Washer Method: Revolving Around Other Axes
Help Questions
AP Calculus AB › Washer Method: Revolving Around Other Axes
Select the correct washer-method integral for revolving the region between $y=\sin x$ and $y=\cos x$ on $0,\tfrac{\pi}{4}$ about $y=-2$.
$V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\cos x)^2-(\sin x)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\sin x+2)^2-(\cos x+2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\cos x-2)^2-(\sin x-2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\cos x+2)^2-(\sin x)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{\pi/4}\left[(\cos x+2)^2-(\sin x+2)^2\right]dx$
Explanation
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.
Which integral gives the volume when the region between $y=x^2$ and $y=2x$ on $0,2$ is revolved about $y=3$?
$V=\pi\displaystyle\int_{0}^{2}\left[(3-2x)^2-(3-x^2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(3-2x)^2-(x^2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(3-x^2)^2-(3-2x)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(2x)^2-(x^2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(2x-3)^2-(x^2-3)^2\right]dx$
Explanation
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.
Which washer-method integral gives the volume when the region between $y=x^3$ and $y=x$ on $0,1$ is revolved about $y=-2$?
$V=\pi\displaystyle\int_{0}^{1}\left[(x)^2-(x^3)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{1}\left[(x-2)^2-(x^3-2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{1}\left[(x^3+2)^2-(x+2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{1}\left[(x+2)^2-(x^3+2)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{1}\left[(x+2)^2-(x^3)^2\right]dx$
Explanation
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.
Select the correct setup for the volume when the region between $y=\sqrt{4-x^2}$ and $y=0$ on $0,2$ is revolved about $y=1$.
$V=\pi\displaystyle\int_{0}^{2}\left[(\sqrt{4-x^2})^2-(0)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(1-\sqrt{4-x^2})^2-(1-0)^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(1-0)^2-(1-\sqrt{4-x^2})^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(1-0)^2-(\sqrt{4-x^2})^2\right]dx$
$V=\pi\displaystyle\int_{0}^{2}\left[(\sqrt{4-x^2}-1)^2-(0-1)^2\right]dx$
Explanation
This problem requires the washer method with revolution about y = 1, above the semicircle region. When revolving about y = 1, the outer radius is the distance from y = 1 to the x-axis: (1 - 0) = 1, and the inner radius is the distance from y = 1 to the semicircle: (1 - √(4-x²)). The washer integral is V = π∫[(1 - 0)² - (1 - √(4-x²))²]dx = π∫[1² - (1 - √(4-x²))²]dx. Choice B incorrectly reverses the subtraction in the outer radius term, writing it as (1 - √(4-x²))² - (1 - 0)². When the axis is above the region, outer radius comes from the bottom boundary and inner radius from the top boundary.
Region bounded by $y=2$ and $y=x^2$ for $0\le x\le\sqrt2$ is revolved about $y=-2$. Which setup is correct?
$V=\pi\int_{0}^{\sqrt2}\big[(x^2+2)^2-(4)^2\big]dx$
$V=\pi\int_{0}^{\sqrt2}\big[2^2-(x^2)^2\big]dx$
$V=\pi\int_{0}^{\sqrt2}\big[(4)-(x^2+2)\big]^2dx$
$V=\pi\int_{0}^{\sqrt2}\big[(2+2)^2-(x^2+2)^2\big]dx$
$V=\pi\int_{0}^{\sqrt2}\big[(-2-2)^2-(-2-x^2)^2\big]dx$
Explanation
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=-2, both radii are curve plus 2 since above, giving $2 + 2 = 4$ and $x^2 + 2$. The outer is 4. This captures the shift. A tempting distractor like option B fails by swapping. A transferable strategy for shifted washer problems is to add shift for below-axis rotations.
The region between $y=\tan x$ and $y=0$ on $0\le x\le\pi/6$ is revolved about $y=3$. Which integral represents the volume?
$V=\pi\int_{0}^{\pi/6}\big[(\tan x-3)^2-(0-3)^2\big]dx$
$V=\pi\int_{0}^{\pi/6}\big[(3-\tan x)^2-(3-0)^2\big]dx$
$V=\pi\int_{0}^{\pi/6}\tan^2x,dx$
$V=\pi\int_{0}^{\pi/6}\big[(3-\tan x)-(3-0)\big]^2dx$
$V=\pi\int_{0}^{\pi/6}\big[(3-0)^2-(3-\tan x)^2\big]dx$
Explanation
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=3, both radii are 3 minus curves since below, giving 3 - 0 and 3 - tan x. The outer is 3. This is correct. A tempting distractor like option A fails by reversing. A transferable strategy for shifted washer problems is to ensure positive by proper order.
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.
$V=\pi\int_{0}^{1}\big[(\sqrt{1-x^2}+1)^2-(0+1)^2\big]dx$
$V=\pi\int_{0}^{1}\big[(\sqrt{1-x^2}+1)-(1)\big]^2dx$
$V=\pi\int_{0}^{1}\big[(\sqrt{1-x^2})^2-0^2\big]dx$
$V=\pi\int_{0}^{1}\big[(-1-\sqrt{1-x^2})^2-(-1-0)^2\big]dx$
$V=\pi\int_{0}^{1}\big[(0+1)^2-(\sqrt{1-x^2}+1)^2\big]dx$
Explanation
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.
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?
$V=\pi\int_{0}^{2}\big[(5-x)^2-(5-(4-x))^2\big]dx$
$V=\pi\int_{0}^{2}\big[(x-5)^2-((4-x)-5)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-(4-x))^2-(5-x)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(4-x)^2-x^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-(4-x))-(5-x)\big]^2dx$
Explanation
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 $y=\frac{x}{2}$ and $y=\frac{x}{4}+1$ for $0\le x\le4$ is revolved about $y=-1$. Which setup is correct?
$V=\pi\int_{0}^{4}\big[(x/2+1)^2-(x/4+2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(x/4+1)^2-(x/2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(x/4+2)^2-(x/2+1)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(-1-x/2)^2-(-1-(x/4+1))^2\big]dx$
$V=\pi\int_{0}^{4}\big[(x/4+2)-(x/2+1)\big]^2dx$
Explanation
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, add 1, outer x/4 +1 +1 = x/4 +2, inner x/2 +1. Yes B. 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?
$V=\pi\int_{0}^{2}\big[(5-\sqrt{x+2})^2-(5-1)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(\sqrt{x+2}-5)^2-(1-5)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-1)^2-(5-\sqrt{x+2})^2\big]dx$
$V=\pi\int_{0}^{2}\big[(\sqrt{x+2})^2-1^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-\sqrt{x+2})-(5-1)\big]^2dx$
Explanation
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.