"Gabriel's Horn" - A 3-D cone formed by rotating a curve

[brotherbobby]

Homework Statement

Consider the three dimensional cone formed by rotating the graph of $f(x) = \frac{1}{x}$ for $x \ge 1$ about the $x$ axis. The graph of $f(x)$ is shown below.

Calculate :

(a) The $\mathbf{\text{volume}}$ of the cone,

(b) The $\mathbf{\text{surface area}}$ of the cone. $\\[10pt]$

Relevant Equations

1. Let $f$ be continuous on $[a,b]$, and let $R$ be the region bounded by the graph of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$. The $\textbf{volume}$ $V$ of the solid of revolution generated by revolving $R$ about the $x$-axis is $$V = \pi \int_a^b [f(x)]^2 dx$$

2. If $f$ is smooth and $f(x)\ge 0$ on $[a,b]$ , then the $\textbf{area S of the surface}$ generated by revolving the graph of $f$ about the $x$-axis is $$ S = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} dx$$ $\\[10pt]$

Problem statement : We have the graph of the function $f(x)$ shown to the right. The function $f(x) = \frac{1}{x}$ and the domain of $x \in [1,\infty)$. We have to find the volume and surface area of the 3-D "cone" formed by rotating the function about the $x$ axis. $\\[10pt]$Attempt : I find the volume of the surface of revolution first.

1. Volume : Using Relevant Equation 1 above, we have $$V = \pi \int_a^b [f(x)]^2 dx = \pi \int_1^{\infty} \frac{1}{x^2} dx = \pi \left[\frac{1}{x}\right]_{\infty}^1 \Rightarrow \; \boxed{\mathbf{V = \pi}}$$

In the website, there is a multiple choice list with the correct answer. I have selected mine as $\mathbf{\pi}$ but I need to complete the second part of the problem for it to let me know whether am right or wrong on both. The two parts come as one whole problem.
2. Surface area : (This is where I am in trouble)

Using the Relavant Equation 2 above, we have $$S = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} dx $$
Now since $f(x) = \frac{1}{x}\Rightarrow f'(x) = - \frac{1}{x^2}$, the above simplifies to,
$$S = 2\pi \int_1^{\infty} \frac{1}{x} \sqrt{1+\frac{1}{x^4}}dx = 2\pi \int_1^{\infty} \frac{\sqrt{x^4+1}}{x^3} dx$$

Let's change variables from $x\rightarrow \theta : x^2 = \tan\theta\Rightarrow 2xdx = \sec^2\theta d\theta$. Rearranging variables and limits, we have $$S = \cancel{2}\pi\int_{\pi/4}^{\pi/2} \frac{\sec \theta}{x^3} \frac{\sec^2 \theta d\theta}{\cancel{2}x} = \pi \int_{\pi/4}^{\pi/2} \frac{\sec^3 \theta d\theta}{\tan^2 \theta} = \pi \int_{\pi/4}^{\pi/2} \frac{(1+\tan^2 \theta)\sec\theta}{\tan^2\theta}d\theta = \pi \left[ \int_{\pi/4}^{\pi/2}\frac{\sec\theta}{\tan^2 \theta} + \int_{\pi/4}^{\pi/2} \sec\theta \right]$$
Multiplying the top and bottom of the first term with $\cos^2\theta$, we have $$S = \pi \int_{\pi/4}^{\pi/2} \frac{\cos\theta}{\sin^2\theta}d\theta + \pi \int_{\pi/4}^{\pi/2} \sec\theta d\theta$$
Let's call the first integral as $$I = \pi \int_{\pi/4}^{\pi/2} \frac{\cos\theta}{\sin^2\theta}d\theta$$ Effecting a change of variables from $\theta \rightarrow z : z = \sin\theta \Rightarrow dz = \cos\theta d\theta$, we have $$I = \pi \int_{1/\sqrt{2}}^{1} \frac{dz}{z^2} = \pi \left[ \frac{1}{z}\right]_{1}^{1/\sqrt{2}} = \pi(\sqrt 2-1)$$

The second integral is : $$\pi \int_{\pi/4}^{\pi/2} \sec\theta d\theta = \ln{|\sec\theta+\tan\theta|} _{\pi/4}^{\pi/2}$$
which is divergent for the upper limit of $\pi/2$

These are the allowed answers on the website - none of which match mine.A help or hint would be welcome.

 

[Mark44]

Homework Statement:: Consider the three dimensional cone formed by rotating the graph of $f(x) = \frac{1}{x}$ for $x \ge 1$ about the $x$ axis. The graph of $f(x)$ is shown below.

Calculate :

(a) The $\mathbf{\text{volume}}$ of the cone,

(b) The $\mathbf{\text{surface area}}$ of the cone. $\\[10pt]$
Relevant Equations:: 1. Let $f$ be continuous on $[a,b]$, and let $R$ be the region bounded by the graph of $f$, the $x$-axis, and the vertical lines $x = a$ and $x = b$. The $\textbf{volume}$ $V$ of the solid of revolution generated by revolving $R$ about the $x$-axis is $$V = \pi \int_a^b [f(x)]^2 dx$$

2. If $f$ is smooth and $f(x)\ge 0$ on $[a,b]$ , then the $\textbf{area S of the surface}$ generated by revolving the graph of $f$ about the $x$-axis is $$ S = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} dx$$ $\\[10pt]$

View attachment 283119Problem statement : We have the graph of the function $f(x)$ shown to the right. The function $f(x) = \frac{1}{x}$ and the domain of $x \in [1,\infty)$. We have to find the volume and surface area of the 3-D "cone" formed by rotating the function about the $x$ axis. $\\[10pt]$Attempt : I find the volume of the surface of revolution first.

