Washer Method: Other Axes - Calculus 2

Let R be the region bounded by the curves $y=4-x^2$ and $y=x+2$ . Which integral gives the volume of the solid formed by revolving R about the line $y=5$ ?
Answer: $\pi \int_{-2}^1 [(3-x)^2 - (1+x^2)^2] dx$ First, find the points of intersection: $4-x^2 = x+2 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0$ . The bounds are $x=-2$ and $x=1$ . In this interval, $4-x^2 \ge x+2$ . The axis of revolution $y=5$ is above the region. The outer radius $R(x)$ is the distance from the axis to the lower curve $y=x+2$ , so $R(x) = 5-(x+2) = 3-x$ . The inner radius $r(x)$ is the distance from the axis to the upper curve $y=4-x^2$ , so $r(x) = 5-(4-x^2) = 1+x^2$ . The volume integral is $\pi \int_{-2}^1 [R(x)^2 - r(x)^2] dx = \pi \int_{-2}^1 [(3-x)^2 - (1+x^2)^2] dx$ .
Let R be the region enclosed by the graphs of $y = x^2$ and $y = 2x$ . Which integral represents the volume of the solid generated by revolving R about the line $y = -1$ ?
Answer: $\pi \int_0^2 [ (2x+1)^2 - (x^2+1)^2 ] dx$ The curves intersect where $x^2 = 2x$ , which gives $x=0$ and $x=2$ . For $x \in [0, 2]$ , $2x \ge x^2$ . The axis of revolution is $y=-1$ . The outer radius $R(x)$ is the distance from the axis to the outer curve $y=2x$ , so $R(x) = 2x - (-1) = 2x+1$ . The inner radius $r(x)$ is the distance from the axis to the inner curve $y=x^2$ , so $r(x) = x^2 - (-1) = x^2+1$ . The volume is given by the washer method formula, $V = \pi \int_a^b [R(x)^2 - r(x)^2] dx$ , which results in $\pi \int_0^2 [ (2x+1)^2 - (x^2+1)^2 ] dx$ .
The region bounded by $y=x$ , $y=2-x$ , and $y=0$ is revolved about the line $x=3$ . Find the volume of the solid generated.
Answer: $4\pi$ The region is a triangle with vertices at (0,0), (2,0), and (1,1). Revolving around the vertical line $x=3$ is best done by integrating with respect to $y$ . The y-bounds are from 0 to 1. We need to express x in terms of y: $x=y$ and $x=2-y$ . For $y \in [0,1]$ , the right boundary is $x=2-y$ and the left boundary is $x=y$ . The axis $x=3$ is to the right of the region. The outer radius is $R(y) = 3 - y$ (distance to the left boundary). The inner radius is $r(y) = 3 - (2-y) = 1+y$ (distance to the right boundary). The volume is $V = \pi \int_0^1 [ (3-y)^2 - (1+y)^2 ] dy = \pi \int_0^1 [(9-6y+y^2)-(1+2y+y^2)] dy = \pi \int_0^1 [8-8y] dy = \pi [8y-4y^2]_0^1 = \pi(8-4) = 4\pi$ .
Let R be the region enclosed by $y = \ln(x)$ , $y=0$ , and $x=e$ . Which definite integral represents 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 $y=\ln(x)$ and $y=0$ . The intersection is at $\ln(x)=0$ , which is $x=1$ . The bounds of integration are from $x=1$ to $x=e$ . The axis of revolution is $y=-2$ , which is below the region. The outer radius $R(x)$ is the distance from the axis to the upper curve $y=\ln(x)$ , so $R(x) = \ln(x) - (-2) = \ln(x)+2$ . The inner radius $r(x)$ is the distance from the axis to the lower curve $y=0$ , so $r(x) = 0 - (-2) = 2$ . The volume is $V = \pi \int_1^e [R(x)^2 - r(x)^2] dx = \pi \int_1^e [(\ln(x)+2)^2 - 2^2] dx$ .
Let R be the region enclosed by $y=x^2$ and $y=4$ . Which integral gives the volume of the solid generated by revolving R about the line $y=4$ ?
