Washer Method: Revolving Around x/y Axes
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AP Calculus AB › Washer Method: Revolving Around x/y Axes
Select the washer-method integral for revolving the region between $y=e^{2x}+3$ and $y=x+2$ on $0,1$ about the x-axis.
$\pi\int_{0}^{1}\left[(e^{2x}+3)^2+(x+2)^2\right]dx$
$\pi\int_{0}^{1}\left[(e^{2x}+3)^2-(x+2)^2\right]dx$
$\pi\int_{0}^{1}(e^{2x}+3)^2dx$
$\pi\int_{0}^{1}\left[(x+2)^2-(e^{2x}+3)^2\right]dx$
$\pi\int_{0}^{1}\left[(e^{2x}+3)-(x+2)\right]^2dx$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the x-axis, the outer radius is the y-value of the upper curve, and the inner radius is the y-value of the lower curve. Here, $y=e^{2x}$+3 is above y=x+2 on [0,1], so the outer radius is $e^{2x}$+3 and the inner radius is x+2. Thus, the volume integral is π ∫ from 0 to 1 of $[(e^{2x}$$+3)^2$ - $(x+2)^2$] dx. A tempting distractor like choice D squares the difference, incorrectly formulating the volume. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Find the washer-method integral when the region between $x=\tfrac{2}{y}+5$ and $x=\tfrac{y}{2}+2$ on $1,2$ is revolved about the y-axis.
$\pi\int_{1}^{2}\left[\left(\tfrac{y}{2}+2\right)^2-\left(\tfrac{2}{y}+5\right)^2\right]dy$
$\pi\int_{1}^{2}\left(\tfrac{2}{y}+5\right)^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$
Explanation
The washer method is used to find the volume of a solid of revolution with a hole, formed by revolving a region between two curves around an axis. In this case, revolving around the y-axis, the radii are the x-values of the curves. To determine which is the outer and inner radius, compare the two functions: 2/y + 5 > y/2 + 2 for y in [1,2], so outer radius is 2/y + 5 and inner is y/2 + 2. The volume integral is then π times the integral from 1 to 2 of [(2/y + $5)^2$ - (y/2 + $2)^2$] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, not correctly forming the washer area. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up π ∫ $(outer^2$ - $inner^2$) d(variable perpendicular to axis), with limits where the region exists.
The region between $y=2x+4$ and $y=x^3+1$ for $0\le x\le1$ is rotated about the $x$-axis; select the washer setup.
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2+(x^3+1)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)-(x^3+1)\big]^2dx$
$\pi\displaystyle\int_{0}^{1}(2x+4)^2dx$
$\pi\displaystyle\int_{0}^{1}\big[(x^3+1)^2-(2x+4)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2-(x^3+1)^2\big]dx$
Explanation
This problem uses the washer method for rotation about the x-axis with linear and cubic functions. We compare y-values to determine which curve forms the outer radius. At x = 0: y = 2(0) + 4 = 4, while y = 0³ + 1 = 1, so the linear function is above. At x = 1: y = 2(1) + 4 = 6, while y = 1³ + 1 = 2, confirming y = 2x + 4 remains above throughout [0,1]. For rotation about the x-axis, the outer radius is (2x + 4) and the inner radius is (x³ + 1), yielding the washer integral π∫₀¹[(2x + 4)² - (x³ + 1)²]dx. Option B incorrectly reverses the subtraction order, which would produce a negative volume. The washer method checklist: identify upper curve (larger y-values), square both functions, subtract inner² from outer², multiply by π.
Which integral gives the washer-method volume when the region between $y=2x+3$ and $y=\tfrac{x}{2}+1$ on $0,4$ is revolved about the x-axis?
