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Compute the area and volume of solids of revolution of e-2x

by RoRi

Define the function f(x) = e^{-2x} for all x \in \mathbb{R}. Let

    \begin{align*} S(t) &= \text{the ordinate set of } f \text{ on } [0,t], \quad t> 0.\\ A(t) &= \text{the area of } S(t).\\ V(t) &= \text{the volume of the solid obtained by rotating } S(t) \text{ about the $x$-axis}.\\ W(t) &= \text{the volume of the solid obtained by rotating } S(t) \text{ about the $y$=axis}. \end{align*}

Compute

  1. A(t);
  2. V(t);
  3. W(t);
  4. \lim_{t \to 0} \frac{V(t)}{A(t)}.
  1. The area of the ordinate set on [0,t] is given by the integral,

        \[ A(t) = \int_0^t f(x) \, dx = \int_0^t e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \Bigr \rvert_0^t = \frac{1}{2}(1 - e^{-2t}). \]

  2. The volume of the solid of revolution obtained by rotating f(x) about the x-axis is

        \begin{align*} V(t) &= \pi \int_0^t (f(x))^2 \, dx \\[9pt] &= \pi \int_0^t e^{-4x} \, dx \\[9pt] &= -\frac{\pi}{4} e^{-4x} \Bigr \rvert_0^t \\[9pt] &= \frac{\pi}{4} (1 - e^{-4t}). \end{align*}

  3. To compute the volume of the solid of revolution obtained by rotating f about the y-axis we first find x as a function of y.

        \[ f(x) = y = e^{-2x} \implies x = -\frac{\log y}{2}. \]

    Since f(t) = e^{-2t}, the integral is then from e^{-2t} to 1 and we have

        \begin{align*} W(t) &= \pi \int_{e^{-2t}}^1 \frac{-\log y}{2} \, dy \\[9pt] &= -\frac{\pi}{2} \int_{e^{-2t}}^1 \log y \, dy \\[9pt] &= -\frac{\pi}{2} ( y \log y - y)\Bigr \rvert_{e^{-2t}}^1 \\[9pt] &= -\frac{\pi}{2} (-1 - e^{-2t} (-2t) - e^{-2t}) \\[9pt] &= \frac{\pi}{2} (1 - e^{-2t}(2t+1)). \end{align*}

  4. Finally, using parts (c) and (d) we can compute the limit,

        \begin{align*} \lim_{t \to 0} \frac{V(t)}{A(t)} &= \lim_{t \to 0} \frac{\frac{\pi}{4} (1-e^{-4t})}{\frac{1}{2} (1-e^{-2t})} \\[9pt] &= \lim_{t \to 0} \frac{\pi (1-e^{-4t})}{2(1-e^{-2t})} \\[9pt] &= \lim_{t \to 0} \frac{\pi (e^{4t} - 1)}{2e^{2t}(e^{2t} - 1)} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \frac{(e^{2t}+1)(e^{2t}-1)}{e^{2t}(e^{2t}-1)} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \frac{e^{2t}+1}{e^{2t}} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \left( 1 + \frac{1}{e^{2t}} \right) \\[9pt] &= \pi. \end{align*}

3 comments

  1. Artem says:

    The solution (c) is incorrect: you need to use the integral of log^2(y). This will not directly take one to the ending answer. After the integral is computed (can be easily done with integration by parts), you have to add \pi * t^2 e^{-2t} to the result – this is the portion of the solid of revolution, that lies below the y = e^{-2t}, it is a square, which revolves around the y-axis. This will give the answer in Apostol. If sth, one can plot this function and immediately see this.

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