Washer Method: Revolving Around x/y Axes
Help Questions
AP Calculus BC › Washer Method: Revolving Around x/y Axes
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?
$\pi\int_{0}^{2}\big[(x)^2-(x/2)^2\big]dx$
$\pi\int_{0}^{4}\big[(x/2)^2-(x)^2\big]dx$
$\pi\int_{0}^{4}\big[x-x/2\big]^2dx$
$\pi\int_{0}^{4}\big[(x)-(x/2)\big]dx$
$\pi\int_{0}^{4}\big[(x)^2-(x/2)^2\big]dx$
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.
Which integral gives the volume when the region between $y=e^x$ and $y=1$ on $0,\ln 3$ is revolved about the $x$-axis?
$\pi\int_{0}^{\ln 3}\big[(1)^2-(e^x)^2\big]dx$
$\pi\int_{0}^{3}\big[(e^y)^2-(1)^2\big]dy$
$\pi\int_{0}^{\ln 3}\big[(e^x)-(1)\big]dx$
$\pi\int_{0}^{\ln 3}\big[(e^x)^2-(1)^2\big]dx$
$\pi\int_{0}^{\ln 3}\big[e^x-1\big]^2dx$
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.
What washer-method integral gives the volume when the region between $y=e^x$ and $y=1$ on $0,\ln 2$ is revolved about the $x$-axis?
$\pi\displaystyle\int_{0}^{\ln 2}\big[(e^x-1)^2\big]dx$
$\pi\displaystyle\int_{0}^{\ln 2}\big[(1)^2-(e^x)^2\big]dx$
$\pi\displaystyle\int_{0}^{\ln 2}\big[(e^x)^2-(1)^2\big]dx$
$\pi\displaystyle\int_{1}^{2}\big[(y)^2-(1)^2\big]dy$
$\pi\displaystyle\int_{0}^{2}\big[(e^x)^2-(1)^2\big]dx$
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, 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.
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?
$\pi\displaystyle\int_{0}^{2}\big[(2y-y^2)^2\big]dy$
$\pi\displaystyle\int_{0}^{2}\big[(y^2)^2-(2y)^2\big]dy$
$\pi\displaystyle\int_{0}^{2}\big[(2y)^2-(y^2)^2\big]dy$
$\pi\displaystyle\int_{0}^{2}\big[(2y)^2-(y^2)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(2-y)^2-(2-y^2)^2\big]dy$
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.
What washer integral gives the volume when the region between $x=1$ and $x=1+y^2$ for $-1\le y\le 1$ is revolved about the $y$-axis?
$\pi\displaystyle\int_{-1}^{1}\big[(1+y^2-1)^2\big]dy$
$\pi\displaystyle\int_{-1}^{1}\big[(1)^2-(1+y^2)^2\big]dy$
$\pi\displaystyle\int_{0}^{1}\big[(1+y^2)^2-(1)^2\big]dy$
$\pi\displaystyle\int_{-1}^{1}\big[(y^2)^2-(1)^2\big]dy$
$\pi\displaystyle\int_{-1}^{1}\big[(1+y^2)^2-(1)^2\big]dy$
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.
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?
$\pi\displaystyle\int_{0}^{2}\big[(2-x)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(3-x)^2-(1)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(1)^2-(3-x)^2\big]dx$
$\pi\displaystyle\int_{1}^{3}\big[(3-x)^2-(1)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(3-x-1)^2-(0)^2\big]dx$
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.
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?
$\pi\displaystyle\int_{0}^{2}\big[(2x-x^2)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(x^2)^2-(2x)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(2x)^2-(x^2)^2\big]dx$
$\pi\displaystyle\int_{0}^{2}\big[(2x)^2-(x^2)^2\big]dy$
$\pi\displaystyle\int_{0}^{2}\big[(2-x^2)^2-(2-2x)^2\big]dx$
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=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.
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?
$\pi\displaystyle\int_{0}^{1}\big[(1)^2-(x^2)^2\big]dx$
$\pi\displaystyle\int_{-1}^{1}\big[(x^2)^2-(1)^2\big]dx$
$\pi\displaystyle\int_{-1}^{1}\big[(1)^2-(x^2)^2\big]dx$
$\pi\displaystyle\int_{-1}^{1}\big[(1-x^2)^2-(0)^2\big]dx$
$\pi\displaystyle\int_{-1}^{1}\big[(1-x^2)^2\big]dx$
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.
Which integral sets up the volume when the region between $x=4-y^2$ and $x=0$ for $-2\le y\le 2$ is revolved about the $y$-axis?
$\pi\displaystyle\int_{-2}^{2}\big[(4-y^2)^2-(0)^2\big]dy$
$\pi\displaystyle\int_{-2}^{2}\big[(4-y^2-0)^2\big]dy$
$\pi\displaystyle\int_{-2}^{2}\big[(y^2)^2-(4)^2\big]dy$
$\pi\displaystyle\int_{0}^{4}\big[(4-y^2)^2-(0)^2\big]dy$
$\pi\displaystyle\int_{-2}^{2}\big[(0)^2-(4-y^2)^2\big]dy$
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 [-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.
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.
$V=\pi\displaystyle\int_{0}^{\ln 3}\big((e^x)^2-1^2\big),dx$
$V=\pi\displaystyle\int_{0}^{\ln 3}\big(e^x-1\big)^2,dx$
$V=\pi\displaystyle\int_{0}^{\ln 3}\big(1^2-(e^x)^2\big),dx$
$V=\pi\displaystyle\int_{0}^{3}\big((e^x)^2-1^2\big),dx$
$V=\pi\displaystyle\int_{1}^{3}\big((\ln y)^2-0^2\big),dy$
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 $y=e^x$, 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.