Washer Method: Revolving Around Other Axes - AP Calculus AB

Region bounded by $y=e^{-x}$ and $y=0$ for $0\le x\le1$ is revolved about $y=2$ . Which setup is correct?
Answer: $V=\pi\int_{0}^{1}\big[(2-0)^2-(2-e^{-x})^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 2. To adjust, subtract y from 2, outer 2-0=2, inner 2 - $e^{-x}$ . Correct for axis above. A tempting distractor is choice A, swapping. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
Select the correct washer-method integral for revolving the region between $y=\sin x$ and $y=\cos x$ on $[0,\tfrac{\pi}{4}]$ about $y=-2$ .
Answer: $V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\cos x+2)^2-(\sin x+2)^2\right]dx$ This problem requires the washer method with revolution about y = -2, a horizontal line below the region. When revolving about y = -2, we calculate distances by subtracting the axis value: the outer radius is (cos x - (-2)) = cos x + 2 and the inner radius is (sin x - (-2)) = sin x + 2. Since cos x ≥ sin x on [0, π/4], cos x + 2 gives the larger radius, so the integral is V = π∫[(cos x + 2)² - (sin x + 2)²]dx. Choice C incorrectly uses the original functions without the +2 shift, forgetting to account for the axis at y = -2. The transferable principle is to always add |k| to both functions when revolving about y = k < 0.
Region bounded by $y=1-x$ and $y=1-x^2$ for $0\le x\le1$ is revolved about $y=3$ . Choose the correct volume setup.
Answer: $V=\pi\int_{0}^{1}\big[(3-(1-x))^2-(3-(1-x^2))^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 3. To adjust for the shifted axis, subtract each y-value from 3, making the outer radius 3 - (1 - x) for the upper curve. The inner radius is 3 - (1 - x²) for the lower curve. These adjustments handle the axis above. A tempting distractor is choice A, which swaps radii. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
Which washer-method integral gives the volume when the region between $y=x^3$ and $y=x$ on $[0,1]$ is revolved about $y=-2$ ?
Answer: $V=\pi\displaystyle\int_{0}^{1}\left[(x+2)^2-(x^3+2)^2\right]dx$ This problem involves the washer method with revolution about y = -2, below the region. When revolving about y = -2, we calculate distances by subtracting the axis value: since x > x³ on [0,1], the outer radius is (x - (-2)) = x + 2 and the inner radius is (x³ - (-2)) = x³ + 2. The washer integral is V = π∫[(x + 2)² - (x³ + 2)²]dx from 0 to 1. Choice E incorrectly adds 2 only to the linear function, forgetting to shift x³ as well, which would create an impossible geometry. The fundamental rule is that both functions must be shifted by the same amount when revolving about a translated axis.
For $0\le x\le1$ , region between $y=2-x$ and $y=1$ is revolved about $y=-2$ . Which integral represents the volume?
Answer: $V=\pi\int_{0}^{1}\big[(2-x+2)^2-(1+2)^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -2. To adjust for the shifted axis, add 2 to each y-value, making the outer radius (2 - x) + 2 = 4 - x for the upper curve. The inner radius is 1 + 2 = 3 for the lower curve. These adjustments reflect distances from below. A tempting distractor is choice B, which reverses the order. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
The region between $y=4$ and $y=\sqrt{x}$ for $0\le x\le16$ is revolved about $y=6$ . Which integral gives the volume?
Answer: $V=\pi\int_{0}^{16}\big[(6-\sqrt{x})^2-(6-4)^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 6. To adjust for the shifted axis, subtract y from 6, making outer 6 - √x for the lower curve. Inner is 6 - 4 = 2 for the upper. These handle axis above. A tempting distractor is choice A, swapping order. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
The region between $y=\frac{1}{x}$ and $y=\frac{1}{4}$ for $1\le x\le4$ is revolved about $y=-1$ . Which integral gives the volume?
Answer: $V=\pi\int_{1}^{4}\big[(1/x+1)^2-(1/4+1)^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=-1, both radii are curve plus 1 since above, giving 1/x + 1 and 1/4 + 1. The outer is 1/x + 1. This handles the shift. A tempting distractor like option B fails by reversing, negative volume. A transferable strategy for shifted washer problems is to add shift when axis below and ensure larger outer.
