Washer Method: Revolving Around Other Axes - AP Calculus BC

What integral gives the volume when the region between $y=x^2$ and $y=2x$ on $[0,2]$ is revolved about $y=-1$ ?
Answer: $\pi\displaystyle\int_{0}^{2}\big[(2x+1)^2-(x^2+1)^2\big]\,dx$ This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis $y = -1$ . To adjust for this axis, add 1 to each y-value to get the distances, since the curves are above the axis. The outer radius is $2x + 1$ from the upper curve, and the inner radius is $x^2 + 1$ from the lower curve. Therefore, the integral is $\pi \int_{0}^{2} [(2x + 1)^2 - (x^2 + 1)^2] \, dx$ . A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand and incorrect volume. In general, when revolving around $y = k$ , compute radii as $|y - k|$ for each curve and identify the outer as the larger distance and inner as the smaller.
What washer-method integral represents the volume when the region between $y=\cos x$ and $y=\sin x$ on $[0,\pi/4]$ rotates about $y=-1$ ?
Answer: $V=\pi\displaystyle\int_{0}^{\pi/4}\Big[(\cos x+1)^2-(\sin x+1)^2\Big]dx$ This problem involves the washer method with rotation about y = -1. When rotating about y = -1, we measure distances upward from this axis: the outer radius is from y = -1 to y = cos x, giving R = cos x - (-1) = cos x + 1, and the inner radius is from y = -1 to y = sin x, giving r = sin x - (-1) = sin x + 1. The volume integral is π∫[0,π/4][(cos x + 1)² - (sin x + 1)²]dx, since cos x ≥ sin x on [0,π/4]. Choice E incorrectly uses 1 - sin x and 1 - cos x, which would apply if rotating about y = 1 with the region above. For rotation about y = -1, add 1 to each function value to get the correct radii.
What integral setup gives the volume when the region between $y=x^2$ and $y=4$ is revolved about $y=1$ ?
Answer: $V=\pi\displaystyle\int_{-2}^{2}\Big[(4-1)^2-(x^2-1)^2\Big]dx$ This problem requires the washer method with rotation about the horizontal line $y = 1$ . When rotating about $y = 1$ , we measure distances from this axis: the outer radius is from $y = 1$ to $y = 4$ , giving $R = 4 - 1 = 3$ , and the inner radius is from $y = 1$ to $y = x^2$ , giving $r = x^2 - 1$ . The volume integral becomes $\pi \int_{-2}^{2} [(4-1)^2 - (x^2-1)^2] \, dx$ , where we integrate from $x = -2$ to $x = 2$ (the intersection points of $y = x^2$ and $y = 4$ ). Choice B incorrectly uses the original y-values without adjusting for the shifted axis. The key strategy is to subtract the axis value from each y-coordinate to find the correct radii.
What integral gives the volume when the region between $y=\ln(x+1)$ and $y=0$ , $0\le x\le e-1$ , is revolved about $y=-1$ ?
Answer: $\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)^2-(1)^2\big]\,dx$ This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = -1. To adjust for this axis, the radii are the distances from the curves to y = -1, which means adding 1 to each y-value. The outer radius is from the upper curve y = ln(x + 1) to the axis, giving ln(x + 1) + 1. The inner radius is from the lower curve y = 0 to the axis, giving 1. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.
Find the volume setup when the region between $y=x^2$ and $y=4$ for $0\le x\le2$ is revolved about $y=5$ .
Answer: $V=\pi\displaystyle\int_{0}^{2}\big[(5-x^2)^2-(5-4)^2\big]dx$ This problem requires the washer method with revolution about the horizontal line y = 5, which is above the bounded region. When revolving about y = 5, we measure distances from this axis: the outer radius is from y = 5 down to the parabola y = x², giving R = 5 - x², and the inner radius is from y = 5 down to the line y = 4, giving r = 5 - 4 = 1. The washer method formula becomes V = π∫[R² - r²]dx = π∫[(5-x²)² - 1²]dx from 0 to 2. Choice A incorrectly uses (4-5)² = 1 as the outer radius and (x²-5)² as the inner radius, reversing the distance calculations and producing negative radii. The key strategy when revolving about y = k is to always compute radii as |k - y(x)|, ensuring positive distances from the axis to each curve.