Answer: $\pi \int_{-2}^2 (4-x^2)^2 dx$ When finding volumes of revolution, you need to identify the axis of rotation and choose the appropriate method. Since we're revolving around the horizontal line $y = 4$ , the washer method works well here. First, visualize the region R. The parabola $y = x^2$ intersects the line $y = 4$ when $x^2 = 4$ , so at $x = -2$ and $x = 2$ . The region is bounded above by $y = 4$ and below by $y = x^2$ . When revolving around $y = 4$ , each vertical slice at position $x$ creates a disk with radius equal to the distance from the parabola to the line of rotation. This distance is $4 - x^2$ (the upper curve minus the lower curve). Using the disk method, the volume is $\pi \int_{-2}^{2} (4-x^2)^2 \, dx$ , which matches option D. Option A uses the wrong setup entirely - it integrates with respect to $y$ from 0 to 4, but the region doesn't extend from $y = 0$ to $y = 4$ . Option B also integrates with respect to $y$ and uses $(\sqrt{y})^2 = y$ as the radius, which ignores the axis of rotation at $y = 4$ . Option C attempts the washer method but incorrectly uses $16 - (x^2)^2$ . The outer radius should be the distance from $y = 4$ to the axis (which is 0), but since we're rotating around $y = 4$ , this approach is fundamentally flawed. Strategy tip: Always identify your axis of rotation first, then determine what function represents the radius from each point in your region to that axis.
Let R be the region in the first quadrant bounded by $y = \sqrt{x}$ , the x-axis, and the line $x=4$ . Which integral gives the volume of the solid formed by revolving R about the line $x = 5$ ?
Answer: $\pi \int_0^2 [ (5-y^2)^2 - 1^2 ] dy$ To revolve around a vertical axis $x=5$ , we must integrate with respect to $y$ . The region is bounded by $x=y^2$ , $x=4$ , and $y=0$ . The y-values range from 0 to $\sqrt{4}=2$ . The axis $x=5$ is to the right of the region. The outer radius $R(y)$ is the distance from the axis $x=5$ to the farther curve $x=y^2$ , so $R(y) = 5 - y^2$ . The inner radius $r(y)$ is the distance from the axis to the nearer curve $x=4$ , so $r(y) = 5 - 4 = 1$ . The volume is $V = \pi \int_0^2 [ (5-y^2)^2 - 1^2 ] dy$ .
The region R is bounded by $y=\sec(x)$ , $y=2$ , $x=-\pi/3$ , and $x=\pi/3$ . Which integral gives the volume of the solid generated by revolving R about the line $y=3$ ?
Answer: $\pi \int_{-\pi/3}^{\pi/3} [(3-\sec(x))^2 - 1^2] dx$ On the interval $[-\pi/3, \pi/3]$ , $\sec(x)$ ranges from 1 to 2. The region R is bounded above by $y=2$ and below by $y=\sec(x)$ . The axis of revolution $y=3$ is above the region. The outer radius $R(x)$ is the distance from the axis to the lower curve $y=\sec(x)$ , so $R(x) = 3 - \sec(x)$ . The inner radius $r(x)$ is the distance from the axis to the upper curve $y=2$ , so $r(x) = 3 - 2 = 1$ . The volume is $V = \pi \int_{-\pi/3}^{\pi/3} [ (3-\sec(x))^2 - 1^2 ] dx$ .
A region is bounded by $y = \sqrt{x}$ , $y = x$ , and revolves around the line $x = 4$ . If we set up the integral using the washer method with respect to $y$ , what are the correct outer and inner radii?
Answer: Outer radius: $4 - y^2$ , Inner radius: $4 - y$ for $0 \leq y \leq 1$ The curves $y = \sqrt{x}$ and $y = x$ intersect where $\sqrt{x} = x$ , giving $x = x^2$ , so $x(1-x) = 0$ , yielding $x = 0$ and $x = 1$ . For $0 \leq y \leq 1$ : from $y = \sqrt{x}$ , we get $x = y^2$ ; from $y = x$ , we get $x = y$ . Since $y^2 \leq y$ in this interval, $y^2$ is the left boundary and $y$ is the right boundary. When revolving around $x = 4$ , the outer radius (farther from axis) is $4 - y^2$ and the inner radius is $4 - y$ . Choice A reverses the radii. Choice C incorrectly adds instead of subtracting from 4. Choice D gives negative radii, which is impossible.
The region bounded by $y = \frac{1}{x}$ , $y = \frac{1}{4}$ , and $x = 1$ is revolved around the line $y = \frac{1}{4}$ . Which statement correctly describes this solid of revolution?
Answer: It can be found using the disk method with radius $r(x) = \frac{1}{x} - \frac{1}{4}$ integrated from $x = 1$ to $x = 4$ The curves intersect when $\frac{1}{x} = \frac{1}{4}$ , giving $x = 4$ . The region is bounded by $x = 1$ , $x = 4$ , $y = \frac{1}{4}$ , and $y = \frac{1}{x}$ . Since $\frac{1}{x} \geq \frac{1}{4}$ for $1 \leq x \leq 4$ , the region lies entirely above the axis of rotation $y = \frac{1}{4}$ . When revolving around $y = \frac{1}{4}$ , each cross-section is a disk with radius equal to the distance from $y = \frac{1}{x}$ to $y = \frac{1}{4}$ , which is $\frac{1}{x} - \frac{1}{4}$ . We integrate from $x = 1$ to $x = 4$ . Choice B incorrectly suggests a washer when there's no hole. Choice C is wrong because the axis forms a boundary, not an interior line. Choice D has incorrect integration bounds.
