There is a short cut to this volume calculation

Geometry Level 4

A line segment is given by $\mathbf{p}(t) = \mathbf{p}_1 (1 - t) + \mathbf{p}_2 t , \hspace{6pt} t \in [0, 1]$ , where $\mathbf{p}_1 = (1, -3, -4)$ and $\mathbf{p}_2 = (1, 3, 4)$ . The line segment is rotated about the $z$ -axis generating a truncated hyperboloid of one sheet. Find the volume of this hyperboloid between $z =-4$ and $z = 4$ .

If the volume can be expressed as $n \pi$ , then enter $n$ as your answer.

The answer is 32.

1 solution

Hosam Hajjir
Jul 23, 2019

The volume can be found by direct evaluation of the $z$ -cross-sectional area, then integrating it between $z = -4$ and $z = 4$ . But there is a short cut which follows because the lateral surface is a ruled surface, meaning that for every point on it there is a straight line segment that lies on that surface. And this means we can use the following formula:

$V = \dfrac{h}{6} ( B_1 + B_2 + 4 B_3 )$

where $B_1$ and $B_2$ are the areas of the top and bottom bases, and $B_3$ is the area of the cross section half way between the two-bases. In our case, $B_1 = B_2 = (1^2 + 3^2) \pi = 10 \pi$ and $B_3 = (1)^2 \pi = \pi$ , and $h = 8$ .

Hence,

$V = \dfrac{8}{6} ( 24 \pi ) = 32 \pi$

Therefore the answer is $\boxed{ 32 }$

Nice problem and solution! This website gives the volume of a one-sheeted hyperboloid as $V = \frac{1}{3}\pi h(2a^2 + r^2)$ , where $h$ is the height, $a$ is the radius at the center of the hyperboloid, and $r$ is the radius at the top and bottom of the hyperboloid, which is basically the same formula you used if $B_1 = B_2$ . (In this problem, $a^2 = 1$ , $r^2 = 1^2 + 3^2 = 10$ , and $h = 8$ .)

David Vreken - 1 year, 10 months ago

Thanks for the information.

Hosam Hajjir - 1 year, 10 months ago

There is a short cut to this volume calculation

The answer is 32.

1 solution

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