Select the washer integral for the region between $y=1$ and $y=e^{-x}$ on $0\le x\le1$ , revolved about $y=2$ .
Answer: $V=\pi\displaystyle\int_{0}^{1}\big[(2-e^{-x})^2-(2-1)^2\big]dx$ This problem involves the washer method with revolution about y = 2, where the region is between y = 1 and y = $e^{-x}$ on [0,1]. Since $e^{-x}$ decreases from 1 to 1/e on this interval and stays below y = 1, and y = 2 is above both curves, the outer radius is R = 2 - $e^{-x}$ and the inner radius is r = 2 - 1 = 1. The volume setup is V = π∫[(2- $e^{-x}$ )² - 1²]dx from 0 to 1. Choice A incorrectly uses (2-1)² as the outer radius and (2- $e^{-x}$ )² as the inner radius, reversing the proper order. When revolving about an axis above the region, identify the outer radius as the distance to the curve farthest below the axis (smallest y-value).
Which integral gives the volume when the region between $y=1-x^2$ and $y=0$ on $[-1,1]$ rotates about $y=-3$ ?
Answer: $V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2+3)^2-(0+3)^2\Big]dx$ This problem uses the washer method with rotation about y = -3. When rotating about y = -3 (below the region), we measure distances upward: the outer radius is from y = -3 to y = 1-x², giving R = (1-x²) - (-3) = 1-x² + 3 = 4-x², and the inner radius is from y = -3 to y = 0, giving r = 0 - (-3) = 3. The volume integral is π∫[-1,1][(1-x²+3)² - (0+3)²]dx = π∫[-1,1][(4-x²)² - 9]dx. Choice D incorrectly subtracts 3 instead of adding it when adjusting for the negative axis value. When the axis of rotation has a negative y-value, add its absolute value to all y-coordinates.
Which integral represents the volume when the region between $y=\sin x$ and $y=0$ on $[0,\pi]$ is revolved about $y=-2$ ?
Answer: $\pi\displaystyle\int_{0}^{\pi}\big[(\sin x+2)^2-(2)^2\big]dx$ This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis $y = -2$ . To adjust for this axis, add 2 to each y-value to get the distances, since the curves are above the axis. The outer radius is $\sin x + 2$ from the upper curve, and the inner radius is $0 + 2 = 2$ from the lower curve. Therefore, the integral is $\pi \int_0^\pi [(\sin x + 2)^2 - (2)^2] \, dx$ . A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand. In general, when revolving around $y = k$ , compute radii as $|y - k|$ for each curve and identify the outer as the larger distance and inner as the smaller.
Which integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$ on $0\le x\le4$ is revolved about $y=-2$ ?
Answer: $V=\pi\displaystyle\int_{0}^{4}\big[(\sqrt{x}+2)^2-(0+2)^2\big]dx$ This problem involves the washer method with revolution about y = -2, a horizontal line below the region between y = √x and y = 0. When revolving about y = -2, the outer radius extends from y = -2 up to y = √x, giving R = √x - (-2) = √x + 2, while the inner radius extends from y = -2 up to y = 0, giving r = 0 - (-2) = 2. The volume integral is V = π∫[(√x + 2)² - 2²]dx from 0 to 4. Choice E incorrectly subtracts 2 from both functions instead of adding 2, which would only be correct if revolving about y = 2. When the axis of revolution y = k is below the region (k < 0), add |k| to each y-value to find the correct radius.
Choose the correct volume setup when the region between $y=x^2+1$ and $y=2x+1$ rotates about $y=0$ .
