1 S = 1 + dx y dx = 2n ---Select--- v 1 + 1 dx x + 1 00 = 2n dx ---Select--- v Rather than trying to evaluate this integral, note the following. V x4 + 1 ? v Vx4 = x for x > 0 Thus, if the area is finite, we have the following. x4 + 1 o y2 dx = 2n S = 2n dx ? v 2n dx /1 Ji But we know that this integral ---Select--- v so the area S is infinite. 118
1 S = 1 + dx y dx = 2n ---Select--- v 1 + 1 dx x + 1 00 = 2n dx ---Select--- v Rather than trying to evaluate this integral, note the following. V x4 + 1 ? v Vx4 = x for x > 0 Thus, if the area is finite, we have the following. x4 + 1 o y2 dx = 2n S = 2n dx ? v 2n dx /1 Ji But we know that this integral ---Select--- v so the area S is infinite. 118
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![The surface formed by rotating the curve \( y = \frac{1}{x} \), \( x \geq 1 \), about the x-axis is known as Gabriel's horn. Show that the surface area is infinite (although the enclosed volume is finite).
**Diagram Description:**
The image shows a 3D representation of Gabriel's horn, which is a shape formed by rotating the curve \( y = \frac{1}{x} \) around the x-axis. The horn extends infinitely to the right and flares outward from the point \( x = 1 \) as it approaches the y-axis, creating an infinitely long, yet finite-volume shape.
**Mathematical Explanation:**
\[ S = \left( \, \int_1^{\infty} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \right) \]
\[ = 2\pi \int_1^{\infty} \left( \frac{1}{x} \right) \sqrt{1 + \frac{1}{x^4}} \, dx \]
\[ = 2\pi \int_1^{\infty} \frac{\sqrt{x^4 + 1}}{x^3} \, dx \]
**Note:**
Rather than trying to evaluate this integral, note the following:
\[ \sqrt{x^4 + 1} \geq \sqrt{x^4} = x^2 \quad \text{for } x > 0 \]
This inequality indicates that the integrand behaves in a manner that leads to an infinite surface area as \( x \) approaches infinity, while the volume remains finite.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff01e8a57-3c53-4370-8dfc-cbadc7875ad2%2F5780d981-ee23-4721-ad04-cca1addd7a3f%2Fixz8ttf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The surface formed by rotating the curve \( y = \frac{1}{x} \), \( x \geq 1 \), about the x-axis is known as Gabriel's horn. Show that the surface area is infinite (although the enclosed volume is finite).
**Diagram Description:**
The image shows a 3D representation of Gabriel's horn, which is a shape formed by rotating the curve \( y = \frac{1}{x} \) around the x-axis. The horn extends infinitely to the right and flares outward from the point \( x = 1 \) as it approaches the y-axis, creating an infinitely long, yet finite-volume shape.
**Mathematical Explanation:**
\[ S = \left( \, \int_1^{\infty} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \right) \]
\[ = 2\pi \int_1^{\infty} \left( \frac{1}{x} \right) \sqrt{1 + \frac{1}{x^4}} \, dx \]
\[ = 2\pi \int_1^{\infty} \frac{\sqrt{x^4 + 1}}{x^3} \, dx \]
**Note:**
Rather than trying to evaluate this integral, note the following:
\[ \sqrt{x^4 + 1} \geq \sqrt{x^4} = x^2 \quad \text{for } x > 0 \]
This inequality indicates that the integrand behaves in a manner that leads to an infinite surface area as \( x \) approaches infinity, while the volume remains finite.
![### Understanding Solid of Revolution: The Gabriel's Horn
The image illustrates a solid of revolution known as Gabriel's Horn, formed by rotating the curve \( y = \frac{1}{x} \) around the x-axis for \( x \geq 1 \).
#### Surface Area Calculation
To determine the surface area \( S \) of this solid, we use the formula:
\[
S = 2\pi \int_{1}^{\infty} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
After substituting \( y = \frac{1}{x} \) and simplifying, we have:
\[
S = 2\pi \int_{1}^{\infty} \left( \sqrt{1 + \frac{1}{x^4}} \right) \, dx
\]
This transforms into:
\[
S = 2\pi \int_{1}^{\infty} \frac{\sqrt{x^4 + 1}}{x^3} \, dx
\]
#### Integral Approximation
Rather than evaluating this complicated integral directly, we approximate:
\[
\sqrt{x^4 + 1} \approx x^2 \quad \text{for} \quad x > 0
\]
This approximation leads to:
\[
S \approx 2\pi \int_{1}^{\infty} \frac{x^2}{x^3} \, dx = 2\pi \int_{1}^{\infty} \frac{1}{x} \, dx
\]
#### Conclusion
The integral:
\[
\int_{1}^{\infty} \frac{1}{x} \, dx
\]
is known to be divergent, implying that the surface area \( S \) of Gabriel's Horn is infinite. Despite its finite volume, the infinite surface area poses an interesting paradox.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff01e8a57-3c53-4370-8dfc-cbadc7875ad2%2F5780d981-ee23-4721-ad04-cca1addd7a3f%2F8a2g1j8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Solid of Revolution: The Gabriel's Horn
The image illustrates a solid of revolution known as Gabriel's Horn, formed by rotating the curve \( y = \frac{1}{x} \) around the x-axis for \( x \geq 1 \).
#### Surface Area Calculation
To determine the surface area \( S \) of this solid, we use the formula:
\[
S = 2\pi \int_{1}^{\infty} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
After substituting \( y = \frac{1}{x} \) and simplifying, we have:
\[
S = 2\pi \int_{1}^{\infty} \left( \sqrt{1 + \frac{1}{x^4}} \right) \, dx
\]
This transforms into:
\[
S = 2\pi \int_{1}^{\infty} \frac{\sqrt{x^4 + 1}}{x^3} \, dx
\]
#### Integral Approximation
Rather than evaluating this complicated integral directly, we approximate:
\[
\sqrt{x^4 + 1} \approx x^2 \quad \text{for} \quad x > 0
\]
This approximation leads to:
\[
S \approx 2\pi \int_{1}^{\infty} \frac{x^2}{x^3} \, dx = 2\pi \int_{1}^{\infty} \frac{1}{x} \, dx
\]
#### Conclusion
The integral:
\[
\int_{1}^{\infty} \frac{1}{x} \, dx
\]
is known to be divergent, implying that the surface area \( S \) of Gabriel's Horn is infinite. Despite its finite volume, the infinite surface area poses an interesting paradox.
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