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AP Calculus AB Flashcards: Washer Method Revolving Around Other Axes

Study Washer Method Revolving Around Other Axes in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Washer Method Revolving Around Other Axes, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus AB Flashcards: Washer Method Revolving Around Other Axes

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QUESTION

How to set R(x)R(x)R(x) if revolving around y=0y = 0y=0, y=x2y = x^2y=x2?

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ANSWER

R(x)=x2R(x) = x^2R(x)=x2. Distance from y=x2y = x^2y=x2 to the xxx-axis (y=0y = 0y=0) is x2x^2x2.

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Flashcard 1: How to set R(x)R(x)R(x) if revolving around y=0y = 0y=0, y=x2y = x^2y=x2?

Answer: R(x)=x2R(x) = x^2R(x)=x2. Distance from y=x2y = x^2y=x2 to the xxx-axis (y=0y = 0y=0) is x2x^2x2.

Flashcard 2: How do you find the inner radius for revolution around yyy-axis?

Answer: Distance from the yyy-axis to the nearest curve. Take the minimum xxx-coordinate of the curves being revolved.

Flashcard 3: Identify the axis of revolution in y=f(x)y = f(x)y=f(x) revolved around y=cy = cy=c.

Answer: y=cy = cy=c. The horizontal line about which the region is rotated.

Flashcard 4: Identify r(x)r(x)r(x) if y=xy = xy=x revolves around y=−1y = -1y=−1.

Answer: r(x)=x+1r(x) = x + 1r(x)=x+1. Distance from y=xy = xy=x to horizontal line y=−1y = -1y=−1 is x−(−1)=x+1x - (-1) = x + 1x−(−1)=x+1.

Flashcard 5: Which axis is used in x=g(y)x = g(y)x=g(y) revolved around x=cx = cx=c?

Answer: x=cx = cx=c. The vertical line about which the region is rotated.

Flashcard 6: Find r(x)r(x)r(x) if revolving around y=2y = 2y=2, g(x)=xg(x) = xg(x)=x.

Answer: r(x)=2−xr(x) = 2 - xr(x)=2−x. Distance from y=xy = xy=x to the line y=2y = 2y=2 is 2−x2 - x2−x.

Flashcard 7: State the limits of integration if revolving around y=cy=cy=c for x=f(y)x = f(y)x=f(y).

Answer: From y=ay = ay=a to y=by = by=b. Integration bounds match the yyy-values where the region is defined.

Flashcard 8: Determine R(y)R(y)R(y) for revolution around x=−1x = -1x=−1, x=y2+1x = y^2 + 1x=y2+1.

Answer: R(y)=y2+2R(y) = y^2 + 2R(y)=y2+2. Distance from x=y2+1x = y^2 + 1x=y2+1 to line x=−1x = -1x=−1 is (y2+1)−(−1)=y2+2(y^2 + 1) - (-1) = y^2 + 2(y2+1)−(−1)=y2+2.

Flashcard 9: Find R(x)R(x)R(x) if revolving around y=2y = 2y=2, f(x)=x2f(x) = x^2f(x)=x2.

Answer: R(x)=2−x2R(x) = 2 - x^2R(x)=2−x2. Distance from y=x2y = x^2y=x2 to the line y=2y = 2y=2 is 2−x22 - x^22−x2.

Flashcard 10: Identify R(x)R(x)R(x) if y=x2y = x^2y=x2 revolves around y=−1y = -1y=−1.

Answer: R(x)=x2+1R(x) = x^2 + 1R(x)=x2+1. Distance from y=x2y = x^2y=x2 to horizontal line y=−1y = -1y=−1 is x2−(−1)=x2+1x^2 - (-1) = x^2 + 1x2−(−1)=x2+1.

Flashcard 11: What is the area of the washer if R=4R = 4R=4 and r=2r = 2r=2?

Answer: π(16−4)\pi (16 - 4)π(16−4). Area of washer is π(R2−r2)=π(16−4)=12π\pi(R^2 - r^2) = \pi(16 - 4) = 12\piπ(R2−r2)=π(16−4)=12π.

Flashcard 12: Determine r(y)r(y)r(y) for revolution around x=−1x = -1x=−1, x=y+1x = y + 1x=y+1.

Answer: r(y)=y+2r(y) = y + 2r(y)=y+2. Distance from x=y+1x = y + 1x=y+1 to line x=−1x = -1x=−1 is (y+1)−(−1)=y+2(y + 1) - (-1) = y + 2(y+1)−(−1)=y+2.

Flashcard 13: What does R(x)R(x)R(x) represent in the washer method?

Answer: The outer radius function. The distance from the axis of revolution to the farther boundary curve.

Flashcard 14: State the formula for volume using the washer method.

Answer: V=π∫ab [(R(x))2−(r(x))2] dxV = \pi \int_a^b \, [(R(x))^2 - (r(x))^2] \, dxV=π∫ab​[(R(x))2−(r(x))2]dx. Standard washer method formula with outer radius R(x)R(x)R(x) and inner radius r(x)r(x)r(x).

