Washer Method: Revolving Around x/y Axes - AP Calculus AB

Which integral gives the washer-method volume when the region between $y=\sin x+4$ and $y=\cos x+2$ on $[0,\tfrac{\pi}{2}]$ is revolved about the x-axis?
Answer: $\pi\displaystyle\int_{0}^{\pi/2}\big((\sin x+4)^2-(\cos x+2)^2\big)\,dx$ This problem requires the washer method for revolving around the x-axis. For washers around the x-axis, we use π∫[R²(x) - r²(x)]dx where R(x) is the outer radius and r(x) is the inner radius. We must determine which curve is farther from the x-axis on [0, π/2]. At x = 0: sin(0) + 4 = 4 and cos(0) + 2 = 3, so sin x + 4 is farther. At x = π/2: sin(π/2) + 4 = 5 and cos(π/2) + 2 = 2, confirming sin x + 4 remains the outer curve. Therefore, the integral is π∫₀^(π/2)[(sin x + 4)² - (cos x + 2)²]dx. Choice A incorrectly reverses the order, subtracting the larger radius squared from the smaller. Key insight: always verify which curve is the outer radius by checking specific x-values in the interval.
Select the washer-method setup for revolving the region between $y=\tfrac{1}{x}+4$ and $y=x+1$ on $[1,3]$ about the x-axis.
Answer: $\pi\int_{1}^{3}\left[\left(\tfrac{1}{x}+4\right)^2-(x+1)^2\right]dx$ 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, $y=\tfrac{1}{x}+4$ is above $y=x+1$ on $[1,3]$ , so the outer radius is $\tfrac{1}{x}+4$ and the inner radius is $x+1$ . Thus, the volume integral is $\pi \int_{1}^{3} \left[ \left( \tfrac{1}{x} + 4 \right)^2 - (x + 1)^2 \right] dx$ . A tempting distractor like choice D squares the difference, incorrectly adapting the formula. 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 integral gives the washer-method volume when the region between $y=\sqrt{1-x}+2$ and $y=\tfrac{1}{2}x+1$ on $[0,1]$ is revolved about the x-axis?
Answer: $\pi\displaystyle\int_{0}^{1}\big((\sqrt{1-x}+2)^2-(\tfrac{1}{2}x+1)^2\big)\,dx$ This problem applies the washer method for revolving around the x-axis. The washer formula is π∫[R²(x) - r²(x)]dx, where R(x) is the outer radius (distance from x-axis to farther curve) and r(x) is the inner radius. To identify the outer curve, we evaluate: at x = 0, y = √(1-0) + 2 = 3 and y = 0/2 + 1 = 1; at x = 1, y = √(1-1) + 2 = 2 and y = 1/2 + 1 = 1.5. Since √(1-x) + 2 > x/2 + 1 throughout [0,1], the curve y = √(1-x) + 2 forms the outer radius. The correct integral is π∫₀¹[(√(1-x) + 2)² - (x/2 + 1)²]dx. Choice C incorrectly squares the difference of the functions instead of taking the difference of their squares. Washer method summary: identify axis, determine outer curve, use difference of squares in the integrand.
What washer-method integral represents the volume when the region between $y=2x+5$ and $y=x^2+3$ on $[0,1]$ is revolved about the x-axis?
Answer: $\pi\displaystyle\int_{0}^{1}\big((2x+5)^2-(x^2+3)^2\big)\,dx$ This problem applies the washer method for revolution around the x-axis. The washer method formula is π∫[R²(x) - r²(x)]dx, where R(x) is the outer radius (farther from x-axis) and r(x) is the inner radius (nearer to x-axis). To identify which curve is outer, we check values: at x = 0, y = 2x + 5 = 5 and y = x² + 3 = 3; at x = 1, y = 2(1) + 5 = 7 and y = 1² + 3 = 4. Since 2x + 5 > x² + 3 throughout [0,1], the line y = 2x + 5 forms the outer radius. The correct integral is π∫₀¹[(2x + 5)² - (x² + 3)²]dx. Choice E incorrectly adds the squared radii instead of subtracting them. Washer method checklist: (1) identify axis of rotation, (2) determine which curve is farther from that axis, (3) use difference of squares, not sum.
Which washer-method integral represents revolving the region between $x=\cos y+6$ and $x=y+3$ on $[0,1]$ about the y-axis?
Answer: $\pi\int_{0}^{1}\left[(\cos y+6)^2-(y+3)^2\right]dy$ 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: $\cos y + 6 > y + 3$ for y in [0,1], so outer radius is $\cos y + 6$ and inner is $ y + 3 $. The volume integral is then $\pi \int_{0}^{1} [(\cos y + 6)^2 - (y + 3)^2] \, dy$ . A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, not matching 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 $\pi \int (\text{outer}^2 - \text{inner}^2) \, d(\text{variable perpendicular to axis})$ , with limits where the region exists.