Let $R$ be bounded by $y=4-x$ and $y=x$ on $0 \leq x \leq 2$ and revolved about $y=5$ . Which setup gives the volume?
Answer: $V=\pi\int_{0}^{2}\big[(5-x)^2-(5-(4-x))^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around $y=5$ , both radii are $5 - x$ and $5 - (4 - x)$ since below, giving $5 - x$ and $5 - (4 - x)$ . The outer is $5 - x$ as lower curve is farther. This adjustment maintains correct distances. A tempting distractor like option B fails by reversing order, yielding negative integrand. A transferable strategy for shifted washer problems is to subtract curves from axis when region is below and verify positive difference.
Region bounded by $y=3$ and $y=x^2$ for $-1\le x\le1$ is revolved about $y=5$ . Which setup is correct?
Answer: $V=\pi\int_{-1}^{1}\big[(5-x^2)^2-(5-3)^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=5, both radii are 5 minus curves since below, giving 5 - x² and 5 - 3 = 2. The outer is 5 - x². This gives correct washers. A tempting distractor like option A fails by swapping order, negative integrand. A transferable strategy for shifted washer problems is to consistently subtract curves from axis when below.
Find the washer-method setup for the volume when the region between $y=x^2$ and $y=4$ on $[ -2,2 ]$ is revolved about $y=1$ .
Answer: $V=\pi\int_{-2}^{2}\Big[(4-1)^2-(x^2-1)^2\Big]dx$ This problem requires the washer method with rotation about the horizontal line y = 1, which is not the x-axis. When revolving about y = 1, we measure distances from this axis: the outer radius is from y = 1 up to the constant y = 4, giving R = 4 - 1 = 3, while the inner radius is from y = 1 up to the parabola y = x², giving r = x² - 1. The washer formula becomes V = π∫[R² - r²]dx = π∫[(4-1)² - (x²-1)²]dx from -2 to 2. Choice B incorrectly uses 4² - (x²)², failing to shift both functions by subtracting 1 from their y-values. The key strategy for washers about y = k is to replace every y-value with (y - k) to measure distances from the new axis.
For $0\le x\le1$ , the region between $y=\sqrt{1-x^2}$ and $y=0$ is revolved about $y=-1$ . Choose the volume integral.
Answer: $V=\pi\int_{0}^{1}\big[(\sqrt{1-x^2}+1)^2-(0+1)^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=-1, both radii are curve plus 1 since above, giving √(1-x²) + 1 and 1. The outer is the varying one. This adjustment is key for shifted axis. A tempting distractor like option B fails by reversing, negative volume. A transferable strategy for shifted washer problems is to adjust by the shift and determine order by distances.
For $0\le x\le2$ , region between $y=\ln(x+1)$ and $y=x/2$ is revolved about $y=3$ . Choose the correct volume integral.
Answer: $V=\pi\int_{0}^{2}\big[(3-x/2)^2-(3-\ln(x+1))^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 3. To adjust, subtract y from 3, outer 3 - (x/2), inner 3 - ln(x+1). But since lower is smaller y, larger distance. Yes B. A tempting distractor is choice A, swapping. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
For $0\le x\le\pi$ , the region between $y=\sin x$ and $y=0$ is revolved about $y=1$ . Which integral represents the volume?
Answer: $V=\pi\int_{0}^{\pi}\big[(1-0)^2-(1-\sin x)^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=1, both radii are 1 minus curves since below, giving 1 - 0 and 1 - sin x. The outer is 1. This works as region touches but doesn't cross. A tempting distractor like option A fails by reversing order. A transferable strategy for shifted washer problems is to use axis minus curve and split if crossing.
For $0\le x\le1$ , region between $y=2x$ and $y=x^3$ is revolved about $y=-1$ . Which integral represents the volume?
