Calculus 2 Quiz: Washer Method Other Axes
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Washer Method Other AxesQuestion 1 of 20

The curves y=2x−x2y = 2x - x^2 and y=0y = 0 bound a region that is revolved around the line y=3y = 3. The volume of the solid formed is:

46π3\frac{46\pi}{3}
92π15\frac{92\pi}{15}
152π15\frac{152\pi}{15}
256π15\frac{256\pi}{15}
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Calculus 2 Quiz

Calculus 2 Quiz: Washer Method Other Axes

Practice Washer Method Other Axes in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Washer Method Other Axes, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The curves y=2x−x2y = 2x - x^2 and y=0y = 0 bound a region that is revolved around the line y=3y = 3. The volume of the solid formed is:

  1. 46π3\frac{46\pi}{3}
  2. 92π15\frac{92\pi}{15}
  3. 152π15\frac{152\pi}{15}
  4. 256π15\frac{256\pi}{15} (correct answer)
Explanation: 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−x2=02x - x^2 = 0, so x(2−x)=0x(2-x) = 0, giving x=0x = 0 and x=2x = 2. The parabola y=2x−x2y = 2x - x^2 opens downward with vertex at (1,1)(1,1), so it's above y=0y = 0 between these points. Since we're rotating around y=3y = 3, each cross-section perpendicular to the x-axis forms a washer. The outer radius is the distance from y=3y = 3 down to y=0y = 0, which is R=3−0=3R = 3 - 0 = 3. The inner radius is the distance from y=3y = 3 down to the parabola y=2x−x2y = 2x - x^2, which is r=3−(2x−x2)=3−2x+x2r = 3 - (2x - x^2) = 3 - 2x + x^2. The volume is V=π∫02[R2−r2]dx=π∫02[9−(3−2x+x2)2]dxV = \pi \int_0^2 [R^2 - r^2] dx = \pi \int_0^2 [9 - (3 - 2x + x^2)^2] dx. Expanding: (3−2x+x2)2=9−12x+6x2+4x2−4x3+x4=9−12x+10x2−4x3+x4(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=π∫02[12x−10x2+4x3−x4]dx=π[6x2−10x33+x4−x55]02=π[24−803+16−325]=256π15V = \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.

Question 2

Let R be the region bounded by y=1/x2y = 1/x^2, y=0y=0, x=1x=1, and x=3x=3. The region is revolved about the line y=1y=1. Which integral gives the volume of the resulting solid?

  1. π∫13[(1−1/x2)2]dx\pi \int_1^3 [ (1 - 1/x^2)^2 ] dx
  2. π∫13[12−(1−1/x2)2]dx\pi \int_1^3 [ 1^2 - (1 - 1/x^2)^2 ] dx (correct answer)
  3. π∫13[(1/x2)2]dx\pi \int_1^3 [ (1/x^2)^2 ] dx
  4. π∫13[(1−1/x2)2−12]dx\pi \int_1^3 [ (1 - 1/x^2)^2 - 1^2 ] dx
Explanation: The region is bounded above by y=1/x2y=1/x^2 and below by y=0y=0. The axis of revolution y=1y=1 is above the region. The outer radius R(x)R(x) is the distance from the axis to the lower boundary y=0y=0, so R(x)=1−0=1R(x) = 1-0 = 1. The inner radius r(x)r(x) is the distance from the axis to the upper boundary y=1/x2y=1/x^2, so r(x)=1−1/x2r(x) = 1 - 1/x^2. The volume is V=π∫13[R(x)2−r(x)2]dx=π∫13[12−(1−1/x2)2]dxV = \pi \int_1^3 [R(x)^2 - r(x)^2] dx = \pi \int_1^3 [ 1^2 - (1 - 1/x^2)^2 ] dx.

Question 3

Let R be the region enclosed by the graphs of y=x2y = x^2 and y=2xy = 2x. Which integral represents the volume of the solid generated by revolving R about the line y=−1y = -1?

