Washer Method: Other Axes - Calculus 1

Let R be the region bounded by $y=e^x$ , $y=e$ , and the y-axis. Which integral represents the volume of the solid formed by revolving R about the line $x=-1$ ?
Answer: $\pi \int_{1}^{e} [(\ln y+1)^2 - 1^2] \,dy$ When finding the volume of a solid of revolution using the washer method, you need to carefully identify the outer and inner radii based on the axis of rotation and set up your integral with the correct variable. First, sketch the region R. It's bounded by $y = e^x$ , the horizontal line $y = e$ , and the y-axis ( $x = 0$ ). These curves intersect when $e^x = e$ , so $x = 1$ . The region extends from $y = 1$ to $y = e$ . Since you're rotating about the vertical line $x = -1$ , it's most efficient to integrate with respect to $y$ . For any horizontal slice at height $y$ , you need the distances from $x = -1$ to the boundaries of the region. The right boundary is $x = \ln y$ (solving $y = e^x$ for $x$ ), so its distance from $x = -1$ is $\ln y - (-1) = \ln y + 1$ . The left boundary is $x = 0$ (the y-axis), so its distance from $x = -1$ is $0 - (-1) = 1$ . Using the washer method, the volume is $\pi \int_{1}^{e} [(\text{outer radius})^2 - (\text{inner radius})^2] \,dy$ . The outer radius is $\ln y + 1$ and the inner radius is $1$ , giving us answer C. Answer A uses the wrong variable ( $x$ instead of $y$ ) and has the radii backwards. Answer B incorrectly makes $1$ the outer radius and $\ln y + 1$ the inner radius. Answer D forgets to add 1 to $\ln y$ when calculating the distance from $x = -1$ . Strategy tip: Always sketch the region and axis of rotation first, then choose your integration variable based on which gives simpler radius expressions.
Let R be the region bounded by $y=x^2$ and $y=2x$ . Which integral represents the volume of the solid obtained by revolving R about the line $y=5$ ?
Answer: $\pi \int_{0}^{2} [(5-x^2)^2 - (5-2x)^2] \,dx$ When finding volumes of revolution using the washer method, you need to identify the outer and inner radii from the axis of rotation to each bounding curve, then apply the formula $V = \pi \int [R_{outer}^2 - R_{inner}^2] \, dx$ . First, find where the curves intersect by solving $x^2 = 2x$ , giving $x = 0$ and $x = 2$ . Between these points, $y = 2x$ lies above $y = x^2$ . When revolving about the horizontal line $y = 5$ , the distance from this axis to each curve determines the radii. For any $x$ in $[0,2]$ : - Distance to $y = x^2$ : $5 - x^2$ (this is the outer radius since $x^2 < 5$ ) - Distance to $y = 2x$ : $5 - 2x$ (this is the inner radius since $2x < 5$ and $2x > x^2$ ) The volume integral becomes $\pi \int_0^2 [(5-x^2)^2 - (5-2x)^2] \, dx$ , which is choice B. Choice A incorrectly swaps the outer and inner radii, treating $5-2x$ as outer and $5-x^2$ as inner. This happens when you don't carefully determine which curve is farther from the axis of rotation. Choice C uses integration with respect to $y$ , which requires different bounds and curve expressions, but the setup shown doesn't correctly represent this region. Choice D attempts to use just the difference between curves without accounting for revolution about $y = 5$ , missing the washer method entirely. Key strategy: Always sketch the region and axis of rotation. Identify which curve is farther from the axis (outer radius) and closer (inner radius) before setting up your integral.
Let R be the region in the first quadrant bounded by $y=1$ , $y=2$ , the y-axis, and the curve $x = 4/y^2$ . Find the volume of the solid generated when R is revolved about the line $x=-3$ .
