AP Calculus BC 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 BC.

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

Explanation: This problem requires the washer method for revolving around the x-axis. The washer method uses π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. Both y = x and y = x/2 are lines through the origin, but y = x has slope 1 while y = x/2 has slope 1/2. Since x > x/2 for all x > 0, the line y = x is farther from the x-axis throughout [0,4], making it the outer radius. The line y = x/2 is closer to the x-axis, making it the inner radius. The correct integral is π∫₀⁴[(x)² - (x/2)²]dx = π∫₀⁴[x² - x²/4]dx. Choice C incorrectly uses (x - x/2)² = (x/2)², which squares the difference instead of taking the difference of squares. When using washers, always square the radii separately—this gives the true area of the washer cross-section.

Question 2

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the x-axis. To identify the outer and inner radii, note that for each x in [0,2], the outer radius is the distance from the x-axis to the farther curve y=2x, while the inner radius is to the closer curve y=x². Since y=2x is above y=x² in this interval, the outer radius is 2x and the inner is x². Squaring these radii and subtracting gives the integrand (2x)^2 - (x²)^2. A tempting distractor like choice B swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 3

1. $V=\pi\displaystyle\int_{0}^{1}\big(x^2-(x^2)^2\big)\,dx$ (correct answer)
2. $V=\pi\displaystyle\int_{0}^{1}\big((x^2)^2-x^2\big)\,dx$
3. $V=\pi\displaystyle\int_{0}^{1}\big(x-(x^2)\big)\,dx$
4. $V=\pi\displaystyle\int_{0}^{1}\big((x^2)^2-x^2\big)\,dy$
5. $V=\pi\displaystyle\int_{-1}^{1}\big(x^2-(x^2)^2\big)\,dx$

Explanation: The washer method is essential for calculating the volume of a solid of revolution with a hole when revolving a region around the x-axis or y-axis. To apply it here, identify the outer radius as the distance from the axis to the farther curve, which is $y=x$, so radius $x$, and the inner radius as the distance to $y=x^2$, so radius $x^2$. Since the revolution is around the x-axis, the cross-sections are washers with area $\pi(\text{outer}^2 - \text{inner}^2) = \pi(x^2 - x^4)$. Integrate this from $x=0$ to $x=1$, the given bounds, to find the total volume. A tempting distractor like choice B swaps inner and outer radii, leading to a negative integrand and incorrect volume. Always decide between washer and disc by checking for a hole: use washer if there's an inner and outer radius, disc if the inner radius is zero.

Question 4

1. $V=\pi\displaystyle\int_{-1}^{1}\big(1^2-(1+y^2)^2\big)\,dy$
2. $V=\pi\displaystyle\int_{-1}^{1}\big((1+y^2)^2-1^2\big)\,dy$ (correct answer)
3. $V=\pi\displaystyle\int_{0}^{1}\big((1+y^2)^2-1^2\big)\,dy$
4. $V=\pi\displaystyle\int_{-1}^{1}\big((1+y^2)-1\big)^2\,dy$
5. $V=\pi\displaystyle\int_{-1}^{1}\big(1-(1+y^2)\big)\,dy$

Explanation: The washer method is essential for calculating the volume of a solid of revolution with a hole when revolving a region around the x-axis or y-axis. To apply it here, identify the outer radius as the distance from the axis to the farther curve, which is x=1+y², so radius 1+y², and the inner radius as the distance to x=1, so radius 1. Since the revolution is around the y-axis, the cross-sections are washers with area π(outer² - inner²) = π((1+y²)² - 1). Integrate this from y=-1 to y=1, the given bounds, to find the total volume. A tempting distractor like choice C integrates only from 0 to 1, ignoring symmetry and yielding half the actual volume. Always decide between washer and disc by checking for a hole: use washer if there's an inner and outer radius, disc if the inner radius is zero.

