AP Calculus AB Quiz: Washer Method Revolving Around Xy Axes

What this quiz covers

This quiz focuses on Washer Method Revolving Around Xy Axes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\pi\displaystyle\int_{0}^{1}\big[(\tfrac12x+2)^2-(\ln(x+1)+3)^2\big]dx$
2. $\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)^2-(\tfrac12x+2)^2\big]dx$ (correct answer)
3. $\pi\displaystyle\int_{0}^{1}(\ln(x+1)+3)^2dx$
4. $\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)-(\tfrac12x+2)\big]^2dx$
5. $\pi\displaystyle\int_{0}^{1}\big[(\ln(x+1)+3)^2+(\tfrac12x+2)^2\big]dx$

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.

Question 2

1. $\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2-(x^3+1)^2\big]dx$ (correct answer)
2. $\pi\displaystyle\int_{0}^{1}\big[(x^3+1)^2-(2x+4)^2\big]dx$
3. $\pi\displaystyle\int_{0}^{1}\big[(2x+4)-(x^3+1)\big]^2dx$
4. $\pi\displaystyle\int_{0}^{1}(2x+4)^2dx$
5. $\pi\displaystyle\int_{0}^{1}\big[(2x+4)^2+(x^3+1)^2\big]dx$

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.

Question 3

1. $\pi\int_{0}^{\pi/2}\left[(\cos x+1)^2-(\sin x+3)^2\right]dx$
2. $\pi\int_{0}^{\pi/2}(\sin x+3)^2dx$
3. $\pi\int_{0}^{\pi/2}\left[(\sin x+3)^2-(\cos x+1)^2\right]dx$ (correct answer)
4. $\pi\int_{0}^{\pi/2}\left[(\sin x+3)-(\cos x+1)\right]^2dx$
5. $\pi\int_{0}^{\pi/2}\left[(\sin x+3)^2+(\cos x+1)^2\right]dx$

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=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.

Question 4

1. $\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)^2-\left(\sqrt{y+1}+2\right)^2\right]dy$ (correct answer)
2. $\pi\int_{0}^{3}\left[\left(\sqrt{y+1}+2\right)^2-\left(\tfrac{y}{2}+4\right)^2\right]dy$
3. $\pi\int_{0}^{3}\left(\tfrac{y}{2}+4\right)^2dy$
4. $\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)-\left(\sqrt{y+1}+2\right)\right]^2dy$
5. $\pi\int_{0}^{3}\left[\left(\tfrac{y}{2}+4\right)^2+\left(\sqrt{y+1}+2\right)^2\right]dy$

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.

Question 5

1. $\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})^2-(2+\tfrac{y}{4})^2\right]dy$ (correct answer)
2. $\pi\int_{0}^{4}\left[(2+\tfrac{y}{4})^2-(6-\tfrac{y}{2})^2\right]dy$
3. $\pi\int_{0}^{4}(6-\tfrac{y}{2})^2dy$
4. $\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})-(2+\tfrac{y}{4})\right]^2dy$
5. $\pi\int_{0}^{4}\left[(6-\tfrac{y}{2})^2+(2+\tfrac{y}{4})^2\right]dy$

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: 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.

Question 6

1. $\pi\int_{1}^{2}\left[(y+2)^2-\left(\tfrac{1}{y}+4\right)^2\right]dy$
2. $\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)^2-(y+2)^2\right]dy$ (correct answer)
3. $\pi\int_{1}^{2}\left(\tfrac{1}{y}+4\right)^2dy$
4. $\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)-(y+2)\right]^2dy$
5. $\pi\int_{1}^{2}\left[\left(\tfrac{1}{y}+4\right)^2+(y+2)^2\right]dy$

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.

Question 7

1. $\pi\int_{1}^{2}\left[\left(\tfrac{y}{2}+2\right)^2-\left(\tfrac{2}{y}+5\right)^2\right]dy$
2. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$ (correct answer)
3. $\pi\int_{1}^{2}\left(\tfrac{2}{y}+5\right)^2dy$
4. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
5. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$

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.

Question 8

1. $\pi\int_{0}^{3}\left[\left(\tfrac{x}{2}+1\right)^2-\left(\sqrt{x+1}+2\right)^2\right]dx$
2. $\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)^2-\left(\tfrac{x}{2}+1\right)^2\right]dx$ (correct answer)
3. $\pi\int_{0}^{3}\left(\sqrt{x+1}+2\right)^2dx$
4. $\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)-\left(\tfrac{x}{2}+1\right)\right]^2dx$
5. $\pi\int_{0}^{3}\left[\left(\sqrt{x+1}+2\right)^2+\left(\tfrac{x}{2}+1\right)^2\right]dx$

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.

Question 9

1. $\pi\int_{0}^{1}\left[(x+2)^2-(e^{2x}+3)^2\right]dx$
2. $\pi\int_{0}^{1}\left[(e^{2x}+3)^2-(x+2)^2\right]dx$ (correct answer)
3. $\pi\int_{0}^{1}(e^{2x}+3)^2dx$
4. $\pi\int_{0}^{1}\left[(e^{2x}+3)-(x+2)\right]^2dx$
5. $\pi\int_{0}^{1}\left[(e^{2x}+3)^2+(x+2)^2\right]dx$

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.