$\pi\int_{0}^{4}\left[(2x+3)-\left(\tfrac{x}{2}+1\right)\right]^2dx$
$\pi\int_{0}^{4}\left[(2x+3)^2-\left(\tfrac{x}{2}+1\right)^2\right]dx$
$\pi\int_{0}^{4}(2x+3)^2dx$
$\pi\int_{0}^{4}\left[(2x+3)^2+\left(\tfrac{x}{2}+1\right)^2\right]dx$
$\pi\int_{0}^{4}\left[\left(\tfrac{x}{2}+1\right)^2-(2x+3)^2\right]dx$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the x-axis, the outer radius is the y-value of the upper curve, and the inner radius is the y-value of the lower curve. Here, y=2x+3 is above y=(x/2)+1 on [0,4], so the outer radius is 2x+3 and the inner radius is (x/2)+1. Thus, the volume integral is π ∫ from 0 to 4 of $[(2x+3)^2$ - $((x/2)+1)^2$] dx. A tempting distractor like choice B reverses the order, resulting in negative volume. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Select the washer-method setup when the region between $y=\sqrt{x}+3$ and $y=\tfrac12 x+1$ on $1,9$ is revolved about the x-axis.
$\pi\int_{1}^{9}\left[\left(\tfrac12 x+1\right)^2-\left(\sqrt{x}+3\right)^2\right]dx$
$\pi\int_{1}^{9}\left(\sqrt{x}+3\right)^2dx$
$\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)^2+\left(\tfrac12 x+1\right)^2\right]dx$
$\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)^2-\left(\tfrac12 x+1\right)^2\right]dx$
$\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)-\left(\tfrac12 x+1\right)\right]^2dx$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the x-axis, the outer radius is the y-value of the upper curve, and the inner radius is the y-value of the lower curve. Here, y=√x+3 is above y=(1/2)x+1 on [1,9], so the outer radius is √x+3 and the inner radius is (1/2)x+1. Thus, the volume integral is π ∫ from 1 to 9 of $[(√x+3)^2$ - $((1/2)x+1)^2$] dx. A tempting distractor like choice D squares the difference, which misapplies the washer formula and resembles an incorrect area computation. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Find the washer-method setup when the region between $y=\cos x+4$ and $y=\sin x+2$ on $0,\tfrac{\pi}{2}$ is revolved about the x-axis.
$\pi\int_{0}^{\pi/2}(\cos x+4)^2dx$
$\pi\int_{0}^{\pi/2}\left[(\cos x+4)-(\sin x+2)\right]^2dx$
$\pi\int_{0}^{\pi/2}\left[(\cos x+4)^2+(\sin x+2)^2\right]dx$
$\pi\int_{0}^{\pi/2}\left[(\cos x+4)^2-(\sin x+2)^2\right]dx$
$\pi\int_{0}^{\pi/2}\left[(\sin x+2)^2-(\cos x+4)^2\right]dx$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the x-axis, the outer radius is the y-value of the upper curve, and the inner radius is the y-value of the lower curve. Here, y=cos x+4 is above y=sin x+2 on [0,π/2], so the outer radius is cos x+4 and the inner radius is sin x+2. Thus, the volume integral is π ∫ from 0 to π/2 of [(cos $x+4)^2$ - (sin $x+2)^2$] dx. A tempting distractor like choice B reverses the subtraction, producing negative values. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Which washer-method setup gives the volume when the region between $x=\ln y+3$ and $x=\tfrac{y}{2}+1$ on $1,3$ is revolved about the y-axis?
$\pi\int_{1}^{3}\left[\left(\ln y+3\right)-\left(\tfrac{y}{2}+1\right)\right]^2dy$
$\pi\int_{1}^{3}(\ln y+3)^2dy$
$\pi\int_{1}^{3}\left[\left(\ln y+3\right)^2-\left(\tfrac{y}{2}+1\right)^2\right]dy$
$\pi\int_{1}^{3}\left[\left(\tfrac{y}{2}+1\right)^2-\left(\ln y+3\right)^2\right]dy$
$\pi\int_{1}^{3}\left[\left(\ln y+3\right)^2+\left(\tfrac{y}{2}+1\right)^2\right]dy$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the y-axis, the outer radius is the larger x-value, and the inner radius is the smaller x-value. Here, x=ln y+3 is to the right of x=(y/2)+1 on [1,3], so the outer radius is ln y+3 and the inner radius is (y/2)+1. Thus, the volume integral is π ∫ from 1 to 3 of [(ln $y+3)^2$ - $((y/2)+1)^2$] dy. A tempting distractor like choice A reverses the order, leading to negative volume. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Find the washer-method setup when the region between $x=\sin y+4$ and $x=\cos y+2$ on $0,\tfrac{\pi}{2}$ is revolved about the y-axis.