  1. π∫02[(2x+1)2−(x2+1)2]dx\pi \int_0^2 [ (2x+1)^2 - (x^2+1)^2 ] dx (correct answer)
  2. π∫02[(x2+1)2−(2x+1)2]dx\pi \int_0^2 [ (x^2+1)^2 - (2x+1)^2 ] dx
  3. π∫02[(2x−x2)2]dx\pi \int_0^2 [ (2x - x^2)^2 ] dx
  4. π∫02[(2x)2−(x2)2]dx\pi \int_0^2 [ (2x)^2 - (x^2)^2 ] dx
Explanation: The curves intersect where x2=2xx^2 = 2x, which gives x=0x=0 and x=2x=2. For x∈[0,2]x \in [0, 2], 2x≥x22x \ge x^2. The axis of revolution is y=−1y=-1. The outer radius R(x)R(x) is the distance from the axis to the outer curve y=2xy=2x, so R(x)=2x−(−1)=2x+1R(x) = 2x - (-1) = 2x+1. The inner radius r(x)r(x) is the distance from the axis to the inner curve y=x2y=x^2, so r(x)=x2−(−1)=x2+1r(x) = x^2 - (-1) = x^2+1. The volume is given by the washer method formula, V=π∫ab[R(x)2−r(x)2]dxV = \pi \int_a^b [R(x)^2 - r(x)^2] dx, which results in π∫02[(2x+1)2−(x2+1)2]dx\pi \int_0^2 [ (2x+1)^2 - (x^2+1)^2 ] dx.

Question 4

Let R be the region enclosed by y=ln⁡(x)y = \ln(x), y=0y=0, and x=ex=e. Which definite integral represents the volume of the solid generated by revolving R about the line y=−2y=-2?

  1. π∫1e[(ln⁡(x)+2)2−22]dx\pi \int_1^e [(\ln(x)+2)^2 - 2^2] dx (correct answer)
  2. π∫1e[(ln⁡(x))2−(−2)2]dx\pi \int_1^e [(\ln(x))^2 - (-2)^2] dx
  3. π∫01[(e+2)2−(ey+2)2]dy\pi \int_0^1 [(e+2)^2 - (e^y+2)^2] dy
  4. π∫1e[(ln⁡(x)−2)2−4]dx\pi \int_1^e [(\ln(x)-2)^2 - 4] dx
Explanation: The region is bounded by y=ln⁡(x)y=\ln(x) and y=0y=0. The intersection is at ln⁡(x)=0\ln(x)=0, which is x=1x=1. The bounds of integration are from x=1x=1 to x=ex=e. The axis of revolution is y=−2y=-2, which is below the region. The outer radius R(x)R(x) is the distance from the axis to the upper curve y=ln⁡(x)y=\ln(x), so R(x)=ln⁡(x)−(−2)=ln⁡(x)+2R(x) = \ln(x) - (-2) = \ln(x)+2. The inner radius r(x)r(x) is the distance from the axis to the lower curve y=0y=0, so r(x)=0−(−2)=2r(x) = 0 - (-2) = 2. The volume is V=π∫1e[R(x)2−r(x)2]dx=π∫1e[(ln⁡(x)+2)2−22]dxV = \pi \int_1^e [R(x)^2 - r(x)^2] dx = \pi \int_1^e [(\ln(x)+2)^2 - 2^2] dx.

Question 5

The region bounded by the parabolas x=2y−y2x=2y-y^2 and x=y2−2yx=y^2-2y is revolved about the line x=3x=3. Which of the following integrals gives the volume of the solid?