Question 10

1. $\pi\int_{0}^{\pi/4}\left[(x+2)^2-(\tan x+4)^2\right]dx$
2. $\pi\int_{0}^{\pi/4}(\tan x+4)^2dx$
3. $\pi\int_{0}^{\pi/4}\left[(\tan x+4)^2-(x+2)^2\right]dx$ (correct answer)
4. $\pi\int_{0}^{\pi/4}\left[(\tan x+4)-(x+2)\right]^2dx$
5. $\pi\int_{0}^{\pi/4}\left[(\tan x+4)^2+(x+2)^2\right]dx$

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.

Question 11

1. $\pi\int_{0}^{\pi/2}\left[(\cos x+4)^2-(\sin x+2)^2\right]dx$ (correct answer)
2. $\pi\int_{0}^{\pi/2}\left[(\sin x+2)^2-(\cos x+4)^2\right]dx$
3. $\pi\int_{0}^{\pi/2}(\cos x+4)^2dx$
4. $\pi\int_{0}^{\pi/2}\left[(\cos x+4)-(\sin x+2)\right]^2dx$
5. $\pi\int_{0}^{\pi/2}\left[(\cos x+4)^2+(\sin x+2)^2\right]dx$

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=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.

Question 12

1. $\pi\int_{0}^{\pi/2}\big[(\cos x+4)^2-2^2\big]dx$ (correct answer)
2. $\pi\int_{0}^{\pi/2}\big[2^2-(\cos x+4)^2\big]dx$
3. $\pi\int_{0}^{\pi/2}(\cos x+4)^2\,dx$
4. $\pi\int_{0}^{\pi/2}2^2\,dx$
5. $\int_{0}^{\pi/2}\big[(\cos x+4)-2\big]^2dx$

Explanation: 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.

Question 13

1. $\pi\int_{0}^{\pi}\left[\left(\tfrac{y}{2}+2\right)^2-(\sin y+5)^2\right]dy$
2. $\pi\int_{0}^{\pi}(\sin y+5)^2dy$
3. $\pi\int_{0}^{\pi}\left[(\sin y+5)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$ (correct answer)
4. $\pi\int_{0}^{\pi}\left[(\sin y+5)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
5. $\pi\int_{0}^{\pi}\left[(\sin y+5)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$

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: sin y + 5 > y/2 + 2 for y in [0,π], so outer radius is sin y + 5 and inner is y/2 + 2. The volume integral is then π times the integral from 0 to π of [(sin 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, leading to an erroneous result. 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.

Question 14

1. $\pi\int_{0}^{1}\left[(3y+1)^2-(e^y+2)^2\right]dy$
2. $\pi\int_{0}^{1}\left[(e^y+2)^2-(3y+1)^2\right]dy$ (correct answer)
3. $\pi\int_{0}^{1}(e^y+2)^2dy$
4. $\pi\int_{0}^{1}\left[(e^y+2)-(3y+1)\right]^2dy$
5. $\pi\int_{0}^{1}\left[(e^y+2)^2+(3y+1)^2\right]dy$

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=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.

Question 15

1. $\pi\int_{1}^{9}\left[\left(\tfrac12 x+1\right)^2-\left(\sqrt{x}+3\right)^2\right]dx$
2. $\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)^2-\left(\tfrac12 x+1\right)^2\right]dx$ (correct answer)
3. $\pi\int_{1}^{9}\left(\sqrt{x}+3\right)^2dx$
4. $\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)-\left(\tfrac12 x+1\right)\right]^2dx$
5. $\pi\int_{1}^{9}\left[\left(\sqrt{x}+3\right)^2+\left(\tfrac12 x+1\right)^2\right]dx$

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=√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.

Question 16

1. $\pi\int_{1}^{2}\left[\left(\tfrac{x}{2}+2\right)^2-\left(\tfrac{2}{x}+5\right)^2\right]dx$
2. $\pi\int_{1}^{2}\left(\tfrac{2}{x}+5\right)^2dx$
3. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{x}+5\right)^2-\left(\tfrac{x}{2}+2\right)^2\right]dx$ (correct answer)
4. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{x}+5\right)-\left(\tfrac{x}{2}+2\right)\right]^2dx$
5. $\pi\int_{1}^{2}\left[\left(\tfrac{2}{x}+5\right)^2+\left(\tfrac{x}{2}+2\right)^2\right]dx$

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: 2/x + 5 > x/2 + 2 on [1,2], so outer radius is 2/x + 5, inner is x/2 + 2. Thus, the integral is π ∫ from 1 to 2 of [(2/x + 5)^2 - (x/2 + 2)^2] dx. A tempting distractor is choice D, which squares the difference, failing to distinguish between washer and other integration methods. 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.

