APEX Calculus

We find \(\fp(x)= \frac32x^{1/2}\text{;}\) note that on \([0,4]\text{,}\) \(f\) is differentiable and \(\fp\) is also continuous. Using the formula, we find the arc length \(L\) as

A graph of \(f\) is given in Figure 7.4.6.

This function was chosen specifically because the resulting integral can be evaluated exactly. We begin by finding \(\fp(x) = x/4-1/x\text{.}\) The arc length is

A graph of \(f\) is given in Figure 7.4.8; the portion of the curve measured in this problem is in bold.

This is somewhat of a mathematical curiosity; in Example 5.4.14 we found the area under one “hump” of the sine curve is 2 square units; now we are measuring its arc length.

The setup is straightforward: \(f(x) = \sin(x)\) and \(\fp(x) = \cos(x)\text{.}\) Thus

This integral cannot be evaluated in terms of elementary functions so we will approximate it with Simpson's Method with \(n=4\text{.}\)

Figure 7.4.10 gives \(\sqrt{1+\cos^2(x) }\) evaluated at 5 evenly spaced points in \([0,\pi]\text{.}\) Simpson's Rule then states that

Using a computer with \(n=100\) the approximation is \(L\approx 3.8202\text{;}\) our approximation with \(n=4\) is quite good.

The setup is relatively straightforward. Using Theorem 7.4.13, we have the surface area \(SA\) is:

The integration step above is nontrivial, utilizing the integration method of Trigonometric Substitution from Section 6.4.

It is interesting to see that the surface area of a solid, whose shape is defined by a trigonometric function, involves both a square root and an inverse hyperbolic trigonometric function.

1. The integral is straightforward to setup:

   \begin{align*} SA \amp = 2\pi\int_0^1 x^2\sqrt{1+(2x)^2}\, dx.\\ \end{align*}

   Like the integral in Example 7.4.14, this requires Trigonometric Substitution.

   \begin{align*} \amp = \left.\frac{\pi}{32}\left(2(8x^3+x)\sqrt{1+4x^2}-\sinh^{-1}(2x)\right)\right|_0^1\\ \amp =\frac{\pi}{32}\left(18\sqrt{5}-\sinh^{-1}(2) \right)\\ \amp \approx 3.81\,\text{units}^2\text{.} \end{align*}

   The solid formed by revolving \(y=x^2\) around the \(x\)-axis is graphed in Figure 7.4.17.(a).

2. Since we are revolving around the \(y\)-axis, the “radius” of the solid is not \(f(x)\) but rather \(x\text{.}\) Thus the integral to compute the surface area is:

   \begin{align*} SA \amp = 2\pi\int_0^1x\sqrt{1+(2x)^2}\, dx.\\ \end{align*}

   This integral can be solved using substitution. Set \(u=1+4x^2\text{;}\) the new bounds are \(u=1\) to \(u=5\text{.}\) We then have

   \begin{align*} \amp = \frac{\pi}4\int_1^5 \sqrt{u}\, du\\ \amp = \left.\frac{\pi}{4}\frac23 u^{3/2}\right|_1^5\\ \amp = \frac{\pi}6\left(5\sqrt{5}-1\right)\\ \amp \approx 5.33\,\text{units}^2\text{.} \end{align*}

   The solid formed by revolving \(y=x^2\) about the \(y\)-axis is graphed in Figure 7.4.17.(b).

To compute the volume it is natural to use the Disk Method. We have:

Gabriel's Horn has a finite volume of \(\pi\) cubic units. Since we have already seen that regions with infinite length can have a finite area, this is not too difficult to accept.

We now consider its surface area. The integral is straightforward to setup:

By Key Idea 6.8.17, the improper integral on the left diverges. Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel's Horn has infinite surface area.

Hence the “paradox”: we can fill Gabriel's Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it.

Somehow this paradox is striking when we think about it in terms of volume and area. However, we have seen a similar paradox before, as referenced above. We know that the area under the curve \(y=1/x^2\) on \([1,\infty)\) is finite, yet the shape has an infinite perimeter. Strange things can occur when we deal with the infinite.