$\pi\int_{0}^{\pi/2}\left[(\sin y+4)^2+(\cos y+2)^2\right]dy$
$\pi\int_{0}^{\pi/2}\left[(\cos y+2)^2-(\sin y+4)^2\right]dy$
$\pi\int_{0}^{\pi/2}\left[(\sin y+4)^2-(\cos y+2)^2\right]dy$
$\pi\int_{0}^{\pi/2}\left[(\cos y+2)-(\sin y+4)\right]^2dy$
$\pi\int_{0}^{\pi/2}(\cos y+2)^2dy$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the y-axis, the outer radius is the larger x-value, and the inner radius is the smaller x-value. Here, x=sin y+4 is to the right of x=cos y+2 on [0,π/2], so the outer radius is sin y+4 and the inner radius is cos y+2. Thus, the volume integral is π ∫ from 0 to π/2 of [(sin $y+4)^2$ - (cos $y+2)^2$] dy. A tempting distractor like choice C reverses the subtraction, yielding a negative result. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Select the washer-method integral for revolving the region between $x=e^y+2$ and $x=3y+1$ on $0,1$ about the y-axis.
$\pi\int_{0}^{1}\left[(e^y+2)^2-(3y+1)^2\right]dy$
$\pi\int_{0}^{1}(e^y+2)^2dy$
$\pi\int_{0}^{1}\left[(e^y+2)^2+(3y+1)^2\right]dy$
$\pi\int_{0}^{1}\left[(3y+1)^2-(e^y+2)^2\right]dy$
$\pi\int_{0}^{1}\left[(e^y+2)-(3y+1)\right]^2dy$
Explanation
The washer method is used to find the volume of a solid formed by revolving a region between two curves around an axis, creating cross-sections that are washers. For revolution about the y-axis, the outer radius is the larger x-value, and the inner radius is the smaller x-value. Here, $x=e^y$+2 is to the right of x=3y+1 on [0,1], so the outer radius is $e^y$+2 and the inner radius is 3y+1. Thus, the volume integral is π ∫ from 0 to 1 of $[(e^y$$+2)^2$ - $(3y+1)^2$] dy. A tempting distractor like choice D uses the square of the difference, which doesn't correspond to the washer area. For washer setups, checklist: confirm the axis, select integration variable, set limits, identify outer (farther) and inner (closer) distances to axis, and subtract inner squared from outer squared.
Find the washer-method setup when the region between $y=\tfrac{x}{3}+4$ and $y=\tfrac{x^2}{9}+2$ on $0,3$ is revolved about the x-axis.
$\pi\int_{0}^{3}\left[\left(\tfrac{x^2}{9}+2\right)^2-\left(\tfrac{x}{3}+4\right)^2\right]dx$
$\pi\int_{0}^{3}\left[\left(\tfrac{x}{3}+4\right)-\left(\tfrac{x^2}{9}+2\right)\right]^2dx$
$\pi\int_{0}^{3}\left[\left(\tfrac{x}{3}+4\right)^2+\left(\tfrac{x^2}{9}+2\right)^2\right]dx$
$\pi\int_{0}^{3}\left(\tfrac{x}{3}+4\right)^2dx$
$\pi\int_{0}^{3}\left[\left(\tfrac{x}{3}+4\right)^2-\left(\tfrac{x^2}{9}+2\right)^2\right]dx$
Explanation
The washer method calculates volumes when revolving regions between curves around an axis, creating washers with outer and inner radii. Here, revolving around the x-axis, the radii are the y-values of the curves. Compare the functions: x/3 + 4 > $x^2$/9 + 2 on [0,3], so outer radius is x/3 + 4, inner is $x^2$/9 + 2. Thus, the integral is π ∫ from 0 to 3 of [(x/3 + $4)^2$ - $(x^2$/9 + $2)^2$] dx. A tempting distractor is choice D, which squares the difference, incorrectly applying a disk-like formula to the height. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up π ∫ $(outer^2$ - $inner^2$) d(variable perpendicular to axis), with limits where the region exists.