  1. π∫02[(2y−y2)2−(y2−2y)2]dy\pi \int_0^2 [ (2y-y^2)^2 - (y^2-2y)^2 ] dy
  2. π∫02[(3−(2y−y2))2−(3−(y2−2y))2]dy\pi \int_0^2 [ (3 - (2y-y^2))^2 - (3 - (y^2-2y))^2 ] dy
  3. π∫−11[((2y−y2)−(y2−2y))2]dy\pi \int_{-1}^1 [ ((2y-y^2)-(y^2-2y))^2 ] dy
  4. π∫02[(3−(y2−2y))2−(3−(2y−y2))2]dy\pi \int_0^2 [ (3 - (y^2-2y))^2 - (3 - (2y-y^2))^2 ] dy (correct answer)
Explanation: When you encounter a volume problem involving revolution around a vertical or horizontal line, you need to set up the washer method carefully, paying close attention to which curve is farther from the axis of rotation. First, find where the parabolas intersect by solving 2y−y2=y2−2y2y - y^2 = y^2 - 2y, which gives 4y=2y24y = 2y^2, so y=0y = 0 and y=2y = 2. Notice that x=2y−y2x = 2y - y^2 can be rewritten as x=−(y−1)2+1x = -(y-1)^2 + 1, a parabola opening left with vertex at (1,1)(1,1). Meanwhile, x=y2−2y=(y−1)2−1x = y^2 - 2y = (y-1)^2 - 1 opens right with vertex at (−1,1)(-1,1). Between y=0y = 0 and y=2y = 2, the first parabola lies to the right of the second. When revolving around x=3x = 3, you use the washer method: V=π∫02[R2−r2]dyV = \pi \int_0^2 [R^2 - r^2] dy, where RR is the outer radius and rr is the inner radius. The outer radius is the distance from x=3x = 3 to the rightmost curve: R=3−(2y−y2)R = 3 - (2y - y^2). The inner radius is the distance to the leftmost curve: r=3−(y2−2y)r = 3 - (y^2 - 2y). Choice A uses the disk method incorrectly without accounting for the axis of rotation. Choice B has the radii backwards—it treats the leftmost curve as the outer radius. Choice C attempts to use a single radius, missing the washer structure entirely. Choice D correctly identifies the outer and inner radii. Study tip: Always sketch the region and identify which curve is farther from the axis of rotation—this determines your outer radius in the washer method.

Question 6

The region bounded by the parabola x=y2x=y^2 and the line x=4x=4 is revolved about the line x=5x=5. What is the volume of the resulting solid?

  1. 256π5\frac{256\pi}{5}
  2. 512π15\frac{512\pi}{15}
  3. 832π15\frac{832\pi}{15} (correct answer)
  4. 416π15\frac{416\pi}{15}
Explanation: Revolving around a vertical axis x=5x=5 requires integrating with respect to yy. The intersections of x=y2x=y^2 and x=4x=4 are at y=±2y=\pm 2. The axis x=5x=5 is to the right of the region. The outer radius is the distance from the axis to the farther curve x=y2x=y^2, so R(y)=5−y2R(y) = 5 - y^2. The inner radius is the distance from the axis to the nearer curve x=4x=4, so r(y)=5−4=1r(y) = 5 - 4 = 1. The volume is V=π∫−22[(5−y2)2−12]dy=π∫−22[y4−10y2+24]dyV = \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=2π[y55−10y33+24y]02=2π(325−803+48)=2π(96−400+72015)=2π(41615)=832π15V = 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}.

Question 7

A solid is formed by revolving the region R, bounded by y=f(x)y=f(x) and y=g(x)y=g(x) on [a,b][a,b] where f(x)≥g(x)≥0f(x) \ge g(x) \ge 0, about the line y=ky=k. If k<0k < 0, which expression represents the volume of the solid?

  1. π∫ab(f(x)−g(x))2dx\pi \int_a^b (f(x)-g(x))^2 dx
  2. π∫ab[(k−f(x))2−(k−g(x))2]dx\pi \int_a^b [ (k-f(x))^2 - (k-g(x))^2 ] dx
  3. π∫ab[(f(x)+k)2−(g(x)+k)2]dx\pi \int_a^b [ (f(x)+k)^2 - (g(x)+k)^2 ] dx
  4. π∫ab[(f(x)−k)2−(g(x)−k)2]dx\pi \int_a^b [ (f(x)-k)^2 - (g(x)-k)^2 ] dx (correct answer)
Explanation: 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=ky = k where k<0k < 0, this axis lies below the x-axis. Given that f(x)≥g(x)≥0f(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)(x, y) to the line y=ky = k is ∣y−k∣=y−k|y - k| = y - k (since y>ky > k). For the washer method, volume equals π∫ab[R2−r2]dx\pi \int_a^b [R^2 - r^2] dx, where RR is the outer radius and rr is the inner radius. Since f(x)≥g(x)f(x) \geq g(x), the outer radius is f(x)−kf(x) - k and the inner radius is g(x)−kg(x) - k. This gives us π∫ab[(f(x)−k)2−(g(x)−k)2]dx\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=ky = k. Answer B incorrectly uses (k−f(x))(k - f(x)) and (k−g(x))(k - g(x)), which would be negative distances since k<0k < 0 and the functions are positive. Answer C uses (f(x)+k)(f(x) + k) terms, which doesn't correctly represent the distance to the axis of rotation. Remember: when rotating about y=ky = k, the radius from any point to the axis is always the absolute value of the difference between the y-coordinate and k.

