9.3 To estimate the surface area and volume of a wine bottle, the radius of the bottle is measured at different heights. The surface area, S, and volume, V, can be determined by: S = 2n["rdz and v = n["r²dz Use the data given below to determine the volume and surface area of the vase: 0 2 4 6 16 z (cm) r (cm) 10 z (cm) 20 r (cm) 8.9 4.7 10 11.9 12.4 13 13.5 13.8 14.1 13.6 12.1 8 12 14 18 11 22 24 26 28 30 32 34 36 4.1 3.5 3.0 2.4 1.9 1.2 1.0 (a) Use the composite rectangle method. (b) Use the composite trapezoidal method. (c) Use the composite Simpson's 3/8 method.
9.3 To estimate the surface area and volume of a wine bottle, the radius of the bottle is measured at different heights. The surface area, S, and volume, V, can be determined by: S = 2n["rdz and v = n["r²dz Use the data given below to determine the volume and surface area of the vase: 0 2 4 6 16 z (cm) r (cm) 10 z (cm) 20 r (cm) 8.9 4.7 10 11.9 12.4 13 13.5 13.8 14.1 13.6 12.1 8 12 14 18 11 22 24 26 28 30 32 34 36 4.1 3.5 3.0 2.4 1.9 1.2 1.0 (a) Use the composite rectangle method. (b) Use the composite trapezoidal method. (c) Use the composite Simpson's 3/8 method.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please see attached image. I only need parts (b) and (c). Thank you!
![**9.3 Estimating Surface Area and Volume of a Wine Bottle**
To estimate the surface area and volume of a wine bottle, the radius of the bottle is measured at different heights. The surface area, \( S \), and volume, \( V \), can be determined by the following integrals:
\[ S = 2\pi \int_{0}^{L} r \, dz \]
\[ V = \pi \int_{0}^{L} r^2 \, dz \]
To determine the volume and surface area of the vase, use the data given below:
| \( z \) (cm) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
|--------------|----|----|----|----|----|----|----|----|----|----|
| \( r \) (cm) | 10 | 11 | 11.9 | 12.4 | 13 | 13.5 | 13.8 | 14.1 | 13.6 | 12.1 |
| \( z \) (cm) | 20 | 22 | 24 | 26 | 28 | 30 | 32 | 34 | 36 | |
|--------------|----|----|----|----|----|----|----|----|----|----|
| \( r \) (cm) | 8.9 | 4.7 | 4.1 | 3.5 | 3.0 | 2.4 | 1.9 | 1.2 | 1.0 | |
**Methods to Use:**
- (a) Use the composite rectangle method.
- (b) Use the composite trapezoidal method.
- (c) Use the composite Simpson’s 3/8 method.
**Diagram Explanation:**
The diagram on the right is a cross-sectional view of the wine bottle, illustrating the varying radius \( r \) as a function of height \( z \). The diagram shows that the bottle has a wider middle section and tapers off towards the top and bottom. The axes are labeled to correspond with the \( z \) and \( r \) dimensions as provided in the table.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2Fed9b86e6-5392-4dff-9625-8415eb7eefe0%2F1fa0ium_processed.png&w=3840&q=75)
Transcribed Image Text:**9.3 Estimating Surface Area and Volume of a Wine Bottle**
To estimate the surface area and volume of a wine bottle, the radius of the bottle is measured at different heights. The surface area, \( S \), and volume, \( V \), can be determined by the following integrals:
\[ S = 2\pi \int_{0}^{L} r \, dz \]
\[ V = \pi \int_{0}^{L} r^2 \, dz \]
To determine the volume and surface area of the vase, use the data given below:
| \( z \) (cm) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
|--------------|----|----|----|----|----|----|----|----|----|----|
| \( r \) (cm) | 10 | 11 | 11.9 | 12.4 | 13 | 13.5 | 13.8 | 14.1 | 13.6 | 12.1 |
| \( z \) (cm) | 20 | 22 | 24 | 26 | 28 | 30 | 32 | 34 | 36 | |
|--------------|----|----|----|----|----|----|----|----|----|----|
| \( r \) (cm) | 8.9 | 4.7 | 4.1 | 3.5 | 3.0 | 2.4 | 1.9 | 1.2 | 1.0 | |
**Methods to Use:**
- (a) Use the composite rectangle method.
- (b) Use the composite trapezoidal method.
- (c) Use the composite Simpson’s 3/8 method.
**Diagram Explanation:**
The diagram on the right is a cross-sectional view of the wine bottle, illustrating the varying radius \( r \) as a function of height \( z \). The diagram shows that the bottle has a wider middle section and tapers off towards the top and bottom. The axes are labeled to correspond with the \( z \) and \( r \) dimensions as provided in the table.
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