Answer: $\pi \int_{1}^{2} [(\frac{4}{y^2}+3)^2 - 3^2] \,dy$ When finding volumes of revolution about vertical lines, you need to carefully set up the washer method and choose the right variable of integration based on how the region is naturally described. Since the region R is bounded by horizontal lines ( $y = 1$ and $y = 2$ ) and the curve is given as $x = \frac{4}{y^2}$ , integrating with respect to $y$ from 1 to 2 is the natural choice. When revolving about the vertical line $x = -3$ , each horizontal slice creates a washer. For any $y$ -value between 1 and 2, the outer radius extends from $x = -3$ to the curve $x = \frac{4}{y^2}$ , giving $R_{outer} = \frac{4}{y^2} - (-3) = \frac{4}{y^2} + 3$ . The inner radius extends from $x = -3$ to the y-axis ( $x = 0$ ), giving $R_{inner} = 0 - (-3) = 3$ . The washer area is $\pi(R_{outer}^2 - R_{inner}^2) = \pi[(\frac{4}{y^2} + 3)^2 - 3^2]$ , confirming answer C. Answer A incorrectly uses $x$ as the variable of integration, which doesn't match the natural bounds of the region. Answer B has the radii reversed, treating the smaller radius as outer and larger as inner, which would give a negative volume. Answer D omits the distance to the axis of rotation entirely, using $\frac{4}{y^2}$ as the outer radius instead of $\frac{4}{y^2} + 3$ . Study tip: Always sketch the region and axis of rotation first. Choose your integration variable based on the natural bounds, and remember that distances to the axis of rotation determine your radii, not the original coordinates.
Let R be the region enclosed by the parabola $x = (y-1)^2$ and the line $x=1$ . Which integral gives the volume of the solid generated by revolving R about the line $x=2$ ?
Answer: $\pi \int_{0}^{2} [(2-(y-1)^2)^2 - 1^2] \,dy$ When finding volumes of revolution using the washer method, you need to identify the outer and inner radii of the washers formed when slicing perpendicular to the axis of rotation. First, let's understand the region R. The parabola $x = (y-1)^2$ opens rightward with vertex at (0,1), and the line $x = 1$ is vertical. These curves intersect when $(y-1)^2 = 1$ , giving $y-1 = ±1$ , so $y = 0$ and $y = 2$ . The region extends from $y = 0$ to $y = 2$ . When revolving about the vertical line $x = 2$ , we use horizontal slicing (integrating with respect to $y$ ). At any height $y$ , the distance from the axis $x = 2$ to the parabola $x = (y-1)^2$ is $2 - (y-1)^2$ (outer radius), and the distance to the line $x = 1$ is $2 - 1 = 1$ (inner radius). The washer method formula gives us: $V = \pi \int_{0}^{2} [(2-(y-1)^2)^2 - 1^2] \,dy$ Option A incorrectly adds the parabola's $x$ -coordinate to 2 instead of subtracting, and uses 3 as the inner radius. Option B has the outer and inner radii backwards and gets a negative volume. Option C attempts to integrate with respect to $x$ , but the bounds and setup are incorrect for this geometry. Study tip: Always sketch the region and identify which boundary is farther from the axis of rotation—that determines your outer radius. The washer method is outer radius squared minus inner radius squared.
Let R be the region in the first quadrant bounded by $y = x^3$ , $y = 8$ , and the y-axis. Which integral represents the volume of the solid obtained by rotating R about the vertical line $x = 3$ ?
Answer: $\pi \int_{0}^{8} [3^2 - (3-\sqrt[3]{y})^2] \,dy$ The revolution is about a vertical line, so we integrate with respect to $y$ . The region is bounded by $x = \sqrt[3]{y}$ on the right and $x=0$ on the left, from $y=0$ to $y=8$ . The axis of revolution $x=3$ is to the right of the region. The outer radius $R(y)$ is the distance from $x=3$ to the farther boundary $x=0$ , so $R(y) = 3 - 0 = 3$ . The inner radius $r(y)$ is the distance from $x=3$ to the closer boundary $x=\sqrt[3]{y}$ , so $r(y) = 3 - \sqrt[3]{y}$ . The volume is $\pi \int_{0}^{8} [R(y)^2 - r(y)^2] \,dy = \pi \int_{0}^{8} [3^2 - (3-\sqrt[3]{y})^2] \,dy$ .
