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Consider the region bounded by the graphs of y = x², y = 30, and x = 11 in the first quadrant. What is the volume of solid obtained by rotating this region about the line x = 5? Select the correct answer below: 30 af ((√6y-16)² - 121) dy T 0 30 7 / (( √6y - 5)² - 36) dy 121 ((√6y-6) ² - 121) dy ((√6y-11)² -36) dy 30 O */ 121 6 ○x 30 Content attribution FEEDBACK

Consider the region bounded by the graphs of y = x², y = 30, and x = 11 in the first quadrant. What is the volume of solid obtained by rotating this region about the line x = 5? Select the correct answer below: 30 af ((√6y-16)² - 121) dy T 0 30 7 / (( √6y - 5)² - 36) dy 121 ((√6y-6) ² - 121) dy ((√6y-11)² -36) dy 30 O */ 121 6 ○x 30 Content attribution FEEDBACK

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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## Volume of Solid of Revolution ### Problem Statement Consider the region bounded by the graphs of \( y = \frac{1}{6} x^2 \), \( y = 30 \), and \( x = 11 \) in the first quadrant. What is the volume of the solid obtained by rotating this region about the line \( x = 5 \)? **Select the correct answer below:** - \( \pi \int_{0}^{30} \left( (\sqrt{6y} - 16)^2 - 121 \right) dy \) - \( \pi \int_{0}^{30} \left( (\sqrt{6y} - 5)^2 - 36 \right) dy \) - \( \pi \int_{0}^{ \frac{121}{6} } \left( (\sqrt{6y} - 6)^2 - 121 \right) dy \) - \( \pi \int_{0}^{ \frac{121}{6} } \left( (\sqrt{6y} - 11)^2 - 36 \right) dy \) ### Antiderivative Calculation Determine the antiderivative given above. Do not include the constant “+C” in your answer. \[ \int 6x^3 \left( -3x^4 + 3 \right)^6 dx \] **Provide your answer below:** \[ \text{Answer:} \ \_\_\_\_\_\_\_\_\_ \]
Transcribed Image Text:## Volume of Solid of Revolution ### Problem Statement Consider the region bounded by the graphs of \( y = \frac{1}{6} x^2 \), \( y = 30 \), and \( x = 11 \) in the first quadrant. What is the volume of the solid obtained by rotating this region about the line \( x = 5 \)? **Select the correct answer below:** - \( \pi \int_{0}^{30} \left( (\sqrt{6y} - 16)^2 - 121 \right) dy \) - \( \pi \int_{0}^{30} \left( (\sqrt{6y} - 5)^2 - 36 \right) dy \) - \( \pi \int_{0}^{ \frac{121}{6} } \left( (\sqrt{6y} - 6)^2 - 121 \right) dy \) - \( \pi \int_{0}^{ \frac{121}{6} } \left( (\sqrt{6y} - 11)^2 - 36 \right) dy \) ### Antiderivative Calculation Determine the antiderivative given above. Do not include the constant “+C” in your answer. \[ \int 6x^3 \left( -3x^4 + 3 \right)^6 dx \] **Provide your answer below:** \[ \text{Answer:} \ \_\_\_\_\_\_\_\_\_ \]
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