The base of a three-dimensional figure is bound by the line y on the interval [0, 4]. Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 4. √y What is the volume of the figure? 3. 5+3NTRA 4 O 3- 2- 1 कृतल पंप -2 3 O V 1 2 3 4 5 6 7 o V= = OV = V = V V = & G w wo 3²3 65 65
The base of a three-dimensional figure is bound by the line y on the interval [0, 4]. Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 4. √y What is the volume of the figure? 3. 5+3NTRA 4 O 3- 2- 1 कृतल पंप -2 3 O V 1 2 3 4 5 6 7 o V= = OV = V = V V = & G w wo 3²3 65 65
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 4T
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![### Volume of a Solid with Rectangular Cross-Sections
The base of a three-dimensional figure is bound by the line \( x = \sqrt{y} \) on the interval \([0, 4]\). Vertical cross-sections that are perpendicular to the y-axis are rectangles with height equal to 4.
**Problem:** What is the volume of the figure?
#### Given Information:
- **Function describing the boundary:** \( x = \sqrt{y} \)
- **Interval for \( y \):** [0, 4]
- **Height of the rectangles:** 4
#### Graphical Representation:
There is a graph illustrating the scenario, which includes:
- **Axes:** The x-axis and y-axis are marked.
- **Curve:** The curve representing \( x = \sqrt{y} \) is plotted.
- **Shaded Region:** The area under the curve from \( y = 0 \) to \( y = 4 \) is shaded, indicating the base of the three-dimensional figure.
#### Mathematical Setup:
To find the volume, we can use the concept of integrating cross-sectional areas perpendicular to the y-axis over the given interval.
The width of each rectangle at a particular \( y \)-value is determined by the function \( x = \sqrt{y} \), and the height is given as 4.
Volume, \( V \), can thus be computed as:
\[ V = \int_{0}^{4} 4 \cdot \text{Width at y} \, dy \]
Since the width is \( 2x = 2\sqrt{y} \):
\[ V = \int_{0}^{4} 4 \cdot 2\sqrt{y} \, dy \]
\[ V = 8 \int_{0}^{4} \sqrt{y} \, dy \]
Solving the integral:
\[ \int_{0}^{4} y^{1/2} \, dy = \left[\frac{2}{3} y^{3/2}\right]_{0}^{4} \]
\[ \left[\frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2}\right] \]
\[ \frac{2}{3} (8) - 0 \]
\[ \frac{16}{3} \]
Multiplying by 8:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb111f9a2-d607-44d0-9cf1-983102308b09%2F30330041-45ad-46c2-b7f8-91f5aaaff605%2Fzrfks1n_processed.png&w=3840&q=75)
Transcribed Image Text:### Volume of a Solid with Rectangular Cross-Sections
The base of a three-dimensional figure is bound by the line \( x = \sqrt{y} \) on the interval \([0, 4]\). Vertical cross-sections that are perpendicular to the y-axis are rectangles with height equal to 4.
**Problem:** What is the volume of the figure?
#### Given Information:
- **Function describing the boundary:** \( x = \sqrt{y} \)
- **Interval for \( y \):** [0, 4]
- **Height of the rectangles:** 4
#### Graphical Representation:
There is a graph illustrating the scenario, which includes:
- **Axes:** The x-axis and y-axis are marked.
- **Curve:** The curve representing \( x = \sqrt{y} \) is plotted.
- **Shaded Region:** The area under the curve from \( y = 0 \) to \( y = 4 \) is shaded, indicating the base of the three-dimensional figure.
#### Mathematical Setup:
To find the volume, we can use the concept of integrating cross-sectional areas perpendicular to the y-axis over the given interval.
The width of each rectangle at a particular \( y \)-value is determined by the function \( x = \sqrt{y} \), and the height is given as 4.
Volume, \( V \), can thus be computed as:
\[ V = \int_{0}^{4} 4 \cdot \text{Width at y} \, dy \]
Since the width is \( 2x = 2\sqrt{y} \):
\[ V = \int_{0}^{4} 4 \cdot 2\sqrt{y} \, dy \]
\[ V = 8 \int_{0}^{4} \sqrt{y} \, dy \]
Solving the integral:
\[ \int_{0}^{4} y^{1/2} \, dy = \left[\frac{2}{3} y^{3/2}\right]_{0}^{4} \]
\[ \left[\frac{2}{3} (4)^{3/2} - \frac{2}{3} (0)^{3/2}\right] \]
\[ \frac{2}{3} (8) - 0 \]
\[ \frac{16}{3} \]
Multiplying by 8:
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