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Find a function whose ordinate set generates a solid of revolution with volume x2 f(x)

by RoRi

Let f(x) be a nonnegative, differentiable function whose graph passes through both points (0,0) and \left( 1, \frac{2}{\pi}\right). For every real number x > 0, the ordinate set of f on the interval [0,x] generates a solid of revolution when rotated about the x-axis whose volume is given by

    \[ x^2 f(x). \]

Find the formula for the function f(x).

Incomplete.

Find a function which divides a rectangle into pieces with given properties

by RoRi

Consider a curve whose Cartesian equation is given by y = f(x), and which passes through the origin. A rectangular region is drawn with one corner at the origin, and the other corner on the curve of the graph of f(x). The curve f(x) then divides the rectangle into two pieces A and B. These two pieces of the rectangle then generate solids of revolution when rotated about the x-axis. If the volume of one solid of revolution is always n times the volume of the other solid of revolution, find the equation for f(x).

Incomplete.

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 function given the volume of the solid of revolution it generates

by RoRi

Let f(x) be a function continuous on an interval [0,a]. The volume of the solid of revolution obtained by rotating f about the x-axis on the interval [0,a] is given by

    \[ V = a^2 + a \]

for every a > 0. Find a formula for the function f.

Using the formula for the volume of the solid of revolution generated by a function on an interval we know

    \[ V = \pi \int_0^a (f(t))^2 \, dt \quad \implies \quad a^2 + a = \pi \int_0^a (f(t))^2 \, dt. \]

Now we differentiate both sides of this equation using the fundamental theorem of calculus on the right-hand side,

    \begin{align*} x^2 + x = \pi \int_0^x (f(t))^2 \, dt && \implies && 2x + 1 = \pi (f(x))^2 \\[9pt] && \implies && f(x) = \sqrt{\frac{2x+1}{\pi}}. \end{align*}

Find the slope and area under the graph for a given function

by RoRi

Let

    \[ f(x) = \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \qquad \text{for} \quad x> 0. \]

  1. Determine the slope of the graph of f at the point with x-coordinate 1.
  2. Find the volume of the solid of revolution formed by rotating the region between the graph of f(x) and the interval [1,4] about the x-axis.
  1. To take this derivative, using logarithmic differentiation will be easier,

        \begin{align*} \log (f(x)) &= \log \left( \sqrt{ \frac{4x+2}{x(x+1)(x+2)}} \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log (x(x+1)(x+2)) \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log x - \log (x+1) - \log (x+2) \right). \end{align*}

    Then differentiating both sides we have,

        \begin{align*} &&\frac{f'(x)}{f(x)} &= \frac{1}{2} \left( \frac{4}{4x+2} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right) \\[9pt] \implies && f'(x) &= \frac{1}{2} \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \left( \frac{2}{2x+1} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right). \end{align*}

    So, to find the slope at the point with x = 1 we evaluate,

        \[ f'(1) = \frac{1}{2} \left( \frac{2}{3} - 1 - \frac{1}{2} - \frac{1}{3} \right) = -\frac{7}{12}. \]

  2. First, the integral to compute the volume of the solid of revolution is,

        \begin{align*} V &= \pi \int_1^4 (f(x))^2 \, dx \\[9pt] &= \int_1^4 \frac{\pi(4x+2)}{x(x+1)(x+2)} \, dx. \end{align*}

    To evaluate this we use the partial fraction decomposition,

        \[ \frac{2x+1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}. \]

    This gives us the equation

        \[ A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = 2x+1. \]

    Evaluating at x = 0, x = -1, and x = -2 we obtain

        \[ A = \frac{1}{2}, \quad B = 1, \quad C = -\frac{3}{2}. \]

    Therefore, we have

        \begin{align*} V &= 2 \pi \int_1^4 \left( \frac{1}{2x} + \frac{1}{x+1} - \frac{3}{2(x+2)} \right) \, dx \\[9pt] &= 2 \pi \left( \frac{1}{2} \int_1^4 \frac{1}{x} \,dx + \int_1^4 \frac{1}{x+1} \, dx - \frac{3}{2} \int_1^4 \frac{1}{x+2} \, dx \right) \\[9pt] &= \pi \log x \Bigr \rvert_1^4 + 2 \pi \log |x+1| \Bigr \rvert_1^4 - 3 \pi \log |x+2| \Bigr \rvert_1^4 \\[9pt] &= \pi \log 4 + 2 \pi (\log 5 - \log 2) - 3 \pi (\log 6 - \log 3) \\[9pt] &= 2 \pi \log 2 + 2 \pi \log 5 - 2 \pi \log 2 - 3 \pi \log (2 \cdot 3) + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 - 3 \pi \log 3 + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 \\[9pt] & = \pi (\log 25 - \log 8) \\[9pt] & = \pi \log \frac{25}{8}. \end{align*}

Compute the volume of the solid of revolution generated by a region bounded by inequalities

by RoRi

Define a region to be the points (x,y) such that

    \[ 0 \leq x \leq 2, \qquad \frac{1}{4} x^2 \leq y \leq 1. \]

