Washer Method: Revolving Around x/y Axes
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AP Calculus AB › Washer Method: Revolving Around x/y Axes
Rotate the region bounded by $y=\ln(x+1)+3$ and $y=\tfrac12x+2$ on $0\le x\le1$ about the $x$-axis; choose the setup.
$\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)-(\tfrac12x+2)\big]^2dx$
$\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)^2-(\tfrac12x+2)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}(\ln(x+1)+3)^2dx$
$\pi\displaystyle\int_{0}^{1}\big[(\tfrac12x+2)^2-(\ln(x+1)+3)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)^2+(\tfrac12x+2)^2\big]dx$
Explanation
This problem applies the washer method for rotation about the x-axis with logarithmic and linear functions. To determine outer and inner radii, we compare y-values on [0,1]. At x = 0: y = ln(1) + 3 = 0 + 3 = 3, while y = ½(0) + 2 = 2, so ln(x+1) + 3 is above. At x = 1: y = ln(2) + 3 ≈ 3.69, while y = ½(1) + 2 = 2.5, confirming ln(x+1) + 3 remains above. Since ln(x+1) + 3 > ½x + 2 throughout [0,1], the outer radius is (ln(x+1) + 3) and inner radius is (½x + 2). The washer integral is π∫₀¹[(ln(x+1) + 3)² - (½x + 2)²]dx. Option A reverses the subtraction, yielding negative volume. Key point: even with transcendental functions, the washer method follows the same pattern of outer² minus inner².
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.
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2-(x^3+1)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}\big[(x^3+1)^2-(2x+4)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)-(x^3+1)\big]^2dx$
$\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2+(x^3+1)^2\big]dx$
$\pi\displaystyle\int_{0}^{1}(2x+4)^2dx$
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.
Choose the washer-method integral for the volume when the region between $y=\sin x+3$ and $y=\cos x+1$ on $0,\tfrac{\pi}{2}$ is revolved about the x-axis.
$\pi\int_{0}^{\pi/2}\left[(\sin x+3)^2+(\cos x+1)^2\right]dx$
$\pi\int_{0}^{\pi/2}\left[(\sin x+3)-(\cos x+1)\right]^2dx$
$\pi\int_{0}^{\pi/2}\left[(\sin x+3)^2-(\cos x+1)^2\right]dx$
$\pi\int_{0}^{\pi/2}\left[(\cos x+1)^2-(\sin x+3)^2\right]dx$
$\pi\int_{0}^{\pi/2}(\sin x+3)^2dx$
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=sin x+3 is above y=cos x+1 on [0,π/2], so the outer radius is sin x+3 and the inner radius is cos x+1. Thus, the volume integral is π ∫ from 0 to π/2 of [(sin $x+3)^2$ - (cos $x+1)^2$] dx. A tempting distractor like choice A reverses the order, producing a negative integrand unsuitable for 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 $x=\sqrt{y+1}+2$ and $x=\tfrac{y}{2}+4$ on $0,3$ is revolved about the y-axis.
$\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)^2-\left(\sqrt{y+1}+2\right)^2\right]dy$
$\pi\int_{0}^{3}\left(\tfrac{y}{2}+4\right)^2dy$
$\pi\int_{0}^{3}\left[\left(\sqrt{y+1}+2\right)^2-\left(\tfrac{y}{2}+4\right)^2\right]dy$
$\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)-\left(\sqrt{y+1}+2\right)\right]^2dy$
$\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)^2+\left(\sqrt{y+1}+2\right)^2\right]dy$
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: y/2 + 4 > √(y + 1) + 2 for y in [0,3], so outer radius is y/2 + 4 and inner is √(y + 1) + 2. The volume integral is then π times the integral from 0 to 3 of [(y/2 + $4)^2$ - (√(y + 1) + $2)^2$] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, failing because it does not correctly model the washer's cross-sectional 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.
Select the washer-method setup when the region between $x=6-\tfrac{y}{2}$ and $x=2+\tfrac{y}{4}$ on $0,4$ is revolved about the y-axis.