Which washer-method integral represents revolving the region between $y=\ln(x+1)+3$ and $y=\tfrac{x}{2}+1$ on $[0,2]$ about the x-axis?
Answer: $\pi\int_{0}^{2}\left[\left(\ln(x+1)+3\right)^2-\left(\tfrac{x}{2}+1\right)^2\right]dx$ 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, $y=\ln(x+1)+3$ is above $y=\tfrac{x}{2}+1$ on $[0,2]$ , so the outer radius is $\ln(x+1)+3$ and the inner radius is $\tfrac{x}{2}+1$ . Thus, the volume integral is $\pi \int_{0}^{2} \left[ \left( \ln(x+1)+3 \right)^2 - \left( \tfrac{x}{2}+1 \right)^2 \right] dx$ . A tempting distractor like choice A subtracts incorrectly, leading to negative integrand. 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 integral gives the volume when the region between $y=\sqrt{x}+3$ and $y=x+1$ on $[0,1]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{0}^{1}\big[(\sqrt{x}+3)^2-(x+1)^2\big]dx$ This problem uses the washer method for rotation about the x-axis. To determine outer and inner radii, we compare the y-values of both functions throughout the interval. At x = 0: y = √x + 3 = 3 and y = x + 1 = 1; at x = 1: y = √x + 3 = 4 and y = x + 1 = 2. Since √x + 3 > x + 1 for all x in [0,1], the function y = √x + 3 forms the outer radius and y = x + 1 forms the inner radius. The washer volume formula gives π∫[(√x + 3)² - (x + 1)²]dx. Choice B incorrectly reverses the order, which would give a negative result. Remember the washer method checklist: identify rotation axis, find which function is farther from axis, square both distances, then subtract inner² from outer².
For the region between $y=x+3$ and $y=x^2+1$ on $0\le x\le1$ , rotated about the $x$ -axis, which washer integral gives the volume?
Answer: $\pi\displaystyle\int_{0}^{1}\big[(x+3)^2-(x^2+1)^2\big]\,dx$ This problem requires the washer method to find the volume when rotating the region between two curves about the x-axis. When rotating about the x-axis, we need to identify which function is farther from the axis (outer radius) and which is closer (inner radius). At x = 0: y = x + 3 gives 3, while y = x² + 1 gives 1, so the linear function y = x + 3 is above. At x = 1: y = x + 3 gives 4, while y = x² + 1 gives 2, confirming y = x + 3 remains above throughout [0,1]. The washer formula is π∫[R²(x) - r²(x)]dx where R is the outer radius and r is the inner radius, giving π∫₀¹[(x+3)² - (x²+1)²]dx. Option A incorrectly reverses the order, which would give a negative volume. Remember: for rotation about the x-axis, always square the function values to get radii, then subtract inner from outer.
Which washer-method setup gives the volume when the region between $y=3x+4$ and $y=x+2$ on $[0,1]$ is revolved about the x-axis?
Answer: $\pi\displaystyle\int_{0}^{1}\big((3x+4)^2-(x+2)^2\big)\,dx$ This problem requires the washer method for revolving around the x-axis. The washer method formula is π∫[R²(x) - r²(x)]dx, where R(x) is the outer radius (farther from x-axis) and r(x) is the inner radius. To determine which curve is outer, we check: at x = 0, y = 3(0) + 4 = 4 and y = 0 + 2 = 2; at x = 1, y = 3(1) + 4 = 7 and y = 1 + 2 = 3. Since 3x + 4 > x + 2 for all x in [0,1], the line y = 3x + 4 is the outer curve. Therefore, the integral is π∫₀¹[(3x + 4)² - (x + 2)²]dx. Choice D incorrectly squares the difference of functions (3x+4) - (x+2) = 2x+2, then squares that result. Washer method checklist: identify outer curve, square each radius separately, then subtract inner² from outer².
Which integral gives the washer-method volume when the region between $x=\tfrac{3}{2}y+2$ and $x=\tfrac{1}{2}y+1$ on $[0,4]$ is revolved about the y-axis?
Answer: $\pi\int_{0}^{4}\left[\left(\tfrac{3}{2}y+2\right)^2-\left(\tfrac{1}{2}y+1\right)^2\right]dy$ 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: $(\tfrac{3}{2}y + 2) > (\tfrac{1}{2}y + 1)$ for y in [0,4], so outer radius is $(\tfrac{3}{2}y + 2)$ and inner is $(\tfrac{1}{2}y + 1)$ . The volume integral is then $\pi$ times the integral from 0 to 4 of $[((\tfrac{3}{2}y + 2)^2 - ((\tfrac{1}{2}y + 1)^2)]$ dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, not reflecting the washer's area formula. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up $\pi \int$ (outer $^2$ - inner $^2$ ) d(variable perpendicular to axis), with limits where the region exists.
