Home » Volume

Tag: Volume

Compute the volume of a tetrahedron with given vertices

by RoRi

Consider the tetrahedron with vertices at the origin and at the points where the plane

    \[ x + 2y + 3z = 6 \]

intersects the coordinate axes. Compute the volume of this tetrahedron.

First, the intercepts of the plane are given by (6,0,0), (0,3,0), (0,0,2). Then from a previous exercise (Section 13.14, Exercise #13) we know that the volume of a tetrahedron with vertices A,B,C,D is

    \[ V = \frac{1}{6} | (B-A) \cdot (C -A) \times (D-A) |. \]

Letting A = (0,0,0), \ B = (6,0,0), \ C = (0,3,0), \ D = (0,0,2) we have

    \[ V = \frac{1}{6} | (6,0,0) \cdot (0,3,0) \times (0,0,2) | = \frac{1}{6} | (6,0,0) \cdot (6,0,0) | = 6. \]

Prove a vector formula for the volume of a tetrahedron

by RoRi
  1. Consider a tetrahedron with vertices A,B,C,D. Prove that the volume of the tetrahedron is given by the formula

        \[ V = \frac{1}{6} |(B-A) \cdot (C -A) \times (D-A)|. \]

  2. Compute the volume in the case that

        \[ A = (1,1,1), \quad B = (0,0,2), \quad C = (0,3,0), \quad D = (4,0,0). \]

  1. Proof. We know the volume of a tetrahedron is given by \frac{1}{3} A_0 h (where A_0 denotes the altitude of the tetrahedron). We know (page 490 of Apostol) that the volume of the parallelepiped with base formed by vector A,B and height formed by vector C is given by A \times B \cdot C. In this case we have that the base of the tetrahedron is formed by the vectors (B-A) and (C-A), and the height is formed by the vector (D-A). Further, we know that the area of the base described by the vectors (B-A) and (C-A) is one half that of the parallelepiped whose base is given by vectors A and B (since the base of the parallelepiped described by vectors A and B is a rectangle, and the base of the tetrahedron is the triangle formed by cutting this rectangle along the diagonal). Therefore we have

        \begin{align*} V &= \frac{1}{3} \left( \frac{1}{2} \right) (B -A)\cdot (C-A) \times (D - A) \\ &= \frac{1}{6} \big( (B-A) \cdot (C-A) \times (D-A) \big). \qquad \blacksquare \end{align*}

  2. Using the formula in part (a) with the given values of A,B,C,D we have

        \begin{align*} V &= \frac{1}{6} \big( (-1,-1,1) \cdot (-1,2,-1) \times (3,-1,-1) \big) \\ &= \frac{1}{6} \big( (-1,-1,1) \cdot (-2-1, -3-1, 1-6) \big) \\ &= \frac{1}{6} \big( (-1,-1,1) \cdot (-3,-4,-5) \big) \\ &= \frac{1}{6} (3 + 4 - 5) \\ &= \frac{1}{3}. \end{align*}

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 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*}

Find a function given the volume of a solid determined by the function

by RoRi

Given a solid with base defined by the ordinate set of a continuous function f on the interval [1,a]. The cross sections take perpendicular to [1,a] are in the shape of squares. Find the function f(a) if the volume of the solid is

    \[ \frac{1}{3} a^3 \log^2 a - \frac{2}{9} a^3 \log a + \frac{2}{27} a^3 - \frac{2}{27} \qquad \text{for all } a \geq 1. \]

The volume of the solid is equal to the integral,

    \[ V = \int_1^a (f(x))^2 \, dx \]

since the cross sections are in the shape of squares and the length of the base is f(x). So the cross sectional area at each point from 1 to a is (f(x))^2, and then the volume is obtained by integrating these areas over [1,a]. Setting this equal to the given formula for the volume,

    \begin{align*} &&\int_1^a (f(x))^2 \, dx &= \frac{1}{3}a^3 \log^2 a - \frac{2}{9} a^2 \log a + \frac{2}{27} a^3 - \frac{2}{27} \\ \implies && \frac{dV}{da} = (f(a))^2 &= a^2 \log^2 a + \frac{2}{3}a^2 \log a - \frac{2}{3} a^2 \log a - \frac{2}{9}a^2 + \frac{2}{9}a^2 \\ &&&= a^2 \log^2 a \\[10pt] \implies && f(a) &= a \log a. \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 solid with given properties

by RoRi

Given a solid with circle base of radius 2 and cross sections which are equilateral triangles, compute the volume of the solid.

We may describe the top half of the circular base of the solid by the equation

    \[ f(x) = \sqrt{4-x^2}. \]

Thus, the length of the base of any equilateral triangular cross section is

    \[ 2 \sqrt{4-x^2} \qquad -2 \leq x \leq 2. \]

Since these are equilateral triangles with side length 2 \sqrt{4-x^2}, the area is given by

    \[ A(x) = \sqrt{3}(4-x^2). \]

Then we compute the volume,

    \begin{align*} V &= \int_{-2}^2 \sqrt{3}(4-x^2) \, dx \\ &= 4\sqrt{3}(4) - \sqrt{3} \left( \left. \frac{x^3}{3} \right|_{-2}^2 \right) \\ &= 16 \sqrt{3} - \frac{16}{3} \sqrt{3} \\ &= \frac{32 \sqrt{3}}{3}. \end{align*}

(Note: Apostol gives the solution \frac{16 \sqrt{3}}{3} in the back of the book, but I keep getting \frac{32 \sqrt{3}}{3}, as does Edwin in the comments. I’m marking this as an error in the book for now. If you see where my solution is wrong and Apostol is correct please leave a comment and let me know.)