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Find the surface area of the portion of the part of the paraboloid z = 7 - 2x² – 2y² that lies above the plane z = −1. O A. π/24(√65 - 1) B. π/24 (65³/2) ● C. π/24(65³/2 - 1) O D. π/24(1-1/√/65)

Find the surface area of the portion of the part of the paraboloid z = 7 - 2x² – 2y² that lies above the plane z = −1. O A. π/24(√65 - 1) B. π/24 (65³/2) ● C. π/24(65³/2 - 1) O D. π/24(1-1/√/65)

Calculus: Early Transcendentals
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:** Find the surface area of the portion of the part of the paraboloid \( z = 7 - 2x^2 - 2y^2 \) that lies above the plane \( z = -1 \). **Answer Choices:** - A. \(\frac{\pi}{24}(\sqrt{65} - 1)\) - B. \(\frac{\pi}{24}(65^{3/2})\) - C. \(\frac{\pi}{24}(65^{3/2} - 1)\) - D. \(\frac{\pi}{24}(1 - \frac{1}{\sqrt{65}})\) The correct answer is choice C.
Transcribed Image Text:**Problem Statement:** Find the surface area of the portion of the part of the paraboloid \( z = 7 - 2x^2 - 2y^2 \) that lies above the plane \( z = -1 \). **Answer Choices:** - A. \(\frac{\pi}{24}(\sqrt{65} - 1)\) - B. \(\frac{\pi}{24}(65^{3/2})\) - C. \(\frac{\pi}{24}(65^{3/2} - 1)\) - D. \(\frac{\pi}{24}(1 - \frac{1}{\sqrt{65}})\) The correct answer is choice C.
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