Flashcard 15: What does R(x)R(x)R(x) represent in the washer method?

Answer: The outer radius function. The distance from the axis of revolution to the farther boundary curve.

Flashcard 16: State the formula for volume using the washer method.

Answer: V=π∫ab [(R(x))2−(r(x))2] dxV = \pi \int_a^b \, [(R(x))^2 - (r(x))^2] \, dxV=π∫ab​[(R(x))2−(r(x))2]dx. Standard washer method formula with outer radius R(x)R(x)R(x) and inner radius r(x)r(x)r(x).

Flashcard 17: How to set R(x)R(x)R(x) if revolving around y=0y = 0y=0, y=x2y = x^2y=x2?

Answer: R(x)=x2R(x) = x^2R(x)=x2. Distance from y=x2y = x^2y=x2 to the xxx-axis (y=0y = 0y=0) is x2x^2x2.

Flashcard 18: Identify r(x)r(x)r(x) if y=xy = xy=x revolves around y=−1y = -1y=−1.

Answer: r(x)=x+1r(x) = x + 1r(x)=x+1. Distance from y=xy = xy=x to horizontal line y=−1y = -1y=−1 is x−(−1)=x+1x - (-1) = x + 1x−(−1)=x+1.

Flashcard 19: Determine r(y)r(y)r(y) for revolution around x=−1x = -1x=−1, x=y+1x = y + 1x=y+1.

Answer: r(y)=y+2r(y) = y + 2r(y)=y+2. Distance from x=y+1x = y + 1x=y+1 to line x=−1x = -1x=−1 is (y+1)−(−1)=y+2(y + 1) - (-1) = y + 2(y+1)−(−1)=y+2.

Flashcard 20: What is the area of the washer if R=4R = 4R=4 and r=2r = 2r=2?

Answer: π(16−4)\pi (16 - 4)π(16−4). Area of washer is π(R2−r2)=π(16−4)=12π\pi(R^2 - r^2) = \pi(16 - 4) = 12\piπ(R2−r2)=π(16−4)=12π.

Flashcard 21: Identify R(x)R(x)R(x) if y=x2y = x^2y=x2 revolves around y=−1y = -1y=−1.

Answer: R(x)=x2+1R(x) = x^2 + 1R(x)=x2+1. Distance from y=x2y = x^2y=x2 to horizontal line y=−1y = -1y=−1 is x2−(−1)=x2+1x^2 - (-1) = x^2 + 1x2−(−1)=x2+1.

Flashcard 22: Which axis is used in x=g(y)x = g(y)x=g(y) revolved around x=cx = cx=c?

Answer: x=cx = cx=c. The vertical line about which the region is rotated.

Flashcard 23: What does r(x)r(x)r(x) represent in the washer method?

Answer: The inner radius function. The distance from the axis of revolution to the nearer boundary curve.

Flashcard 24: Identify the axis of revolution in y=f(x)y = f(x)y=f(x) revolved around y=cy = cy=c.

Answer: y=cy = cy=c. The horizontal line about which the region is rotated.

Flashcard 25: How do you find the outer radius for revolution around yyy-axis?

Answer: Distance from the yyy-axis to the farthest curve. Take the maximum xxx-coordinate of the curves being revolved.

Flashcard 26: How do you find the inner radius for revolution around yyy-axis?

Answer: Distance from the yyy-axis to the nearest curve. Take the minimum xxx-coordinate of the curves being revolved.

Flashcard 27: Find R(x)R(x)R(x) if revolving around y=2y = 2y=2, f(x)=x2f(x) = x^2f(x)=x2.

Answer: R(x)=2−x2R(x) = 2 - x^2R(x)=2−x2. Distance from y=x2y = x^2y=x2 to the line y=2y = 2y=2 is 2−x22 - x^22−x2.

Flashcard 28: Determine R(y)R(y)R(y) for revolution around x=−1x = -1x=−1, x=y2+1x = y^2 + 1x=y2+1.

Answer: R(y)=y2+2R(y) = y^2 + 2R(y)=y2+2. Distance from x=y2+1x = y^2 + 1x=y2+1 to line x=−1x = -1x=−1 is (y2+1)−(−1)=y2+2(y^2 + 1) - (-1) = y^2 + 2(y2+1)−(−1)=y2+2.

Flashcard 29: State the limits of integration if revolving around y=cy=cy=c for x=f(y)x = f(y)x=f(y).

Answer: From y=ay = ay=a to y=by = by=b. Integration bounds match the yyy-values where the region is defined.

Flashcard 30: Find r(x)r(x)r(x) if revolving around y=2y = 2y=2, g(x)=xg(x) = xg(x)=x.

Answer: r(x)=2−xr(x) = 2 - xr(x)=2−x. Distance from y=xy = xy=x to the line y=2y = 2y=2 is 2−x2 - x2−x.