Sketch this region, and compute the volume of the solid of revolution under the following revolutions:

  1. Revolution about the x-axis.
  2. Revolution about the y-axis.
  3. Revolution about the vertical line through the point (2,0).
  4. Revolution about the horizontal line through the point (0,1).
First, the sketch of the region is:

Rendered by QuickLaTeX.com

  1. We compute the volume from revolving about the x-axis as follows,

        \begin{align*} V &= \pi \int_0^2 \left( 1 - \left( \frac{x^2}{4} \right)^2 \right) \, dx \\ &= 2 \pi - \frac{\pi}{16} \int_0^2 x^4 \, dx \\ &= 2 \pi - \frac{2 \pi}{5} \\ &= \frac{8 \pi}{5}. \end{align*}

  2. First, we find an equation for y = \frac{1}{4} x^2 in terms of x.

        \[ y = \frac{1}{4} x^2 \quad \implies \quad |x| = 2 \sqrt{y}. \]

    Then, we compute the volume as follows from revolving about the y-axis as follows,

        \begin{align*} V &= \pi \int_0^1 (2 \sqrt{y})^2 \, dy \\ &= \pi \int_0^1 4y \, dy \\ &= 2 \pi. \end{align*}

  3. We compute the volume as follows from revolving about the vertical line x=2 as follows,

        \begin{align*} V &= \pi \int_0^1 4 \, dy - \pi \int_0^1 (2 - 2 \sqrt{y})^2 \, dy \\ &= 4 \pi - \pi \int_0^1 (4 - 8 \sqrt{y} + 4 y) \, dy \\ &= 4 \pi - \pi \left( 4 - \frac{16}{3} + 2 \right) \\ &= 4 \pi - \frac{2 \pi}{3} \\ &= \frac{10 \pi}{3}. \end{align*}

  4. We compute the volume as follows from revolving about the horizontal line y = 1 as follows,

        \begin{align*} V &= \pi \int_0^2 \left( \frac{1}{4} x^2 - 1 \right)^2 \, dx \\ &= \pi \int_0^2 \left( \frac{x^4}{16} - \frac{x^2}{2} + 1 \right) \, dx \\ &= \pi \left( \frac{2}{5} - \frac{4}{3} + 2 \right) \\ &= \frac{16 \pi}{15}. \end{align*}

Compute the volume of a sphere with a cylindrical hole of length 2h removed

by RoRi

Given a solid sphere, drill a cylindrical hole of length 2h through the center of the sphere. Prove that the volume of the resulting ring is \pi a h^3, where a \in \mathbb{Q}.

Proof. Let r_s denote the radius of the sphere, and r_c denote the radius of the cylindrical hole. Then the volume of the ring is the volume of the solid of revolution formed by rotating the area between the functions

    \[ f(x) = \sqrt{r_s^2 - x^2}, \qquad g(x) = r_c, \qquad -h \leq x \leq h \]

about the x-axis. Since the length of the hole is 2h, we know f(h) = g(h); thus, r_s^2 - h^2 = r_c^2 \implies r_s^2 - r_c^2 = h^2. So, we have,

    \begin{align*} V &= \pi \int_{-h}^h \left( (\sqrt{r_s^2 - x^2})^2 - r_c^2 \right) \, dx \\ &= \pi \int_{-h}^h (r_s^2 - r_c^2 - x^2) \, dx \\ &= \pi h^2 (2h) - pi \left. \frac{x^3}{3} \right|_{-h}^h \\ &= 2 \pi h^3 - \frac{2}{3} \pi h^3 \\ &= \frac{4}{3} \pi h^3 \\ &= \pi a h^3 & \text{where } a = \frac{4}{3} \in \mathbb{Q}. \qquad \blacksquare \end{align*}

Volume of cylindrical hole removed from a sphere

by RoRi

Given a solid sphere of radius 2r, what is the volume of material from a hole of radius r through the center of the sphere.

First, the volume of a sphere of radius 2r is given by

    \[ V_S = \frac{4}{3} \pi (2r)^3 = \frac{32}{3} \pi r^3. \]

Then, the volume of a sphere with a hole drilled in it is the volume of the solid of revolution generated by the region between f(x) = \sqrt{4r^2 - x^2} and g(x) = r from -\sqrt{3} r to \sqrt{3} r. Denoting this volume by V_T we then have,

    \begin{align*} V_T &= \pi \int_{-\sqrt{3} r}^{\sqrt{3} r} (4r^2 - x^2 - r^2) \, dx \\ &= \pi \left( \int_{-\sqrt{3} r}^{\sqrt{3}r} (3r^2 - x^2) \, dx \right) \\ &= 6 \sqrt{3} \pi r^3 - 2 \sqrt{3} \pi r^3 \\ &= 4 \sqrt{3} \pi r^3. \end{align*}

Thus, the volume of the material removed from the sphere by drilling a hole in it is given by

    \[ \frac{32}{3} \pi r^3 - 4\sqrt{3} \pi r^3 = \left( \frac{32}{3} - 4 \sqrt{3} \right) \pi r^3. \]