Question 8

Let R be the region enclosed by the graphs of y=sin⁡(x)y = \sin(x) and y=cos⁡(x)y = \cos(x) for 0≤x≤π40 \le x \le \frac{\pi}{4}. Which integral represents the volume of the solid generated when R is revolved about the line y=2y = 2?

  1. π∫0π/4[(2−cos⁡(x))2−(2−sin⁡(x))2]dx\pi \int_0^{\pi/4} [ (2 - \cos(x))^2 - (2 - \sin(x))^2 ] dx
  2. π∫0π/4[(2−sin⁡(x))2−(2−cos⁡(x))2]dx\pi \int_0^{\pi/4} [ (2 - \sin(x))^2 - (2 - \cos(x))^2 ] dx (correct answer)
  3. π∫0π/4[(cos⁡(x)−sin⁡(x))2]dx\pi \int_0^{\pi/4} [ (\cos(x) - \sin(x))^2 ] dx
  4. π∫0π/4[(sin⁡(x)+2)2−(cos⁡(x)+2)2]dx\pi \int_0^{\pi/4} [ (\sin(x) + 2)^2 - (\cos(x) + 2)^2 ] dx
Explanation: In the interval [0,π/4][0, \pi/4], cos⁡(x)≥sin⁡(x)\cos(x) \ge \sin(x). The axis of revolution y=2y=2 is above this region. The outer radius R(x)R(x) is the distance from the axis to the farther curve, y=sin⁡(x)y=\sin(x), so R(x)=2−sin⁡(x)R(x) = 2 - \sin(x). The inner radius r(x)r(x) is the distance from the axis to the nearer curve, y=cos⁡(x)y=\cos(x), so r(x)=2−cos⁡(x)r(x) = 2 - \cos(x). The volume is V=π∫0π/4[R(x)2−r(x)2]dx=π∫0π/4[(2−sin⁡(x))2−(2−cos⁡(x))2]dxV = \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.

Question 9

A region is bounded by x=y2−4x = y^2 - 4 and x=0x = 0. When revolved around the line x=−5x = -5, what is the correct integral setup for the volume?

  1. π∫−22[(5)2−(5−y2+4)2]dy\pi \int_{-2}^{2} [(5)^2 - (5 - y^2 + 4)^2] dy
  2. π∫−22[(y2+1)2−(5)2]dy\pi \int_{-2}^{2} [(y^2 + 1)^2 - (5)^2] dy
  3. π∫−22[(9−y2)2−(5)2]dy\pi \int_{-2}^{2} [(9 - y^2)^2 - (5)^2] dy
  4. π∫−22[(5)2−(y2+1)2]dy\pi \int_{-2}^{2} [(5)^2 - (y^2 + 1)^2] dy (correct answer)
Explanation: When finding volumes of revolution around vertical lines, you need to carefully identify the outer and inner radii from the axis of rotation to each boundary curve. This problem uses the washer method since you're revolving around an external axis. First, find where the curves intersect. Setting x=y2−4=0x = y^2 - 4 = 0 gives y2=4y^2 = 4, so y=±2y = \pm 2. The region exists between these y-values where the parabola x=y2−4x = y^2 - 4 lies to the left of the line x=0x = 0. When revolving around x=−5x = -5, calculate the distances from this axis to each boundary:
  • Distance to the right boundary (x=0x = 0): 0−(−5)=50 - (-5) = 5 (outer radius)
  • Distance to the left boundary (x=y2−4x = y^2 - 4): (y2−4)−(−5)=y2+1(y^2 - 4) - (-5) = y^2 + 1 (inner radius)
The washer method formula is V=π∫ab[Router2−Rinner2]dyV = \pi \int_{a}^{b} [R_{outer}^2 - R_{inner}^2] dy, giving us π∫−22[52−(y2+1)2]dy\pi \int_{-2}^{2} [5^2 - (y^2 + 1)^2] dy. Choice A incorrectly writes the inner radius as 5−y2+4=9−y25 - y^2 + 4 = 9 - y^2, confusing the distance calculation. Choice B reverses the outer and inner radii, which would give a negative volume. Choice C uses 9−y29 - y^2 as the outer radius, stemming from the same distance miscalculation as choice A. Study tip: Always sketch the region and axis of rotation first. For washers, the radius is always the positive distance from the axis to each curve, and outer minus inner prevents negative volumes.