Let R be the region enclosed by the graphs of $y = \sqrt{x}$ and $y = x^2$ . Which integral gives the volume of the solid generated when R is revolved about the line $y = -1$ ?
Answer: $\pi \int_{0}^{1} [(\sqrt{x}+1)^2 - (x^2+1)^2] \,dx$ The region R is bounded by $y = \sqrt{x}$ (top curve) and $y = x^2$ (bottom curve) from $x=0$ to $x=1$ . When revolving around $y = -1$ , the cross-sections are washers. The outer radius $R(x)$ is the distance from the axis $y=-1$ to the top curve $y=\sqrt{x}$ , so $R(x) = \sqrt{x} - (-1) = \sqrt{x}+1$ . The inner radius $r(x)$ is the distance from $y=-1$ to the bottom curve $y=x^2$ , so $r(x) = x^2 - (-1) = x^2+1$ . The volume is given by the washer method formula $V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \,dx$ . Therefore, the integral is $\pi \int_{0}^{1} [(\sqrt{x}+1)^2 - (x^2+1)^2] \,dx$ .
Let $R$ be the region in the first quadrant enclosed by the graphs of $y=x$ and $y=\sqrt[3]{x}$ . Which definite integral represents the volume of the solid generated when $R$ is revolved about the line $y=-2$ ?
Answer: $\pi \int_{0}^{1} [(\sqrt[3]{x}+2)^2 - (x+2)^2] \,dx$ When finding the volume of a solid of revolution about a horizontal line, you need to use the washer method if the region has both an upper and lower boundary. The key insight is measuring distances from each boundary curve to the axis of rotation. First, let's identify the region R. The curves $y = x$ and $y = \sqrt[3]{x}$ intersect when $x = \sqrt[3]{x}$ , which occurs at $x = 0$ and $x = 1$ . Between these points, $\sqrt[3]{x} > x$ (since the cube root function grows faster than the linear function for $0 < x < 1$ ), so $y = \sqrt[3]{x}$ is the upper boundary. When revolving about $y = -2$ , each cross-section perpendicular to the x-axis forms a washer. The outer radius extends from $y = -2$ to the upper curve $y = \sqrt[3]{x}$ , giving distance $\sqrt[3]{x} - (-2) = \sqrt[3]{x} + 2$ . The inner radius extends from $y = -2$ to the lower curve $y = x$ , giving distance $x - (-2) = x + 2$ . The washer method formula is $V = \pi \int_a^b [R_{outer}^2 - R_{inner}^2] \, dx$ , so we get $\pi \int_0^1 [(\sqrt[3]{x} + 2)^2 - (x + 2)^2] \, dx$ , which is choice D. Choice A uses the difference of functions, not the washer method. Choice B incorrectly swaps which radius is outer versus inner. Choice C ignores the axis of rotation entirely, treating it as if revolving about the x-axis. Strategy tip: Always sketch the region and identify which curve is farther from the axis of rotation—that determines your outer radius in the washer method.
Let R be the region bounded by the graphs of $y = e^x$ , $y = 1$ , and $x = 2$ . Which of the following integrals represents the volume of the solid formed by revolving R about the line $y = -2$ ?
Answer: $\pi \int_{0}^{2} [(e^x+2)^2 - 3^2] \,dx$ The region is bounded above by $y=e^x$ and below by $y=1$ . The intersection of $y=e^x$ and $y=1$ is at $x=0$ , so the region extends from $x=0$ to $x=2$ . The axis of revolution is $y=-2$ . The outer radius is the distance from the axis to the outer curve: $R(x) = e^x - (-2) = e^x + 2$ . The inner radius is the distance from the axis to the inner curve: $r(x) = 1 - (-2) = 3$ . The volume is $V = \pi \int_{0}^{2} [R(x)^2 - r(x)^2] \,dx = \pi \int_{0}^{2} [(e^x+2)^2 - 3^2] \,dx$ .
Let R be the region in the first quadrant bounded by $y=2^x$ , $y=8$ , and the y-axis. What is the volume of the solid generated by revolving R about the line $y=10$ ?