A solid is formed by revolving the region R, bounded by $y=f(x)$ and $y=g(x)$ on $[a,b]$ where $f(x) \ge g(x) \ge 0$ , about the line $y=k$ . If $k < 0$ , which expression represents the volume of the solid?
Answer: $\pi \int_a^b [ (f(x)-k)^2 - (g(x)-k)^2 ] dx$ When you encounter a solid of revolution problem where the region is bounded by two functions and rotated about a horizontal line, you need to visualize the washer method. The key insight is determining the correct radii based on the axis of rotation. Since we're rotating about the line $y = k$ where $k < 0$ , this axis lies below the x-axis. Given that $f(x) \geq g(x) \geq 0$ , both functions are above the x-axis, making them even further from the axis of rotation. The distance from any point $(x, y)$ to the line $y = k$ is $|y - k| = y - k$ (since $y > k$ ). For the washer method, volume equals $\pi \int_a^b [R^2 - r^2] dx$ , where $R$ is the outer radius and $r$ is the inner radius. Since $f(x) \geq g(x)$ , the outer radius is $f(x) - k$ and the inner radius is $g(x) - k$ . This gives us $\pi \int_a^b [(f(x) - k)^2 - (g(x) - k)^2] dx$ , which is answer D. Answer A represents the volume when rotating about the x-axis, ignoring the offset from $y = k$ . Answer B incorrectly uses $(k - f(x))$ and $(k - g(x))$ , which would be negative distances since $k < 0$ and the functions are positive. Answer C uses $(f(x) + k)$ terms, which doesn't correctly represent the distance to the axis of rotation. Remember: when rotating about $y = k$ , the radius from any point to the axis is always the absolute value of the difference between the y-coordinate and k.
Let R be the region enclosed by the graphs of $y = \sin(x)$ and $y = \cos(x)$ for $0 \le x \le \frac{\pi}{4}$ . Which integral represents the volume of the solid generated when R is revolved about the line $y = 2$ ?
Answer: $\pi \int_0^{\pi/4} [ (2 - \sin(x))^2 - (2 - \cos(x))^2 ] dx$ In the interval $[0, \pi/4]$ , $\cos(x) \ge \sin(x)$ . The axis of revolution $y=2$ is above this region. The outer radius $R(x)$ is the distance from the axis to the farther curve, $y=\sin(x)$ , so $R(x) = 2 - \sin(x)$ . The inner radius $r(x)$ is the distance from the axis to the nearer curve, $y=\cos(x)$ , so $r(x) = 2 - \cos(x)$ . The volume is $V = \pi \int_0^{\pi/4} [R(x)^2 - r(x)^2] dx = \pi \int_0^{\pi/4} [ (2 - \sin(x))^2 - (2 - \cos(x))^2 ] dx$ .
Let R be the region bounded by $y=6-x$ , $y=1$ , and $x=0$ . Which integral represents the volume of the solid generated by revolving R about the line $x=7$ ?
Answer: $\pi \int_1^6 [7^2 - (1+y)^2] dy$ To revolve around a vertical axis $x=7$ , we integrate with respect to $y$ . The region's vertices are at (0,6), (0,1), and (5,1). The y-bounds are from 1 to 6. The region is bounded on the right by $x=6-y$ and on the left by $x=0$ . The axis $x=7$ is to the right of the region. The outer radius $R(y)$ is the distance from the axis to the farther boundary $x=0$ , so $R(y) = 7-0=7$ . The inner radius $r(y)$ is the distance from the axis to the nearer boundary $x=6-y$ , so $r(y) = 7-(6-y) = 1+y$ . The volume is $V = \pi \int_1^6 [R(y)^2 - r(y)^2] dy = \pi \int_1^6 [7^2 - (1+y)^2] dy$ .
Let R be the region bounded by $y = e^x$ , $y=1$ , and $x=2$ . Find the volume of the solid generated when R is revolved about the line $y=5$ .