Question 10

A region is bounded by y=xy = \sqrt{x}, y=xy = x, and revolves around the line x=4x = 4. If we set up the integral using the washer method with respect to yy, what are the correct outer and inner radii?

  1. Outer radius: 4−y4 - y, Inner radius: 4−y24 - y^2 for 0≤y≤10 \leq y \leq 1
  2. Outer radius: 4−y24 - y^2, Inner radius: 4−y4 - y for 0≤y≤10 \leq y \leq 1 (correct answer)
  3. Outer radius: 4+y4 + y, Inner radius: 4+y24 + y^2 for 0≤y≤10 \leq y \leq 1
  4. Outer radius: y2−4y^2 - 4, Inner radius: y−4y - 4 for 0≤y≤10 \leq y \leq 1
Explanation: The curves y=xy = \sqrt{x} and y=xy = x intersect where x=x\sqrt{x} = x, giving x=x2x = x^2, so x(1−x)=0x(1-x) = 0, yielding x=0x = 0 and x=1x = 1. For 0≤y≤10 \leq y \leq 1: from y=xy = \sqrt{x}, we get x=y2x = y^2; from y=xy = x, we get x=yx = y. Since y2≤yy^2 \leq y in this interval, y2y^2 is the left boundary and yy is the right boundary. When revolving around x=4x = 4, the outer radius (farther from axis) is 4−y24 - y^2 and the inner radius is 4−y4 - y. Choice A reverses the radii. Choice C incorrectly adds instead of subtracting from 4. Choice D gives negative radii, which is impossible.

Question 11

The region in the first quadrant bounded by the curves y=x2y=x^2 and y=x4y=x^4 is rotated about the line y=1y=1. What is the volume of the solid generated?

  1. 2π15\frac{2\pi}{15}
  2. 4π21\frac{4\pi}{21}
  3. 8π45\frac{8\pi}{45} (correct answer)
  4. 2π9\frac{2\pi}{9}
Explanation: The curves intersect at x=0x=0 and x=1x=1. In the interval [0,1][0,1], x2≥x4x^2 \ge x^4. The axis of revolution y=1y=1 is above the region. The outer radius is R(x)=1−x4R(x) = 1 - x^4 (distance to the lower curve). The inner radius is r(x)=1−x2r(x) = 1 - x^2 (distance to the upper curve). The volume is V=π∫01[(1−x4)2−(1−x2)2]dx=π∫01[(1−2x4+x8)−(1−2x2+x4)]dx=π∫01[x8−3x4+2x2]dx=π[x99−3x55+2x33]01=π(19−35+23)=π(5−27+3045)=8π45V = \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}.

Question 12

Let R be the region bounded by y=exy = e^x, y=1y=1, and x=2x=2. Find the volume of the solid generated when R is revolved about the line y=5y=5.

  1. π∫02[(5−1)2−(5−ex)2]dx\pi \int_0^2 [ (5-1)^2 - (5-e^x)^2 ] dx (correct answer)
  2. π∫02[(5−ex)2−(5−1)2]dx\pi \int_0^2 [ (5-e^x)^2 - (5-1)^2 ] dx
  3. π∫1e2[22−(ln⁡y)2]dy\pi \int_1^{e^2} [ 2^2 - (\ln y)^2 ] dy
  4. π∫02[(ex−5)2−(1−5)2]dx\pi \int_0^2 [ (e^x-5)^2 - (1-5)^2 ] dx
Explanation: The region is integrated with respect to xx from x=0x=0 (where y=exy=e^x intersects y=1y=1) to x=2x=2. The axis of revolution y=5y=5 is above the region. The outer radius R(x)R(x) is the distance from the axis to the farther curve y=1y=1, so R(x)=5−1=4R(x) = 5-1=4. The inner radius r(x)r(x) is the distance from the axis to the nearer curve y=exy=e^x, so r(x)=5−exr(x) = 5-e^x. The volume is V=π∫02[R(x)2−r(x)2]dx=π∫02[42−(5−ex)2]dxV = \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.