Answer: $\pi \int_{0}^{3} [(10-2^x)^2 - 2^2] \,dx$ When finding volumes of revolution, you need to carefully identify which method to use and set up the integral correctly based on the axis of rotation and the region's boundaries. First, let's understand the region R. It's bounded by $y = 2^x$ , $y = 8$ , and the y-axis in the first quadrant. The curves intersect when $2^x = 8$ , so $x = 3$ . The region extends from $x = 0$ to $x = 3$ . Since we're revolving around the horizontal line $y = 10$ , we'll use the washer method with vertical slices. At any x-value, we have an outer radius (from $y = 10$ to the farther boundary) and an inner radius (from $y = 10$ to the closer boundary). The outer boundary is $y = 8$ , giving outer radius $10 - 8 = 2$ . The inner boundary is $y = 2^x$ , giving inner radius $10 - 2^x$ . The volume is $\pi \int_{0}^{3} [(10-2^x)^2 - 2^2] \,dx$ , which is answer B. Answer A incorrectly switches the outer and inner radii, which would give a negative volume since $2^2 < (10-2^x)^2$ throughout the interval. Answer C attempts integration with respect to y, but makes errors in the setup of the radii and limits. Answer D uses the wrong outer boundary, treating $y = 8$ as if it were at distance 8 from the axis of rotation instead of distance 2. Strategy tip: Always sketch the region and identify which boundary is farther from the axis of rotation. The larger radius goes first in the washer formula: $\pi \int [R^2 - r^2] \,dx$ .
Let R be the region bounded by $y=1/x$ , $y=0$ , $x=1$ , and $x=4$ . Which expression represents the volume of the solid generated when R is revolved about the line $x=-2$ ?
Answer: $\pi \int_{0}^{1/4} (6^2 - 3^2) \,dy + \pi \int_{1/4}^{1} [(\frac{1}{y}+2)^2 - 3^2] \,dy$ Revolving around a vertical line $x=-2$ requires integration with respect to $y$ . The region must be described with $x$ as a function of $y$ . The boundary $y=1/x$ becomes $x=1/y$ . The y-values range from 0 to 1. The shape of the region changes at $y=1/4$ (the y-value when x=4). For $y \in [0, 1/4]$ , the region is bounded by $x=1$ and $x=4$ . For $y \in [1/4, 1]$ , the region is bounded by $x=1$ and $x=1/y$ . The axis is $x=-2$ . For the first integral ( $y \in [0, 1/4]$ ), $R(y) = 4 - (-2) = 6$ and $r(y) = 1 - (-2) = 3$ . For the second integral ( $y \in [1/4, 1]$ ), $R(y) = (1/y) - (-2) = 1/y + 2$ and $r(y) = 1 - (-2) = 3$ . The total volume is the sum of the volumes of these two parts.
Let $R$ be the region in the first quadrant bounded by the graph of $y=\sqrt{x}$ , the x-axis, and the line $x=4$ . Which integral represents the volume of the solid formed by revolving $R$ about the line $y=3$ ?
Answer: $\pi \int_{0}^{4} [3^2 - (3-\sqrt{x})^2] \,dx$ The region is bounded by $y=\sqrt{x}$ and $y=0$ from $x=0$ to $x=4$ . The axis of revolution is the horizontal line $y=3$ , which is above the region. Using the washer method, we integrate with respect to $x$ . The outer radius $R(x)$ is the distance from the axis of revolution $y=3$ to the farther function, which is $y=0$ . So, $R(x) = 3 - 0 = 3$ . The inner radius $r(x)$ is the distance from the axis of revolution $y=3$ to the closer function, which is $y=\sqrt{x}$ . So, $r(x) = 3 - \sqrt{x}$ . The volume is given by $V = \pi \int_{0}^{4} [R(x)^2 - r(x)^2] \,dx = \pi \int_{0}^{4} [3^2 - (3-\sqrt{x})^2] \,dx$ .
The region R is enclosed by the graph of $y = |x|$ and the line $y=3$ . Find the volume of the solid generated when R is revolved about the line $y=4$ .
