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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*}

Find a curve bisecting the curves x2 and x2/2

by RoRi

Consider the figure

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We say the curve C bisects the region between C_1 and C_2 in area if for every point P on the curve C the area of regions A and B are equal. Given the equations

    \[ C: \ y = x^2 \qquad C_1: \ y = \frac{1}{2}x^2 \]

find an equation for C_2 such that the curve C bisects the region between C_1 and C_2 in area.

First, we can calculate the area of the region A. This is the difference in the integrals from 0 to the x-coordinate of P of x^2 and \frac{1}{2}x^2. Since P lies on the curve C defined by the equation y = x^2 we may write P = (t,t^2) for some t \geq 0. Then the area of A is given by

    \begin{align*} \int_0^t x^2 \, dx - \int_0^t \frac{x^2}{2} \, dx &= \int_0^t \left( x^2 - \frac{x^2}{2} \right) \, dx \\ &= \int_0^t \frac{x^2}{2} \, dx \\ &= \left. \frac{x^3}{6} \right|_0^t \\ &= \frac{t^3}{6}. \end{align*}

Now, we make the assumption that the equation for C_2 is of the form kx^2 for some positive real number k. (I don’t know a good way to justify this assumption other than it’s the most obvious first thing to try, and it happens to work.) Then to find the area of region B first we find equations for the curves C and C_2 in terms of y (so that we may integrate along the y-axis which is somewhat easier). So we have

    \begin{align*} C: \ y &= x^2 &\implies && x = \sqrt{y} \\ C_2: \ y &= kx^2 & \implies && x = \frac{1}{\sqrt{k}} \sqrt{y}. \end{align*}

Then, we integrate along the y-axis from 0 to t^2 (since this is the y-coordinate of P) the difference between these two curves. The area of B is then

    \begin{align*} \int_0^{t^2}\left( \sqrt{y} - \frac{1}{\sqrt{k}} \sqrt{y} \right) \, dy &= \left( 1 - \frac{1}{\sqrt{k}} \right) \int_0^{t^2} \sqrt{y} \, dy \\ &= \left(1 - \frac{1}{\sqrt{k}} \right) \left( \frac{2}{3} y^{\frac{3}{2}} \Big \rvert_0^{t^2} \right) \\ &= \left( 1 - \frac{1}{\sqrt{k}} \right) \left( \frac{2t^3}{3} \right) \\ &= \frac{2t^3}{3} - \frac{2t^3}{3 \sqrt{k}}. \end{align*}

Now we set the areas of the regions equal and solve for k,

    \begin{align*} \frac{t^3}{6} = \frac{2t^3}{3} - \frac{2t^3}{3 \sqrt{k}} && \implies && \frac{1}{6} &= \frac{2}{3} - \frac{2}{3 \sqrt{k}} \\ && \implies && \frac{2}{3 \sqrt{k}} &= \frac{1}{2} \\ && \implies && 4 &= 3 \sqrt{k} \\ && \implies && k &= \frac{16}{9}. \end{align*}

Therefore, the equation describing C_2 is given by

    \[ y = \frac{16}{9} x^2. \]