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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### Problem Statement:
**4-** Find the volume of the solid obtained by rotating about the x-axis the region under the curve \( y = \sqrt{5x} \) from 0 to 4.
---
To solve this problem, we use the method of disks (also known as the method of washers). The volume \( V \) of the solid of revolution formed by rotating a function \( y = f(x) \) around the x-axis from \( x = a \) to \( x = b \) is given by the following integral:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
In this case, \( f(x) = \sqrt{5x} \), \( a = 0 \), and \( b = 4 \). Thus, the volume can be computed as:
\[ V = \pi \int_{0}^{4} (\sqrt{5x})^2 \, dx \]
Simplifying the integrand:
\[ V = \pi \int_{0}^{4} 5x \, dx \]
Next, we compute the integral:
\[ V = 5\pi \int_{0}^{4} x \, dx \]
\[ V = 5\pi \left[ \frac{x^2}{2} \right]_{0}^{4} \]
\[ V = 5\pi \left( \frac{4^2}{2} - \frac{0^2}{2} \right) \]
\[ V = 5\pi \left( \frac{16}{2} \right) \]
\[ V = 5\pi \times 8 \]
\[ V = 40\pi \]
Therefore, the volume of the solid obtained by rotating the region under the curve \( y = \sqrt{5x} \) from 0 to 4 about the x-axis is \( 40\pi \) cubic units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98ea0ed6-be6f-4556-80a5-46c7ae10c3b9%2F61268117-3cc2-4dbd-8017-b51024a1aa57%2Focmqane_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Problem Statement:
**4-** Find the volume of the solid obtained by rotating about the x-axis the region under the curve \( y = \sqrt{5x} \) from 0 to 4.
---
To solve this problem, we use the method of disks (also known as the method of washers). The volume \( V \) of the solid of revolution formed by rotating a function \( y = f(x) \) around the x-axis from \( x = a \) to \( x = b \) is given by the following integral:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
In this case, \( f(x) = \sqrt{5x} \), \( a = 0 \), and \( b = 4 \). Thus, the volume can be computed as:
\[ V = \pi \int_{0}^{4} (\sqrt{5x})^2 \, dx \]
Simplifying the integrand:
\[ V = \pi \int_{0}^{4} 5x \, dx \]
Next, we compute the integral:
\[ V = 5\pi \int_{0}^{4} x \, dx \]
\[ V = 5\pi \left[ \frac{x^2}{2} \right]_{0}^{4} \]
\[ V = 5\pi \left( \frac{4^2}{2} - \frac{0^2}{2} \right) \]
\[ V = 5\pi \left( \frac{16}{2} \right) \]
\[ V = 5\pi \times 8 \]
\[ V = 40\pi \]
Therefore, the volume of the solid obtained by rotating the region under the curve \( y = \sqrt{5x} \) from 0 to 4 about the x-axis is \( 40\pi \) cubic units.
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