All flashcards
Flashcard 1: What change is made to the disc method when revolving around a horizontal line y=c?
Answer: Use ∣f(x)−c∣ as radius. Distance from curve to line y=c becomes the new radius.
Flashcard 2: How is the disc method implemented for a solid of revolution about y=b?
Answer: Adjust radius: ∣g(y)−b∣. Distance from curve to horizontal line y=b becomes radius.
Flashcard 3: What is the effect of changing the axis of rotation to y=c?
Answer: Radius becomes ∣f(x)−c∣. Shift from x-axis changes radius to distance from line.
Flashcard 4: How do you adjust the disc method formula for revolving around the y-axis?
Answer: Use x=f(y) instead of y=f(x). Switch to integrating with respect to y and use horizontal cross-sections.
Flashcard 5: What is the role of the function g(y) in the disc method when using the y-axis?
Answer: It defines the radius for integration. Function g(y) gives horizontal distance from y-axis.
Flashcard 6: What adjustment is made to the disc method for a non-axis vertical line?
Answer: Adjust radius: ∣g(y)−line∣. Distance from function to vertical line becomes radius.
Flashcard 7: How does the function f(x) affect the volume when using the disc method?
Answer: It determines the radius. Function values determine cross-sectional radii at each point.
Flashcard 8: Describe how to use the disc method when revolving around a non-axis line.
Answer: Adjust radius as ∣f(x)−line∣. The distance from curve to the line becomes the radius.
Flashcard 9: What is the role of the limits of integration in the disc method?
Answer: They define the bounds of integration. Limits specify the interval over which to integrate.
Flashcard 10: What is the effect of changing the axis of rotation to x=c?
Answer: Radius becomes ∣g(y)−c∣. Shift from y-axis changes radius to distance from line.
Flashcard 11: How do you determine the limits of integration for revolving around the y-axis?
Answer: Use the interval [c,d] of y. Integration bounds match the range of the function.
Flashcard 12: What change is made to the disc method when revolving around a vertical line x=c?
Answer: Use ∣g(y)−c∣ as radius. Distance from curve to line x=c becomes the new radius.
Flashcard 13: Identify the expression for the radius in the disc method revolving around the x-axis.
Answer: Radius = f(x). The function value gives the distance from x-axis to curve.
Flashcard 14: How do you determine the limits of integration for revolving around the x-axis?
Answer: Use the interval [a,b] of x. Integration bounds match the domain of the function.
Flashcard 15: What adjustment is made to the disc method for a non-axis horizontal line?
Answer: Adjust radius: ∣f(x)−line∣. Distance from function to horizontal line becomes radius.
Flashcard 16: Identify the expression for the radius in the disc method revolving around the y-axis.
Answer: Radius = g(y). The function g(y) gives distance from y-axis to curve.
Flashcard 17: Describe how to use the disc method when revolving around a non-axis line.
Answer: Adjust radius as ∣f(x)−line∣. The distance from curve to the line becomes the radius.
Flashcard 18: What adjustment is made to the disc method for a non-axis horizontal line?
Answer: Adjust radius: ∣f(x)−line∣. Distance from function to horizontal line becomes radius.
Flashcard 19: What is the effect of changing the axis of rotation to x=c?
Answer: Radius becomes ∣g(y)−c∣. Shift from y-axis changes radius to distance from line.
Flashcard 20: What is the role of the limits of integration in the disc method?
Answer: They define the bounds of integration. Limits specify the interval over which to integrate.
Flashcard 21: What change is made to the disc method when revolving around a horizontal line y=c?
Answer: Use ∣f(x)−c∣ as radius. Distance from curve to line y=c becomes the new radius.
Flashcard 22: How does the function f(x) affect the volume when using the disc method?
Answer: It determines the radius. Function values determine cross-sectional radii at each point.
Flashcard 23: What adjustment is made to the disc method for a non-axis vertical line?
Answer: Adjust radius: ∣g(y)−line∣. Distance from function to vertical line becomes radius.
Flashcard 24: Identify the expression for the radius in the disc method revolving around the y-axis.
Answer: Radius = g(y). The function g(y) gives distance from y-axis to curve.
Flashcard 25: How do you determine the limits of integration for revolving around the x-axis?
Answer: Use the interval [a,b] of x. Integration bounds match the domain of the function.
Flashcard 26: How is the disc method implemented for a solid of revolution about y=b?
Answer: Adjust radius: ∣g(y)−b∣. Distance from curve to horizontal line y=b becomes radius.
Flashcard 27: How do you adjust the disc method formula for revolving around the y-axis?
Answer: Use x=f(y) instead of y=f(x). Switch to integrating with respect to y and use horizontal cross-sections.
Flashcard 28: Identify the expression for the radius in the disc method revolving around the x-axis.
Answer: Radius = f(x). The function value gives the distance from x-axis to curve.
Flashcard 29: What change is made to the disc method when revolving around a vertical line x=c?
Answer: Use ∣g(y)−c∣ as radius. Distance from curve to line x=c becomes the new radius.
Flashcard 30: How do you determine the limits of integration for revolving around the y-axis?
Answer: Use the interval [c,d] of y. Integration bounds match the range of the function.