Answer: $\pi \int_{-3}^{3} [(4-|x|)^2 - 1^2] \,dx$ The region is bounded by $y=|x|$ below and $y=3$ above. The intersection points are $x=\pm 3$ . The axis of revolution is $y=4$ , which is above the region. The outer radius is the distance from the axis to the farther curve, $y=|x|$ : $R(x) = 4 - |x|$ . The inner radius is the distance from the axis to the closer curve, $y=3$ : $r(x) = 4-3=1$ . The volume integral is $\pi \int_{-3}^{3} [(4-|x|)^2 - 1^2] \,dx$ . Since the integrand is an even function, this is equivalent to $2\pi \int_{0}^{3} [(4-x)^2 - 1^2] \,dx$ . Both B and A are correct expressions, but B is the most direct setup. Often in such cases, one form is preferred. Let's check if A is fully equivalent. Yes. Let's make B the primary correct answer as it is the most direct translation of the problem. A is also correct due to symmetry.
Let $R$ be the region enclosed by the graphs of the line $y=x-1$ and the parabola $y^2=2x+6$ . Which of the following definite integrals represents the volume of the solid generated when $R$ is revolved about the line $x=10$ ?
Answer: $\pi \int_{-2}^{4} \left[ \left(13 - \frac{1}{2}y^2\right)^2 - (9-y)^2 \right] \,dy$ To revolve around a vertical line $x=10$ , we must integrate with respect to $y$ . First, express $x$ in terms of $y$ : $x=y+1$ and $x=\frac{1}{2}y^2-3$ . Find intersection points: $y+1 = \frac{1}{2}y^2-3 \implies y^2-2y-8=0 \implies (y-4)(y+2)=0$ . So, the limits of integration are $y=-2$ to $y=4$ . The axis of revolution $x=10$ is to the right of the region. The outer radius $R(y)$ is the distance from $x=10$ to the farther curve (the left curve), which is $x=\frac{1}{2}y^2-3$ . So, $R(y) = 10 - (\frac{1}{2}y^2-3) = 13 - \frac{1}{2}y^2$ . The inner radius $r(y)$ is the distance from $x=10$ to the closer curve (the right curve), which is $x=y+1$ . So, $r(y) = 10 - (y+1) = 9-y$ . The volume is $V = \pi \int_{-2}^{4} [R(y)^2 - r(y)^2] \,dy = \pi \int_{-2}^{4} [ (13 - \frac{1}{2}y^2)^2 - (9-y)^2 ] \,dy$ .
The region R is bounded by $y = \arctan(x)$ , $y=\pi/4$ , and the y-axis. Find the volume of the solid generated by revolving R about the line $y=-1$ .
Answer: $\pi \int_{0}^{1} [(\pi/4+1)^2 - (\arctan(x)+1)^2] \,dx$ When finding volumes of revolution, you need to carefully identify the region, the axis of rotation, and choose the appropriate method. This problem involves rotating about a horizontal line that's not a boundary of the region, which suggests using the washer method. First, sketch the region R. It's bounded by $y = \arctan(x)$ , the horizontal line $y = \pi/4$ , and the y-axis. Since $\arctan(1) = \pi/4$ , these curves intersect at $(1, \pi/4)$ , so the region extends from $x = 0$ to $x = 1$ . When revolving about $y = -1$ , each vertical slice creates a washer. The outer radius extends from $y = -1$ to the top boundary $y = \pi/4$ , giving $R_{outer} = \pi/4 - (-1) = \pi/4 + 1$ . The inner radius extends from $y = -1$ to the curve $y = \arctan(x)$ , giving $R_{inner} = \arctan(x) - (-1) = \arctan(x) + 1$ . The washer method formula is $V = \pi \int_a^b [R_{outer}^2 - R_{inner}^2] \, dx$ , which gives us $\pi \int_0^1 [(\pi/4 + 1)^2 - (\arctan(x) + 1)^2] \, dx$ . Choice A uses the disk method incorrectly and measures distances between curves rather than from the axis. Choice B has the radii reversed, giving negative area. Choice C attempts the shell method with $y$ as the variable, but uses incorrect bounds and setup. Key strategy: Always identify your axis of rotation first, then determine whether each cross-section is a disk or washer. Draw distances from the axis to boundaries to find your radii correctly.