Question 13

The region bounded by the parabola x=y2+1x = y^2 + 1 and the line x=2x=2 is revolved about the line x=−1x = -1. What is the volume of the resulting solid?

  1. 64π15\frac{64\pi}{15}
  2. 104π15\frac{104\pi}{15} (correct answer)
  3. 128π15\frac{128\pi}{15}
  4. 56π15\frac{56\pi}{15}
Explanation: To revolve around the vertical axis x=−1x=-1, we integrate with respect to yy. The intersections occur when y2+1=2y^2+1=2, so y2=1y^2=1 and y=±1y=\pm 1. The axis x=−1x=-1 is to the left of the region. The outer radius is R(y)=2−(−1)=3R(y) = 2 - (-1) = 3. The inner radius is r(y)=(y2+1)−(−1)=y2+2r(y) = (y^2+1) - (-1) = y^2+2. The volume is V=π∫−11[32−(y2+2)2]dy=π∫−11[9−(y4+4y2+4)]dy=π∫−11[−y4−4y2+5]dyV = \pi \int_{-1}^1 [3^2 - (y^2+2)^2] dy = \pi \int_{-1}^1 [9 - (y^4+4y^2+4)] dy = \pi \int_{-1}^1 [-y^4-4y^2+5] dy. Since the integrand is even, this is 2π[−y55−4y33+5y]01=2π(−15−43+5)=2π(−3−20+7515)=2π(5215)=104π152\pi [-\frac{y^5}{5} - \frac{4y^3}{3} + 5y]_0^1 = 2\pi(-\frac{1}{5}-\frac{4}{3}+5) = 2\pi(\frac{-3-20+75}{15}) = 2\pi(\frac{52}{15}) = \frac{104\pi}{15}.

Question 14

Let R be the region bounded by the curves y=4−x2y=4-x^2 and y=x+2y=x+2. Which integral gives the volume of the solid formed by revolving R about the line y=5y=5?

  1. π∫−21[(1+x2)2−(3−x)2]dx\pi \int_{-2}^1 [(1+x^2)^2 - (3-x)^2] dx
  2. π∫−21[((4−x2)−(x+2))2]dx\pi \int_{-2}^1 [((4-x^2)-(x+2))^2] dx
  3. π∫−21[(3−x)2−(1+x2)2]dx\pi \int_{-2}^1 [(3-x)^2 - (1+x^2)^2] dx (correct answer)
  4. π∫−21[(5−(4−x2))2−(5−(x+2))2]dx\pi \int_{-2}^1 [(5-(4-x^2))^2 - (5-(x+2))^2] dx
Explanation: First, find the points of intersection: 4−x2=x+2⇒x2+x−2=0⇒(x+2)(x−1)=04-x^2 = x+2 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0. The bounds are x=−2x=-2 and x=1x=1. In this interval, 4−x2≥x+24-x^2 \ge x+2. The axis of revolution y=5y=5 is above the region. The outer radius R(x)R(x) is the distance from the axis to the lower curve y=x+2y=x+2, so R(x)=5−(x+2)=3−xR(x) = 5-(x+2) = 3-x. The inner radius r(x)r(x) is the distance from the axis to the upper curve y=4−x2y=4-x^2, so r(x)=5−(4−x2)=1+x2r(x) = 5-(4-x^2) = 1+x^2. The volume integral is π∫−21[R(x)2−r(x)2]dx=π∫−21[(3−x)2−(1+x2)2]dx\pi \int_{-2}^1 [R(x)^2 - r(x)^2] dx = \pi \int_{-2}^1 [(3-x)^2 - (1+x^2)^2] dx.

Question 15

The region enclosed by y=1/xy=1/x, y=1y=1, y=4y=4, and the y-axis is revolved about the line x=−1x=-1. What is the volume of the solid generated?

