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

Which integral sets up the volume when the region between $y=3-x$ and $y=1$ on $[0,2]$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{2}\big[(3-x)^2-(1)^2\big]dx$ 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.
Revolve the region between $x=1$ and $x=1+y^2$ about the $y$ -axis for $-1\le y\le1$ ; choose the correct washer setup.
Answer: $V=\pi\displaystyle\int_{-1}^{1}\big((1+y^2)^2-1^2\big)\,dy$ 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.
Rotate the region between $y=\sin x$ and $y=0$ on $0\le x\le\pi$ about the $x$ -axis; choose the correct setup.
Answer: $V=\pi\displaystyle\int_{0}^{\pi}\big[(\sin x)^2-0^2\big]dx$ This problem uses the washer method for rotating about the x-axis, though it simplifies to the disk method since the inner boundary is y = 0 (the x-axis itself). The sine curve y = sin x lies above y = 0 throughout [0, π], making sin x the outer radius and 0 the inner radius. The washer formula V = π∫[R²(x) - r²(x)]dx gives V = π∫₀^π[(sin x)² - 0²]dx. Choice A shows π∫(sin x - 0)dx = π∫sin x dx, which is missing the squaring entirely and would give the area under the curve times π, not the volume of revolution. The fundamental rule: volumes of revolution always involve squared radii, whether using disks (R²) or washers (R² - r²).
Which integral sets up the volume when the region between $y=x+1$ and $y=3-x$ on $[0,1]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{0}^{1}\big[(3-x)^2-(x+1)^2\big]dx$ This problem uses the washer method for revolving around the x-axis. For the washer method, we use π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. To find which function is farther from the x-axis on [0,1], evaluate at a test point: at x = 0, y = x + 1 gives 1 and y = 3 - x gives 3; at x = 1, y = x + 1 gives 2 and y = 3 - x gives 2. Since 3 - x ≥ x + 1 on [0,1], the line y = 3 - x is the outer radius and y = x + 1 is the inner radius. The correct integral is π∫₀¹[(3-x)² - (x+1)²]dx. Choice D incorrectly squares the difference (3-x) - (x+1) = 2-2x instead of taking the difference of the squares. Remember: the washer method requires the difference of squared radii, not the square of the difference of radii.
Rotate the region between $y=3$ and $y=x^2$ from $x=-\sqrt3$ to $x=\sqrt3$ about the $x$ -axis; choose the setup.
Answer: $V=\pi\displaystyle\int_{-\sqrt3}^{\sqrt3}\big[3^2-(x^2)^2\big]dx$ This problem uses the washer method for rotating about the x-axis. To identify the outer radius, note that y = 3 is a horizontal line while y = x² is a parabola opening upward with vertex at the origin. At x = ±1, we have y = 1 for the parabola and y = 3 for the line, and at x = ±√3, both curves meet at y = 3. Since 3 > x² for all x in (-√3, √3), the outer radius is R(x) = 3 and the inner radius is r(x) = x². The washer method gives V = π∫₋√₃^√₃[3² - (x²)²]dx = π∫₋√₃^√₃[9 - x⁴]dx. Choice D incorrectly uses (3 - x²)², which squares the difference rather than taking the difference of squares, fundamentally changing the solid's volume. Remember: washers always use R² - r², representing the area of an annular cross-section.
What washer integral gives the volume when the region between $y=2x$ and $y=x^2$ on $[0,2]$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{2}\big[(2x)^2-(x^2)^2\big]dx$ 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.
Revolve the region between $y=2x$ and $y=x^2$ about the $x$ -axis from $x=0$ to $x=2$ ; pick the correct setup.
Answer: $V=\pi\displaystyle\int_{0}^{2}\big((2x)^2-(x^2)^2\big)\,dx$ 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.
What washer-method integral gives the volume when the region between $y=\sin x$ and $y=\cos x$ on $[0,\tfrac{\pi}{4}]$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{\pi/4}\big[(\cos x)^2-(\sin x)^2\big]dx$ 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, π/4], the outer radius is the distance from the x-axis to the farther curve y=cos x, while the inner radius is to the closer curve y=sin x. Since y=cos x is above y=sin x in this interval, the outer radius is cos x and the inner is sin x. Squaring these radii and subtracting gives the integrand (cos x)^2 - (sin x)^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.
Which integral sets up the volume when the region between $x=y^2$ and $x=2y$ for $0\le y\le 2$ is revolved about the $y$ -axis?
Answer: $\pi\displaystyle\int_{0}^{2}\big[(2y)^2-(y^2)^2\big]dy$ 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.
Which integral gives the volume when the region between $y=\ln x$ and $y=0$ on $[1,e]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{1}^{e}\big[(\ln x)^2-(0)^2\big]dx$ This problem uses the washer method for revolving around the x-axis. For revolution around the x-axis, we use π∫[R(x)]² - [r(x)]² dx, where R(x) is the outer radius and r(x) is the inner radius. On the interval [1,e], ln(1) = 0 and ln(e) = 1, so ln x goes from 0 to 1. Since ln x ≥ 0 on [1,e], the curve y = ln x is above the x-axis (y = 0), making ln x the outer radius and 0 the inner radius. The correct integral is π∫₁ᵉ[(ln x)² - (0)²]dx = π∫₁ᵉ(ln x)²dx. Choice A incorrectly has (0)² - (ln x)², which would give a negative integrand since it subtracts the larger squared value. When setting up washers, always ensure the outer radius squared comes first in the subtraction to avoid negative volumes.