The region bounded by the parabola $y=x^2$ and the line $y=4$ is revolved about the line $y=k$ , where $k$ is a constant and $k > 4$ . The volume of the resulting solid is $V(k)$ . Which of the following expressions represents $V(k)$ ?
Answer: $\pi \int_{-2}^{2} [ (k-x^2)^2 - (k-4)^2 ] \,dx$ When you encounter a solid of revolution problem, you need to carefully set up the washer method formula, paying close attention to which function is farther from and closer to the axis of rotation. Since we're revolving about the line $y = k$ where $k > 4$ , we're rotating above the entire region. The region is bounded by $y = x^2$ (parabola) and $y = 4$ (horizontal line), with intersection points at $x = -2$ and $x = 2$ . For the washer method, the volume formula is $V = \pi \int_a^b [R^2 - r^2] \, dx$ , where $R$ is the outer radius and $r$ is the inner radius. Since $k > 4$ , the distance from the axis $y = k$ to any point on the parabola $y = x^2$ is $k - x^2$ , and the distance to the line $y = 4$ is $k - 4$ . Because $x^2 \leq 4$ in our region, we have $k - x^2 \geq k - 4$ , making $k - x^2$ the outer radius and $k - 4$ the inner radius. Therefore, $V(k) = \pi \int_{-2}^{2} [(k-x^2)^2 - (k-4)^2] \, dx$ , which is choice B. Choice A incorrectly swaps the outer and inner radii. Choice C ignores the axis of rotation entirely, using a disk method setup as if rotating about the x-axis. Choice D uses the wrong inner radius, substituting $16$ instead of $(k-4)^2$ . Strategy tip: Always identify which boundary is farther from the rotation axis to determine your outer radius correctly. Draw a quick sketch to visualize the setup before writing your integral.
Let $R$ be the region enclosed by the graphs of $y=x^2+1$ and $y=x+3$ . Which of the following integrals gives the volume of the solid generated when $R$ is revolved about the line $y=-1$ ?
Answer: $\pi \int_{-1}^{2} [(x+4)^2 - (x^2+2)^2] \,dx$ First, find the points of intersection by setting the functions equal: $x^2+1 = x+3 \implies x^2-x-2=0 \implies (x-2)(x+1)=0$ . The points of intersection are $x=-1$ and $x=2$ . In the interval $[-1, 2]$ , the line $y=x+3$ is above the parabola $y=x^2+1$ . The solid is generated by revolving the region about the horizontal line $y=-1$ . The method of washers is appropriate. The outer radius $R(x)$ is the distance from the axis of revolution $y=-1$ to the outer curve $y=x+3$ , so $R(x) = (x+3) - (-1) = x+4$ . The inner radius $r(x)$ is the distance from the axis of revolution to the inner curve $y=x^2+1$ , so $r(x) = (x^2+1) - (-1) = x^2+2$ . The volume $V$ is given by the integral $V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \,dx$ . Substituting the radii and limits gives $V = \pi \int_{-1}^{2} [(x+4)^2 - (x^2+2)^2] \,dx$ .
The region R is bounded by $y = \ln(x)$ , the x-axis, and the line $x = e$ . Find the volume of the solid generated by revolving R about the line $y = -2$ .
Answer: $\pi \int_{1}^{e} [(\ln(x)+2)^2 - 2^2] \,dx$ The region is bounded by $y=\ln(x)$ above and $y=0$ below. It starts where $\ln(x)=0$ , which is $x=1$ , and ends at $x=e$ . The axis of revolution is $y=-2$ . The outer radius is $R(x) = \ln(x) - (-2) = \ln(x) + 2$ . The inner radius is $r(x) = 0 - (-2) = 2$ . The volume is $\pi \int_{1}^{e} [R(x)^2 - r(x)^2] \,dx = \pi \int_{1}^{e} [(\ln(x)+2)^2 - 4] \,dx$ . Distractor D represents the common error of treating this as a disk method problem around y=0 and also using the $(R-r)^2$ fallacy, since $((\ln x+2)-2)^2 = (\ln x)^2$ .
