a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve. c. Use technology to find the surface area numerically. xy = 3, 2

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...
Question
**Problem Statement:**

a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.

b. Graph the curve.

c. Use technology to find the surface area numerically.

Given:
\[ xy = 3, \quad 2 \leq y \leq 8; \quad y\text{-axis} \]

---

**Instructions:**

a. **Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer.)**

- Choice A: \[ S = \int_{2}^{8} \ \square \ dx \]
  
- Choice B: \[ S = \int_{2}^{8} \ \square \ dy \]

b. **Choose the correct answer below.**

- Graph A: A curve on the x-y plane with coordinates starting around (0,10) and moving towards (1,0).
- Graph B: A curve on the x-y plane with coordinates starting around (0,2) and moving towards (10,0).
- Graph C: A curve on the x-y plane with coordinates starting around (2,-10) and moving towards (0,-2).
- Graph D: A curve on the x-y plane with coordinates starting around (-2,0) and moving upwards to (0,10).

c. **The area of the surface is \(\square\) square units. (Do not round until the final answer. Then round to the nearest hundredth as needed.)**

---

**Graph Descriptions:**

- **Graph A** shows a decreasing function, which starts at a higher y-value and approaches the x-axis.
- **Graph B** depicts a downward-sloping curve, starting at a low x-value and stretching out towards a higher x-value.
- **Graph C** features an inverted curve in the negative region, arching upwards.
- **Graph D** illustrates a curve starting from a negative x-value and moving upwards towards the origin.
Transcribed Image Text:**Problem Statement:** a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve. c. Use technology to find the surface area numerically. Given: \[ xy = 3, \quad 2 \leq y \leq 8; \quad y\text{-axis} \] --- **Instructions:** a. **Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer.)** - Choice A: \[ S = \int_{2}^{8} \ \square \ dx \] - Choice B: \[ S = \int_{2}^{8} \ \square \ dy \] b. **Choose the correct answer below.** - Graph A: A curve on the x-y plane with coordinates starting around (0,10) and moving towards (1,0). - Graph B: A curve on the x-y plane with coordinates starting around (0,2) and moving towards (10,0). - Graph C: A curve on the x-y plane with coordinates starting around (2,-10) and moving towards (0,-2). - Graph D: A curve on the x-y plane with coordinates starting around (-2,0) and moving upwards to (0,10). c. **The area of the surface is \(\square\) square units. (Do not round until the final answer. Then round to the nearest hundredth as needed.)** --- **Graph Descriptions:** - **Graph A** shows a decreasing function, which starts at a higher y-value and approaches the x-axis. - **Graph B** depicts a downward-sloping curve, starting at a low x-value and stretching out towards a higher x-value. - **Graph C** features an inverted curve in the negative region, arching upwards. - **Graph D** illustrates a curve starting from a negative x-value and moving upwards towards the origin.
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