Revolve the region between $y=e^x$ and $y=1$ about the $x$ -axis for $0\le x\le \ln 3$ ; choose the washer setup.
Answer: $V=\pi\displaystyle\int_{0}^{\ln 3}\big((e^x)^2-1^2\big)\,dx$ 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.
Which integral sets up the volume when the region between $y=1$ and $y=x^2$ on $[-1,1]$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{-1}^{1}\big[(1)^2-(x^2)^2\big]dx$ 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.
What is the correct washer-method integral for the volume when the region between $y=x$ and $y=x^2$ on $[0,1]$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{1}\big[(x)^2-(x^2)^2\big]\,dx$ 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 [0,1], the outer radius is the distance from the x-axis to the farther curve y=x, while the inner radius is to the closer curve y=x². Since y=x is above y=x² in this interval, the outer radius is x and the inner is x². Squaring these radii and subtracting gives the integrand $(x)^2$ - $(x²)^2$ . 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.
Rotate the region between $y=x+1$ and $y=1$ on $0\le x\le3$ about the $x$ -axis; select the correct washer integral.
Answer: $V=\pi\displaystyle\int_{0}^{3}\big[(x+1)^2-1^2\big]dx$ This problem requires the washer method since we're rotating the region between two curves about the x-axis. The line y = x + 1 starts at y = 1 when x = 0 and rises to y = 4 when x = 3, while y = 1 is constant, so y = x + 1 is always above y = 1 on [0,3]. This makes the outer radius R(x) = x + 1 and the inner radius r(x) = 1. Applying the washer method formula V = π∫[R²(x) - r²(x)]dx gives V = π∫₀³[(x+1)² - 1²]dx. Choice C shows (x+1-1)² = x², which would give the volume of a solid with radius x, completely ignoring the hole created by the inner boundary y = 1. The decision rule: use washers (R² - r²) when rotating a region between two curves, creating a solid with a hole.
Rotate the region between $x=y+2$ and $x=2y$ on $0\le y\le2$ about the $y$ -axis; select the correct setup.
Answer: $V=\pi\displaystyle\int_{0}^{2}\big[(y+2)^2-(2y)^2\big]dy$ This problem uses the washer method for rotation about the y-axis. We need to determine which curve is farther from the y-axis: at y = 1, x = y + 2 = 3 while x = 2y = 2, and at y = 2, x = y + 2 = 4 while x = 2y = 4 (they meet at the endpoint). Since y + 2 ≥ 2y when y ≤ 2, the line x = y + 2 forms the outer radius throughout [0, 2]. The washer formula V = π∫[R²(y) - r²(y)]dy gives V = π∫₀²[(y+2)² - (2y)²]dy. Choice C shows (2y - (y+2))² = (-y - 2)² = (y + 2)², which would give an incorrect volume by squaring the (negative) difference. When setting up washers, always ensure outer radius minus inner radius, then square each separately.
Which integral gives the volume when the region between $y=4-x^2$ and $y=x$ on $[0,2]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{0}^{2}\big[(4-x^2)^2-(x)^2\big]\,dx$ 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.
Revolve the region between $x=y$ and $x=2$ about the $y$ -axis for $0\le y\le2$ ; select the correct washer setup.
Answer: $V=\pi\displaystyle\int_{0}^{2}\big(2^2-y^2\big)\,dy$ 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=2, so radius 2, and the inner radius as the distance to x=y, so radius y. Since the revolution is around the y-axis, the cross-sections are washers with area π(outer² - inner²) = π(4 - y²). Integrate this from y=0 to y=2, the given bounds, to find the total volume. A tempting distractor like choice A swaps inner and outer radii, producing a negative integrand and invalid 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.
Which integral sets up the volume when the region between $y=x$ and $y=\tfrac{x}{2}$ on $[0,4]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{0}^{4}\big[(x)^2-(x/2)^2\big]dx$ 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.
Which integral sets up the volume when the region between $y=2x$ and $y=x^2$ on $[0,2]$ is revolved about the $x$ -axis?
Answer: $\pi\int_{0}^{2}\big[(2x)^2-(x^2)^2\big]dx$ This problem uses the washer method for revolving around the x-axis. For the washer method, 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 on [0,2], note that at x = 0 both functions equal 0, but at x = 1, y = 2x gives 2 while y = x² gives 1. The functions intersect where 2x = x², which gives x = 0 and x = 2. Since 2x ≥ x² on [0,2], the line y = 2x is the outer radius and y = x² is the inner radius. The correct integral is π∫₀²[(2x)² - (x²)²]dx = π∫₀²[4x² - x⁴]dx. Choice C incorrectly uses (2x - x²)², which squares the difference rather than taking the difference of squares. The washer method requires the difference of squared radii to find the area of the annular cross-section.
Revolve the region between $y=x$ and $y=x^2$ about the $x$ -axis for $0 \le x \le 1$ ; which washer integral is correct?
Answer: $V=\pi\displaystyle\int_{0}^{1}\big(x^2-(x^2)^2\big)\,dx$ 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.

Start this quiz