Let R be the region in the first quadrant enclosed by the graphs of $x = y^2$ and $x = 2y$ . Which integral gives the volume of the solid generated when R is revolved about the line $x = -1$ ?
Answer: $\pi \int_{0}^{2} [(2y+1)^2 - (y^2+1)^2] \,dy$ The curves intersect when $y^2 = 2y$ , which gives $y=0$ and $y=2$ . Since the revolution is about a vertical line, we integrate with respect to $y$ . For $y \in [0, 2]$ , $2y \ge y^2$ , so $x=2y$ is the right curve and $x=y^2$ is the left curve. The axis is $x=-1$ . The outer radius $R(y)$ is the distance from $x=-1$ to the right curve $x=2y$ , so $R(y) = 2y - (-1) = 2y+1$ . The inner radius $r(y)$ is the distance from $x=-1$ to the left curve $x=y^2$ , so $r(y) = y^2 - (-1) = y^2+1$ . The volume is $\pi \int_{0}^{2} [(2y+1)^2 - (y^2+1)^2] \,dy$ .
The region enclosed by the parabola $x=y^2$ and the line $x=y+2$ is revolved about the line $x=-1$ . Which integral represents the volume of the resulting solid?
Answer: $\pi \int_{-1}^{2} [(y+3)^2 - (y^2+1)^2] \,dy$ When finding volumes of revolution about vertical or horizontal lines that aren't the coordinate axes, you need to use the washer method with careful attention to the radii from each curve to the axis of revolution. First, find where the curves intersect by solving $y^2 = y + 2$ , which gives $y^2 - y - 2 = 0$ , so $(y-2)(y+1) = 0$ . The curves intersect at $y = -1$ and $y = 2$ . Since we're revolving about the vertical line $x = -1$ , we integrate with respect to $y$ . For any horizontal slice at height $y$ , we need the distances from each curve to the line $x = -1$ : - Distance from parabola $x = y^2$ to $x = -1$ : $y^2 - (-1) = y^2 + 1$ - Distance from line $x = y + 2$ to $x = -1$ : $(y + 2) - (-1) = y + 3$ The line $x = y + 2$ is farther from the axis of revolution (forms the outer radius), while the parabola forms the inner radius. Using the washer method: $V = \pi \int_{-1}^{2} [R^2 - r^2] \,dy = \pi \int_{-1}^{2} [(y+3)^2 - (y^2+1)^2] \,dy$ . Answer A incorrectly uses $(y^2)^2$ instead of $(y^2+1)^2$ for the inner radius. Answer B switches the outer and inner radii, which would give a negative volume under the integral. Answer D attempts to use a single function approach, missing that this is a washer (not disk) problem. Strategy tip: Always sketch the region and identify which curve is farther from the axis of revolution—that becomes your outer radius in the washer formula.
Let R be the region bounded by $y = \sin(x)$ and $y = \cos(x)$ over the interval $[0, \pi/4]$ . What is the volume of the solid generated by revolving R about the line $y = 2$ ?
Answer: $\pi \int_{0}^{\pi/4} [(2-\sin(x))^2 - (2-\cos(x))^2] \,dx$ On the interval $[0, \pi/4]$ , $\cos(x) \ge \sin(x)$ . The axis of revolution $y=2$ is above the region. The outer radius $R(x)$ is the distance from the axis $y=2$ to the farther curve, which is $y=\sin(x)$ . Thus, $R(x) = 2 - \sin(x)$ . The inner radius $r(x)$ is the distance from the axis $y=2$ to the closer curve, which is $y=\cos(x)$ . Thus, $r(x) = 2 - \cos(x)$ . The volume is $V = \pi \int_{0}^{\pi/4} [R(x)^2 - r(x)^2] \,dx = \pi \int_{0}^{\pi/4} [(2-\sin(x))^2 - (2-\cos(x))^2] \,dx$ .

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