1. Volume : Using Relevant Equation 1 above, we have $$V = \pi \int_a^b [f(x)]^2 dx = \pi \int_1^{\infty} \frac{1}{x^2} dx = \pi \left[\frac{1}{x}\right]_{\infty}^1 \Rightarrow \; \boxed{\mathbf{V = \pi}}$$View attachment 283121In the website, there is a multiple choice list with the correct answer. I have selected mine as $\mathbf{\pi}$ but I need to complete the second part of the problem for it to let me know whether am right or wrong on both. The two parts come as one whole problem.
2. Surface area : (This is where I am in trouble)

Using the Relavant Equation 2 above, we have $$S = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} dx $$
Now since $f(x) = \frac{1}{x}\Rightarrow f'(x) = - \frac{1}{x^2}$, the above simplifies to,
$$S = 2\pi \int_1^{\infty} \frac{1}{x} \sqrt{1+\frac{1}{x^4}}dx = 2\pi \int_1^{\infty} \frac{\sqrt{x^4+1}}{x^3} dx$$

Let's change variables from $x\rightarrow \theta : x^2 = \tan\theta\Rightarrow 2xdx = \sec^2\theta d\theta$. Rearranging variables and limits, we have $$S = \cancel{2}\pi\int_{\pi/4}^{\pi/2} \frac{\sec \theta}{x^3} \frac{\sec^2 \theta d\theta}{\cancel{2}x} = \pi \int_{\pi/4}^{\pi/2} \frac{\sec^3 \theta d\theta}{\tan^2 \theta} = \pi \int_{\pi/4}^{\pi/2} \frac{(1+\tan^2 \theta)\sec\theta}{\tan^2\theta}d\theta = \pi \left[ \int_{\pi/4}^{\pi/2}\frac{\sec\theta}{\tan^2 \theta} + \int_{\pi/4}^{\pi/2} \sec\theta \right]$$
Multiplying the top and bottom of the first term with $\cos^2\theta$, we have $$S = \pi \int_{\pi/4}^{\pi/2} \frac{\cos\theta}{\sin^2\theta}d\theta + \pi \int_{\pi/4}^{\pi/2} \sec\theta d\theta$$
Let's call the first integral as $$I = \pi \int_{\pi/4}^{\pi/2} \frac{\cos\theta}{\sin^2\theta}d\theta$$ Effecting a change of variables from $\theta \rightarrow z : z = \sin\theta \Rightarrow dz = \cos\theta d\theta$, we have $$I = \pi \int_{1/\sqrt{2}}^{1} \frac{dz}{z^2} = \pi \left[ \frac{1}{z}\right]_{1}^{1/\sqrt{2}} = \pi(\sqrt 2-1)$$
View attachment 283125
The second integral is : $$\pi \int_{\pi/4}^{\pi/2} \sec\theta d\theta = \ln{|\sec\theta+\tan\theta|} _{\pi/4}^{\pi/2}$$
which is divergent for the upper limit of $\pi/2$

These are the allowed answers on the website - none of which match mine.A help or hint would be welcome.

These questions are well beyond precalculus, so I have moved the thread to the appropriate forum section.

For the second integral (surface area), saying that it is divergent isn't as meaningful as saying that its value is infinite.

This question commonly appears in calculus textbooks. Its purpose is to show the "paradox" of a solid figure with finite volume, but infinite surface area. Intuitively, one might think that the horn could be filled with a finite volume of paint, but that amount of paint would not suffice to cover the horn with paint, either the inside or the outside.

 

[brotherbobby]

These questions are well beyond precalculus, so I have moved the thread to the appropriate forum section.

For the second integral (surface area), saying that it is divergent isn't as meaningful as saying that its value is infinite.

This question commonly appears in calculus textbooks. Its purpose is to show the "paradox" of a solid figure with finite volume, but infinite surface area. Intuitively, one might think that the horn could be filled with a finite volume of paint, but that amount of paint would not suffice to cover the horn with paint, either the inside or the outside.

Yes @Mark44 thank you. It has been a learning. I was puzzled about its finite volume but infinite surface area. It is certainly non-intuitive.

 

[LCKurtz]

Of course, if you use KNMP paint, there is no paradox.

Definition: The xy plane is painted if for any $r > 0$, the circular area at the origin of radius $r$ is covered.

Theorem: A single drop of KNMP paint is sufficient to paint the $xy$ plane.
Proof: Suppose a drop of KNMP paint has volume $V$. A drop is placed on the horizontal $xy$ plane at the origin and allowed to spread. Given $r>0$, when the drop has spread to cover the circle, it will have thickness $T = \frac V {\pi r^2} > 0$, meaning it hasn't finished spreading. Consequently the plane is painted. Q.E.D.

Note: You don't see this demonstrated very often because KNMP paint is expensive and hard to get.

Note: KNMP is the brand name for "Kurtz's Non-Molecular Paint".