  1. π(34+2ln⁡4)\pi (\frac{3}{4} + 2\ln 4) (correct answer)
  2. π(74+2ln⁡4)\pi (\frac{7}{4} + 2\ln 4)
  3. π(34+ln⁡4)\pi (\frac{3}{4} + \ln 4)
  4. π(154−2ln⁡4)\pi (\frac{15}{4} - 2\ln 4)
Explanation: To revolve about the vertical axis x=−1x=-1, we integrate with respect to yy. The function is x=1/yx=1/y. The y-bounds are given as 1 to 4. The region is bounded by x=1/yx=1/y and x=0x=0 (the y-axis). The axis x=−1x=-1 is to the left of the region. The outer radius is R(y)=(1/y)−(−1)=1/y+1R(y) = (1/y) - (-1) = 1/y+1. The inner radius is r(y)=0−(−1)=1r(y) = 0 - (-1) = 1. The volume is V=π∫14[(1/y+1)2−12]dy=π∫14[1/y2+2/y]dy=π[−1/y+2ln⁡∣y∣]14=π[(−1/4+2ln⁡4)−(−1+2ln⁡1)]=π[3/4+2ln⁡4]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 ].

Question 16

The region bounded by y=xy=x, y=2−xy=2-x, and y=0y=0 is revolved about the line x=3x=3. Find the volume of the solid generated.

  1. 2π2\pi
  2. 8π3\frac{8\pi}{3}
  3. 4π4\pi (correct answer)
  4. 16π3\frac{16\pi}{3}
Explanation: The region is a triangle with vertices at (0,0), (2,0), and (1,1). Revolving around the vertical line x=3x=3 is best done by integrating with respect to yy. The y-bounds are from 0 to 1. We need to express x in terms of y: x=yx=y and x=2−yx=2-y. For y∈[0,1]y \in [0,1], the right boundary is x=2−yx=2-y and the left boundary is x=yx=y. The axis x=3x=3 is to the right of the region. The outer radius is R(y)=3−yR(y) = 3 - y (distance to the left boundary). The inner radius is r(y)=3−(2−y)=1+yr(y) = 3 - (2-y) = 1+y (distance to the right boundary). The volume is V=π∫01[(3−y)2−(1+y)2]dy=π∫01[(9−6y+y2)−(1+2y+y2)]dy=π∫01[8−8y]dy=π[8y−4y2]01=π(8−4)=4πV = \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.

Question 17

The region R is bounded by y=2xy=2^x, y=8y=8, and x=0x=0. Which integral gives the volume of the solid formed by revolving R about the line y=10y=10?

  1. π∫03[(10−8)2−(10−2x)2]dx\pi \int_0^3 [ (10-8)^2 - (10-2^x)^2 ] dx
  2. π∫03[(10−2x)2−(10−8)2]dx\pi \int_0^3 [ (10-2^x)^2 - (10-8)^2 ] dx (correct answer)
  3. π∫18[(log⁡2y)2]dy\pi \int_1^8 [ (\log_2 y)^2 ] dy
  4. π∫03[(8−2x)2]dx\pi \int_0^3 [ (8-2^x)^2 ] dx
Explanation: The intersection of y=2xy=2^x and y=8y=8 is at 2x=82^x=8, so x=3x=3. The integration is from x=0x=0 to x=3x=3. In this region, y=8y=8 is the upper curve and y=2xy=2^x is the lower curve. The axis of revolution y=10y=10 is above the region. The outer radius R(x)R(x) is the distance from the axis to the lower curve y=2xy=2^x, so R(x)=10−2xR(x) = 10-2^x. The inner radius r(x)r(x) is the distance from the axis to the upper curve y=8y=8, so r(x)=10−8=2r(x) = 10-8=2. The volume is V=π∫03[(10−2x)2−22]dxV = \pi \int_0^3 [ (10-2^x)^2 - 2^2 ] dx.

Question 18

Let R be the region in the first quadrant bounded by y=arctan⁡(x)y = \arctan(x), y=π/4y=\pi/4, and the y-axis. Find the setup for the volume of the solid generated by revolving R about the line x=−2x=-2.