Question 5

1. $V=\pi\displaystyle\int_{0}^{2}\big((x^2)^2-(2x)^2\big)\,dx$
2. $V=\pi\displaystyle\int_{0}^{2}\big((2x)^2-(x^2)^2\big)\,dx$ (correct answer)
3. $V=\pi\displaystyle\int_{0}^{4}\big((2\sqrt{y})^2-y^2\big)\,dy$
4. $V=\pi\displaystyle\int_{0}^{2}\big(2x-x^2\big)\,dx$
5. $V=\pi\displaystyle\int_{0}^{1}\big((2x)^2-(x^2)^2\big)\,dx$

Explanation: The washer method is essential for calculating the volume of a solid of revolution with a hole when revolving a region around the x-axis or y-axis. To apply it here, identify the outer radius as the distance from the axis to the farther curve, which is y=2x, so radius 2x, and the inner radius as the distance to y=x², so radius x². Since the revolution is around the x-axis, the cross-sections are washers with area π(outer² - inner²) = π(4x² - x⁴). Integrate this from x=0 to x=2, the bounds where the curves intersect, to find the total volume. A tempting distractor like choice A swaps inner and outer radii, resulting in a negative value that doesn't make sense for volume. Always decide between washer and disc by checking for a hole: use washer if there's an inner and outer radius, disc if the inner radius is zero.

Question 6

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

Explanation: This problem requires the washer method since we're 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 = 1, we have y = 2(1) = 2 for the line and y = 1² = 1 for the parabola, so the line y = 2x is above y = x² throughout [0,2], making 2x the outer radius and x² the inner radius. The washer method formula is V = π∫[R²(x) - r²(x)]dx, where R is the outer radius and r is the inner radius, giving us V = π∫₀²[(2x)² - (x²)²]dx. Choice C incorrectly squares the difference of the functions rather than taking the difference of the squares, which would give the volume of a solid disk with varying radius. Remember: washers require difference of squares (R² - r²), not square of differences.

Question 7

1. $V=\pi\displaystyle\int_{0}^{4}\big[(\sqrt{y})^2-0^2\big]dy$ (correct answer)
2. $V=\pi\displaystyle\int_{0}^{4}\big[0^2-(\sqrt{y})^2\big]dy$
3. $V=\pi\displaystyle\int_{0}^{4}\big[(\sqrt{y})-0\big]^2dy$
4. $V=\pi\displaystyle\int_{0}^{4}\big[(\sqrt{y})^2-0\big]dy$
5. $V=\pi\displaystyle\int_{0}^{2}\big[(\sqrt{y})^2-0^2\big]dy$

Explanation: This problem involves the washer method for rotating about the y-axis, though it simplifies to the disk method since one boundary is the y-axis itself. The curve x = √y starts at (0,0) and reaches (2,4), always lying to the right of x = 0 (the y-axis) for y ∈ [0,4]. This makes the outer radius R(y) = √y and the inner radius r(y) = 0. The washer formula V = π∫[R²(y) - r²(y)]dy becomes V = π∫₀⁴[(√y)² - 0²]dy = π∫₀⁴[y - 0]dy. Choice C incorrectly uses (√y - 0)² = y, which happens to give the same integrand but conceptually misapplies the disk method formula. When one boundary is the axis of rotation, washers reduce to disks, but maintain the R² - r² structure for consistency.

Question 8

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

Explanation: This problem involves the washer method for rotating a region about the x-axis, though it simplifies to the disk method since one boundary is the x-axis itself. The parabola y = 4 - x² is above y = 0 throughout [-2, 2], making it the outer radius R(x) = 4 - x², while the x-axis gives inner radius r(x) = 0. The washer formula V = π∫[R²(x) - r²(x)]dx becomes V = π∫₋₂²[(4-x²)² - 0²]dx. Choice B incorrectly uses (4-x² - 0)² = (4-x²)², which would be appropriate for the disk method but here we already have the correct disk method setup since r = 0. When one boundary is the axis of rotation itself, the washer method reduces to the disk method, but the formula structure remains V = π∫[R² - 0²]dx.