Answer: $V=\pi\displaystyle\int_{0}^{2}\Big[(2x+1)^2-(x^2+1)^2\Big]dx$ This problem requires the washer method with rotation about $y = 0$ (the x-axis). When rotating about $y = 0$ , the radii are simply the y-values themselves: we need to determine which curve is farther from the x-axis. At $x = 0$ , $y = 2(0) + 1 = 1$ and $y = 0^2 + 1 = 1$ (they start equal), but at $x = 2$ , $y = 2(2) + 1 = 5$ and $y = 2^2 + 1 = 5$ (they end equal), while at $x = 1$ , $y = 2(1) + 1 = 3$ and $y = 1^2 + 1 = 2$ , showing that $2x + 1 > x^2 + 1$ on $(0,2)$ . Therefore, the outer radius is $R = 2x + 1$ and inner radius is $r = x^2 + 1$ , giving $\pi \int_{0}^{2} [(2x+1)^2 - (x^2+1)^2] \, dx$ . Choice A reverses these radii. When rotating about the x-axis, no adjustment is needed—just use the y-values directly as radii.
What integral gives the volume when the region between $y=x^3$ and $y=x$ , $0\le x\le1$ , is revolved about $y=2$ ?
Answer: $\pi\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big]\,dx$ This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 2. To adjust for this axis, the radii are the distances from the curves to y = 2, which means subtracting each y-value from 2 since the region is below. The outer radius is from the lower curve y = x³ to the axis, giving 2 - x³. The inner radius is from the upper curve y = x to the axis, giving 2 - x. A tempting distractor is choice A, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.
Which integral sets up the volume when the region between $y=\ln x$ and $y=0$ on $[1,e]$ is revolved about $y=-1$ ?
Answer: $\pi\displaystyle\int_{1}^{e}\big[(\ln x+1)^2-(1)^2\big]dx$ This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y = -1. To adjust for this axis, add 1 to each y-value to get the distances, since the curves are above the axis. The outer radius is ln x + 1 from the upper curve, and the inner radius is 0 + 1 = 1 from the lower curve. Therefore, the integral is π ∫ from 1 to e of [(ln x + 1)² - (1)²] dx. A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand. In general, when revolving around y = k, compute radii as |y - k| for each curve and identify the outer as the larger distance and inner as the smaller.
What integral gives the volume when the region between $y=e^x$ and $y=1$ , $0\le x\le1$ , is revolved about $y=0$ ?
Answer: $\pi\int_{0}^{1}\big[(e^x)^2-(1)^2\big]\,dx$ This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 0. To adjust for this axis, the radii are the distances from the curves to y = 0, which is simply the y-value since the region is above. The outer radius is from the upper curve y = $e^x$ to the axis, giving $e^x$ . The inner radius is from the lower curve y = 1 to the axis, giving 1. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.
Which integral represents the volume when the region between $y=x^3$ and $y=x$ on $[0,1]$ is revolved about $y=-3$ ?
Answer: $\pi\displaystyle\int_{0}^{1}\big[(x+3)^2-(x^3+3)^2\big]dx$ This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y = -3. To adjust for this axis, add 3 to each y-value to get the distances, since the curves are above the axis. The outer radius is x + 3 from the upper curve, and the inner radius is x³ + 3 from the lower curve. Therefore, the integral is π ∫ from 0 to 1 of [(x + 3)² - (x³ + 3)²] dx. A tempting distractor is choice B, which swaps the inner and outer radii, resulting in a negative integrand. In general, when revolving around y = k, compute radii as |y - k| for each curve and identify the outer as the larger distance and inner as the smaller.
Which integral represents the volume when the region between $y=\sqrt{x}$ and $y=0$ on $[0,4]$ rotates about $y=-2$ ?
Answer: $V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}+2)^2-(0+2)^2\Big]dx$ This problem uses the washer method with rotation about the line $y = -2$ . When rotating about $y = -2$ , we calculate distances from this axis: the outer radius extends from $y = -2$ to $y = \sqrt{x}$ , giving $R = \sqrt{x} - (-2) = \sqrt{x} + 2$ , and the inner radius extends from $y = -2$ to $y = 0$ , giving $r = 0 - (-2) = 2$ . The volume integral is $\pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - (0 + 2)^2]\, dx = \pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - 4]\, dx$ . Choice B incorrectly subtracts 2 instead of adding it when adjusting for the axis below the region. Remember that when the axis is below the region, you add the absolute value of the axis position to find radii.
