AP Calculus BC Flashcards: Washer Method Revolving Around Other Axes

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 BC.

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.

Flashcard 1: Calculate the volume for $y = x^2$ and $y = x$ revolved around $y = 1$ from $x=0$ to $x=1$.

Answer: $V = \pi \int_{0}^{1} [(1-x^2)^2 - (1-x)^2] \ dx$. Both curves shifted by subtracting from 1 for revolution around $y=1$.

Flashcard 2: Determine the outer radius for $x = 2$ revolved around $x = -3$.

Answer: $R(y) = |2 - (-3)| = 5$. Distance from vertical line $x=2$ to axis $x=-3$ is $|2-(-3)|=5$.

Flashcard 3: Identify the axis of revolution for $y = f(x)$ revolved around $y = 2$.

Answer: $y = 2$. The axis of revolution is the horizontal line around which the region rotates.

Flashcard 4: What does $R(x)$ represent in the washer method formula?

Answer: The outer radius of the washer. $R(x)$ is the distance from the axis of revolution to the farther curve.

Flashcard 5: What is the modified formula for revolving around a horizontal line $y = k$?

Answer: $V = \pi \int_{a}^{b} [(R(x)-k)^2 - (r(x)-k)^2] \ dx$. Shifts both radii by subtracting the axis position $k$ from each radius function.

Flashcard 6: What is the modified formula for revolving around a vertical line $x = k$?

Answer: $V = \pi \int_{c}^{d} [(R(y)-k)^2 - (r(y)-k)^2] \ dy$. Integrates with respect to $y$ and shifts radii by the vertical axis position $k$.

Flashcard 7: How do you determine the outer radius $R(x)$ for $y = f(x)$ revolved around $y = k$?

Answer: $R(x) = |f(x) - k|$. Distance from curve $f(x)$ to horizontal axis $y = k$ using absolute value.

Flashcard 8: How do you determine the inner radius $r(x)$ for $y = g(x)$ revolved around $y = k$?

Answer: $r(x) = |g(x) - k|$. Distance from curve $g(x)$ to horizontal axis $y = k$ using absolute value.

Flashcard 9: Determine the outer radius for $x = 3$ revolving around $x = -1$.

Answer: $R(y) = |3 - (-1)| = 4$. Distance from vertical line $x=3$ to axis $x=-1$ is $|3-(-1)|=4$.

Flashcard 10: State the bounds $a$ and $b$ for $y = f(x)$ from $x = 1$ to $x = 3$.

Answer: $a = 1$, $b = 3$. Integration bounds match the given $x$-interval from 1 to 3.

Flashcard 11: For $x = 4$ revolving around $x = 0$, what is the outer radius?

Answer: $R(y) = |4 - 0| = 4$. Distance from vertical line $x=4$ to y-axis is simply the absolute value 4.

Flashcard 12: When is the washer method preferred over the disk method?

Answer: When there is an inner radius creating a hole. Washer method applies when the solid has a hollow center or gap.

Flashcard 13: Determine the outer radius for $x = 2$ revolved around $x = -3$.

Answer: $R(y) = |2 - (-3)| = 5$. Distance from vertical line $x=2$ to axis $x=-3$ is $|2-(-3)|=5$.

Flashcard 14: In the washer method, how do you find the volume if $R(x) = x+1$ and $r(x) = x$?

Answer: $V = \pi \int_{a}^{b} [(x+1)^2 - x^2] \ dx$. Standard washer formula with given outer and inner radius functions.

Flashcard 15: What modification is needed for a vertical axis of revolution?

Answer: Integrate with respect to $y$ instead of $x$. Vertical axes require integrating with respect to $y$ instead of $x$.

Flashcard 16: Find the volume for $y = \sqrt{x}$ and $y = x$ revolved around $y = 3$ from $x=0$ to $x=1$.

Answer: $V = \pi \int_{0}^{1} [(3-\sqrt{x})^2 - (3-x)^2] \ dx$. Both curves shifted by subtracting from 3 for revolution around $y=3$.

Flashcard 17: Identify the outer radius for $x = 5$ revolved around $x = 0$.

Answer: $R(y) = |5 - 0| = 5$. Distance from vertical line $x=5$ to y-axis ($x=0$) is simply 5.

Flashcard 18: Determine the outer radius for $y = f(x)$ revolved around $x = k$.

Answer: Calculate $R(y) = |f(y) - k|$. Distance from curve to vertical axis using absolute value for outer radius.

Flashcard 19: What happens to $R(x)$ if the axis of revolution is above the graph?

Answer: Increase $R(x)$ by the vertical distance. When axis is above the curve, add the vertical distance to get the radius.

Flashcard 20: Identify the inner radius for $y = 2x$ revolved around $y = -2$.

Answer: $r(x) = |2x - (-2)|$. Distance from line $y=2x$ to axis $y=-2$ using absolute value formula.

Flashcard 21: If $R(x) = 5$ and $r(x) = 2$, what is the area of one washer?

Answer: $A = \pi (5^2 - 2^2)$. Area of washer equals $\pi$ times outer radius squared minus inner radius squared.

Flashcard 22: What effect does a horizontal shift have on the washer method formula?

Answer: Adjust radii by the shift value in the formula. Horizontal shifts affect the distance calculations in the radius formulas.

Flashcard 23: Find the inner radius for $y = f(x)$ revolving around $x = c$ and $y = g(x)$.

Answer: $r(y) = |g(y) - c|$. Distance from inner curve $g(y)$ to vertical axis $x=c$ using absolute value.

Flashcard 24: What adjustment is made in the washer method when revolving around $x = k$?

Answer: Use $R(y)$ and $r(y)$, integrate with respect to $y$. Switch to functions of $y$ and integrate with respect to $dy$ for vertical axes.

Flashcard 25: State the corrected formula for revolving around a vertical line $x = k$.

Answer: $V = \pi \int_{c}^{d} [(R(y)-k)^2 - (r(y)-k)^2] \ dy$. General formula for revolution around any vertical line at $x=k$.

Flashcard 26: State the formula for the volume of a solid using the washer method.

Answer: $V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \ dx$. Standard washer method formula with outer radius squared minus inner radius squared.

Flashcard 27: How do you modify $R(x)$ and $r(x)$ for a shift along the y-axis?

Answer: Adjust by the shift value in the $y$ direction. Add or subtract the shift distance from each radius function accordingly.

Flashcard 28: Find the volume of $y = x^2$ and $y = 4$ revolved around $y = 5$ from $x=0$ to $x=2$.

Answer: $V = \pi \int_{0}^{2} [(5-x^2)^2 - (5-4)^2] \ dx$. Outer radius is $5-x^2$, inner radius is $5-4=1$ for revolution around $y=5$.

Flashcard 29: Identify the formula for the solid's volume when $y = 3x$ and $y = 0$ are revolved around $y = -1$.

Answer: $V = \pi \int_{0}^{1} [(3x+1)^2 - 1^2] dx$. Both curves shifted by adding 1 for revolution around $y=-1$.

Flashcard 30: For $R(x) = x^2$ and $r(x) = x$, what is the volume formula?

Answer: $V = \pi \int_{a}^{b} [x^4 - x^2] dx$. Outer radius squared is $x^4$, inner radius squared is $x^2$.