Answer: $\pi \int_0^2 [ (5-1)^2 - (5-e^x)^2 ] dx$ The region is integrated with respect to $x$ from $x=0$ (where $y=e^x$ intersects $y=1$ ) to $x=2$ . The axis of revolution $y=5$ is above the region. The outer radius $R(x)$ is the distance from the axis to the farther curve $y=1$ , so $R(x) = 5-1=4$ . The inner radius $r(x)$ is the distance from the axis to the nearer curve $y=e^x$ , so $r(x) = 5-e^x$ . The volume is $V = \pi \int_0^2 [R(x)^2 - r(x)^2] dx = \pi \int_0^2 [ 4^2 - (5-e^x)^2 ] dx$ . Choice D is equivalent to B, which incorrectly swaps the radii resulting in a negative volume.
Let R be the region in the first quadrant bounded by $y = \arctan(x)$ , $y=\pi/4$ , and the y-axis. Find the setup for the volume of the solid generated by revolving R about the line $x=-2$ .
Answer: $\pi \int_0^{\pi/4} [ (\tan(y)+2)^2 - 2^2 ] dy$ When you're finding volumes of revolution about a line that's not a coordinate axis, you need to carefully set up the washer method and choose the right variable for integration. First, sketch the region R bounded by $y = \arctan(x)$ , $y = \pi/4$ , and the y-axis. This region extends from $y = 0$ to $y = \pi/4$ , with the right boundary being $x = \tan(y)$ (the inverse of $y = \arctan(x)$ ). When revolving about the vertical line $x = -2$ , it's natural to integrate with respect to $y$ . For each horizontal slice at height $y$ , the distance from $x = -2$ to the y-axis is 2 (inner radius), and the distance from $x = -2$ to the curve $x = \tan(y)$ is $\tan(y) + 2$ (outer radius). Using the washer method, the volume is $\pi \int_0^{\pi/4} [(\text{outer radius})^2 - (\text{inner radius})^2] \, dy = \pi \int_0^{\pi/4} [(\tan(y) + 2)^2 - 2^2] \, dy$ , which is choice D. Choice A incorrectly reverses the radii, subtracting the outer radius squared from the inner radius squared. Choice B uses integration with respect to $x$ , which complicates the setup since the axis of revolution is vertical. Choice C forgets about the washer method entirely, only accounting for the outer radius without subtracting the inner radius. Strategy tip: When revolving about a line parallel to the y-axis, integrate with respect to $y$ and carefully identify which boundary is closer to (inner radius) and farther from (outer radius) the axis of revolution.
The region R is bounded by $y=2^x$ , $y=8$ , and $x=0$ . Which integral gives the volume of the solid formed by revolving R about the line $y=10$ ?
Answer: $\pi \int_0^3 [ (10-2^x)^2 - (10-8)^2 ] dx$ The intersection of $y=2^x$ and $y=8$ is at $2^x=8$ , so $x=3$ . The integration is from $x=0$ to $x=3$ . In this region, $y=8$ is the upper curve and $y=2^x$ is the lower curve. The axis of revolution $y=10$ is above the region. The outer radius $R(x)$ is the distance from the axis to the lower curve $y=2^x$ , so $R(x) = 10-2^x$ . The inner radius $r(x)$ is the distance from the axis to the upper curve $y=8$ , so $r(x) = 10-8=2$ . The volume is $V = \pi \int_0^3 [ (10-2^x)^2 - 2^2 ] dx$ .
The region in the first quadrant bounded by the curves $y=x^2$ and $y=x^4$ is rotated about the line $y=1$ . What is the volume of the solid generated?
Answer: $\frac{8\pi}{45}$ The curves intersect at $x=0$ and $x=1$ . In the interval $[0,1]$ , $x^2 \ge x^4$ . The axis of revolution $y=1$ is above the region. The outer radius is $R(x) = 1 - x^4$ (distance to the lower curve). The inner radius is $r(x) = 1 - x^2$ (distance to the upper curve). The volume is $V = \pi \int_0^1 [ (1-x^4)^2 - (1-x^2)^2 ] dx = \pi \int_0^1 [(1-2x^4+x^8) - (1-2x^2+x^4)] dx = \pi \int_0^1 [x^8 - 3x^4 + 2x^2] dx = \pi [\frac{x^9}{9} - \frac{3x^5}{5} + \frac{2x^3}{3}]_0^1 = \pi (\frac{1}{9} - \frac{3}{5} + \frac{2}{3}) = \pi (\frac{5-27+30}{45}) = \frac{8\pi}{45}$ .
The region bounded by the parabola $x=y^2$ and the line $x=4$ is revolved about the line $x=5$ . What is the volume of the resulting solid?
