Errata Archives - Page 2 of 2

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.)

For $p \in \mathbb{Z}_{>0}$ define a step function on the interval $[0,p]$ by

$s(x) = \begin{cases} (-1)^n \cdot n & \text{if } n \leq x < n+1, \ n=0,1, \ldots, p-1 \\ 0 & \text{if } x =p. \end{cases}$

Then, define

$f(p) = \int_0^p s(x) \, dx.$

Calculate $f(3), f(4), f(f(3))$ .
Find all values of $p$ such that $|f(p)| = 7$ .

We calculuate:
$\begin{align*} f(3) &= \int_0^3 s(x) \, dx = (-1)^0 0 (1-0) + (-1)^1 (1) (2-1) + (-1)^2 (2)(3-2) = 0 - 1 + 2 &= \phantom{-}1.\\ f(4) &= \int_0^4 s(x) \, dx = \int_0^3 s(x) \, dx + \int_3^4 s(x) \, dx = 1 + (-1)^3 (3) (4-3) &= -2. \\ f(f(3)) &= f(1) = \int_0^1 s(x) \, dx = (-1)^0 0 (1-0) &= \phantom{-}0. \end{align*}$
$p = 14,15$ .

Note: There is an error in the book. The answers in the back of the book claim that $f(4) = -1$ , which is incorrect.

Stumbling Robot

Tag: Errata

Compute the volume of a solid with given properties

Calculate the values of an integral of a step function