Which washer integral gives the volume when the region between $y=x$ and $y=0$ on $[0,3]$ is revolved about $y=4$ ?
Answer: $V=\pi\displaystyle\int_{0}^{3}\Big[(4-0)^2-(4-x)^2\Big]dx$ This problem uses the washer method with rotation about y = 4. Since y = 4 is above the triangular region between y = x and y = 0, we measure distances downward: the outer radius extends from y = 4 to y = 0, giving R = 4 - 0 = 4, and the inner radius extends from y = 4 to y = x, giving r = 4 - x. The volume integral is π∫[0,3][(4-0)² - (4-x)²]dx = π∫[0,3][16 - (4-x)²]dx. Choice A reverses the radii, which would result in a negative integrand. When rotating about an axis above the region, the constant boundary (here y = 0) gives the larger radius.
What integral represents the volume when the region between $y=\cos x$ and $y=0$ on $[0,\tfrac{\pi}{2}]$ is revolved about $y=1$ ?
Answer: $\pi\displaystyle\int_{0}^{\pi/2}\big[(1-0)^2-(1-\cos x)^2\big]dx$ This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y = 1. To adjust for this axis, subtract each y-value from 1 to get the distances, since the curves are below or at the axis. The outer radius is 1 - 0 = 1 from the lower curve, and the inner radius is 1 - cos x from the upper curve. Therefore, the integral is π ∫ from 0 to π/2 of [(1 - 0)² - (1 - cos x)²] dx. A tempting distractor is choice A, which swaps the inner and outer radii, resulting in a negative integrand in some parts. In general, when revolving around y = k, compute radii as |y - k| for each curve and identify the outer as the larger distance and inner as the smaller.
What integral gives the volume when the region between $y=\sin x$ and $y=0$ , $0\le x\le\pi$ , is revolved about $y=1$ ?
Answer: $\pi\int_{0}^{\pi}\big[(1-0)^2-(1-\sin x)^2\big]\,dx$ This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 1. To adjust for this axis, the radii are the distances from the curves to y = 1, which means subtracting each y-value from 1 since the region is below or touching. The outer radius is from the lower curve y = 0 to the axis, giving 1. The inner radius is from the upper curve y = sin x to the axis, giving 1 - sin x. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.
Which integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$ on $[0,4]$ is revolved about $y=-2$ ?
Answer: $V=\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big]dx$ This problem involves the washer method with revolution about y = -2, a horizontal line below the x-axis. For revolution about y = -2, the outer radius extends from y = -2 up to y = √x, giving R = √x - (-2) = √x + 2, and the inner radius extends from y = -2 up to y = 0, giving r = 0 - (-2) = 2. The washer formula becomes π∫[(√x + 2)² - (2)²]dx from 0 to 4. Choice D incorrectly subtracts 2 from the functions instead of adding 2, which would give negative radii. When the axis of revolution is below the region, add the absolute value of the axis to all y-values to find positive radii.
Choose the correct volume integral for revolving the region between $y=\sin x$ and $y=0$ on $[0,\pi]$ about $y=2$ .
Answer: $V=\pi\displaystyle\int_{0}^{\pi}\Big[(2-0)^2-(2-\sin x)^2\Big]dx$ This problem requires the washer method with rotation about y = 2. Since y = 2 is above the region bounded by y = sin x and y = 0 on [0,π], we measure distances downward: the outer radius is from y = 2 to y = 0, giving R = 2 - 0 = 2, and the inner radius is from y = 2 to y = sin x, giving r = 2 - sin x. The volume integral is π∫[0,π][(2-0)² - (2-sin x)²]dx = π∫[0,π][4 - (2-sin x)²]dx. Choice A incorrectly orders the radii, placing the smaller radius as the outer one. Always ensure the outer radius is larger than the inner radius in the washer method.

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