Question 9

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

Explanation: This problem requires the washer method for revolution around the x-axis. The washer method uses π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. On [0, ln 3], we need to determine which function is farther from the x-axis. Since e⁰ = 1 and e^(ln 3) = 3, the exponential function y = e^x starts at 1 and increases to 3, while y = 1 is constant. Therefore, e^x ≥ 1 throughout the interval, making e^x the outer radius and 1 the inner radius. The correct integral is π∫₀^(ln 3)[($e^x$)² - (1)²]dx. Choice D incorrectly uses ($e^x$ - 1)², squaring the difference instead of taking the difference of squares. When using washers, always square each function separately—this represents the areas of the outer and inner circles whose difference gives the washer area.

Question 10

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the x-axis. To identify the outer and inner radii, note that for each x in [0, ln2], the outer radius is the distance from the x-axis to the farther curve y=e^x, while the inner radius is to the closer curve y=1. Since y=e^x is above y=1 in this interval, the outer radius is e^x and the inner is 1. Squaring these radii and subtracting gives the integrand (e^x)^2 - (1)^2. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 11

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the y-axis. To identify the outer and inner radii, note that for each y in [0,2], the outer radius is the distance from the y-axis to the farther curve x=2y, while the inner radius is to the closer curve x=y². Since x=2y is to the right of x=y² in this interval, the outer radius is 2y and the inner is y². Squaring these radii and subtracting gives the integrand (2y)^2 - (y²)^2. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 12

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the y-axis. To identify the outer and inner radii, note that for each y in [-1,1], the outer radius is the distance from the y-axis to the farther curve x=1+y², while the inner radius is to the closer curve x=1. Since x=1+y² is to the right of x=1 in this interval, the outer radius is 1+y² and the inner is 1. Squaring these radii and subtracting gives the integrand (1+y²)^2 - (1)^2. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 13

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the x-axis. To identify the outer and inner radii, note that for each x in $[0,2]$, the outer radius is the distance from the x-axis to the farther curve $y=3-x$, while the inner radius is to the closer curve $y=1$. Since $y=3-x$ is above $y=1$ in this interval, the outer radius is $3-x$ and the inner is $1$. Squaring these radii and subtracting gives the integrand $(3-x)^2 - (1)^2$. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 14

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the x-axis. To identify the outer and inner radii, note that for each x in [-1,1], the outer radius is the distance from the x-axis to the farther curve y=1, while the inner radius is to the closer curve y=x². Since y=1 is above y=x² in this interval, the outer radius is 1 and the inner is x². Squaring these radii and subtracting gives the integrand (1)^2 - (x²)^2. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 15

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

Explanation: This problem requires the washer method to find the volume of a solid formed by revolving a region around the y-axis. To identify the outer and inner radii, note that for each y in [-2,2], the outer radius is the distance from the y-axis to the farther curve x=4-y², while the inner radius is to the closer curve x=0. Since x=4-y² is to the right of x=0 in this interval, the outer radius is 4-y² and the inner is 0. Squaring these radii and subtracting gives the integrand (4-y²)^2 - (0)^2. A tempting distractor like choice A swaps the radii, which would yield a negative volume and is incorrect. Always ensure the outer radius is larger than the inner to avoid this error. Use the washer method when the region is between two curves, creating a hole in the solid, and the disk method when it's bounded by one curve and the axis.

Question 16

1. $V=\pi\displaystyle\int_{0}^{\ln 3}\big(1^2-(e^x)^2\big)\,dx$
2. $V=\pi\displaystyle\int_{1}^{3}\big((\ln y)^2-0^2\big)\,dy$
3. $V=\pi\displaystyle\int_{0}^{\ln 3}\big((e^x)^2-1^2\big)\,dx$ (correct answer)
4. $V=\pi\displaystyle\int_{0}^{3}\big((e^x)^2-1^2\big)\,dx$
5. $V=\pi\displaystyle\int_{0}^{\ln 3}\big(e^x-1\big)^2\,dx$

Explanation: The washer method is essential for calculating the volume of a solid of revolution with a hole when revolving a region around the x-axis or y-axis. To apply it here, identify the outer radius as the distance from the axis to the farther curve, which is y=e^x, so radius e^x, and the inner radius as the distance to y=1, so radius 1. Since the revolution is around the x-axis, the cross-sections are washers with area π(outer² - inner²) = π(($e^x$)² - 1). Integrate this from x=0 to x=ln 3, the given bounds, to find the total volume. A tempting distractor like choice A swaps inner and outer radii, leading to a negative value that can't represent volume. Always decide between washer and disc by checking for a hole: use washer if there's an inner and outer radius, disc if the inner radius is zero.