$\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})^2-(2+\tfrac{y}{4})^2\right]dy$
$\pi\int_{0}^{4}\left[(2+\tfrac{y}{4})^2-(6-\tfrac{y}{2})^2\right]dy$
$\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})^2+(2+\tfrac{y}{4})^2\right]dy$
$\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})-(2+\tfrac{y}{4})\right]^2dy$
$\pi\int_{0}^{4}(6-\tfrac{y}{2})^2dy$
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: 6 - y/2 > 2 + y/4 for y in [0,4], so outer radius is 6 - y/2 and inner is 2 + y/4. The volume integral is then π times the integral from 0 to 4 of [(6 - $y/2)^2$ - (2 + $y/4)^2$] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, misapplying the formula for volume. 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.
Select the washer-method setup for revolving the region between $x=\tfrac{1}{y}+4$ and $x=y+2$ on $1,2$ about the y-axis.
$\pi\int_{1}^{2}\left(\tfrac{1}{y}+4\right)^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)^2-(y+2)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)-(y+2)\right]^2dy$
$\pi\int_{1}^{2}\left[(y+2)^2-\left(\tfrac{1}{y}+4\right)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)^2+(y+2)^2\right]dy$
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=(1/y)+4 is to the right of x=y+2 on [1,2], so the outer radius is (1/y)+4 and the inner radius is y+2. Thus, the volume integral is π ∫ from 1 to 2 of $[((1/y)+4)^2$ - $(y+2)^2$] dy. A tempting distractor like choice D squares the difference, which doesn't fit the washer model. 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.
$\pi\int_{1}^{2}\left[\left(\tfrac{y}{2}+2\right)^2-\left(\tfrac{2}{y}+5\right)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
$\pi\int_{1}^{2}\left(\tfrac{2}{y}+5\right)^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$
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.
Find the washer-method integral when the region between $y=\sqrt{x+1}+2$ and $y=\tfrac{x}{2}+1$ on $0,3$ is revolved about the x-axis.
$\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)-\left(\tfrac{x}{2}+1\right)\right]^2dx$
$\pi\int_{0}^{3}\left(\sqrt{x+1}+2\right)^2dx$
$\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)^2+\left(\tfrac{x}{2}+1\right)^2\right]dx$
$\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)^2-\left(\tfrac{x}{2}+1\right)^2\right]dx$
$\pi\int_{0}^{3}\left[\left(\tfrac{x}{2}+1\right)^2-\left(\sqrt{x+1}+2\right)^2\right]dx$
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 + 1) + 2 > x/2 + 1 on [0,3], so outer radius is √(x + 1) + 2, inner is x/2 + 1. Thus, the integral is π ∫ from 0 to 3 of [(√(x + 1) + $2)^2$ - (x/2 + $1)^2$] dx. A tempting distractor is choice D, which squares the difference, mistaking the washer method for a disk with radius equal to the height difference, ignoring the varying distances from the axis. 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.
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.
$\pi\int_{0}^{1}\left[(x+2)^2-(e^{2x}+3)^2\right]dx$
$\pi\int_{0}^{1}(e^{2x}+3)^2dx$
$\pi\int_{0}^{1}\left[(e^{2x}+3)^2+(x+2)^2\right]dx$
$\pi\int_{0}^{1}\left[(e^{2x}+3)-(x+2)\right]^2dx$
$\pi\int_{0}^{1}\left[(e^{2x}+3)^2-(x+2)^2\right]dx$
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, $y=e^{2x}$+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.
Select the washer-method integral when the region between $y=\tan x+4$ and $y=x+2$ on $0,\tfrac{\pi}{4}$ is revolved about the x-axis.
$\pi\int_{0}^{\pi/4}\left[(\tan x+4)^2-(x+2)^2\right]dx$
$\pi\int_{0}^{\pi/4}\left[(\tan x+4)^2+(x+2)^2\right]dx$
$\pi\int_{0}^{\pi/4}(\tan x+4)^2dx$
$\pi\int_{0}^{\pi/4}\left[(\tan x+4)-(x+2)\right]^2dx$
$\pi\int_{0}^{\pi/4}\left[(x+2)^2-(\tan x+4)^2\right]dx$
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: tan x + 4 > x + 2 on [0, π/4], so outer radius is tan x + 4, inner is x + 2. Thus, the integral is π ∫ from 0 to π/4 of [(tan x + $4)^2$ - (x + $2)^2$] dx. A tempting distractor is choice D, which squares the difference, mistaking it for a different volume method like cylindrical shells. 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.