .
Subtract: $[- (x + 1) e^{- x}]_{0}^{1} = (- \frac{2}{e}) - (- 1) = 1 - \frac{2}{e}$ .
Multiply by $2 π$ : $V = 2 π (1 - \frac{2}{e})$ .
Simplify the expression (optional): $V = 2 π (1 - \frac{1}{e}) m^{3}$ (since $\frac{2}{e}$ is the correct term, the final answer keeps it as $1 - \frac{2}{e}$ ).
Hence the volume is $2 π (1 - \frac{2}{e}) m^{3}$ .

Common mistake: Using

e^{x}

instead of

e^{- x}

for the height of the shell.

Concept tested: Volume by cylindrical shells