A region bounded by f(y) = √√√y, y = 4, and x = 0 is shown. What is the volume of the solid formed by revolving the region about the y-axis? 8 7 6 5 4 3 2 1 -2 -11 1 2 3 4 5 6 7 8 ○ 24π 32 3 T 26 3 8π π X
A region bounded by f(y) = √√√y, y = 4, and x = 0 is shown. What is the volume of the solid formed by revolving the region about the y-axis? 8 7 6 5 4 3 2 1 -2 -11 1 2 3 4 5 6 7 8 ○ 24π 32 3 T 26 3 8π π X
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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![**Calculating the Volume of a Solid of Revolution**
**Problem Statement:**
A region bounded by the function \( f(y) = \sqrt{y} \), \( y = 4 \), and \( x = 0 \) is shown. What is the volume of the solid formed by revolving the region about the y-axis?
**Diagram Explanation:**
The provided graph is a coordinate plane with axes labeled as \(X\) and \(Y\). A curve representing the function \( f(y) = \sqrt{y} \) is drawn, starting from \( x = 0 \) and increasing as \( y \) increases. The shaded region is bounded by \( x = 0 \) (y-axis), the curve \( f(y) = \sqrt{y} \), and the line \( y = 4 \). The area of interest is highlighted with diagonal lines.
**Solution Options:**
- \( 24\pi \)
- \( \dfrac{32\pi}{3} \)
- \( \dfrac{26\pi}{3} \)
- \( 8\pi \)
To find the volume of the solid of revolution when the region is revolved about the y-axis, we use the formula for the volume of a solid of revolution:
\[ V = \pi \int_{a}^{b} (R(y))^2 \, dy \]
Here the function \( f(y) = \sqrt{y} \) is our radius \( R(y) = \sqrt{y} \), and the limits of integration are from \( y = 0 \) to \( y = 4 \).
\[ V = \pi \int_{0}^{4} (\sqrt{y})^2 \, dy \]
\[ V = \pi \int_{0}^{4} y \, dy \]
We now evaluate this integral:
\[ V = \pi \left[ \frac{y^2}{2} \right]_{0}^{4} \]
\[ V = \pi \left( \frac{4^2}{2} - \frac{0^2}{2} \right) \]
\[ V = \pi \left( \frac{16}{2} \right) \]
\[ V = \pi \times 8 \]
\[ V = 8\pi \]
Thus, the volume of the solid formed by revolving the region about](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6539f692-5656-46fd-b828-f0fe9bb331a2%2Fc262b989-fd7d-4a2d-a409-85854ca23d7d%2Fj08brv9_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the Volume of a Solid of Revolution**
**Problem Statement:**
A region bounded by the function \( f(y) = \sqrt{y} \), \( y = 4 \), and \( x = 0 \) is shown. What is the volume of the solid formed by revolving the region about the y-axis?
**Diagram Explanation:**
The provided graph is a coordinate plane with axes labeled as \(X\) and \(Y\). A curve representing the function \( f(y) = \sqrt{y} \) is drawn, starting from \( x = 0 \) and increasing as \( y \) increases. The shaded region is bounded by \( x = 0 \) (y-axis), the curve \( f(y) = \sqrt{y} \), and the line \( y = 4 \). The area of interest is highlighted with diagonal lines.
**Solution Options:**
- \( 24\pi \)
- \( \dfrac{32\pi}{3} \)
- \( \dfrac{26\pi}{3} \)
- \( 8\pi \)
To find the volume of the solid of revolution when the region is revolved about the y-axis, we use the formula for the volume of a solid of revolution:
\[ V = \pi \int_{a}^{b} (R(y))^2 \, dy \]
Here the function \( f(y) = \sqrt{y} \) is our radius \( R(y) = \sqrt{y} \), and the limits of integration are from \( y = 0 \) to \( y = 4 \).
\[ V = \pi \int_{0}^{4} (\sqrt{y})^2 \, dy \]
\[ V = \pi \int_{0}^{4} y \, dy \]
We now evaluate this integral:
\[ V = \pi \left[ \frac{y^2}{2} \right]_{0}^{4} \]
\[ V = \pi \left( \frac{4^2}{2} - \frac{0^2}{2} \right) \]
\[ V = \pi \left( \frac{16}{2} \right) \]
\[ V = \pi \times 8 \]
\[ V = 8\pi \]
Thus, the volume of the solid formed by revolving the region about
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