Question 17

1. $\pi\int_{0}^{1}\big[(\ln(x+1)+1)^2-(e^x+2)^2\big]dx$
2. $\pi\int_{0}^{1}(e^x+2)^2\,dx$
3. $\pi\int_{0}^{1}\big[(e^x+2)^2-(\ln(x+1)+1)^2\big]dx$ (correct answer)
4. $\int_{0}^{1}\big[(e^x+2)-(\ln(x+1)+1)\big]^2dx$
5. $\pi\int_{0}^{1}(\ln(x+1)+1)^2\,dx$

Explanation: This problem requires the washer method for revolving around the x-axis. To identify outer and inner radii, we compare the functions' y-values on the given interval. At x = 0: e⁰ + 2 = 3 and ln(1) + 1 = 1; at x = 1: e¹ + 2 ≈ 4.72 and ln(2) + 1 ≈ 1.69. Since e^x + 2 > ln(x+1) + 1 throughout [0,1], the function y = e^x + 2 forms the outer radius. The washer volume formula is π∫[(e^x + 2)² - (ln(x+1) + 1)²]dx. Choice A incorrectly places the inner radius first, producing a negative integrand. For washer method success: determine axis of rotation, find which curve is farther from axis, square both radii, then compute π∫[outer² - inner²]dx.

Question 18

1. $\pi\int_{0}^{\pi/2}\big[(\cos x+1)^2-(\sin x+3)^2\big]dx$
2. $\pi\int_{0}^{\pi/2}(\sin x+3)^2\,dx$
3. $\pi\int_{0}^{\pi/2}\big[(\sin x+3)^2-(\cos x+1)^2\big]dx$ (correct answer)
4. $\int_{0}^{\pi/2}\big[(\sin x+3)-(\cos x+1)\big]^2dx$
5. $\pi\int_{0}^{\pi/2}(\cos x+1)^2\,dx$

Explanation: This problem applies the washer method for rotation about the x-axis. We must determine which function creates the outer radius by comparing their distances from the x-axis. At x = 0: sin(0) + 3 = 3 and cos(0) + 1 = 2; at x = π/2: sin(π/2) + 3 = 4 and cos(π/2) + 1 = 1. Throughout [0, π/2], we have sin x + 3 > cos x + 1, making y = sin x + 3 the outer radius. The washer formula π∫[R²(x) - r²(x)]dx gives π∫[(sin x + 3)² - (cos x + 1)²]dx. Choice A incorrectly subtracts the larger value from the smaller, yielding negative volume. The washer setup process: (1) identify rotation axis, (2) compare function values to find outer curve, (3) apply formula π∫[outer² - inner²]dx.

Question 19

1. $\pi\displaystyle\int_{0}^{1}(6-x^2)^2dx$
2. $\pi\displaystyle\int_{0}^{1}\big[(2x+1)^2-(6-x^2)^2\big]dx$
3. $\pi\displaystyle\int_{0}^{1}\big[(6-x^2)^2-(2x+1)^2\big]dx$ (correct answer)
4. $\pi\displaystyle\int_{0}^{1}\big[(6-x^2)-(2x+1)\big]^2dx$
5. $\pi\displaystyle\int_{0}^{1}\big[(6-x^2)^2+(2x+1)^2\big]dx$

Explanation: This problem requires the washer method for rotation about the x-axis with a parabola and linear function. We must identify which curve is farther from the x-axis on [0,1]. At x = 0: y = 6 - 0² = 6, while y = 2(0) + 1 = 1, so the parabola is above. At x = 1: y = 6 - 1² = 5, while y = 2(1) + 1 = 3, confirming y = 6 - x² remains above throughout. For rotation about the x-axis, the outer radius is (6 - x²) and the inner radius is (2x + 1), giving the washer integral π∫₀¹[(6 - x²)² - (2x + 1)²]dx. Option B incorrectly places (2x + 1)² first, reversing the subtraction and producing a negative result. Remember the washer formula structure: π∫[R²(x) - r²(x)]dx where R is the outer (farther) radius.

Question 20

1. $\pi\displaystyle\int_{0}^{2}\big[(5-x)^2+(2+\tfrac12x)^2\big]dx$
2. $\pi\displaystyle\int_{0}^{2}\big[(5-x)^2-(2+\tfrac12x)^2\big]dx$ (correct answer)
3. $\pi\displaystyle\int_{0}^{2}(5-x)^2dx$
4. $\pi\displaystyle\int_{0}^{2}\big[(2+\tfrac12x)^2-(5-x)^2\big]dx$
5. $\pi\displaystyle\int_{0}^{2}\big[(5-x)-(2+\tfrac12x)\big]^2dx$

Explanation: This problem applies the washer method for rotation about the x-axis. We must determine which curve is farther from the x-axis to identify the outer radius. At x = 0: y = 5 - x = 5, while y = 2 + ½x = 2, so y = 5 - x is above. At x = 2: y = 5 - x = 3, while y = 2 + ½x = 3, so the curves meet at the right endpoint. Throughout [0,2], y = 5 - x ≥ y = 2 + ½x, making (5 - x) the outer radius and (2 + ½x) the inner radius. The washer integral becomes π∫₀²[(5 - x)² - (2 + ½x)²]dx. Option D reverses the subtraction order, which would give a negative volume since inner² > outer² is impossible. Key steps: compare y-values, identify outer curve, use π∫[outer² - inner²]dx.