The given curve is rotated about the x-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y. x = In(4y + 1), 0≤ y ≤ 1 (a) Integrate with respect to x. In (5) 27 (In ( 4y + 1)) {√ + 1)) ( √ ¹ + ( + ² + 1)² ) ) ** )dx 1+ 4y (b) Integrate with respect to y. SC X ) dy
The given curve is rotated about the x-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y. x = In(4y + 1), 0≤ y ≤ 1 (a) Integrate with respect to x. In (5) 27 (In ( 4y + 1)) {√ + 1)) ( √ ¹ + ( + ² + 1)² ) ) ** )dx 1+ 4y (b) Integrate with respect to y. SC X ) dy
Calculus: Early Transcendentals
8th Edition
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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![**Surface Area of a Curve Rotated About the x-axis**
In this example, we consider a given curve rotated about the x-axis and setup, but do not evaluate, an integral for the area of the resulting surface. The curve is defined by the equation \( x = \ln(4y + 1) \) for \( 0 \leq y \leq 1 \).
### (a) Integration with Respect to \( x \)
The setup for integrating with respect to \( x \) involves determining the surface area of revolution around the x-axis. The integral can be expressed as:
\[
\int_{0}^{\ln(5)} \left( 2\pi (\ln(4y + 1)) \sqrt{1 + \left(\frac{4}{4y+1}\right)^2}\right) dx
\]
Here:
- The limits of integration are from \( x = 0 \) to \( x = \ln(5) \), corresponding to the given range of \( y \) values.
- The integrand consists of \( 2\pi (\ln(4y + 1)) \) multiplied by the square root of \( 1 + \left(\frac{4}{4y + 1}\right)^2 \).
### (b) Integration with Respect to \( y \)
The setup for integrating with respect to \( y \) to find the surface area is shown, but incomplete in the image. The complete integral should integrate from \( y = 0 \) to \( y = 1 \):
\[
\int_{0}^{1} \left( 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \right) dy
\]
Here:
- The limits of integration are from \( y = 0 \) to \( y = 1 \).
- The integrand is \( 2\pi x \) multiplied by the square root of \( 1 + \left(\frac{dx}{dy}\right)^2 \).
For the given curve:
\[
x = \ln(4y + 1)
\]
we find \( \frac{dx}{dy} = \frac{4}{4y + 1} \), so the surface area integral becomes:
\[
\int_{0}^{1} \left( 2\pi \ln(4y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faea7a391-77a2-4be9-b3a3-98e0f70bf810%2Feb7dfe9f-5325-47b5-bbc6-01725ce2fbc6%2Fokc0br_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Surface Area of a Curve Rotated About the x-axis**
In this example, we consider a given curve rotated about the x-axis and setup, but do not evaluate, an integral for the area of the resulting surface. The curve is defined by the equation \( x = \ln(4y + 1) \) for \( 0 \leq y \leq 1 \).
### (a) Integration with Respect to \( x \)
The setup for integrating with respect to \( x \) involves determining the surface area of revolution around the x-axis. The integral can be expressed as:
\[
\int_{0}^{\ln(5)} \left( 2\pi (\ln(4y + 1)) \sqrt{1 + \left(\frac{4}{4y+1}\right)^2}\right) dx
\]
Here:
- The limits of integration are from \( x = 0 \) to \( x = \ln(5) \), corresponding to the given range of \( y \) values.
- The integrand consists of \( 2\pi (\ln(4y + 1)) \) multiplied by the square root of \( 1 + \left(\frac{4}{4y + 1}\right)^2 \).
### (b) Integration with Respect to \( y \)
The setup for integrating with respect to \( y \) to find the surface area is shown, but incomplete in the image. The complete integral should integrate from \( y = 0 \) to \( y = 1 \):
\[
\int_{0}^{1} \left( 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \right) dy
\]
Here:
- The limits of integration are from \( y = 0 \) to \( y = 1 \).
- The integrand is \( 2\pi x \) multiplied by the square root of \( 1 + \left(\frac{dx}{dy}\right)^2 \).
For the given curve:
\[
x = \ln(4y + 1)
\]
we find \( \frac{dx}{dy} = \frac{4}{4y + 1} \), so the surface area integral becomes:
\[
\int_{0}^{1} \left( 2\pi \ln(4y
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