  1. π∫0π/4[22−(tan⁡(y)+2)2]dy\pi \int_0^{\pi/4} [ 2^2 - (\tan(y)+2)^2 ] dy
  2. π∫01[(π/4−arctan⁡x)2]dx\pi \int_0^1 [ (\pi/4 - \arctan x)^2 ] dx
  3. π∫0π/4[(tan⁡(y))2]dy\pi \int_0^{\pi/4} [ (\tan(y))^2 ] dy
  4. π∫0π/4[(tan⁡(y)+2)2−22]dy\pi \int_0^{\pi/4} [ (\tan(y)+2)^2 - 2^2 ] dy (correct answer)
Explanation: 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 = \arctan(x), y=π/4y = \pi/4, and the y-axis. This region extends from y=0y = 0 to y=π/4y = \pi/4, with the right boundary being x=tan⁡(y)x = \tan(y) (the inverse of y=arctan⁡(x)y = \arctan(x)). When revolving about the vertical line x=−2x = -2, it's natural to integrate with respect to yy. For each horizontal slice at height yy, the distance from x=−2x = -2 to the y-axis is 2 (inner radius), and the distance from x=−2x = -2 to the curve x=tan⁡(y)x = \tan(y) is tan⁡(y)+2\tan(y) + 2 (outer radius). Using the washer method, the volume is π∫0π/4[(outer radius)2−(inner radius)2] dy=π∫0π/4[(tan⁡(y)+2)2−22] dy\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 xx, 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 yy and carefully identify which boundary is closer to (inner radius) and farther from (outer radius) the axis of revolution.

Question 19

A solid is generated by rotating the region bounded by y=xy=x, x=0x=0, and y=2y=2 about the line y=3y=3. What is the volume of the solid?

  1. 4π4\pi
  2. 16π3\frac{16\pi}{3}
  3. 20π3\frac{20\pi}{3} (correct answer)
  4. 8π8\pi
Explanation: The region is a triangle bounded by y=xy=x, the y-axis (x=0x=0), and the horizontal line y=2y=2. We integrate with respect to x from 0 to 2. The region is bounded above by y=2y=2 and below by y=xy=x. The axis of revolution is y=3y=3, which is above the region. The outer radius is the distance from the axis to the lower curve, R(x)=3−xR(x) = 3 - x. The inner radius is the distance from the axis to the upper curve, r(x)=3−2=1r(x) = 3 - 2 = 1. The volume is V=π∫02[(3−x)2−12]dx=π[−(3−x)33−x]02=π[(−133−2)−(−333−0)]=π[−13−2+9]=π[7−13]=20π3V = \pi \int_0^2 [ (3-x)^2 - 1^2 ] dx = \pi [-\frac{(3-x)^3}{3} - x]_0^2 = \pi [(-\frac{1^3}{3} - 2) - (-\frac{3^3}{3} - 0)] = \pi [-\frac{1}{3} - 2 + 9] = \pi [7 - \frac{1}{3}] = \frac{20\pi}{3}.

Question 20

The region R is bounded by y=sec⁡(x)y=\sec(x), y=2y=2, x=−π/3x=-\pi/3, and x=π/3x=\pi/3. Which integral gives the volume of the solid generated by revolving R about the line y=3y=3?

  1. π∫−π/3π/3[(3−sec⁡(x))2−12]dx\pi \int_{-\pi/3}^{\pi/3} [(3-\sec(x))^2 - 1^2] dx (correct answer)
  2. π∫−π/3π/3[12−(3−sec⁡(x))2]dx\pi \int_{-\pi/3}^{\pi/3} [1^2 - (3-\sec(x))^2] dx
  3. π∫−π/3π/3[(2−sec⁡(x))2]dx\pi \int_{-\pi/3}^{\pi/3} [(2-\sec(x))^2] dx
  4. π∫12[(sec⁡−1y)2]dy\pi \int_{1}^{2} [(\sec^{-1}y)^2] dy
Explanation: On the interval [−π/3,π/3][-\pi/3, \pi/3], sec⁡(x)\sec(x) ranges from 1 to 2. The region R is bounded above by y=2y=2 and below by y=sec⁡(x)y=\sec(x). The axis of revolution y=3y=3 is above the region. The outer radius R(x)R(x) is the distance from the axis to the lower curve y=sec⁡(x)y=\sec(x), so R(x)=3−sec⁡(x)R(x) = 3 - \sec(x). The inner radius r(x)r(x) is the distance from the axis to the upper curve y=2y=2, so r(x)=3−2=1r(x) = 3 - 2 = 1. The volume is V=π∫−π/3π/3[(3−sec⁡(x))2−12]dxV = \pi \int_{-\pi/3}^{\pi/3} [ (3-\sec(x))^2 - 1^2 ] dx.