Answer: $V=\pi\int_{0}^{1}\big[(2x+1)^2-(x^3+1)^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -1. To adjust for the shifted axis, add 1 to y, outer 2x +1 for upper. Inner $x^3$ +1 for lower. Adjustments correct. A tempting distractor is choice B, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
Region between $y=3x+2$ and $y=x^2+2$ on $0\le x\le3$ is revolved about $y=2$ . The volume is
Answer: $\pi\int_{0}^{3}\left[(3x+2-2)^2-(x^2+2-2)^2\right]dx$ This problem uses the washer method with revolution around y = 2, requiring distance measurements from this horizontal line. For the line y = 3x + 2, the distance to y = 2 is |(3x + 2) - 2| = |3x| = 3x (since x ≥ 0). For the parabola y = x² + 2, the distance to y = 2 is |(x² + 2) - 2| = |x²| = x². Since 3x ≥ x² on [0,3], the outer radius is 3x and the inner radius is x². The volume integral is π∫₀³[(3x)² - (x²)²]dx = π∫₀³[9x² - x⁴]dx. The expression (3x+2-2)² - (x²+2-2)² correctly shows the subtraction but simplifies to the cleaner form. When setting up washers around y = k, subtract k from each y-expression before squaring to find the radius squared.
The region bounded by $y=\sqrt{4x}+1$ and $y=x+1$ for $0\le x\le4$ is revolved about $y=1$ . Volume equals
Answer: $\pi\int_{0}^{4}\left[(\sqrt{4x}+1-1)^2-(x+1-1)^2\right]dx$ This problem requires the washer method with revolution around y = 1, adjusting both functions for the shifted axis. The curve y = √(4x) + 1 gives a radius of (√(4x) + 1) - 1 = √(4x) = 2√x from y = 1. The line y = x + 1 gives a radius of (x + 1) - 1 = x from y = 1. Since √(4x) ≥ x on [0,4] (checking: at x = 4, we have 4 ≥ 4), the outer radius is √(4x) and inner radius is x. The volume integral is π∫₀⁴[(√(4x))² - x²]dx = π∫₀⁴[4x - x²]dx. The expression (√(4x)+1-1)² - (x+1-1)² correctly shows the adjustment process. When setting up washers around y = k, subtract k from each function before squaring to find radius squared terms.
Which integral gives the volume when the region between $y=x^2$ and $y=2x$ on $[0,2]$ is revolved about $y=3$ ?
Answer: $V=\pi\displaystyle\int_{0}^{2}\left[(3-2x)^2-(3-x^2)^2\right]dx$ This problem involves the washer method with revolution about the horizontal line y = 3, which is above the region. When revolving about y = 3, we need to find distances from this axis: since y = 3 is above both curves, the outer radius is (3 - 2x) and the inner radius is (3 - x²). The washer method integral is V = π∫[(3 - 2x)² - (3 - x²)²]dx from 0 to 2. Choice C incorrectly uses the original functions without adjusting for the shifted axis, missing the crucial step of subtracting each function from 3. For revolution about a line y = k above the region, remember to use (k - f(x)) for each radius.
Region bounded by $y=e^x$ and $y=1$ for $0\le x\le\ln 3$ is revolved about $y=4$ . Which is the correct setup?
Answer: $V=\pi\int_{0}^{\ln 3}\big[(4-1)^2-(4-e^x)^2\big]dx$ The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=4, both radii are 4 minus curves since below, giving 4 - 1 and 4 - $e^x$ . The outer is constant 3. This captures the washer correctly. A tempting distractor like option A fails by reversing, negative integrand. A transferable strategy for shifted washer problems is to use axis minus curve for below-axis regions and outer as farther.
Region bounded by $y=2$ and $y=\sin x$ for $0\le x\le\pi$ is revolved about $y=4$ . Which setup is correct?
Answer: $V=\pi\int_{0}^{\pi}\big[(4-\sin x)^2-(4-2)^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 4. To adjust, subtract y from 4, outer 4 - sin x for lower, inner 4-2=2. Yes. A tempting distractor is choice A, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.
For $0 \le x \le 2$ , region between $y=\sqrt{x+2}$ and $y=1$ is revolved about $y=5$ . Which integral represents the volume?
Answer: $V=\pi\int_{0}^{2}\big[(5-1)^2-(5-\sqrt{x+2})^2\big]dx$ This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically $y = 5$ . To adjust, subtract y from 5, outer $5-1=4$ , inner $5 - \sqrt{x+2}$ . Yes B. A tempting distractor is choice A, swapping. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate $\pi$ times ( $\text{outer}^2 - \text{inner}^2$ ) dx.

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