Which washer-method setup gives the volume when the region between $y=\sqrt{x}+3$ and $y=\tfrac{x}{2}+1$ on $[0,4]$ is revolved about the x-axis?
Answer: $\pi\displaystyle\int_{0}^{4}\big((\sqrt{x}+3)^2-(\tfrac{x}{2}+1)^2\big)\,dx$ This problem uses the washer method for revolution around the x-axis. When revolving around the x-axis, the washer formula is π∫[R²(x) - r²(x)]dx, where R(x) is the outer radius and r(x) is the inner radius. We need to determine which curve is farther from the x-axis on [0,4]. At x = 0: y = √x + 3 = 3 and y = x/2 + 1 = 1, so √x + 3 is farther. At x = 4: y = √4 + 3 = 5 and y = 4/2 + 1 = 3, confirming that √x + 3 remains the outer curve. Therefore, the integral is π∫₀⁴[(√x + 3)² - (x/2 + 1)²]dx. Choice D incorrectly squares the difference (outer - inner)² instead of taking the difference of squares. For washers: identify which curve is farther from the axis of rotation, then use (farther)² - (nearer)².
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.
Answer: $\pi\int_{0}^{\pi/4}\left[(\tan x+4)^2-(x+2)^2\right]dx$ 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.
Select the washer-method integral when the region between $y=\tfrac{3}{2}x+2$ and $y=\tfrac{1}{2}x+1$ on $[0,4]$ is revolved about the x-axis.
Answer: $\pi\int_{0}^{4}\left[\left(\tfrac{3}{2}x+2\right)^2-\left(\tfrac{1}{2}x+1\right)^2\right]dx$ 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: (3/2)x + 2 > (1/2)x + 1 on [0,4], so outer radius is (3/2)x + 2, inner is (1/2)x + 1. Thus, the integral is π ∫ from 0 to 4 of [ $((3/2)x + 2)^2$ - $((1/2)x + 1)^2$ ] dx. A tempting distractor is choice D, which squares the difference, erroneously simplifying the washer to a single radius. 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 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.
Answer: $\pi\int_{0}^{\pi/2}\left[(\cos x+4)^2-(\sin x+2)^2\right]dx$ 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.
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.
Answer: $\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})^2-(2+\tfrac{y}{4})^2\right]dy$ 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.
Set up the washer-method integral for volume when the region between $y=\cos x+4$ and $y=2$ on $[0,\pi/2]$ is revolved about the $x$ -axis.
Answer: $\pi\int_{0}^{\pi/2}\big[(\cos x+4)^2-2^2\big]dx$ This problem uses the washer method for revolving around the x-axis. We must compare the distances of both functions from the x-axis. The function y = cos x + 4 varies: at x = 0, cos(0) + 4 = 5; at x = π/2, cos(π/2) + 4 = 4. The horizontal line y = 2 is constant. Since cos x ∈ [0,1] on [0,π/2], we have cos x + 4 ∈ [4,5], which means cos x + 4 > 2 throughout the interval. Therefore, y = cos x + 4 forms the outer radius and y = 2 forms the inner radius. The washer formula gives π∫[(cos x + 4)² - 2²]dx. Choice B incorrectly subtracts in the wrong order. Remember: for washers, always subtract inner² from outer² to ensure positive volume.
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?
Answer: $\pi\int_{1}^{3}\left[\left(\ln y+3\right)^2-\left(\tfrac{y}{2}+1\right)^2\right]dy$ 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 for revolving the region between $x=y^2+3$ and $x=2y+6$ on $[0,2]$ about the y-axis.
Answer: $\pi\int_{0}^{2}\left[(2y+6)^2-(y^2+3)^2\right]dy$ 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: 2y + 6 > $y^2$ + 3 for y in [0,2], so outer radius is 2y + 6 and inner is $y^2$ + 3. The volume integral is then π times the integral from 0 to 2 of [ $(2y + 6)^2$ - $(y^2 + 3)^2$ ] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, leading to an incorrect volume calculation. 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 for revolving the region between $y=\ln x+4$ and $y=\tfrac{x}{2}+2$ on $[1,4]$ about the x-axis.
Answer: $\pi\int_{1}^{4}\left[\left(\ln x+4\right)^2-\left(\tfrac{x}{2}+2\right)^2\right]dx$ 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=ln x+4 is above y=(x/2)+2 on [1,4], so the outer radius is ln x+4 and the inner radius is (x/2)+2. Thus, the volume integral is π ∫ from 1 to 4 of [ $(ln x+4)^2$ - $((x/2)+2)^2$ ] dx. A tempting distractor like choice A reverses the subtraction, resulting in a negative value that doesn't represent 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.
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.
Answer: $\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)^2-(\tfrac12x+2)^2\big]dx$ 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.

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