Answer: $\frac{832\pi}{15}$ Revolving around a vertical axis $x=5$ requires integrating with respect to $y$ . The intersections of $x=y^2$ and $x=4$ are at $y=\pm 2$ . The axis $x=5$ is to the right of the region. The outer radius is the distance from the axis to the farther curve $x=y^2$ , so $R(y) = 5 - y^2$ . The inner radius is the distance from the axis to the nearer curve $x=4$ , so $r(y) = 5 - 4 = 1$ . The volume is $V = \pi \int_{-2}^2 [ (5-y^2)^2 - 1^2 ] dy = \pi \int_{-2}^2 [y^4 - 10y^2 + 24] dy$ . The integrand is even, so $V = 2\pi [\frac{y^5}{5} - \frac{10y^3}{3} + 24y]_0^2 = 2\pi (\frac{32}{5} - \frac{80}{3} + 48) = 2\pi (\frac{96-400+720}{15}) = 2\pi (\frac{416}{15}) = \frac{832\pi}{15}$ .
The region enclosed by $y=1/x$ , $y=1$ , $y=4$ , and the y-axis is revolved about the line $x=-1$ . What is the volume of the solid generated?
Answer: $\pi (\frac{3}{4} + 2\ln 4)$ To revolve about the vertical axis $x=-1$ , we integrate with respect to $y$ . The function is $x=1/y$ . The y-bounds are given as 1 to 4. The region is bounded by $x=1/y$ and $x=0$ (the y-axis). The axis $x=-1$ is to the left of the region. The outer radius is $R(y) = (1/y) - (-1) = 1/y+1$ . The inner radius is $r(y) = 0 - (-1) = 1$ . The volume is $V = \pi \int_1^4 [ (1/y+1)^2 - 1^2 ] dy = \pi \int_1^4 [1/y^2 + 2/y] dy = \pi [ -1/y + 2\ln|y| ]_1^4 = \pi [ (-1/4 + 2\ln 4) - (-1 + 2\ln 1) ] = \pi [ 3/4 + 2\ln 4 ]$ .
Let R be the region bounded by $y = 1/x^2$ , $y=0$ , $x=1$ , and $x=3$ . The region is revolved about the line $y=1$ . Which integral gives the volume of the resulting solid?
Answer: $\pi \int_1^3 [ 1^2 - (1 - 1/x^2)^2 ] dx$ The region is bounded above by $y=1/x^2$ and below by $y=0$ . The axis of revolution $y=1$ is above the region. The outer radius $R(x)$ is the distance from the axis to the lower boundary $y=0$ , so $R(x) = 1-0 = 1$ . The inner radius $r(x)$ is the distance from the axis to the upper boundary $y=1/x^2$ , so $r(x) = 1 - 1/x^2$ . The volume is $V = \pi \int_1^3 [R(x)^2 - r(x)^2] dx = \pi \int_1^3 [ 1^2 - (1 - 1/x^2)^2 ] dx$ .
The curves $y = 2x - x^2$ and $y = 0$ bound a region that is revolved around the line $y = 3$ . The volume of the solid formed is:
Answer: $\frac{256\pi}{15}$ When you encounter a volume of revolution problem where the axis of rotation is not a coordinate axis, you need to use the washer method with careful attention to how distances are measured from the rotation axis. First, find where the curves intersect: $2x - x^2 = 0$ , so $x(2-x) = 0$ , giving $x = 0$ and $x = 2$ . The parabola $y = 2x - x^2$ opens downward with vertex at $(1,1)$ , so it's above $y = 0$ between these points. Since we're rotating around $y = 3$ , each cross-section perpendicular to the x-axis forms a washer. The outer radius is the distance from $y = 3$ down to $y = 0$ , which is $R = 3 - 0 = 3$ . The inner radius is the distance from $y = 3$ down to the parabola $y = 2x - x^2$ , which is $r = 3 - (2x - x^2) = 3 - 2x + x^2$ . The volume is $V = \pi \int_0^2 [R^2 - r^2] dx = \pi \int_0^2 [9 - (3 - 2x + x^2)^2] dx$ . Expanding: $(3 - 2x + x^2)^2 = 9 - 12x + 6x^2 + 4x^2 - 4x^3 + x^4 = 9 - 12x + 10x^2 - 4x^3 + x^4$ So $V = \pi \int_0^2 [12x - 10x^2 + 4x^3 - x^4] dx = \pi[6x^2 - \frac{10x^3}{3} + x^4 - \frac{x^5}{5}]_0^2 = \pi[24 - \frac{80}{3} + 16 - \frac{32}{5}] = \frac{256\pi}{15}$ Answer D is correct. Choices A, B, and C likely result from calculation errors in expanding the squared term or integrating incorrectly. Study tip: Always double-check your algebra when squaring binomials in washer method problems—small errors multiply through integration.

Start this quiz