Question 17

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

Explanation: This problem requires the washer method to find the volume when revolving around the x-axis. When revolving around the x-axis, we use $\pi \int [R(x)]^2 - [r(x)]^2 \, dx$, where R(x) is the outer radius (distance from x-axis to the farther curve) and r(x) is the inner radius (distance from x-axis to the nearer curve). Since $y = 4 - x^2$ starts at y = 4 when x = 0 and $y = x$ starts at y = 0 when x = 0, the parabola $y = 4 - x^2$ is farther from the x-axis throughout [0,2], making it the outer radius. The line $y = x$ is closer to the x-axis, making it the inner radius. Choice C incorrectly squares the difference of the functions rather than taking the difference of the squares, which would give the wrong volume formula. Remember: for washers, always square each radius separately, then subtract—never square the difference of the radii.

Question 18

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

Explanation: This problem requires the washer method for revolution around the x-axis. The washer method uses π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius (farther from x-axis) and r(x) is the inner radius (closer to x-axis). On the interval [0, π/4], we need to determine which function is farther from the x-axis. At x = 0, cos(0) = 1 and sin(0) = 0, so cosine starts higher. At x = π/4, both equal √2/2, and throughout [0, π/4], cos x ≥ sin x. Therefore, cos x is the outer radius and sin x is the inner radius. Choice C incorrectly uses (cos x - sin x)², which squares the difference rather than taking the difference of squares. The washer method requires squaring each radius separately to find the area of the annular cross-section, not squaring their difference.

Question 19

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

Explanation: This problem uses the washer method for revolving around the x-axis. For the washer method around the x-axis, we need π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. To determine which function is farther from the x-axis, we can check at a test point like x = 0.5: √0.5 ≈ 0.707 while (0.5)² = 0.25, so y = √x is above y = x². This makes √x the outer radius R(x) and x² the inner radius r(x). The correct integral is π∫₀¹[(√x)² - (x²)²]dx = π∫₀¹[x - x⁴]dx. Choice C makes the common error of squaring the difference (√x - x²)² instead of taking the difference of squares. When using the washer method, always remember to square each function separately before subtracting—this gives the area of the washer cross-section.

Question 20

1. $V=\pi\displaystyle\int_{0}^{2}\big((x^2)^2-4^2\big)\,dx$
2. $V=\pi\displaystyle\int_{-2}^{2}\big(4^2-(x^2)^2\big)\,dx$ (correct answer)
3. $V=\pi\displaystyle\int_{0}^{4}\big(2^2-(\sqrt{y})^2\big)\,dy$
4. $V=\pi\displaystyle\int_{0}^{2}\big(4^2-(x^2)^2\big)\,dx$
5. $V=\pi\displaystyle\int_{-2}^{2}\big((x^2)^2-4^2\big)\,dx$

Explanation: The washer method is essential for calculating the volume of a solid of revolution with a hole when revolving a region around the x-axis or y-axis. To apply it here, identify the outer radius as the distance from the axis to the farther curve, which is y=4, so radius 4, and the inner radius as the distance to the closer curve, y=x², so radius x². Since the revolution is around the x-axis, the cross-sections are washers with area π(outer² - inner²) = π(16 - x⁴). Integrate this from x=-2 to x=2, the bounds where the curves intersect, to find the total volume. A tempting distractor like choice D integrates from 0 to 2 without doubling, failing to account for the full symmetric region and thus underestimating the volume. Always decide between washer and disc by checking for a hole: use washer if there's an inner and outer radius, disc if the inner radius is zero.